7.15 Exercise Problems

Q1: A wheel in uniform motion about an axis passing through its centre and perpendicular to its plane is considered to be in mechanical (translational plus rotational) equilibrium because no net external force or torque is required to sustain its motion. However, the particles that constitute the wheel do experience a centripetal acceleration directed towards the centre. How do you reconcile this fact with the wheel being in equilibrium?
How would you set a half-wheel into uniform motion about an axis passing through the centre of mass of the wheel and perpendicular to its plane? Will you require external forces to sustain the motion?

Answer: The centripetal acceleration in a wheel arise due to the internal elastic forces which in pairs cancel each other; being part of a symmetrical system. In a half wheel the distribution of mass about its centre of mass (axis of rotation) is not symmetrical. Therefore, the direction of angular momentum does not coincide with the direction of angular velocity and hence an external torque is required to maintain rotation.

Q2: Find the centre of mass of a uniform (a) half-disc, (b) quarter-disc.

Answer: (a)

Mass of half disc \(=M\)
Surface density \(\sigma=\frac{2 M}{\pi a^2}\)
Mass of the element \((d m)=\sigma d A=\sigma \pi r d r\)
As disc is symmetric about \(\mathrm{x}\)-axis, so \({x}_{CM}=0\)
\(
\begin{aligned}
x_{CM} & =\frac{\int x d m}{\int d m}=\int_{r=0}^a \int_{\theta=0}^\pi r \cos \theta \sigma r d r d \theta=\int_{r=0}^a \int_{\theta=0}^\pi \sigma r d r d \theta \\
& =\frac{\left.\int_0^a r^2 d r \sin \theta\right|_0 ^\pi}{\int_0^a r d r \int_0^\pi d \theta}=0
\end{aligned}
\)
Here \(y\) – coordinate of centre of mass of ring of radius \(r\) is \(y=\frac{2 r}{\pi}\) \(y\)-coordinate of half-disc:
\(y\)-coordinate of half-disc:
\(y_{C M}=\frac{\int y d m}{\int d m}\)
\(\int_{\theta=0}^\pi \int_{r=0}^a r \sin \theta \sigma r d r d \theta = \int_{r=0}^a \int_{\theta=0}^\pi \sigma r d r d \theta\)
\(
=\frac{\int_0^a r^2 d r \int_{\theta=0}^\pi \sin \theta d \theta}{\int_0^a r d r \int_0^\pi d \theta}=\frac{a^3}{3} \frac{[-\cos \theta]_0^\pi}{\left(a^2 / 2\right) \pi}=\frac{a}{3} \frac{4}{\pi}=\frac{4 a}{3 \pi} .
\)
\(
\text { Final Answer: } 0, \frac{4 a}{3 \pi}
\)

(b) 

Mass of quarter disc \(=M\)
Surface charge density be \((\sigma)=\frac{4 M}{\pi a^2}\)
Mass of the element \((d m)=\sigma d A=\sigma\left(\frac{\pi r}{2}\right) d r\)
Here centre of mass of the ring element, \(x=y=\frac{2 r}{\pi}\)
x-coordinate of quarter-disc:
\(
x_{C M}=\frac{\int x d m}{\int d m}
\)
\(
x_{C M}=\frac{\int_0^a\left(\frac{2 r}{\pi}\right) \sigma\left(\frac{\pi r}{2}\right) d r}{\int_0^a\left(\sigma\left(\frac{\pi r}{2}\right) d r\right)}
\)
\(
\left(\frac{2}{\pi}\right) \frac{\int_0^a\left(r^2 d r\right)}{\int_0^a(r d r)}
\)
\(
x_{C M}=\frac{4 a}{3 \pi}
\)
\(
\text { Same procedure will follow for } y \text {-coordinate of mass of disc }
\)
\(
y_{C M}=\frac{\int y d m}{\int d m}=\frac{\int_0^a\left(\frac{2 r}{\pi}\right) \sigma\left(\frac{\pi r}{2}\right) d r}{\int_0^a\left(\sigma\left(\frac{\pi r}{2}\right) d r\right)}=\frac{4 a}{3 \pi}
\)
\(
\text { Final Answer: } \frac{4 a}{3 \pi} \frac{4 a}{3 \pi}
\)

Q3: Two discs of moments of inertia \(I_1\) and \(I_2\) about their respective axes (normal to the disc and passing through the centre), and rotating with angular speed \(\omega_1\) and \(\omega_2\) are brought into contact face to face with their axes of rotation coincident.
(a) Does the law of conservation of angular momentum apply to the situation? why?
(b) Find the angular speed of the two-disc system.
(c) Calculate the loss in kinetic energy of the system in the process.
(d) Account for this loss.

Answer:
(a) Yes, because there is no net external torque on the system. External forces, gravitation and normal reaction act through the axis of rotation, hence produce no torque.
(b) By angular momentum conservation
\(
\begin{aligned}
& I_{\omega}=I_1 \omega_1+I_2 \omega_2 \\
& \therefore \omega=\frac{I_1 \omega_1+I_2 \omega_2}{I_1+I_2}
\end{aligned}
\)
(c)
\(
\begin{aligned}
& K_f=\frac{1}{2}\left(I_1+I_2\right) \frac{\left(I_1 \omega_1+I_2 \omega_2\right)^2}{\left(I_1+I_2\right)^2}=\frac{1}{2} \frac{\left(I_1 \omega_1+I_2 \omega_2\right)^2}{I_1+I_2} \\
& K_t=\frac{1}{2}\left(I_1 \omega_1^2+I_2 \omega_2^2\right) \\
& \Delta K=K_f-K_i=-\frac{I_1 I_2}{2\left(I_1+I_2\right)}\left(\omega_1-\omega_2\right)^2
\end{aligned}
\)
(d) The loss in kinetic energy is due to the work against the friction between the two discs.

Q4: A disc of radius \(R\) is rotating with an angular speed \(\omega_o\) about a horizontal axis. It is placed on a horizontal table. The coefficient of kinetic friction is \(\mu_k\).
(a) What was the velocity of its centre of mass before being brought in contact with the table?
(b) What happens to the linear velocity of a point on its rim when placed in contact with the table?
(c) What happens to the linear speed of the centre of mass when disc is placed in contact with the table?
(d) Which force is responsible for the effects in (b) and (c)?
(e) What condition should be satisfied for rolling to begin?
(f) Calculate the time taken for the rolling to begin.

Answer: (a) Before being brought in contact with the table the disc was in pure rotational motion hence, \(\mathrm{v}_{\mathrm{CM}}=0\).
(b) When the disc is placed in contact with the table due to friction velocity of a point on the rim decreases.
(c) When the rotating disc is placed in contact with the table due to friction centre of mass acquires some linear velocity.
(d) Friction is responsible for the effects in (b) and (c).
\(
\text { When rolling starts } \mathrm{v}_{\mathrm{CM}}=\omega \mathrm{R} \text {. }
\)
\(
\text { Where } \omega \text { is the angular speed of the disc when rolling just starts. }
\)
(f) Acceleration produced in centre of mass due to friction:
\(
a_{c m}=\frac{F}{m}=\frac{\mu_k m g}{m}=\mu_k g \text {. }
\)
Angular acceleration produced by the torque due to friction,
\(
\begin{aligned}
& \alpha=\frac{\tau}{I}=\frac{\mu_k m g R}{I} \\
& \therefore v_{c m}=u_{c m}+a_{c m} t \Rightarrow v_{c m}=\mu_k g t \\
& \text { and } \omega=\omega_o+\alpha t \Rightarrow \omega=\omega_o-\frac{\mu_k m g R}{I} t
\end{aligned}
\)
For rolling without slipping,
\(
\begin{aligned}
& \frac{v_{a m}}{R}=\omega_o-\frac{\mu_K m g R}{I} t \\
& \frac{\mu_K g t}{R}=\omega_O-\frac{\mu_K m g R}{I} t \\
& t=\frac{R \omega_o}{\mu_k g\left(1+\frac{m R^2}{I}\right)}
\end{aligned}
\)

Q5: Two cylindrical hollow drums of radii \(R\) and \(2 R\), and of a common height \(h\), are rotating with angular velocities ‘ \(\omega\) (anti-clockwise) and \(\omega\) (clockwise), respectively. Their axes, fixed are parallel and in a horizontal plane separated by \((3 R+\delta)\). They are now brought in contact \((\delta \rightarrow 0)\).
(a) Show the frictional forces just after contact.
(b) Identify forces and torques external to the system just after contact.
(c) What would be the ratio of final angular velocities when friction ceases?

Answer: (a)

(b)

\(F^{\prime}=F=F^{\prime \prime}\) where \(F\) and \(F^{\prime \prime}\) and external forces through support.
\(
\mathrm{F}_{\text {net }}=0
\)
External torque \(=F \times 3 R\), anticlockwise.

(c) Let \(\omega_1\) and \(\omega_2\) be final angular velocities (anticlockwise and clockwise respectively)
Finally, there will be no friction.
Hence, \(R \omega_1=2 R \omega_2 \Rightarrow \frac{\omega_1}{\omega_2}=2\)

Q6: A uniform square plate \(\mathrm{S}\) (side c) and a uniform rectangular plate \(\mathrm{R}\) (sides b, a) have identical areas and masses (Fig. 7.11).

Show that \(\text { (i) } I_{x R} / I_{x S}<\text { l; (ii) } I_{y R} / I_{y S}>\text { l; (iii) } I_{z R} / I_{z S}>1 \text {. }\)

Answer: (i) Area of square \(=\) area of rectangle \(\Rightarrow c^2=a b\)
\(
\frac{I_{x R}}{I_{x S}} \times \frac{I_{y R}}{I_{y S}}=\frac{b^2}{c^2} \times \frac{a^2}{c^2}=\left(\frac{a b}{c^2}\right)^2=1
\)
(ii)
\(
\begin{aligned}
& \frac{I_{y R}}{I_{y S}}>\frac{I_{x R}}{I_{x S}} \Rightarrow \frac{I_{y R}}{I_{y S}}>1 \\
& \text { and } \frac{I_{x R}}{I_{x S}}<1
\end{aligned}
\)
(iii)

\(
\begin{aligned}
I_{z r}-I_{Z S} & \propto\left(a^2+b^2-2 c^2\right) \\
& =a^2+b^2-2 a b>0
\end{aligned}
\)
\(
\begin{aligned}
& \therefore\left(I_{\mathrm{zR}}-I_{\mathrm{zS}}\right)>0 \\
& \therefore \frac{I_{\mathrm{z} R}}{I_{\mathrm{zS}}}>1 .
\end{aligned}
\)

Q7: Find the components along the \(x, y, z\) axes of the angular momentum 1 of a particle, whose position vector is \(\mathbf{r}\) with components \(x, y, z\) and momentum is \(\mathrm{p}\) with components \(p_x, p_y\) and \(p_z\). Show that if the particle moves only in the \(x-y\) plane the angular momentum has only a \(z\)-component.

Answer:

\(
\text { Linear momentum of particle, } \vec{p}=p_x \hat{\imath}+p_y \hat{\jmath}+p_z \hat{k}
\)
Position vector of the particle, \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\)
\(
\text { Angular momentum, } \overrightarrow{l}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{p}}
\)
\(
\Longrightarrow l_x^{\hat{\imath}}+l_y \hat{\jmath}+l_z \hat{k}=\hat{i}\left(y p_z-z p_y\right)-\hat{j}\left(x_z-z p_x\right)+\hat{k}\left(x p_y-y p_x\right)
\)
Therefore on comparison of coefficients,
\(
\begin{aligned}
& l_x=y p_z-z p_y \\
& l_y=z p_x-x p_z \\
& l_z=x p_y-y p_x
\end{aligned}
\)
The particle moves in the \(x-y\) plane. Hence the \(z\) component of the position vector and linear momentum vector becomes zero.
\(
\mathrm{z}=\mathrm{p}_{\mathrm{z}}=0
\)
\(
\text { Thus } l_x=0
\)
\(
l_y=0
\)
\(
i_z=x p_y-y p_x
\)
Thus when the particle is confined to move in the \(x\)-y plane, the angular momentum of the particle is along the \(\mathrm{z}\)-direction.

Q8: Two particles, each of mass \(m\) and speed \(v\), travel in opposite directions along parallel lines separated by a distance \(d\). Show that the angular momentum vector of the two-particle system is the same whatever be the point about which the angular momentum is taken.

Answer:

Let at a certain instant two particles be at points \(\mathrm{P}\) and \(\mathrm{Q}\), as shown in the following figure.
Consider a point \(R\), which is at a distance y from point Q, i.e.,
\(
\begin{aligned}
& \mathrm{QR}=\mathrm{y} \\
& \therefore \mathrm{PR}=\mathrm{d}-\mathrm{y}
\end{aligned}
\)
Angular momentum of the system about point \(\mathrm{P}\) :
\(
\mathrm{L}_{\mathrm{P}}=\mathrm{mv} \times 0+\mathrm{mv} \times \mathrm{d}=\mathrm{mvd} . \ldots . .(\mathrm{i})
\)
Angular momentum of the system about point Q:
\(\mathrm{L}_{\mathrm{Q}}=\mathrm{mv} \times \mathrm{d}+\mathrm{mv} \times 0=\mathrm{mvd} . \ldots .(\) ii)
Angular momentum of the system about point R:
\(\mathrm{L}_{\mathrm{R}}=\mathrm{mv} \times(\mathrm{d}-\mathrm{y})+\mathrm{mv} \times \mathrm{y}=\mathrm{mvd} \ldots .(\mathrm{iii})\)
Comparing equations (i), (ii), and (iii), we get:
\(\mathrm{L}_{\mathrm{P}}=\mathrm{L}_{\mathrm{Q}}=\mathrm{L}_{\mathrm{R}} \dots(iv)\)
We infer from equation (iv) that the angular momentum of a system does not depend on the point about which it is taken.

Q9: (a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be \(2 M R^2 / 5\), where \(M\) is the mass of the sphere and \(R\) is the radius of the sphere.
(b) Given the moment of inertia of a disc of mass \(M\) and radius \(R\) about any of its diameters to be \(M R^2 / 4\), find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

Answer:

(a)
The moment of inertia (M.I.) of a sphere about its diameter \(=2 \mathrm{MR}^2 / 5\)
According to the theorem of parallel axes, the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.
The M.I. about a tangent of the sphere \(=2 \mathrm{MR}^2 / 5+\mathrm{MR}^2=7 \mathrm{MR}^2 / 5\)
(b)
The moment of inertia of a disc about its diameter \(=\mathrm{MR}^2 / 4\)
According to the theorem of the perpendicular axis, the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with the perpendicular axis and lying in the plane of the body.
The M.I. of the disc about an axis normal to the disc through the center = \(\mathrm{MR}^2 / 4+\mathrm{MR}^2 / 4=\mathrm{MR}^2 / 2\)
Now, Applying the theorem of parallel axes:
The moment of inertia about an axis normal to the disc and passing through a point on its edge
\(
=\mathrm{MR}^2 / 2+\mathrm{MR}^2=3 \mathrm{MR}^2 / 2
\)

Q10: As shown in Fig. 7.40, the two sides of a step ladder BA and CA are \(1.6 \mathrm{~m}\) long and hinged at \(\mathrm{A}\). A rope DE, \(0.5 \mathrm{~m}\) is tied halfway up. A weight \(40 \mathrm{~kg}\) is suspended from a point F, \(1.2 \mathrm{~m}\) from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. (Take \(g=9.8 \mathrm{~m} / \mathrm{s}^2\) )
(Hint: Consider the equilibrium of each side of the ladder separately.)

Answer:

The forces acting on the ladder are shown in Fig. below. Here, W \(=40 \mathrm{~kg}=40 \times 9.8 \mathrm{~N}=392 \mathrm{~N}, \mathrm{AB}=\mathrm{AC}=1.6 \mathrm{~m}, \mathrm{BD}=1 / 2 \times 1.6 \mathrm{~m}\) \(=0.8 \mathrm{~m}\),
\(
B F=1.2 \mathrm{~m} \text { and DE } 0.5 \mathrm{~m} \text {, }
\)
In figure \(\triangle A D E\) and \(\triangle A B C\) are similar hence
\(
B C=D E \times \frac{A B}{A D}=\frac{0.5 \times 1.6}{0.8}=1.0 \mathrm{~m}
\)
\(
\text { Now considering equilibrium at a point B, }
\)
\(
\therefore W \times(M B)=N_c \times(C B) \ldots \ldots \text { (i) }
\)
\(
\text { But } M B=\frac{K B \times B F}{B A}=\frac{0.5 \times 1.2}{1.6}=0.375 \mathrm{~m}
\)
Substituting this value in (i) we get
\(
\therefore N_C=\frac{W \times(M B)}{C B}=\frac{392 \times .0375}{1}=147 N
\)
Again considering equilibrium at point \(C\) in a similar manner we have
\(
\begin{aligned}
& W \times(M C)=N_B \times(B C) \\
& \therefore N_B=\frac{W \times(M C)}{B C}=\frac{W \times(B C-B M)}{B C} \\
& =(392 \times(1-0.375)) / 1=245 \mathrm{~N}
\end{aligned}
\)
Now, it can be easily shown that tension in the string
\(
\mathrm{T}=\mathrm{N}_{\mathrm{B}}-\mathrm{N}_{\mathrm{C}}=245-147=98 \mathrm{~N}
\)

Q11: Two discs of moments of inertia \(I_1\) and \(I_2\) about their respective axes (normal to the disc and passing through the centre), and rotating with angular speeds \(\omega_1\) and \(\omega_2\) are brought into contact face to face with their axes of rotation coincident. (a) What is the angular speed of the two-disc system? (b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? Take \(\omega_1 \neq \omega_2\).

Answer: (a)Moment of inertia of \(\operatorname{disc} I=I_1\)
Angular speed of \(\operatorname{disc} I=\omega_1\)
Angular speed of disc \(\mathrm{II}=I_2\)
Angular momentum of disc \(\mathrm{II}=\omega_2\)
Angular momentum of disc \(\mathrm{I}=L_1=I_1 \omega_1\)
Angular momentum of disc II, \(L_2=I_2 \omega 2\)
Total initial angular momentum, \(L_1=I_1 \omega_1+I_2 \omega_2\)
When the two discs are joined together, their moments of inertia get added up.
Moment of inertia of the system of two discs, \(I=I_1+I_2\)
Let \(\omega\) be the angular speed of the system
Total final angular momentum, \(L_f=\left(I_1+I_2\right) \omega\)
Using the law of conservation of angular momentum, we have:
\(
\begin{aligned}
& L_i=L_f \\
& I_1 \omega_1+I_2 \omega_2=\left(I_1+I_2\right) \omega \\
& \therefore \omega=\frac{I_1 \omega_1+I_2 \omega_2}{I_1+I_2}
\end{aligned}
\)
(b) Kinetic energy of disc I, \(E_1=\frac{1}{2} I_1 \omega_1^2\)
Kinetic energy of disc II, \(E_2=\frac{1}{2} I_2 \omega_2^2\)
Total initial kinetic energy, \(E_i=\frac{1}{2}\left(I_1 \omega_1^2+I_2 \omega_2^2\right)\)
When the discs are joined, their moments of inertia get added up.
Moment of inertia of the system, \(I=I_1+I_2\)
Angular speed of the system \(=\omega\)
Final kinetic energy \(E_f\) :
\(
\begin{aligned}
& =\frac{1}{2}\left(I_1+I_2\right) \omega^2 \\
& =\frac{1}{2}\left(I_1+I_2\right)\left(\frac{I_1 \omega_1+I_2 \omega_2}{I_1+I_2}\right)^2=\frac{1}{2} \frac{I_1 \omega_1+I_2 \omega_2}{I_1+I_2} \\
& \therefore E_i-E_f
\end{aligned}
\)
\(
\begin{aligned}
& =\frac{1}{2}\left(I_1 \omega_1^2+I_2 \omega_2^2\right)-\frac{\left(I_1 \omega_1+I_1 \omega_2\right)^2}{2\left(I_1+I_2\right)} \\
& =\frac{1}{2} I_1 \omega_1^2+\frac{1}{2} I_2 \omega_2^2-\frac{1}{2} \frac{I_1^2 \omega_1^2}{I_1+I_2}-\frac{1}{2} \frac{I_2^2 \omega_2^2}{2\left(I_1+I_2\right)}-\frac{1}{2} \frac{2 I_1 I_2 \omega_1 \omega_2}{2\left(I_1+I_2\right)} \\
& =\frac{1}{I_1+I_2}\left[\frac{1}{2} I_1^2 \omega_1^2+\frac{1}{2} I_1 I_2 \omega_1^2+\frac{1}{2} I_1 I_2 \omega_2^2+\frac{1}{2} I_2^2 \omega^2-\frac{1}{2} I_1^2 \omega_1^2-\frac{1}{2} I_2^2 \omega_2^2-I_1 I_2 \omega_1 \omega_2\right] \\
& =\frac{I_1 I_2}{2\left(I_1+I_2\right)}\left[\omega_1^2+\omega_2^2-2 \omega_1 \omega_2\right] \\
& =\frac{I_1 I_2\left(\omega_1-\omega_2\right)^2}{2\left(I_1+I_2\right)}
\end{aligned}
\)
All the quantities on RHS are positive
\(
\begin{aligned}
& \therefore E_i-E_f>0 \\
& E_i>E_f
\end{aligned}
\)
The loss of KE can be attributed to the frictional force that comes into play when the two discs come in contact with each other.

Q12: (a) Prove the theorem of perpendicular axes.
(Hint: Square of the distance of a point \((x, y)\) in the \(x-y\) plane from an axis through the origin and perpendicular to the plane is \(\left.x^2+y^2\right)\).
(b) Prove the theorem of parallel axes.
(Hint: If the centre of mass of a system of \(n\) particles is chosen to be the origin \(\left.\sum m_i \mathbf{r}_i=0\right)\)

Answer: 

(a)The theorem of perpendicular axes states that the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with the perpendicular axis and lying in the plane of the body.

A physical body with centre \(\mathrm{O}\) and a point mass \(m\), in the \(x-y\) plane at \((x, y)\) is shown in the following figure.

Moment of inertia about \(x\)-axis, \(I_x=m x^2\)
Moment of inertia about \(y\)-axis, \(l_y=m y^2\)
Moment of inertia about \(z\)-axis, \(I_z=m\left(\sqrt{x^2+y^2}\right)^2\)
\(
\begin{aligned}
& I_x+I_y=m x^2+m y^2 \\
& =m\left(x^2+y^2\right) \\
& =m\left(\sqrt{x^2+y^2}\right)^2 \\
& I_x+I_y=I_z
\end{aligned}
\)
Hence, the theorem is proved.
(b) The theorem of parallel axes states that the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.

Suppose a rigid body is made up of \(n\) particles, having masses \(m_1, m_2, m_3, \ldots\), \(m_n\), at perpendicular distances \(r_1, r_2, r_3, \ldots, r_n\) respectively from the centre of mass \(O\) of the rigid body.
The moment of inertia about axis RS passing through the point \(O\) :
\(
I_{\mathrm{RS}}=\sum_{i=1}^n m_i r_i^2
\)
The perpendicular distance of mass \(m_i\), from the axis \(\mathrm{QP}=a+r_i\) Hence, the moment of inertia about axis QP:
\(
\begin{aligned}
I_{\mathrm{QP}} & =\sum_{i=1}^n m_j\left(a+r_i\right)^2 \\
& =\sum_{i=1}^n m_i\left(a^2+r_i^2+2 a r_i\right)^2 \\
& =\sum_{i=1}^n m_i a^2+\sum_{i=1}^n m_i r_i^2+\sum_{i=1}^n m_i 2 a r_i \\
& =I_{\mathrm{RS}}+\sum_{i=1}^n m_i a^2+2 \sum_{i=1}^n m_i a r_i^2
\end{aligned}
\)
Now, at the centre of mass, the moment of inertia of all the particles about the axis passing through the centre of mass is zero, that is,
\(
\begin{aligned}
& 2 \sum_{i=1}^n m_i a r_i=0 \\
& \because a \neq 0 \\
& \therefore \sum m_i r_i=0
\end{aligned}
\)
Also,
\(
\begin{aligned}
& \sum_{i=1}^n m_i=M ; \quad M=\text { Total mass of the rigid body } \\
& \therefore I_{\mathrm{QP}}=I_{\mathrm{RS}}+M a^2
\end{aligned}
\)
Hence, the theorem is proved.

Q13: Prove the result that the velocity \(v\) of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height \(h\) is given by
\(
v^2=\frac{2 g h}{\left(1+k^2 / R^2\right)}
\)
using dynamical consideration (i.e. by consideration of forces and torques). Note \(k\) is the radius of gyration of the body about its symmetry axis, and \(R\) is the radius of the body. The body starts from rest at the top of the plane.

Answer: 

A body rolling on an inclined plane of height \(h\) is shown in the following figure:

\(m=\) Mass of the body
\(R=\) Radius of the body
\(K=\) Radius of gyration of the body
\(v=\) Translational velocity of the body
\(h=\) Height of the inclined plane
\(\mathrm{g}=\) Acceleration due to gravity
Total energy at the top of the plane, \(E_t=m g h\)
Total energy at the bottom of the plane
\(
E_b=\frac{1}{2} I \omega^2+\frac{1}{2} m v^2
\)
But \(I=m k^2\) and \(\omega=\frac{v}{r}\)
\(
\begin{aligned}
& \therefore E_b=\frac{1}{2}\left(m k^2\right)\left(\frac{v^2}{R_2}\right)+\frac{1}{2} m v^2 \\
& =\frac{1}{2} m v^2 \frac{k^2}{R^2}+\frac{1}{2} m v^2 \\
& =\frac{1}{2} m v^2\left(1+\frac{k^2}{R^2}\right)
\end{aligned}
\)
From the law of conservation of energy, we have:
\(
\begin{aligned}
& E_t=E_b \\
& m g h=\frac{1}{2} m v^2\left(1+\frac{k^2}{R^2}\right) \\
& \therefore v=\frac{2 g h}{1+k^2 / R^2}
\end{aligned}
\)
Hence, the given result is proved.

Q14: Explain why friction is necessary to make the disc in Fig. 7.41 roll in the direction indicated.
(a) Give the direction of frictional force at B, and the sense of frictional torque, before perfect rolling begins.
(b) What is the force of friction after perfect rolling begins?

Answer: (a) Frictional force at B opposes the velocity of B. Therefore, the frictional force is in the same direction as the arrow. The sense of frictional torque is such as to oppose angular motion. \(\omega_0\) and \(\tau\) are both normal to the paper, the first into the paper, and the second coming out of the paper.
(b) Frictional force decreases the velocity of the point of contact B. Perfect rolling ensues when this velocity is zero. Once this is so, the force of friction is zero.

Q15: A solid disc and a ring, both of radius \(10 \mathrm{~cm}\) are placed on a horizontal table simultaneously, with an initial angular speed equal to \(10 \pi \mathrm{rad} \mathrm{s}^{-1}\). Which of the two will start to roll earlier? The coefficient of kinetic friction is \(\mu_{\mathrm{k}}=0.2\).

Answer: Frictional force causes the CM to accelerate from its initial zero velocity. Frictional torque causes retardation in the initial angular speed \(\omega_0\).
The equations of motion are : \(\quad \mu_{\mathrm{k}} m g=m a\)
and \(\mu_{\mathrm{k}} m g R=-I \alpha\),
which yield
\(v=\mu_{\mathrm{k}} g t,
\omega=\omega_{\mathrm{o}}-\mu_{\mathrm{k}} m g R t / I\).
Rolling begins when \(v=R \omega\).
For a ring, \(I=m R^2\), and rolling begins at \(t=\omega_{\mathrm{o}} R / 2 \mu_{\mathrm{k}} g\).
For a disc, \(I=1 / 2 m R^2\) and rolling starts at break line \(t=R \omega_0 / 3 \mu_{\mathrm{k}} g\).
Thus, the disc begins to roll earlier than the ring, for the same \(R\) and \(\omega_0\).
The actual times can be obtained for \(R=10 \mathrm{~cm}, \omega_{\mathrm{o}}=10 \pi \mathrm{rad} \mathrm{s}^{-1}, \mu_{\mathrm{k}}=0.2\)

Q16: Three particles of masses \(0.50 \mathrm{~kg}, 1.0 \mathrm{~kg}\) and \(1.5 \mathrm{~kg}\) are placed at the three corners of a right-angled triangle of sides \(3.0 \mathrm{~cm}, 4.0 \mathrm{~cm}\) and \(5.0 \mathrm{~cm}\) as shown in figure (9-W1). Locate the centre of mass of the system.

Answer: Let us take the \(4.0 \mathrm{~cm}\) line as the \(X\)-axis and the \(3.0 \mathrm{~cm}\) line as the \(Y\)-axis. The coordinates of the three particles are as follows :
\(
\begin{array}{lll}
m & x & y \\
0.50 \mathrm{~kg} & 0 & 0 \\
1.0 \mathrm{~kg} & 4.0 \mathrm{~cm} & 0 \\
1.5 \mathrm{~kg} & 0 & 3.0 \mathrm{~cm}
\end{array}
\)
The \(x\)-coordinate of the centre of mass is
\(
\begin{aligned}
X & =\frac{m_1 x_1+m_2 x_2+m_3 x_3}{m_1+m_2+m_{\mathrm{a}}} \\
& =\frac{(0 \cdot 50 \mathrm{~kg}) \cdot 0+(1 \cdot 0 \mathrm{~kg}) \cdot(4 \cdot 0 \mathrm{~cm})+(1 \cdot 5 \mathrm{~kg}) \cdot 0}{0 \cdot 50 \mathrm{~kg}+1 \cdot 0 \mathrm{~kg}+1 \cdot 5 \mathrm{~kg}}
\end{aligned}
\)
\(
=\frac{4 \mathrm{~kg}-\mathrm{cm}}{3 \mathrm{~kg}}=1.3 \mathrm{~cm} \text {. }
\)
The \(y\)-coordinate of the centre of mass is
\(
\begin{aligned}
Y & =\frac{m_1 y_1+m_2 y_2+m_3 y_3}{m_1+m_2+m_3} \\
& =\frac{(0.50 \mathrm{~kg}) \cdot 0+(1.0 \mathrm{~kg}) \cdot 0+(1.5 \mathrm{~kg})(3.0 \mathrm{~cm})}{0.50 \mathrm{~kg}+1.0 \mathrm{~kg}+1.5 \mathrm{~kg}} \\
& =\frac{4.5 \mathrm{~kg}-\mathrm{cm}}{3 \mathrm{~kg}}=1.5 \mathrm{~cm} .
\end{aligned}
\)
Thus, the centre of mass is \(1.3 \mathrm{~cm}\) right and \(1.5 \mathrm{~cm}\) above the \(0.5 \mathrm{~kg}\) particle.

Q17: Half of the rectangular plate shown in figure (9-W2) is made of a material of density \(\rho_1\) and the other half of density \(\rho_2\). The length of the plate is L. Locate the centre of mass of the plate.

Answer: The centre of mass of each half is located at the geometrical centre of that half. Thus, the left half may be replaced by a point particle of mass \(K \rho_1\) placed at \(C_1\) and the right half may be replaced by a point particle of mass \(K \rho_2\) placed at \(C_2\). This replacement is for the specific purpose of locating the combined centre of mass. Take the middle point of the left edge to be the origin. The \(x\)-coordinate of \(C_1\) is \(L / 4\) and that of \(C_2\) is \(3 L / 4\). Hence, the \(x\)-coordinate of the centre of mass is
\(
\begin{aligned}
X & =\frac{\left(K \rho_1\right) \frac{L}{4}+\left(K \rho_2\right) \frac{3 L}{4}}{K \rho_1+K \rho_2} \\
& =\frac{\left(\rho_1+3 \rho_2\right)}{4\left(\rho_1+\rho_2\right)} L .
\end{aligned}
\)
The combined centre of mass is this much to the right of the assumed origin.

Q18: The density of a linear rod of length \(L\) varies as \(\rho=A+B x\) where \(x\) is the distance from the left end. Locate the centre of mass.

Answer: Let the cross-sectional area be \(\alpha\). The mass of an element of length \(d x\) located at a distance \(x\) away from the left end is \((A+B x) \alpha d x\). The \(x\)-coordinate of the centre of mass is given by
\(
\begin{aligned}
X_{C M} & =\frac{\int x d m}{\int d m}=\frac{\int_0^L x(A+B x) \alpha d x}{\int_0^L(A+B x) \alpha d x} \\
& =\frac{A \frac{L^2}{2}+B \frac{L^3}{3}}{A L+B \frac{L^2}{2}}=\frac{3 A L+2 B L^2}{3(2 A+B L)} .
\end{aligned}
\)

Q19: A cubical block of ice of mass \(m\) and edge \(L\) is placed in a large tray of mass \(M\). If the ice melts, how far does the centre of mass of the system “ice plus tray” come down?

Answer: Consider Figure (9-W4).

Suppose the centre of mass of the tray is a distance \(x_1\) above the origin and that of the ice is a distance \(x_2\) above the origin. The height of the centre of mass of the ice-tray system is
\(
x=\frac{m x_2+M x_1}{m+M} .
\)
When the ice melts, the water of mass \(m\) spreads on the surface of the tray. As the tray is large, the height of the water is negligible. The centre of mass of the water is then on the surface of the tray and is at a distance \(x_2-L / 2\) above the origin. The new centre of mass of the ice-tray system will be at the height
\(
x^{\prime}=\frac{m\left(x_2-\frac{L}{2}\right)+M x_1}{m+M} .
\)
The shift in the centre of mass \(=x-x^{\prime}=\frac{m L}{2(m+M)}\).

Q20: consider a two-particle system with the particles having masses \(m_1\) and \(m_2\). If the first particle is pushed towards the centre of mass through a distance \(d\), by what distance should the second particle be moved so as to keep the centre of mass at the same position?

Answer: Consider Figure (9-W5).

Suppose the distance of \(m_1\) from the centre of mass \(C\) is \(x_1\) and that of \(m_2\) from \(C\) is \(x_2\). Suppose the mass \(m_2\) is moved through a distance \(d^{\prime}\) towards \(C\) so as to keep the centre of mass at \(C\)
Then,
\(
\begin{aligned}
& m_1 x_1=m_2 x_2 \dots(i)\\
& \text { and } \\
& m_1\left(x_1-d\right)=m_2\left(x_2-d^{\prime}\right) \dots(ii) \\
&
\end{aligned}
\)
Subtracting (ii) from (i)
\(
\begin{aligned}
m_1 d & =m_2 d^{\prime} \\
d^{\prime} & =\frac{m_1}{m_2} d .
\end{aligned}
\)

Q21: A body of mass \(25 \mathrm{~kg}\) is subjected to the forces shown in figure (9-W6). Find the acceleration of the centre of mass.

Answer: Take the \(X\) and \(Y\) axes as shown in the figure. The \(x\)-component of the resultant force is
\(
\begin{aligned}
F_x & =-6 \mathrm{~N}+(5 \mathrm{~N}) \cos 37^{\circ}+(6 \mathrm{~N}) \cos 53^{\circ}+(4 \mathrm{~N}) \cos 60^{\circ} \\
& =-6 \mathrm{~N}+(5 \mathrm{~N}) \cdot(4 / 5)+(6 \mathrm{~N}) \cdot(3 / 5)+(4 \mathrm{~N}) \cdot(1 / 2)=3 \cdot 6 \mathrm{~N}
\end{aligned}
\)
Similarly, the \(y\)-component of the resultant force is
\(
\begin{aligned}
F_y & =5 \mathrm{~N} \sin 37^{\circ}-(6 \mathrm{~N}) \sin 53^{\circ}+4 \mathrm{~N} \sin 60^{\circ} \\
& =(5 \mathrm{~N}) \cdot(3 / 5)-(6 \mathrm{~N}) \cdot(4 / 5)+(4 \mathrm{~N}) \cdot(\sqrt{ } 3 / 2)=1 \cdot 7 \mathrm{~N}
\end{aligned}
\)
The magnitude of the resultant force is
\(
F=\sqrt{F_{\mathrm{x}}{ }^2+F_y{ }^2}=\sqrt{(3.6 \mathrm{~N})^2+(1.7 \mathrm{~N})^2} \approx 4.0 \mathrm{~N} \text {. }
\)
The direction of the resultant force makes an angle \(\theta\) with the \(X\)-axis where
\(
\tan \theta=\frac{F_y}{F_x}=\frac{1 \cdot 7}{3 \cdot 6}=0.47 .
\)
The acceleration of the centre of mass is
\(
a_{C M}=\frac{F}{M}=\frac{4 \cdot 0 \mathrm{~N}}{2.5 \mathrm{~kg}}=1.6 \mathrm{~m} / \mathrm{s}^2
\)
in the direction of the resultant force.

Q22: Two blocks of equal mass \(m\) are connected by an unstretched spring and the system is kept at rest on a frictionless horizontal surface. A constant force \(F\) is applied on one of the blocks pulling it away from the other as shown in figure (9-W7). (a) Find the position of the centre of mass at time \(t\). (b) If the extension of the spring is \(x_0\) at time \(t\), find the displacement of the two blocks at this instant.

Answer: (a) The acceleration of the centre of mass is given by
\(
a_{C M}=\frac{F}{M}=\frac{F}{2 m} \text {. }
\)
The position of the centre of mass at time \(t\) is
\(
x=\frac{1}{2} a_{C M} t^2=\frac{F t^2}{4 m} .
\)
(b) Suppose the displacement of the first block is \(x_1\) and that of the second is \(x_2\). As the centre of mass is at \(x\), we should have
\(
x=\frac{m x_1+m x_2}{2 m}
\)
\(
\text { or, } \quad \frac{F t^2}{4 m}=\frac{x_1+x_2}{2}
\)
\(
\text { or, } \quad x_1+x_2=\frac{F t^2}{2 m} \dots(i)
\)
The extension of the spring is \(x_2-x_1\). Therefore, \(x_2-x_1=x_0 \dots(ii)\)
from (i) and (ii), \(x_1=\frac{1}{2}\left(\frac{F t^2}{2 m}-x_0\right)\)
and \(\quad x_2=\frac{1}{2}\left(\frac{F t^2}{2 m}+x_0\right)\).

Q23: A projectile is fired at a speed of \(100 \mathrm{~m} / \mathrm{s}\) at an angle of \(37^{\circ}\) above the horizontal. At the highest point, the projectile breaks into two parts of mass ratio \(1: 3\), the smaller coming to rest. Find the distance from the launching point to the point where the heavier piece lands.

Answer: See figure (9-W8). At the highest point, the projectile has horizontal velocity. The lighter part comes to rest. Hence the heavier part will move hith increased horizontal velocity. In vertical direction, both parts have zero velocity and undergo same acceleration, hence they will cover equal vertical displacements in a given time. Thus, both will hit the ground together. As internal forces do not affect the motion of the centre of mass, the centre of mass hits the ground at the position where the original projectile would have landed. The range of the original projectile is
\(
\begin{aligned}
x_{C M}=\frac{2 u^2 \sin \theta \cos \theta}{g} & =\frac{2 \times 10^4 \times \frac{3}{5} \times \frac{4}{5}}{10} \mathrm{~m} \\
& =960 \mathrm{~m} .
\end{aligned}
\)
The centre of mass will hit the ground at this position. As the smaller block comes to rest after breaking, it falls down vertically and hits the ground at half of the range i.e., at \(x=480 \mathrm{~m}\). If the heavier block hits the ground at \(x_2\), then
\(
\begin{array}{rlrl}
x_{C M} & =\frac{m_1 x_1+m_2 x_2}{m_1+m_2} \\
\text { or, } & 960 \mathrm{~m} & =\frac{\frac{M}{4} \times 480 \mathrm{~m}+\frac{3 M}{4} \times x_2}{M} \\
\text { or, } & x_2 & =1120 \mathrm{~m} .
\end{array}
\)

Q24: A block of mass \(M\) is placed on the top of a bigger block of mass \(10 \mathrm{M}\) as shown in figure (9-W9). All the surfaces are frictionless. The system is released from rest. Find the distance moved by the bigger block at the instant the smaller block reaches the ground.

Answer: If the bigger block moves towards right by a distance \(X\), the smaller block will move towards left by a distance \((2 \cdot 2 \mathrm{~m}-X)\). Taking the two blocks together as the system, there is no horizontal external force on it. The centre of mass, which was at rest initially, will remain at the same horizontal position. Thus,
\(
\begin{aligned}
M(2 \cdot 2 \mathrm{~m}-X) & =10 M X \\
2 \cdot 2 \mathrm{~m} & =11 X \\
X & =0.2 \mathrm{~m} .
\end{aligned}
\)

Q25: The hero of a stunt film fires \(50 \mathrm{~g}\) bullets from a machine gun, each at a speed of \(1.0 \mathrm{~km} / \mathrm{s}\). If he fires 20 bullets in 4 seconds, what average force does he exert against the machine gun during this period?

Answer: The momentum of each bullet \(=(0.050 \mathrm{~kg})(1000 \mathrm{~m} / \mathrm{s})=50 \mathrm{~kg}-\mathrm{m} / \mathrm{s}\).
The gun is imparted this much of momentum by each bullet fired. Thus, the rate of change of momentum of the gun
\(
=\frac{(50 \mathrm{~kg}-\mathrm{m} / \mathrm{s}) \times 20}{4 \mathrm{~s}}=250 \mathrm{~N}
\)
In order to hold the gun, the hero must exert a force of \(250 \mathrm{~N}\) against the gun.

Q26: A block moving horizontally on a smooth surface with a speed of \(20 \mathrm{~m} / \mathrm{s}\) bursts into two equal parts continuing in the same direction. If one of the parts moves at \(30 \mathrm{~m} / \mathrm{s}\), with what speed does the second part move and what is the fractional change in the kinetic energy?

Answer: There is no external force on the block. Internal forces break the block in two parts. The linear momentum of the block before the break should, therefore, be equal to the linear momentum of the two parts after the break. As all the velocities are in same direction, we get,

\(
M(20 \mathrm{~m} / \mathrm{s})=\frac{M}{2}(30 \mathrm{~m} / \mathrm{s})+\frac{M}{2} v
\)
where \(v\) is the speed of the other part. From this equation \(v=10 \mathrm{~m} / \mathrm{s}\). The change in kinetic energy is
\(
\begin{gathered}
\frac{1}{2} \frac{M}{2}(30 \mathrm{~m} / \mathrm{s})^2+\frac{1}{2} \frac{M}{2}(10 \mathrm{~m} / \mathrm{s})^2-\frac{1}{2} M(20 \mathrm{~m} / \mathrm{s})^2 \\
=\frac{M}{2}(450+50-400) \frac{\mathrm{m}^2}{\mathrm{~s}^2}=\left(50 \frac{\mathrm{m}^2}{\mathrm{~s}^2}\right) M .
\end{gathered}
\)
Hence, the fractional change in the kinetic energy
\(
=\frac{M\left(50 \frac{\mathrm{m}^2}{\mathrm{~s}^2}\right)}{\frac{1}{2} M(20 \mathrm{~m} / \mathrm{s})^2}=\frac{1}{4} .
\)

Q27: A car of mass \(M\) is moving with a uniform velocity \(v\) on a horizontal road when the hero of a Hindi film drops himself on it from above. Taking the mass of the hero to be \(m\), what will be the velocity of the car after the event?

Answer: Consider the car plus the hero as the system. In the horizontal direction, there is no external force. Since the hero has fallen vertically, so his initial horizontal momentum \(=0\).

Initial horizontal momentum of the system \(=M v\) towards the right.

Finally, the hero sticks to the roof of the car, so they move with equal horizontal velocity say \(V\). Final horizontal momentum of the system
\(
=(M+m) V
\)
Hence,
\(
M v=(M+m) V
\)
or,
\(
V=\frac{M v}{M+m} .
\)

Q28: A space shuttle, while travelling at a speed of \(4000 \mathrm{~km} / \mathrm{h}\) with respect to the earth, disconnects and ejects a module backward, weighing one-fifth of the residual part. If the shuttle ejects the disconnected module at a speed of \(100 \mathrm{~km} / \mathrm{h}\) with respect to the state of the shuttle before the ejection, find the final velocity of the shuttle.

Answer: Suppose the mass of the shuttle including the module is \(M\). The mass of the module will be \(M / 6\). The total linear momentum before disconnection
\(
=M(4000 \mathrm{~km} / \mathrm{h}) .
\)
The velocity of the ejected module with respect to the earth \(=\) its velocity with respect to the shuttle + the velocity of the shuttle with respect to the earth
\(
=-100 \mathrm{~km} / \mathrm{h}+4000 \mathrm{~km} / \mathrm{h}=3900 \mathrm{~km} / \mathrm{h} \text {. }
\)
If the final velocity of the shuttle is \(V\) then the total final linear momentum
\(
=\frac{5 M}{6} V+\frac{M}{6} \times 3900 \mathrm{~km} / \mathrm{h} \text {. }
\)
By the principle of conservation of linear momentum,
\(
M(4000 \mathrm{~km} / \mathrm{h})=\frac{5 M}{6} V+\frac{M}{6} \times 3900 \mathrm{~km} / \mathrm{h}
\)
or,
\(
V=4020 \mathrm{~km} / \mathrm{h} \text {. }
\)

Q29: A boy of mass \(25 \mathrm{~kg}\) stands on a board of mass \(10 \mathrm{~kg}\) which in turn is kept on a frictionless horizontal ice surface. The boy makes a jump with a velocity component \(5 \mathrm{~m} / \mathrm{s}\) in a horizontal direction with respect to the ice. With what velocity does the board recoil? With what rate are the boy and the board separating from each other?

Answer: Consider the “board + boy” as a system. The external forces on this system are (a) weight of the system and (b) normal contact force by the ice surface. Both these forces are vertical and there is no external force in the horizontal direction. The horizontal component of linear momentum of the “board + boy” system is, therefore, constant.
If the board recoils at a speed \(v\),
\(
\begin{aligned}
& 0=(25 \mathrm{~kg}) \times(5 \mathrm{~m} / \mathrm{s})-(10 \mathrm{~kg}) v \\
& \text { or, } \quad v=12.5 \mathrm{~m} / \mathrm{s} \text {. } \\
&
\end{aligned}
\)
The boy and the board are separating with a rate
\(
5 \mathrm{~m} / \mathrm{s}+12.5 \mathrm{~m} / \mathrm{s}=17.5 \mathrm{~m} / \mathrm{s} \text {. }
\)

Q30: A man of mass \(m\) is standing on a platform of mass \(M\) kept on smooth ice. If the man starts moving on the platform with a speed \(v\) relative to the platform, with what velocity relative to the ice does the platform recoil?

Answer: Consider the situation shown in figure (9-W10).

uppose the man moves at a speed \(w\) towards right and the platform recoils at a speed \(V\) towards left, both relative to the ice. Hence, the speed of the man relative to the platform is \(V+w\). By the question,
\(
V+w=v, \text { or } w=v-V \dots(i)
\)
Taking the platform and the man to be the system, there is no external horizontal force on the system. The linear momentum of the system remains constant. Initially, both the man and the platform were at rest. Thus,
\(
\begin{array}{rlrl}
0 & =M V-m w \\
& \text { or, } \quad M V  =m(v-V) \text { [Using (i)] } \\
V & =\frac{m v}{M+m}
\end{array}
\)

Q31: A ball of mass \(m\), moving with a velocity \(v\) along \(X\)-axis, strikes another ball of mass \(2 m\) kept at rest. The first ball comes to rest after collision and the other breaks into two equal pieces. One of the pieces starts moving along \(Y\)-axis with a speed \(v_1\). What will be the velocity of the other piece?

Answer: The total linear momentum of the balls before the collision is \(m v\) along the \(X\)-axis. After the collision, momentum of the first ball \(=0\), momentum of the first piece \(=m v_1\) along the \(Y\)-axis and momentum of the second piece \(=m v_2\) along its direction of motion where \(v_2\) is the speed of the second piece. These three should add to \(m v\) along the \(X\)-axis, which is the initial momentum of the system.

Taking components along the \(X\)-axis,
\(
m v_2 \cos \theta=m v \dots(i)
\)
and taking components along the \(Y\)-axis,
\(
m v_2 \sin \theta=m v_1 \dots(ii)
\)
From (i) and (ii),
\(
v_2=\sqrt{v^2+v_1^2} \text { and } \tan \theta=v_1 / v .
\)

Q32: A bullet of mass \(50 \mathrm{~g}\) is fired from below into the bob of mass \(450 \mathrm{~g}\) of a long simple pendulum as shown in figure (9-W12). The bullet remains inside the bob and the bob rises through a height of \(1.8 \mathrm{~m}\). Find the speed of the bullet. Take \(g=10 \mathrm{~m} / \mathrm{s}^2\)

Answer: Let the speed of the bullet be \(v\). Let the common velocity of the bullet and the bob, after the bullet is embedded into the bob, is \(V\). By the principle of conservation of linear momentum,
\(
V=\frac{(0.05 \mathrm{~kg}) v}{0.45 \mathrm{~kg}+0.05 \mathrm{~kg}}=\frac{v}{10}
\)
The string becomes loose and the bob will go up with a deceleration of \(g=10 \mathrm{~m} / \mathrm{s}^2\). As it comes to rest at a height of \(1.8 \mathrm{~m}\), using the equation \(v^2=u^2+2 a x\),
\(
\begin{aligned}
1.8 \mathrm{~m} & =\frac{(v / 10)^2}{2 \times 10 \mathrm{~m} / \mathrm{s}^2} \\
v & =60 \mathrm{~m} / \mathrm{s} .
\end{aligned}
\)

Q33: A light spring of spring constant \(k\) is kept compressed between two blocks of masses \(m\) and \(M\) on a smooth horizontal surface (figure 9-W13). When released, the blocks acquire velocities in opposite directions. The spring loses contact with the blocks when it acquires natural length. If the spring was initially compressed through a distance \(x\), find the final speeds of the two blocks.

Answer: Consider the two blocks plus the spring to be the system. No external force acts on this system in the horizontal direction. Hence, the linear momentum will remain constant. As the spring is light, it has no linear momentum. Suppose the block of mass \(M\) moves with a speed \(V\) and the other block with a speed \(v\) after losing contact with the spring. As the blocks are released from rest, the initial momentum is zero. The final momentum is \(M V-m v\) towards the right. Thus,
\(
M V-m v=0 \quad \text { or, } \quad V=\frac{m}{M} v \dots(i)
\)
Initially, the energy of the system \(=\frac{1}{2} k x^2\).
Finally, the energy of the system \(=\frac{1}{2} m v^2+\frac{1}{2} M V^2\).
As there is no friction,
\(
\frac{1}{2} m v^2+\frac{1}{2} M V^2=\frac{1}{2} k x^2 \dots(ii)
\)
Using (i) and (ii),
\(
\begin{aligned}
& m v^2\left(1+\frac{m}{M}\right)=k x^2 \\
& v=\sqrt{\frac{k M}{m(M+m)}} x \\
& V=\sqrt{\frac{k m}{M(M+m)}} x . \\
&
\end{aligned}
\)

Q34: A block of mass \(m\) is connected to another block of mass \(M\) by a massless spring of spring constant \(k\). The blocks are kept on a smooth horizontal plane. Initially, the blocks are at rest and the spring is unstretched when a constant force \(F\) starts acting on the block of mass \(M\) to pull it. Find the maximum extension of the spring.

Answer: Let us take the two blocks plus the spring as the system. The centre of mass of the system moves with an acceleration \(a=\frac{F}{m+M}\). Let us work from a reference frame with its origin at the centre of mass. As this frame is accelerated with respect to the ground we have to apply a pseudo force \(m a\) towards left on the block of mass \(m\) and \(M a\) towards left on the block of mass \(M\). The net external force on \(m\) is
\(
F_1=m a=\frac{m F}{m+M} \text { towards left }
\)
and the net external force on \(M\) is
\(
F_2=F-M a=F-\frac{M F}{m+M}=\frac{m F}{m+M} \text { towards right. }
\)
The situation from this frame is shown in figure (9-W14b). As the centre of mass is at rest in this frame, the blocks move in opposite directions and come to instantaneous rest at some instant. The extension of the spring will be maximum at this instant. Suppose the left block is displaced through a distance \(x_1\) and the right block through a distance \(x_2\) from the initial positions. The total work done by the external forces \(F_1\) and \(F_2\) in this period are
\(
W=F_1 x_1+F_2 x_2=\frac{m F}{m+M}\left(x_1+x_2\right) .
\)
This should be equal to the increase in the potential energy of the spring as there is no change in the kinetic energy. Thus,
\(
\begin{aligned}
\frac{m F}{m+M}\left(x_1+x_2\right) & =\frac{1}{2} k\left(x_1+x_2\right)^2 \\
x_1+x_2 & =\frac{2 m F}{k(m+M)} .
\end{aligned}
\)
This is the maximum extension of the spring.

Q35: The two balls shown in Figure (9-W15) are identical, the first moving at a speed \(v\) towards the right and the second staying at rest. The wall at the extreme right is fixed. Assume all collisions to be elastic. Show that the speeds of the balls remain unchanged after all the collisions have taken place.

Answer: 1st collision: As the balls have equal mass and make elastic collision, the velocities are interchanged.
Hence, after the first collision, the ball \(A\) comes to rest and the ball \(B\) moves towards right at a speed \(v\).
2nd collision : The ball \(B\) moving with a speed \(v\), collides with the wall and rebounds. As the wall is rigid and may be taken to be of infinite mass, momentum conservation gives no useful result. The velocity of separation should be equal to the velocity of approach. Hence, the ball rebounds at the same speed \(v\) towards the left.
3rd collision: The ball \(B\) moving towards left at the speed \(v\) again collides with the ball \(A\) kept at rest. As the masses are equal and the collision is elastic, the velocities are interchanged. Thus, the ball \(B\) comes to rest and the ball \(A\) moves towards the left at a speed \(v\). No further collision takes place. Thus, the speeds of the balls remain the same as their initial values.

Q36: A block of mass \(m\) moving at a velocity \(v\) collides head on with another block of mass \(2 m\) at rest. If the coefficient of restitution is \(1 / 2\), find the velocities of the blocks after the collision.

Answer: Suppose after the collision the block of mass \(m\) moves at a velocity \(u_1\) and the block of mass \(2 m\) moves at a velocity \(u_2\). By conservation of momentum,
\(
m v=m u_1+2 m u_2 \dots(i)
\)
The velocity of separation is \(u_2-u_1\) and the velocity of approach is \(v\).
So,
\(
u_2-u_1=v / 2 \dots(ii)
\)
From (i) and (ii), \(u_1=0\) and \(u_2=v / 2\).

Q37: A block of mass \(1.2 \mathrm{~kg}\) moving at a speed of \(20 \mathrm{~cm} / \mathrm{s}\) collides head-on with a similar block kept at rest. The coefficient of restitution is \(3 / 5\). Find the loss of kinetic energy during the collision.

Answer: Suppose the first block moves at a speed \(v_1\) and the second at \(v_2\) after the collision. Since the collision is head-on, the two blocks move along the original direction of motion of the first block.
By conservation of linear momentum,
\(
(1 \cdot 2 \mathrm{~kg})(20 \mathrm{~cm} / \mathrm{s})=(1 \cdot 2 \mathrm{~kg}) v_1+(1 \cdot 2 \mathrm{~kg}) v_2
\)
or, \(v_1+v_2=20 \mathrm{~cm} / \mathrm{s} \dots(i)\)
The velocity of separation is \(v_2-v_1\) and the velocity of approach is \(20 \mathrm{~cm} / \mathrm{s}\). As the coefficient of restitution is \(3 / 5\), we have,
\(
v_2-v_1=(3 / 5) \times 20 \mathrm{~cm} / \mathrm{s}=12 \mathrm{~cm} / \mathrm{s} . \quad \ldots \text { (ii) }
\)
By (i) and (ii),
\(
v_1=4 \mathrm{~cm} / \mathrm{s} \text { and } v_2=16 \mathrm{~cm} / \mathrm{s} \text {. }
\)
The loss in kinetic energy is
\(
\begin{aligned}
& \frac{1}{2}(1.2 \mathrm{~kg})\left[(20 \mathrm{~cm} / \mathrm{s})^2-(4 \mathrm{~cm} / \mathrm{s})^2-(16 \mathrm{~cm} / \mathrm{s})^2\right] \\
& =(0.6 \mathrm{~kg})\left[0.04 \mathrm{~m}^2 / \mathrm{s}^2-0.0016 \mathrm{~m}^2 / \mathrm{s}^2-0.0256 \mathrm{~m}^2 / \mathrm{s}^2\right] \\
& =(0.6 \mathrm{~kg})\left(0.0128 \mathrm{~m}^2 / \mathrm{s}^2\right)=7.7 \times 10^{-3} \mathrm{~J} .
\end{aligned}
\)

Q38: A ball of mass \(m\) hits the floor with a speed \(v\) making an angle of incidence \(\theta\) with the normal. The coefficient of restitution is e. Find the speed of the reflected ball and the angle of reflection of the ball.

Answer: See figure (9-W16). Suppose the angle of reflection is \(\theta^{\prime}\) and the speed after the collision is \(v^{\prime}\). The floor exerts a force on the ball along the normal during the collision. There is no force parallel to the surface. Thus, the parallel component of the velocity of the ball remains unchanged. This gives
\(
v^{\prime} \sin \theta^{\prime}=v \sin \theta \dots(i)
\)
For the components normal to the floor, the velocity of separation \(=v^{\prime} \cos \theta^{\prime}\) and the velocity of approach \(=v \cos \theta\).
Hence,
\(
v^{\prime} \cos \theta^{\prime}=e v \cos \theta \dots(ii)
\)
From (i) and (ii),
\(
v^{\prime}=v \sqrt{\sin ^2 \theta+e^2 \cos ^2 \theta}
\)
\(
\text { and } \quad \tan \theta^{\prime}=\frac{\tan \theta}{e} \text {. }
\)
\(
\text { For elastic collision, } e=1 \text { so that } \theta^{\prime}=\theta \text { and } v^{\prime}=v \text {. }
\)

Q39: A block of mass \(m\) and a pan of equal mass are connected by a string going over a smooth light pulley as shown in figure (9-W17). Initially, the system is at rest when a particle of mass \(m\) falls on the pan and sticks to it. If the particle strikes the pan with a speed \(v\) find the speed with which the system moves just after the collision.

Answer: Let the required speed be \(V\).
As there is a sudden change in the speed of the block, the tension must change by a large amount during the collision.
Let \(N=\) magnitude of the contact force between the particle and the pan
\(
T=\text { tension in the string }
\)
Consider the impulse imparted to the particle. The force is \(N\) in upward direction and the impulse is \(\int N d t\). This should be equal to the change in its momentum.
\(
\text { Thus, } \quad \int N d t=m v-m V \dots(i)
\)
Similarly considering the impulse imparted to the pan,
\(
\int(N-T) d t=m V \dots(ii)
\)
and that to the block,
\(
\int T d t=m V \dots(iii)
\)
Adding (ii) and (iii),
\(
\int N d t=2 m V \text {. }
\)
Comparing with (i),
\(
m v-m V=2 m V
\)
or,
\(
V=v / 3 .
\)

Q40: Three particles of masses \(1.0 \mathrm{~kg}, 2.0 \mathrm{~kg}\) and \(3.0 \mathrm{~kg}\) are placed at the corners \(A, B\) and \(C\) respectively of an equilateral triangle \(A B C\) of edge \(1 \mathrm{~m}\). Locate the centre of mass of the system.

Answer:

\(
\begin{aligned}
& m_1=1 \mathrm{~kg}, \\
& m_2=2 \mathrm{~kg}, \\
& m_3=3 \mathrm{~kg}
\end{aligned}
\)
\(
\begin{aligned}
& \mathrm{x}_1=0 \\
& \mathrm{x}_2=1, \mathrm{x}_3=\frac{1}{2} \\
& \mathrm{y}_1=0, \mathrm{y}_2=0, \mathrm{y}_3=\frac{\sqrt{3}}{2}
\end{aligned}
\)
The position of centre of mass is
\(
\begin{aligned}
& \text { C. } M=\frac{\left(m_1 x_1+m_2 x_2+m_3 x_3\right)}{m_1+m_2+m_3}, \frac{\left(m_1 y_1+m_2 y_2+m_3 y_3\right)}{m_1+m_2+m_3} \\
& =\left(\frac{(1 \times 0)+(2 \times 1)+\left(3 \times \frac{1}{2}\right)}{1+2+3}, \frac{(1 \times 0)+(2 \times 0)+\left(3 \times \frac{\sqrt{3}}{2}\right)}{1+2+3}\right) \\
& =\left(\frac{7}{12}, \frac{\sqrt{3}}{4}\right) \text { from the point B. }
\end{aligned}
\)

Q41: A uniform disc of radius \(R\) is put over another uniform disc of radius \(2 R\) of the same thickness and density. The peripheries of the two discs touch each other. Locate the centre of mass of the system.

Answer:

Let the centre of the bigger disc be the origin.
\(2 \mathrm{R}=\) Radius of bigger disc
\(R=\) Radius of smaller disc
\(\mathrm{m}_1=\pi \mathrm{R}^2 \times \mathrm{T} \times \rho\)
\(m_2=\pi(2 R)^2\) \(\mathrm{T} \times \rho\)
where \(T=\) Thickness of the two discs
\(\rho=\) Density of the two discs
\(\therefore\) The position of the centre of mass
\(
\begin{aligned}
& \left(\frac{m_1 x_1+m_2 x_2}{m_1+m_2}, \frac{m_1 y_1+m_2 y_2}{m_1+m_2}\right) \\
& x_1=R \quad y_1=0 \\
& x_2=0 \quad y_2=0 \\
& \left(\frac{\pi R^2 T \rho R+0}{\pi R^2 T \rho+\pi(2 R)^2 T \rho}, \frac{0}{m_1+m_2}\right)=\left(\frac{\pi R^2 T \rho R}{5 \pi R^2 T \rho}, 0\right)=\left(\frac{R}{5}, 0\right)
\end{aligned}
\)
At R/5 from the centre of the bigger disc towards the centre of the smaller disc.

Q42: A disc of radius \(R\) is cut out from a larger disc of radius \(2 R\) in such a way that the edge of the hole touches the edge of the disc. Locate the centre of mass of the residual disc.

Answer:

Let ‘ O ‘ be the origin of the system.
\(\mathrm{R}=\) radius of the smaller disc
\(2 \mathrm{R}=\) radius of the bigger disc
The smaller disc is cut out from the bigger disc
As from the figure
\(
\begin{array}{lll}
\mathrm{m}_1=\pi \mathrm{R}^2 \mathrm{~T} \rho & \mathrm{x}_1=\mathrm{R} & \mathrm{y}_1=0 \\
\mathrm{~m}_2=\pi(2 \mathrm{R})^2 \mathrm{~T} \rho & \mathrm{x}_2=0 & \mathrm{y}_2=0
\end{array}
\)
The position of C.M. \(=\left(\frac{-\pi R^2 T \rho R+0}{-\pi R^2 T \rho+\pi(2 R)^2 T \rho R}, \frac{0+0}{m_1+m_2}\right)\)
\(
=\left(\frac{-\pi R^2 T \rho R}{3 \pi R^2 T \rho}, 0\right)=\left(-\frac{R}{3}, 0\right)
\)
C.M. is at R/3 from the centre of the bigger disc away from centre of the hole.

Q43: A square plate of edge \(d\) and a circular disc of diameter \(d\) are placed touching each other at the midpoint of an edge of the plate as shown in figure (9-Q2). Locate the centre of mass of the combination, assuming same mass per unit area for the two plates.

Answer:

Let \(m\) be the mass per unit area.
\(\therefore\) Mass of the square plate \(=M_1=d^2 m\)
Mass of the circular disc \(=M_2=\frac{\pi d^2}{4} m\)
Let the centre of the circular disc be the origin of the system.
\(
\begin{aligned}
& x_1=d, y_1=0 \\
& x_2=0, y_2=0
\end{aligned}
\)
\(\therefore\) Position of centre of mass
\(
=\left(\frac{m_1 x_1+m_2 x_2}{m_1+m_2}, \frac{m_1 y_1+m_2 y_2}{m_1+m_2}\right)
\)
\(
\begin{aligned}
& =\left(\frac{d^2 m d+\pi\left(d^2 / 4\right) m \times 0}{d^2 m+\pi\left(d^2 / 4\right) m}, \frac{0+0}{m_1+m_2}\right) \\
& =\left(\frac{d^3 m}{d^2 m(1+\pi / 4)}, 0\right) \\
& =\left(\frac{4 d}{\pi+4}, 0\right)
\end{aligned}
\)
Hence, the new centre of mass of the system (circular disc plus square plate) lies at distance \(\frac{4 d}{(\pi+4)}\) from the centre of the circular disc, towards the right.

Q44: Calculate the velocity of the centre of mass of the system of particles shown in Figure (9-E3).

Answer:

\(
\mathrm{m}_1=1 \mathrm{~kg}, \text { Velocity, } \vec{v}_1=\left(-1.5 \cos 37^{\circ} \hat{i}-1.5 \sin 37^{\circ} \hat{j}\right)=-1.2 \hat{i}-0.9 \hat{j}
\)
\(
\mathrm{m}_2=1.2 \mathrm{~kg}, \quad \overrightarrow{\mathrm{v}}_2=0.4 \hat{\mathrm{j}}
\)
\(
\mathrm{m}_3=1.5 \mathrm{~kg}, \quad \overrightarrow{\mathrm{v}}_3=-0.8 \hat{\mathrm{i}}+0.6 \hat{\mathrm{j}}
\)
\(
\mathrm{m}_4=0.5 \mathrm{~kg}, \quad \overrightarrow{\mathrm{v}}_4=3 \hat{\mathrm{i}}
\)
\(
\mathrm{m}_5=1 \mathrm{~kg}, \quad \overrightarrow{\mathrm{v}}_5=1.6 \hat{\mathrm{i}}-1.2 \hat{\mathrm{j}}
\)
\(
V_{c m}=\frac{m_1 \vec{v}_1+m_2 \vec{v}_2+m_3 \vec{v}_3+m_4 \vec{v}_4+m_5 \vec{v}_5}{m_1+m_2+m_3+m_4+m_5}
\)
On solving the above equation, we get:
\(V_{c m}\) is \(0.20 \mathrm{~m} / \mathrm{s}\), at \(45^{\circ}\) below the direction, towards right.

Q45: Find the centre of mass of a uniform plate having semicircular inner and outer boundaries of radii \(R_1\) and \(R_2\) (figure 9-E5).

Answer: Let the mass of the plate be \(M\).
Consider a small semicircular portion of mass \(\mathrm{dm}\) and radius \(r\), as shown in Fig.

\(
\mathrm{dm}=\frac{M \pi r d r}{\frac{\pi\left(R_2^2-R_1^2\right)}{2}}=\frac{M}{\frac{\left(R_2^2-R_1^2\right)}{2}} \mathrm{rdr}
\)
The centre of mass is given as:
\(
\begin{aligned}
& y_{c m}=\frac{\int y \mathrm{dm}}{M} \\
& y_{c m}=\int_{R_1}^{R_2}\left(\frac{2 r}{\pi}\right) \cdot \frac{M}{\frac{\left(R_2^2-R_1^2\right)}{2}} \times \frac{\mathrm{rdr}}{M} \\
& =\frac{2}{\pi \frac{\left(R_2^2-R_1^2\right)}{2}} \int_{R_1}^{R_2} r^2 d r \\
& =\frac{2}{\pi \frac{\left(R_2^2-R_1^2\right)}{2}}\left[\frac{\left(R_2^3-R_1^3\right)}{3}\right] \\
& =\frac{4\left(R_1^2+R_1 R_2+R_2^2\right)}{3 \pi\left(R_1+R_2\right)}
\end{aligned}
\)

Q46: A man of mass \(M\) having a bag of mass \(m\) slips from the roof of a tall building of height \(H\) and starts falling vertically (figure 9-E8). When at a height \(h\) from the ground, he notices that the ground below him is pretty hard, but there is a pond at a horizontal distance \(x\) from the line of fall. In order to save himself he throws the bag horizontally (with respect to himself) in the direction opposite to the pond. Calculate the minimum horizontal velocity imparted to the bag so that the man lands in the water. If the man just succeeds to avoid the hard ground, where will the bag land?

Answer:

Mass of \(\operatorname{man}=M\)
Initial velocity of the \(\operatorname{man}=0\)
Mass of bag \(=\mathrm{m}\)
Let the throws the bag towards left with a velocity \(v\) towards left. So, there is no external force in the horizontal direction. The momentum will be conserved. Let he goes right with a velocity \(\mathrm{V}\).
Using the law of conservation of momentum,
\(
\begin{aligned}
& m v=M V \\
& \Rightarrow v=\frac{M V}{m} \ldots(1)
\end{aligned}
\)
Let the total time he takes to reach the ground be
\(
\Rightarrow t_1=\sqrt{\frac{2 H}{g}}
\)
Let the total time he takes to reach the height \(\mathrm{h}\) be
\(
\Rightarrow t_2=\sqrt{\frac{2(H-h)}{g}}
\)
\(\therefore\) The time of flying in covering the remaining height \(h\) is,
\(
\begin{aligned}
& t=t_1-t_2 \\
& \Rightarrow t=\sqrt{\frac{2 H}{g}}-\sqrt{\frac{2(H-h)}{g}} \\
& =\sqrt{\frac{2}{g}}(\sqrt{H}-\sqrt{H-h})
\end{aligned}
\)
During this time, the man covers a horizontal distance \(\mathrm{x}\) and lands in the water.
\(
\begin{aligned}
& \Rightarrow x=V \times t \\
& \Rightarrow V=\frac{x}{t} \\
& \therefore v=\frac{M}{m} \frac{x}{t}[\text { using equation (1)] } \\
& =\frac{M}{m} \frac{x}{\sqrt{\frac{2}{g}}(\sqrt{H}-\sqrt{H-h})}
\end{aligned}
\)
Let the bag lands at a distance \(x\) ‘ towards the left from the actual line of fall.
As there is no external force in the horizontal direction, the \(x\)-coordinate of the centre of mass will remain the same.
\(
\begin{aligned}
& \Rightarrow 0=\frac{M \times(x)+m \times x^{\prime}}{M+m} \\
& \Rightarrow x^{\prime}=-\frac{M}{m} x
\end{aligned}
\)
Therefore, the bag will land at a distance
\(
\frac{M}{m} x
\)

Q47: A block at rest explodes into three equal parts. Two parts start moving along \(X\) and \(Y\) axes respectively with equal speeds of \(10 \mathrm{~m} / \mathrm{s}\). Find the initial velocity of the third part.

Answer: As the block is exploded only because of its internal energy, the net external force on the system is zero. Thus, the centre of mass of does not change.

Let the body was at the origin of the coordinate system during the explosion.
The resultant velocity of two bodies of equal mass moving at a speed of \(10 \mathrm{~m} / \mathrm{s}\) in \(+x\)-axis and \(+y\)-axis direction, is given as:
\(
v=\sqrt{10^2+10^2+2 \cdot 10 \cdot 10 \cos 90^{\circ}}=10 \sqrt{2} \mathrm{~m} / \mathrm{s}, 45^{\circ} \mathrm{w} \cdot \mathrm{r} \cdot \mathrm{t} \mathrm{~x} \text { – axis }
\)
If the centre of mass is at rest, the third part having equal mass as that of the other two masses will move in the opposite direction (i.e. \(135^{\circ}\) w.r.t. \(+x\)-axis) at the same velocity of \(10 \sqrt{2} \mathrm{~m} / \mathrm{s}\).

Q48: Two persons each of mass \(m\) are standing at the two extremes of a railroad car of mass \(M\) resting on a smooth track (figure 9-E10). The person on left jumps to the left with a horizontal speed \(u\) with respect to the state of the car before the jump. Thereafter, the other person jumps to the right, again with the same horizontal speed \(u\) with respect to the state of the car before his jump. Find the velocity of the car after both the persons have jumped off.

Answer: Two persons each of mass \(m\) are standing at the two extremes of a railroad car of mass \(m\) resting on a smooth track.
Case-I:
It is given that:
Mass of each persons \(=m\)
Mass of railroad car \(=\mathrm{M}\)
Let the velocity of the railroad w.r.t. earth, when the man on the left jumps off be V.
By the law of conservation of momentum:
\(
\begin{aligned}
& 0=-m u+(M+m) V \\
& \Rightarrow V=\left(\frac{m u}{M+m}\right), \text { towards right }
\end{aligned}
\)

Case II: When the man on the right jumps, the velocity of it w.r.t the car is \(u\).
\(
\begin{aligned}
& 0=m u-M V^{\prime} \\
& \Rightarrow V^{\prime}=\frac{m u}{M}
\end{aligned}
\)
( \(V^{\prime}\) is the change in velocity of the platform when the platform itself is taken as reference, assuming the car to be at rest.)
\(\therefore\) Net velocity towards the left, (i.e. the velocity of the car)
\(
\begin{aligned}
& V^{\prime}-V=\frac{m u}{M}-\frac{m u}{(M+m)} \\
& =\frac{m M u+m^2 u-M m u}{M(M+m)} \\
& \Rightarrow V^{\prime}-V=\frac{m^2 u}{M(M+m)}
\end{aligned}
\)

Q49: A bullet of mass \(m\) moving at a speed \(v\) hits a ball of mass \(M\) kept at rest. A small part having mass \(m^{\prime}\) breaks from the ball and sticks to the bullet. The remaining ball is found to move at a speed \(v_1\) in the direction of the bullet. Find the velocity of the bullet after the collision.

Answer: Mass of the bullet \(=\mathrm{m}\) and speed \(=v\)
Mass of the ball \(=M\)
\(\mathrm{m}^{\prime}=\) frictional mass from the ball.
The remaining mass of the ball moves with the velocity \(v_1\).
On applying the law of conservation of momentum, we get:
\(
m v+0=\left(m^{\prime}+m\right) v^{\prime}+\left(M-m^{\prime}\right) v_1
\)
where \(v^{\prime}=\) final velocity of the bullet + frictional mass
\(\Rightarrow v^{\prime}=\frac{m v-\left(M+m^{\prime}\right) v_1}{m+m^{\prime}}\)

Q50: Two friends \(A\) and \(B\) (each weighing \(40 \mathrm{~kg}\) ) are sitting on a frictionless platform some distance \(d\) apart. \(A\) rolls a ball of mass \(4 \mathrm{~kg}\) on the platform towards \(B\) which \(B\) catches. Then \(B\) rolls the ball towards \(A\) and \(A\) catches it. The ball keeps on moving back and forth between \(A\) and \(B\). The ball has a fixed speed of \(5 \mathrm{~m} / \mathrm{s}\) on the platform. (a) Find the speed of \(A\) after he rolls the ball for the first time. (b) Find the speed of \(A\) after he catches the ball for the first time. (c) Find the speeds of \(A\) and \(B\) after the ball has made 5 round trips and is held by \(A\). (d) How many times can \(A\) roll the ball? (e) Where is the centre of mass of the system ” \(A+B+\) ball” at the end of the \(n\)th trip?

Answer: It is given that:
Weight of \(A=\) Weight of \(B=40 \mathrm{~kg}\)
Velocity of ball \(=5 \mathrm{~m} / \mathrm{s}\)
(a) Case-1: Total momentum of the man A and the ball remains constant.
\(
\begin{aligned}
& \therefore 0=4 \times 5-40 \times v \\
& \Rightarrow v=0.5 \mathrm{~m} / \mathrm{s}, \text { towards left }
\end{aligned}
\)
(b) Case-2: When B catches the ball, the momentum between B and the ball remains constant.
\(
\begin{aligned}
& \Rightarrow 4 \times 5=44 \mathrm{v} \\
& \Rightarrow v=\left(\frac{20}{44}\right) \mathrm{m} / \mathrm{s}
\end{aligned}
\)
Case-3: When B throws the ball,
On applying the law of conservation of linear momentum, we get:
\(\Rightarrow 44 \times\left(\frac{20}{44}\right)=-4 \times 5+40 \times v\)
\(\Rightarrow v=1 \mathrm{~m} / \mathrm{s}\), towards right)
Case-4: When A catches the ball,
Applying the law of conservation of linear momentum, we get:
\(-4 \times 5+(-0.5) \times 40=44 v\)
\(\Rightarrow v=\frac{40}{44}=\frac{10}{11} \mathrm{~m} / \mathrm{s}\), towards left
(c) Case-5: When A throws the ball,
Applying the law of conservation of linear momentum, we get:
\(
\begin{aligned}
& 44 \times\left(\frac{10}{11}\right)=4 \times 5+40 \times v \\
& \Rightarrow v=\frac{60}{40}=\frac{3}{2} \mathrm{~m} / \mathrm{s} \text { towards left) }
\end{aligned}
\)
Case- 6 : When \(B\) receives the ball,
Applying the law of conservation of linear momentum, we get:
\(
\begin{aligned}
& 40 \times 1+4 \times 5=44 \times v \\
& \Rightarrow v=\frac{60}{44} \mathrm{~m} / \mathrm{s}, \text { towards right }
\end{aligned}
\)
Case-7: When B throws the ball,
On applying the law of conservation of linear momentum, we get:
\(
\Rightarrow v=44 \times\left(\frac{60}{44}\right) \mathrm{m} / \mathrm{s} \text {, towards right }
\)
Case-8: When A catches the ball,
On applying the law of conservation of linear momentum, we get:
\(
\begin{aligned}
& -4 \times 5+40\left(\frac{3}{2}\right)=-44 v \\
& \Rightarrow V=-\frac{40}{44}=\left(\frac{10}{11}\right) \mathrm{m} / \mathrm{s} \text {, towards left }
\end{aligned}
\)
Similarly, after 5 round trips,
The velocity of \(A\) will be \(\left(\frac{50}{11}\right) \mathrm{m} / \mathrm{s}\) and the velocity of \(B\) will be \(5 \mathrm{~m} / \mathrm{s}\).
(d) As after 6 round trips, the velocity of A becomes \(\frac{60}{11}>5 \mathrm{~m} / \mathrm{s}\), it cannot catch the ball. Thus, A can only roll the ball six times.
(e) Let the ball and the body \(\mathrm{A}\) be at the origin, in the initial position.
\(
\begin{aligned}
& \therefore X_c=\frac{40 \times 0+4 \times 0+40 \times d}{40+40+4} \\
& =\frac{10}{21} d
\end{aligned}
\)

Q51: In a gamma decay process, the internal energy of a nucleus of mass \(M\) decreases, a gamma photon of energy \(E\) and linear momentum \(E / c\) is emitted and the nucleus recoils. Find the decrease in internal energy.

Answer:

Let the nucleus recoils with a velocity \(\mathrm{v}\).
Applying the law of conservation of linear momentum, we get:
Linear momentum of recoiled nucleus \(=\) Linear momentum of gamma photon
\(\Rightarrow \mathrm{mv}=\frac{E}{c}\)
\(\therefore v=\frac{E}{m c}\)
Kinetic energy of the recoiled nucleus \(=\frac{1}{2} M v^2\)
\(\Rightarrow K . E .=\frac{1}{2} m\left(\frac{E}{m c}\right)^2=\frac{1}{2} \frac{E^2}{m c^2}\)
Energy limited by Gamma photon \(=\mathrm{E}\).
Decrease in the internal energy = photon energy + the kinetic energy of the recoiled nucleus
\(\Rightarrow\) Decrease in the internal energy \(=E+\frac{E^2}{2 m c^2}\)

Q52: A projectile is fired with a speed \(u\) at an angle \(\theta\) above a horizontal field. The coefficient of restitution of collision between the projectile and the field is \(e\). How far from the starting point, does the projectile makes its second collision with the field?

Answer:

Given:
The initial velocity of the projectile \(=u\)
The angle of projection of the projectile with respect to the ground \(=\theta\)
When the projectile hits the ground for the first time, the velocity remains the same i.e. \(u\). The component of velocity parallel to the ground, \(u \cos \theta\) should remain constant.
However, the vertical component of the projectile undergoes a change after the collision.
\(
\Rightarrow \mathrm{e}=\frac{\mathrm{u} \sin \theta}{\mathrm{v}} \Rightarrow \mathrm{v}=\mathrm{eu} \sin \theta \dots(1)
\)
Therefore, for the second projectile motion, the Velocity of projection ( \(\left.u^{\prime}\right)\) will be,
\(
u^{\prime}=\sqrt{(u \cos \theta)^2+(e u \sin \theta)^2}
\)
Angle of projection, \(\alpha=\tan ^{-1}\left(\frac{e u \sin \theta}{u \cos \theta}\right)\)
\(
\Rightarrow \alpha=\tan ^{-1}(e \tan \theta)
\)
or, \(\tan \alpha=e \tan \theta \ldots(2)\)
Also, \(\mathrm{y}=x \tan \alpha-\left(\frac{g x^2 \sec ^2 \alpha}{2 u^{\prime 2}}\right) \dots(3)\)
Here, \(y=0\)
\(
\begin{aligned}
& \therefore \tan \alpha=e \tan \theta \\
& \Rightarrow \sec ^2 \alpha=1+e^2 \tan ^2 \theta
\end{aligned}
\)
and \(u^{\prime 2}=u^2 \cos ^2 \theta+e^2 u^2 \sin ^2 \theta\)
Putting the above-calculated values in equation (3), we get:
\(
\text { xe } \tan \theta=\frac{g x^2\left(1+e^2 \tan ^2 \theta\right)}{2 u^2\left(\cos ^2 \theta+e^2 \sin ^2 \theta\right)}
\)
or, \(x=\frac{2 e u^2 \tan \theta\left(\cos ^2 \theta+e^2 \sin ^2 \theta\right)}{g\left(1+e^2 \tan ^2 \theta\right)}\)
\(
\begin{aligned}
& \Rightarrow x=\frac{2 e u^2 \tan \theta .}{g} \\
& \Rightarrow x=\frac{e u^2 \sin 2 \theta}{g}
\end{aligned}
\)
Thus, from the starting point, the projectile makes its second collision with the field at a distance,
\(
\begin{aligned}
& x^{\prime}=\frac{u^2 \sin 2 \theta}{g}+\frac{e u^2 \sin 2 \theta}{g} \\
& \Rightarrow x^{\prime}=\frac{u^2 \sin 2 \theta}{g}(1+e)
\end{aligned}
\)

Q53: A ball falls on an inclined plane of inclination \(\theta\) from a height \(h\) above the point of impact and makes a perfectly elastic collision. Where will it hit the plane again?

Answer:

Angle inclination of the plane \(=\theta\)
\(M\) the body falls through a height of \(h\),
The striking velocity of the projectile with the inclined plane \(v=\sqrt{2 \mathrm{gh}}\)
Now, the projectile makes on angle \(\left(90^{\circ}-2 \theta\right)\)
Velocity of projection \(=u=\sqrt{2 g h}\)
Let \(A B=\ell\).
So, \(x=\ell \cos \theta, y=-\ell \sin \theta\)
From the equation of trajectory,
\(
y=x \tan \alpha-\frac{g^2 \sec ^2 \alpha}{2 u^2}
\)
\(
-\ell \sin \theta=\ell \cos \theta \cdot \tan \left(90^{\circ}-2 \theta\right)-\frac{g \times \ell^2 \cos ^2 \theta \sec ^2\left(90^{\circ}-2 \theta\right)}{2 \times 2 g h}
\)
\(
\Rightarrow-\ell \sin \theta=\ell \cos \theta \cdot \cot 2 \theta-\frac{g \ell^2 \cos ^2 \theta \operatorname{cosec}^2 2 \theta}{4 \mathrm{gh}}
\)
\(
\text { So, } \frac{\ell \cos ^2 \theta \operatorname{cosec}^2 2 \theta}{4 h}=\sin \theta+\cos \theta \cot 2 \theta
\)
\(
\Rightarrow \ell=\frac{4 h}{\cos ^2 \theta \operatorname{cosec}^2 2 \theta}(\sin \theta+\cos \theta \cot 2 \theta)=\frac{4 h \times \sin ^2 2 \theta}{\cos ^2 \theta}\left(\sin \theta+\cos \theta \times \frac{\cos 2 \theta}{\sin 2 \theta}\right)
\)
\(
=\frac{4 \mathrm{~h} \times 4 \sin ^2 \theta \cos ^2 \theta}{\cos ^2 \theta}\left(\frac{\sin \theta \times \sin 2 \theta+\cos \theta \cos 2 \theta}{\sin 2 \theta}\right)
\)
\(
=16 \mathrm{~h} \sin ^2 \theta \times \frac{\cos \theta}{2 \sin \theta \cos \theta}=8 \mathrm{~h} \sin \theta
\)

Q54: Solve the previous problem if the coefficient of restitution is \(e\). Use \(\theta=45^{\circ}, e=\frac{3}{4}\) and \(h=5 \mathrm{~m}\).

Answer:

\(
\mathrm{h}=5 \mathrm{~m}, \quad \theta=45^{\circ}, \quad \mathrm{e}=(3 / 4)
\)
Here the velocity with which it would strike \(=v=\sqrt{2 g \times 5}=10 \mathrm{~m} / \mathrm{sec}\)
After the collision, let it make an angle \(\beta\) with horizontal. The horizontal component of velocity \(10 \cos 45^{\circ}\) will remain unchanged and the velocity in the perpendicular direction to the plane after collision will be
\(
\Rightarrow \mathrm{V}_{\mathrm{y}}=\mathrm{e} \times 10 \sin 45^{\circ}
\)
\(
\begin{aligned}
& =(3 / 4) \times 10 \times \frac{1}{\sqrt{2}}=(3.75) \sqrt{2} \mathrm{~m} / \mathrm{sec} \\
& V_x=10 \cos 45^{\circ}=5 \sqrt{2} \mathrm{~m} / \mathrm{sec}
\end{aligned}
\)
\(
\text { So, } u=\sqrt{\mathrm{V}_{\mathrm{x}}^2+\mathrm{V}_{\mathrm{y}}^2}=\sqrt{50+28.125}=\sqrt{78.125}=8.83 \mathrm{~m} / \mathrm{sec}
\)
\(
\text { Angle of reflection from the wall } \beta=\tan ^{-1}\left(\frac{3.75 \sqrt{2}}{5 \sqrt{2}}\right)=\tan ^{-1}\left(\frac{3}{4}\right)=37^{\circ}
\)
\(\Rightarrow\) Angle of projection \(\alpha=90-(\theta+\beta)=90-\left(45^{\circ}+37^{\circ}\right)=8^{\circ}\) Let the distance where it falls \(=\ell\)
\(
\Rightarrow \mathrm{x}=\ell \cos \theta, \mathrm{y}=-\ell \sin \theta
\)
Angle of projection \((\alpha)=8^{\circ}\)
Using equation of trajectory, \(\mathrm{y}=\mathrm{x} \tan \alpha-\frac{\mathrm{gx}^2 \sec ^2 \alpha}{2 \mathrm{u}^2}\)
\(
\begin{aligned}
& \Rightarrow-\ell \sin \theta=\ell \cos \theta \times \tan 8^{\circ}-\frac{g}{2} \times \frac{\ell \cos ^2 \theta \sec ^2 8^{\circ}}{u^2} \\
& \Rightarrow-\sin 45^{\circ}=\cos 45^{\circ}-\tan 8^{\circ}-\frac{10 \cos ^2 45^{\circ} \sec 8^{\circ}}{(8.83)^2}(\ell)
\end{aligned}
\)
Solving the above equation we get, \(\ell=18.5 \mathrm{~m}\).

Q55: Two masses \(m_1\) and \(m_2\) are connected by a spring of spring constant \(k\) and are placed on a frictionless horizontal surface. Initially, the spring is stretched through a distance \(x_0\) when the system is released from rest. Find the distance moved by the two masses before they again come to rest.

Answer:

It is given that two blocks of masses \(\mathrm{m}_1\) and \(\mathrm{m}_2\) are connected with a spring having spring constant \(\mathrm{k}\). Initially, the spring is stretched by a distance \(x_0\).
For the block to come to rest again, Let the distance travelled by \(m_1\) be \(x_1\) (towards right), and that travelled by \(m_2\) be \(x_2\) towards the left.
As no external force acts in the horizontal direction, we can write:
\(
m_1 x_1=m_2 x_2 \dots(1)
\)
As the energy is conserved in the spring, we get:
\(
\begin{aligned}
& \left(\frac{1}{2}\right) k x_0^2=\left(\frac{1}{2}\right) k\left(x_1+x_2-x_0\right)^2 \\
& \Rightarrow x_0=x_1+x_2-x_0 \\
& \Rightarrow x_1+x_2=2 x_0 \ldots(2)
\end{aligned}
\)
\(
\therefore x_1=2 x_0-x_2
\)
Putting this value in equation (1), we get :
\(
\begin{aligned}
& m_1\left(2 x_0-x_0\right)=m_2 x_2 \\
& \Rightarrow 2 m_1 x_0-m_1 x_2=m_2 x_2 \\
& \Rightarrow x_2=\frac{2 m_2}{m_1+m_2} x_0 \\
& \text { Similarly, } x_1=\left(\frac{2 m_2}{m_1+m_2}\right) x_0
\end{aligned}
\)

Q56: Two blocks of masses \(m_1\) and \(m_2\) are connected by a spring of spring constant \(k\) (figure 9-E15). The block of mass \(m_2\) is given a sharp impulse so that it acquires a velocity \(v_0\) towards the right. Find (a) the velocity of the centre of mass, (b) the maximum elongation that the spring will suffer.

Answer: Given,
Velocity of mass, \(m_2=v_0\)
Velocity of mass, \(m_1=0\)

(a) Velocity of centre of mass is given by,
\(
\begin{aligned}
& v_{c m}=\frac{m_1 v_1+m_2 v_2}{m_1+m_2} \\
& \Rightarrow v_{c m}=\frac{m_1 \times 0+m_2 \times v_0}{m_1+m_2} \\
& \Rightarrow v_{c m}=\frac{m_2 v_0}{m_1+m_2}
\end{aligned}
\)

(b) Let the maximum elongation in spring be \(x\).
The spring attains maximum elongation when the velocities of both blocks become equal to the velocity of centre of mass.
i.e. \(v_1=v_2=v_{c m}\)
On applying the law of conservation of energy, we can write:
Change in kinetic energy \(=\) Potential energy stored in spring
\(
\begin{aligned}
& \Rightarrow \frac{1}{2} m_2 v_0^2-\frac{1}{2}\left(m_1+m_2\right)\left(\frac{m_2 v_0}{m_1+m_2}\right)^2=\frac{1}{2} k x^2 \\
& \Rightarrow m_2 v_0^2\left(1-\frac{m_2}{m_1+m_2}\right)=k x^2 \\
& \Rightarrow=V_o\left[\frac{m_1 m_2}{\left(m_1+m_2\right) k}\right]^{\frac{1}{2}}
\end{aligned}
\)

Q57: Consider the situation of the previous problem. Suppose each of the blocks is pulled by a constant force \(F\) instead of any impulse. Find the maximum elongation that the spring will suffer and the distances moved by the two blocks in the process.

Answer:

It is given that the force on both the blocks is \(F\).
Let \(x_1\) and \(x_2\) be the extensions of blocks \(m_1\) and \(m_2\) respectively.
Total work done by the forces on the blocks \(=\mathrm{Fx}_1+\mathrm{Fx}_2 \dots(1)\)
\(\therefore\) Increase in the potential energy of spring \(=\left(\frac{1}{2}\right) K\left(x_1+x_2\right)^2 \ldots(2)\)
Equating the equations (1) and (2), we get:
\(
\Rightarrow\left(x_1+x_2\right)=\left(\frac{2 F}{k}\right) \ldots(3)
\)
As the net external force on the system is zero, the centre of mass does not shift.
\(
\therefore m_1 x_1=m_2 x_2 \ldots \text { (4) }
\)
Using the equations (3) and (4), we get :
\(
\begin{aligned}
& \frac{m_1}{m_2} x_1+x_1=\frac{2 F}{k} \\
& \Rightarrow x_1\left(1+\frac{m_1}{m_2}\right)=\frac{2 F}{k} \\
& \Rightarrow x_1=\frac{2 F m_2}{k\left(m_1+m_2\right)}
\end{aligned}
\)
Similarly,
\(
x_2=\frac{2 F m_1}{k\left(m_1+m_2\right)}
\)

Q58: Consider the situation of the previous problem. Suppose the block of mass \(m_1\) is pulled by a constant force \(F_1\) and the other block is pulled by a constant force \(F_2\). Find the maximum elongation that the spring will suffer.

Answer: Given:
Force on the block of mass, \(m_1=F_1\)
Force on the block of mass, \(m_2=F_2\)

Let the acceleration produced in mass \(m_1\) be \(a_1\).
\(
a_1=\frac{F_1-F_2}{m_1+m_2}
\)
Let the acceleration of mass \(m_2\) be \(a_2\).
\(
a_2=\frac{F_2-F_1}{m_1+m_2}
\)
Due to the force \(F_2\), the mass \(m_1\) experiences a pseudo force.

\(\therefore\) Net force on \(\mathrm{m}_1=F_1+m_1 a_2\)
\(
\begin{aligned}
& F^{\prime}=F_1+m_1 \times \frac{\left(F_2-F_1\right)}{m_1+m_2} \\
& =\frac{m_1 F_1+m_2 F_1+m_1 F_2-m_1 F_1}{m_1+m_2} \\
& =\frac{m_2 F_1+m_1 F_2}{m_1+m_2}
\end{aligned}
\)
Similarly, mass \(m_2\) experiences a pseudo force due to force \(F_1\).

\(
\begin{aligned}
& \therefore \text { Net force on } \mathrm{m}_2=F_2+m_2 a_1 \\
& F \prime \prime=F_2+m_2 \times \frac{\left(F_1-F_2\right)}{m_1+m_2} \\
& =\frac{m_1 F_2+m_2 F_2+m_2 F_1-m_2 F_2}{m_1+m_2} \\
& =\frac{m_1 F_2+m_2 F_1}{m_1+m_2}
\end{aligned}
\)
Let \(\mathrm{m}_1\) be displaced by a distance \(\mathrm{x}_1\) and \(\mathrm{m}_2\) be displaced by a distance \(\mathrm{x}_2\).
Therefore, the maximum elongation of the spring \(=x_1+x_2\)
Work done by the blocks \(=\) Energy stored in the spring
\(
\begin{aligned}
& \Rightarrow \frac{m_2 F_1+m_1 F_2}{m_1+m_2} \times x_1 \times \frac{m_2 F_1+m_1 F_2}{m_1+m_2} \times x_2=\left(\frac{1}{2}\right) k\left(x_1+x_2\right)^2 \\
& \Rightarrow x_1+x_2=\frac{2}{k}\left(\frac{m_1 F_2+m_2 F_1}{m_1+m_2}\right)
\end{aligned}
\)

Q59: The track shown in Figure (9-E16) is frictionless. The block \(B\) of mass \(2 m\) is lying at rest and the block \(A\) of mass \(m\) is pushed along the track with some speed. The collision between \(A\) and \(B\) is perfectly elastic. With what velocity should the block \(A\) be started to get the sleeping man awakened?

Answer: Mass of the block, \(A=m\)
Mass of the block, \(B=2 m\)
Let the initial velocity of block \(A\) be \(u_1\) and the final velocity of block \(A\), when it reaches the block \(B\) be \(v_1\).
Using the work-energy theorem for block A, we can write:
The gain in kinetic energy = Loss in potential energy
\(
\begin{aligned}
& \therefore\left(\frac{1}{2}\right) m v_1^2-\left(\frac{1}{2}\right) m u_1^2=m g h \\
& \Rightarrow v_1^2-u_1^2=2 g h \\
& \Rightarrow v_1=\sqrt{2 g h+u_1^2} \dots(1)
\end{aligned}
\)
Let the block B just manages to reach the man’s head. i.e. the velocity of block \(B\) is zero at that point.
Again, applying the work-energy theorem for block B, we get:
\(
\begin{aligned}
& \left(\frac{1}{2}\right) \times 2 m \times(0)^2-\left(\frac{1}{2}\right) \times 2 m \times v^2=m g h \\
& \Rightarrow v=\sqrt{2 g h}
\end{aligned}
\)
Therefore, before the collision :
Velocity of \(\mathrm{A}, \mathrm{u}_A=v_1\)
Velocity of \(\mathrm{B}, \mathrm{u}_B=0\)
After the collision:
Velocity of \(\mathrm{A}, \mathrm{v}_A=v\) ( say )
Velocity of \(\mathrm{B}, \mathrm{v}_B=\sqrt{2 g h}\)
As the collision is elastic, K.E. and momentum are conserved.
\(
\begin{aligned}
& m v_1+2 m \times 0=m v+2 m \sqrt{2 g h} \\
& \Rightarrow v_1-v=2 \sqrt{2 g h} \ldots(2)
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow\left(\frac{1}{2}\right) m v_1^2+\left(\frac{1}{2}\right) 2 m \times(0)^2=\left(\frac{1}{2}\right) m v^2+\left(\frac{1}{2}\right) 2 m(\sqrt{2 g h})^2 \\
& \Rightarrow v_1^2-v^2=2 \times \sqrt{2 g h} \times \sqrt{2 g h} \ldots(3)
\end{aligned}
\)
Dividing equation (3) by equation (2), we get:
\(
v_1+v=\sqrt{2 g h} \ldots(4)
\)
Adding the equations (4) and (2), we get:
\(
2 v_1=3 \sqrt{2 g h}
\)
Now using equation (1) to substitute the value of \(v_1\), we get:
\(
\begin{aligned}
& \sqrt{2 g h+u^2}=\left(\frac{3}{2}\right) \sqrt{2 g h} \\
& \Rightarrow 2 g h+u^2=\left(\frac{9}{4}\right)(2 g h) \\
& \Rightarrow u=\sqrt{2.5 g h}
\end{aligned}
\)
Block a should be started with a minimum velocity of \(\sqrt{2.5 g h}\)
So the block will travel with a velocity greater than \(2.5 \sqrt{2 \mathrm{gh}}\) so awake the man by \(B\).

Q60: A bullet of mass \(10 \mathrm{~g}\) moving horizontally at a speed of \(50 \sqrt{7} \mathrm{~m} / \mathrm{s}\) strikes a block of mass \(490 \mathrm{~g}\) kept on a frictionless track as shown in figure (9-E17). The bullet remains inside the block and the system proceeds towards the semicircular track of radius \(0.2 \mathrm{~m}\). Where will the block strike the horizontal part after leaving the semicircular track? 

Answer:

The mass of the block is \(490 \mathrm{gm}\),
The mass of the bullet is \(10 \mathrm{gm}\).
Since the bullet is embedded inside the block, it is a plastic collision.
The initial velocity of the bullet is
\(
\mathrm{v}_1=50 \sqrt{7} \mathrm{~m} / \mathrm{s}
\)
The velocity of the block is \(v_2=0\).
Let the final velocity of both is \(\mathrm{v}_{\mathrm{A}}\).
Therefore, by the law of conservation of momentum
\(
m_1 \times v_1+m_2 \times v_2=\left(m_1+m_2\right) V_A
\)
\(
\begin{aligned}
& 10 \times 10^{-3} \times 50 \sqrt{7}+490 \times 10^{-3} \times 0=(490+10) \times 10^{-3} \times \mathrm{v}_{\mathrm{A}} \\
& \mathrm{v}_{\mathrm{A}}=\sqrt{7} \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
When the block losses contact at ‘ \(\mathrm{D}\) ‘ the component mg will act on it.
Angular momentum acting at point \(D\)
\(
\begin{aligned}
& \frac{m\left(v_B\right)^2}{r}=m g \sin \theta \\
& \left(v_B\right)^2=\operatorname{gr} \sin \theta \dots(1)
\end{aligned}
\)
Putting work energy principle
\(
\begin{aligned}
& \frac{1}{2} \mathrm{~m}\left(\mathrm{v}_{\mathrm{B}}\right)^2-\frac{1}{2} \mathrm{~m}\left(\mathrm{v}_{\mathrm{A}}\right)^2=-\mathrm{mgh} \\
& \operatorname{since} \mathrm{h}=(\mathrm{r}+\mathrm{r} \sin \theta) \\
& \frac{1}{2} \mathrm{gr} \sin \theta-\frac{1}{2} \times 7=\mathrm{g}(0.2+0.2 \sin \theta) \\
& 3.5-\frac{1}{2} \times 9.8 \times 0.2 \times \sin \theta=9.8 \times 0.2(1+\sin \theta) \\
& 3.5-0.98 \sin \theta=1.96+1.96 \sin \theta \\
& \sin \theta=\frac{1}{2} \\
& \sin \theta=\sin 30^{\circ} \\
& \theta=30^{\circ}
\end{aligned}
\)
Therefore, the angle of projection will be
\(
90^{\circ}-30^{\circ}=60^{\circ}
\)
Then, the time taken to reach the ground will be \(\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}\)
\(
\sqrt{\frac{2 \times(0.2+0.2 \sin \theta)}{9.8}}=0.247 \mathrm{sec}
\)
Thus, distance travelled in the horizontal direction
\(
\begin{aligned}
& \mathrm{s}=\mathrm{v}_{\mathrm{B}} \times \mathrm{t} \\
& \mathrm{s}=\sqrt{\operatorname{gr} \sin \theta} \times \mathrm{t} \\
& \mathrm{s}=\sqrt{9.8 \times 0.2 \times\left(\frac{1}{2}\right)} \times 0.247 \\
& \mathrm{~s}=0.196 \mathrm{~m}
\end{aligned}
\)
So, the total distance will be
\(
\left(0.2-0.2 \cos 30^{\circ}\right)+0.196=0.22 m
\)
Hence, at the junction of the straight and the curved parts, the block will strike the horizontal part after leaving the semicircular track.

Q61: Two balls having masses \(m\) and \(2 m\) are fastened to two light strings of the same length \(l\) (figure 9-E18). The other ends of the strings are fixed at \(O\). The strings are kept in the same horizontal line and the system is released from rest. The collision between the balls is elastic. (a) Find the velocities of the balls just after their collision. (b) How high will the balls rise after the collision?

Answer: 

Given that two balls having masses \(\mathrm{m}\) and \(2 \mathrm{~m}\) are fastened to two light strings of the same length \(l\),
Let the velocity of \(m\) reaching at lower end is \(\mathrm{V}_1\),
From the work-energy principle
\(
\begin{aligned}
& \frac{1}{2} \mathrm{mv}_1^2-\frac{1}{2} \mathrm{~m} \times 0^2=\mathrm{mgl} \\
& \mathrm{V}_1=\sqrt{2 \mathrm{gl}}
\end{aligned}
\)
Similarly, the velocity of the heavy block will be
\(
V_2=\sqrt{2 g l}
\)
Therefore,
\(
\mathrm{V}_1=-\mathrm{V}_2=\mathrm{u}(\text { say) }
\)
Since both the balls act in opposite directions.
(a) Let the final velocity of \(m\) and \(2 m\) is \(v_1\) and \(v_2\) respectively.
According to the law of conservation of momentum
\(
\begin{aligned}
& m v_1+2 m v_2=m v+2 m v_2 \\
& m u-2 m u=m v_1+2 m v_2 \\
& -u=v_1+2 v_2 \dots(1)
\end{aligned}
\)
Again,
\(
\begin{aligned}
& \mathrm{v}_1-\mathrm{v}_2=-\left(\mathrm{v}_1-\mathrm{v}_2\right) \\
& \mathrm{v}_1-\mathrm{v}_2=-(\mathrm{u}-(-\mathrm{u})) \\
& \mathrm{v}_1-\mathrm{v}_2=-2 \mathrm{u} \dots(2)
\end{aligned}
\)
Subtracting equation (2) from (1)
\(
3 v_2=u
\)
\(
\mathrm{v}_2=\frac{\mathrm{u}}{3}
\)
\(
\mathrm{v}_2=\frac{\sqrt{(2 \mathrm{gl})}}{3}
\)
Putting in equation (2)
\(
\mathrm{v}_1=-2 \mathrm{u}+2 \mathrm{v}_2
\)
\(
\mathrm{v}_1=-2 \sqrt{(2 \mathrm{gl})}+\left(\frac{\sqrt{(2 \mathrm{gl})}}{3}\right)
\)
\(
\mathrm{v}_1=\left(\frac{-6 \sqrt{(2 \mathrm{gl})}}{3}\right)+\left(\frac{\sqrt{(2 \mathrm{gl})}}{3}\right)
\)
\(
\mathrm{v}_1=\left(\frac{-5 \sqrt{(2 \mathrm{gl})}}{3}\right)
\)
\(
\mathrm{v}_1=\left(\frac{-\sqrt{(50 \mathrm{gl})}}{3}\right)
\)
Hence, light ball will have velocity \(\mathrm{v}_1=\left(\frac{-\sqrt{(50 \mathrm{gl})}}{3}\right)\) towards left
And heavy ball will velocity \(\mathrm{V}_2=\frac{\sqrt{(2 \mathrm{gl})}}{3}\).
(b) Let \(h\) be the height gone by heavy ball
Putting work energy principle
\(
\frac{1}{2} \times 2 \mathrm{~m} \times 0^2-\frac{1}{2} \times 2 \mathrm{~m} \times \mathrm{v}_2^2=-2 \mathrm{mgh}
\)
\(
\begin{aligned}
& \frac{1}{2} \times \frac{2 g l}{9}=g h \\
& h=\frac{l}{9}
\end{aligned}
\)
Similarly, \(h^{\prime}\) be the height reached by light ball
\(
\frac{1}{2} \times \mathrm{m} \times 0^2-\frac{1}{2} \times \mathrm{m} \times \mathrm{v}_1^2=\mathrm{mgh}^{\prime}
\)
\(
\frac{1}{2} \times \frac{50 \mathrm{gl}}{9}=\mathrm{gh}^{\prime}
\)
\(
\mathrm{h}^{\prime}=\frac{25 {l}}{9}
\)
So, the balls will rise of following height after the collision

Q62: A block of mass \(m\) is placed on a triangular block of mass \(M\), which in turn is placed on a horizontal surface as shown in figure (9-E21). Assuming frictionless surfaces find the velocity of the triangular block when the smaller block reaches the bottom end.

Answer: 

According to the question, the surface is frictionless. Thus, the block \(\mathrm{m}\) will slide down the inclined plane of mass M.
Acceleration, \(a_1=g \sin \alpha \quad\) (Relative to the inclined plane)
The horizontal component of acceleration \(\mathrm{a}_1\) is given by \(\mathrm{a}_{\mathrm{x}}=\mathrm{g} \sin \alpha \cos \alpha\), for which the block \(\mathrm{M}\) accelerates towards left.
Let the left acceleration be \(\mathrm{a}_2\).
By the concept of centre of mass, we can say that the external force is zero in the horizontal direction.
\(m a_x=(M+m) a_2\)
\(
\Rightarrow a_2=\frac{m a_x}{M+m}=\frac{m g \sin \alpha \cos \alpha}{M+m} \dots(1)
\)
\(
\text { Absolute (resultant) acceleration of } \mathrm{m} \text { on the plane } \mathrm{M} \text {, along the direction of the incline will be }=a=g \sin \alpha-a_2 \cos \alpha
\)
\(
\begin{aligned}
& =g \sin \alpha-\frac{m g \sin \alpha \cos ^2 \alpha}{M+m}=g \sin \alpha\left[1-\frac{m \cos ^2 \alpha}{M+m}\right] \\
& =g \sin \alpha\left[\frac{M+m-m \cos ^2 \alpha}{M+m}\right]
\end{aligned}
\)
So, \(a=g \sin \alpha\left[\frac{M+m \sin ^2 \alpha}{M+m}\right] \dots(2)\)
Let the time taken by the block \(m\) to reach the bottom end be \(t\).
Now,
\(
\begin{aligned}
& s=u t+\left(\frac{1}{2}\right) a t^2 \\
& \Rightarrow \frac{h}{\sin \alpha}=\left(\frac{1}{2}\right) a t^2 \\
& \Rightarrow t=\sqrt{\frac{2 h}{a \sin \alpha}}
\end{aligned}
\)
Thus, the velocity of the bigger block after time \(t\) will be,
\(
v_m=u=a_2 t
\)
\(
=\frac{m g \sin \alpha \cos \alpha}{M+m} \sqrt{\frac{2 h}{a \sin \alpha}}=\left[\frac{2 m^2 g^2 h \sin ^2 \alpha \cos ^2 \alpha}{(M+m)^2 a \sin \alpha}\right]^{1 / 2}
\)
Subtracting the value of a from equation (2), we get:
\(
v_M=\left[\frac{2 m^2 g^2 h \sin ^2 \alpha}{(M+m)^2 \sin \alpha} \times \frac{\cos ^2 \alpha}{g \sin \alpha} \frac{(M+m)}{\left(M+m \sin ^2 \alpha\right)}\right]
\)
\(
\text { or } V_M=\left[\frac{2 m^2 g^2 h \cos ^2 \alpha}{(M+m)\left(M+m \sin ^2 \alpha\right)}\right]^{1 / 2}
\)

Q63: Figure (9-E22) shows a small body of mass \(m\) placed over a larger mass \(M\) whose surface is horizontal near the smaller mass and gradually curves to become vertical. The smaller mass is pushed on the longer one at a speed \(v\) and the system is left to itself. Assume that all the surfaces are frictionless. (a) Find the speed of the larger block when the smaller block is sliding on the vertical part. (b) Find the speed of the smaller mass when it breaks off the larger mass at the height \(h[latex]. (c) Find the maximum height (from the ground) that the smaller mass ascends. (d) Show that the smaller mass will again land on the bigger one. Find the distance traversed by the bigger block during the time when the smaller block was in its flight under gravity.

Answer: 

According to the question, mass [latex]m\) is given with a speed \(v\) over the larger mass \(M\).
(a) When the smaller block travels on the vertical part, let the velocity of the bigger block be \(v_1\), towards the left.
From the law of conservation of momentum (in the horizontal direction), we get:
\(
\begin{aligned}
& m v=(M+m) v_1 \\
& \Rightarrow v_1=\frac{m v}{M+m}
\end{aligned}
\)
(b) When the smaller block breaks off, let its resultant velocity be \(\mathrm{v}_2\).
Using the law of conservation of energy, we get:
\(
\begin{aligned}
& \left(\frac{1}{2}\right) m v^2=\left(\frac{1}{2}\right) M v_1^2+\left(\frac{1}{2}\right) m v_2^2+m g h \\
& \Rightarrow v_2{ }^2=v^2-\frac{M}{m} v_1^2-2 g h \ldots(i) \\
& \Rightarrow v_2{ }^2=v^2\left[1-\frac{M}{m} \cdot \frac{m^2}{(M+m)^2}\right]-2 g h \\
& \Rightarrow v_2=\left[\frac{\left[\left(m^2+M m+m^2\right)\right] v^2}{(M+m)^2}-2 g h\right]^{1 / 2}
\end{aligned}
\)
(c) Vertical component of the velocity \(v_2\) of mass \(m\) is given by,
\(
\begin{aligned}
& v_y{ }^2=v_2{ }^2-v_1{ }^2 \\
& =\frac{\left(M^2+M m+m^2\right)}{(M+m)^2} v^2-2 g h-\frac{m^2 v^2}{(M+m)^2} \\
& \left(\therefore v_1=\frac{m v}{M+m}\right) \\
& \Rightarrow v_y{ }^2=\frac{M^2+M m+m^2-m^2}{(M+m)^2} v^2-2 g h \\
& \Rightarrow v_y{ }^2=\frac{M v^2}{(M+m)}-2 g h \ldots(i i)
\end{aligned}
\)
To determine the maximum height (from the ground),
Let us assume that the body rises to a height \(h_1\) over and above \(h\).
\(
\begin{aligned}
& \text { Now, }\left(\frac{1}{2}\right) m v_y^2=m g h_1 \\
& \Rightarrow h_1=\frac{v_y^2}{2 g} \ldots(i i i) \\
& \therefore \text { Total height, } \mathrm{H}=\mathrm{h}+\mathrm{h}_1 \\
& \Rightarrow H=h+\frac{v_y^2}{2 g} \\
& \Rightarrow H=h+\frac{M v^2}{(M+m) 2 g}-h \\
& \Rightarrow H=\frac{M v^2}{(M+m) 2 g}
\end{aligned}
\)
(d) Because the smaller mass also has a horizontal component of velocity \(\mathrm{V}_1\) at the time it breaks off from \(\mathrm{M}\) (that has a velocity \(\mathrm{v}_1\) ), the block \(\mathrm{m}[latex] will again land on the block [latex]\mathrm{M}\).
The time of flight of block \(m\), after it breaks off, is calculated as:
During the upward motion (BC),
\(
\begin{aligned}
& 0=v_y-g t \\
& \Rightarrow t_1=\frac{v_y}{g}=\frac{1}{g}\left[\frac{M v^2}{(M+m)}-2 g h\right] \ldots
\end{aligned}
\)
Thus, the time for which the smaller block is in flight is given by,
\(
\begin{aligned}
& T=2 t_1 \\
& =\frac{2}{g}\left[\frac{M v^2-2(M+m) g h}{(M+m)}\right]
\end{aligned}
\)
The distance travelled by the bigger block during this time is, \(S=v_1 T\)
\(
=\frac{m v}{M+m} \times \frac{2}{g} \frac{\left[M v^2-2(M+m) g h\right]^{1 / 2}}{(M+m)^{1 / 2}}
\)
\(
S=\frac{2 m v\left[M v^2-2(M+m) g h\right]^{1 / 2}}{g(M+m)^{3 / 2}}
\)

Q64: A body of mass \(m\) makes an elastic collision with another identical body at rest. Show that if the collision is not head-on, the bodies go at the right angle to each other after the collision.

Answer:

According to the question, the collision of the two bodies of mass \(\mathrm{m}\) is not head-on. Thus, the two bodies move in different directions.
Let the velocity vectors of the two bodies after collision be \(v_1\) and \(v_2\).
As the collision in the question is elastic, momentum is conserved.
On applying the law of conservation of momentum in the X-direction, we get:
\(
m u_1+m \times 0=m v_1 \cos \alpha+m v_2 \cos \beta \ldots(i)
\)
On applying the law of conservation of momentum in Y-direction, we get:
\(
\begin{aligned}
& 0=m v_1 \sin \alpha-m v_2 \sin \beta \ldots(i i) \\
& \Rightarrow m v_1 \sin \alpha=m v_2 \sin \beta
\end{aligned}
\)
Now,
\(
\begin{aligned}
& \left(\frac{1}{2}\right) m u_1^2+0=\left(\frac{1}{2}\right) m v_1^2+\left(\frac{1}{2}\right) m \times v_2^2 \\
& \Rightarrow u_1^2=v_1^2+v_2^2 \ldots(i i i)
\end{aligned}
\)
On squaring equation (i), we get:
\(
u_1^2=v_1^2 \cos ^2 \alpha+v_2^2 \cos ^2 \beta+2 v_1 v_2 \cos \alpha \cos \beta
\)
Equating the equations (i) and (iii), we get
\(
\begin{aligned}
& v_1^2+v_2^2=v_1^2 \cos ^2 \alpha+v_2^2 \cos ^2 \beta+2 v_1 v_2 \cos \alpha \cos \beta \\
& \Rightarrow v_1{ }^2 \sin ^2 \alpha+v_2{ }^2 \sin ^2 \beta=2 v_1 v_2 \cos \alpha \cos \beta \\
& \Rightarrow 2 v_1{ }^2 \sin ^2 \alpha=2 \times v_1 \times \frac{v_1 \sin \alpha}{\sin \beta} \cos \alpha \cos \beta \\
& \Rightarrow \sin \alpha \sin \beta=\cos \alpha \cos \beta \\
& \Rightarrow \cos \alpha \cdot \cos \beta-\sin \alpha \cos \beta=0 \\
& \Rightarrow \cos (\alpha+\beta)=0=\cos 90^{\circ} \\
& \Rightarrow \alpha+\beta=90^{\circ}
\end{aligned}
\)

Q65: A small particle travelling with a velocity \(v\) collides elastically with a spherical body of equal mass and of radius \(r\) initially kept at rest. The centre of this spherical body is located a distance \(\rho(<r)\) away from the direction of motion of the particle (figure 9-E23). Find the final velocities of the two particles.

[Hint: The force acts along the normal to the sphere through the contact. Treat the collision as onedimensional for this direction. In the tangential direction, no force acts and the velocities do not change].

Answer: 

Let the mass of both the particle and the spherical body be ‘ \(m\) ‘. The particle velocity ‘ \(\mathrm{v}\) ‘ has two components, \(v \cos \alpha\) normal to the sphere and \(v \sin \alpha\) tangential to the sphere.
After the collision, they will exchange their velocities. So, the spherical body will have a velocity \(v \cos \alpha\) and the particle will not have any component of velocity in this direction.
[The collision will be due to the component \(v \cos \alpha\) in the normal direction. But, the tangential velocity, of the particle \(v \sin \alpha\) will be unaffected]
So, velocity of the sphere \(=v \cos \alpha=\frac{v}{r} \sqrt{r^2-\rho^2}\) [from second figure)] And velocity of the particle \(=v \sin \alpha=\frac{v \rho}{r}\)

Short Answer Questions

Q1: Can the centre of mass of a body be at a point outside the body?

Answer: 
1. The Centre of mass of a body or system of a particle is defined as, a point at which the whole of the mass of the body or all the masses of a system of particles appeared to be concentrated.
2. For Hollow objects like a uniform ring, the center of mass lies outside the object. For a ring, the centre of mass lies at the centre of the circle made by the ring
3. So, we can say that center of mass of an object can lie outside.

Q2: If all the particles of a system lie in \(X-Y\) plane, is it necessary that the centre of mass be in \(X-Y\) plane?

Answer: Yes, if all the particles of a system lie in the \(X-Y\) plane, then it’s necessary that its centre of mass lies in the \(X-Y\) plane.
\(
z_{c m}=\frac{m_1 z_1+m_2 z_2+\ldots m_n z_n}{\sum m}
\)
As all the particles lie in the \(X-Y\) plane, their \(z\)-coordinates are zero.
Therefore, for the whole system, \(z_{\mathrm{cm}}=0\); i.e., its centre of mass lies in the \(X-Y\) plane.

Q3: If all the particle of a system lies in a cube, is it necessary that the centre of mass be in the cube?

Answer: Yes. As a cube is a 3-dimensional body, all the particles of a system lying in a cube lie in the \(x, y[latex], and [latex]z\) plane.
Let the \(i^{\text {th }}\) element of mass \(\Delta m_{\mathrm{i}}\) is located at the point \(\left(x_{\mathrm{i}}, y_{\mathrm{i}}, z_{\mathrm{i}}\right)\).
The coordinates of the centre of mass are given as:
\(
\begin{aligned}
X & =\frac{1}{M} \sum_{i=1}^{i=n}\left(\Delta m_i\right) x_i \\
Y & =\frac{1}{M} \sum_{i=1}^{i=n}\left(\Delta m_i\right) y_i \\
Z & =\frac{1}{M} \sum_{i=1}^{i=n}\left(\Delta m_i\right) z_i
\end{aligned}
\)
\(X, Y\), and \(Z\) lie inside the cube because it is a weighted mean.

Q4: The centre of mass is defined as \(\vec{R}=\frac{1}{M} \sum_i m_i \vec{r}_i\). Suppose we define “centre of charge” as \(\vec{R}_C=\frac{1}{Q} \sum q_i \vec{r}_i\) where \(q_i\) represents the \(i\) th charge placed at \(\vec{r}_i\) and \(Q\) is the total charge of the system.
(a) Can the centre of charge of a two-charge system be outside the line segment joining the charges?
(b) If all the charges of a system are in \(X-Y\) plane, is it necessary that the centre of charge be in \(X-Y\) plane?
(c) If all the charges of a system lie in a cube, is it necessary that the centre of charge be in the cube?

Answer: (a) Yes
Consider a charge distributed in \(X-Y\) plane.

\(
X_{c m}=\frac{-6 q \times 0+q \times 5 a}{-6 q+q}=-a
\)
(b) Yes. Because the \(z\)-coordinates of all the charges are zero, the centre of charge lies in \(X-Y\) plane.
(c) No, it is not necessary that the centre of charge lies in the cube because the charge can be either negative or positive.

Q5: The weight \(M g\) of an extended body is generally shown in a diagram to act through the centre of mass. Does it mean that the earth does not attract other particles?

Answer: The earth does attract each and every particle. The resultant of all these forces of attraction is \({Mg}\) which acts at the center of mass (where \(M\) is the total mass). That is why the weight \({Mg}\) of an extended body is generally shown in a diagram to act through the center of mass.

Q6: A bob suspended from the ceiling of a car which is accelerating on a horizontal road. The bob stays at rest with respect to the car with the string making an angle \(\theta\) with the vertical. The linear momentum of the bob as seen from the road is increasing with time. Is it a violation of the conservation of linear momentum? If not, where is the external force which changes the linear momentum?

Answer: 

Let the acceleration of the car be ‘ \(a\) ‘. The forces on the bob from the frame of reference of the car is shown in the above picture.
Since there is no motion in the vertical direction we can say,
\(\operatorname{Tcos}(e)-m g=0\)
or, \(\operatorname{Tcos}(e)=m g\).
In the horizontal direction, on applying Newton’s second law we get,
\(T \sin (e)=m a\)
where \(a[latex] is the acceleration of the bob in the horizontal direction
with respect to the ground frame. So, we can see that there is a
net non-zero force in the horizontal direction which is responsible
for the acceleration observed.
So, there is no violation of the law of conservation of momentum.

Q7: You are waiting for a train on a railway platform. Your three-year-old niece is standing on your iron trunk containing the luggage. Why does the trunk not recoil as she jumps off on the platform?

Answer: The momentum of the girl and trunk system was zero before she jumped off the trunk. When she jumps off the trunk, she exerts a backward force on the trunk and the trunk exerts a forward force on the girl (Newton’s third law).
The girl moves forward but the trunk does not recoil. The reason is that there is a force of static friction in between the trunk surface and the railway platform. The force exerted by the girl was not enough to overcome this friction and so the trunk does not recoil.
Since there is an external friction force on the system in this case, so linear momentum is not conserved here.

Q8: In a head-on collision between two particles, is it necessary that the particles will acquire a common velocity at least for one instant?

Answer: Yes, the particles acquire a common velocity at least for one instant of time. In the case of elastic collision, the particles separate after this instant and move with different velocities.
In the case of inelastic collision, the particles stick together even after this instant and move with a common velocity.

Q9: A collision experiment is done on a horizontal table kept in an elevator. Do you expect a change in the results if the elevator is accelerated up or down because of the noninertial character of the frame?

Answer: No, there should not be any change in the results.
Explanation:
Since acceleration is in the vertical direction, so there won’t be any effect on motion in the horizontal plane of the table and so the collision experiment results will remain the same.

Q10: Two bodies make an elastic head-on collision on a smooth horizontal table kept in a car. Do you expect a change in the result if the car is accelerated on a horizontal road because of the noninertial character of the frame? Does the equation “Velocity of separation = Velocity of approach” remain valid in an accelerating car? Does the equation “final momentum = initial momentum” remain valid in the accelerating car?

Answer: Yes, we can expect a change in the result of the experiment. If we look at the motion from the frame of reference of the car, the two bodies experience a pseudo force in a direction opposite to the direction of motion of the car. Now since there is a net non-zero force on the two bodies, the momentum of the two-body system is not conserved in this case. So, the equation “final momentum = initial momentum’ does not remain valid.
Also, since the momentum of the two-body system is not conserved the equation “Velocity of separation = Velocity of approach” also does not remain valid.

Q11: If the total mechanical energy of a particle is zero, is its linear momentum necessarily zero? Is it necessarily nonzero?

Answer: Yes, it is possible for the particle to have a non-zero momentum.
Explanation:

Consider the above scenario. The particle is ‘ [latex]h\) ‘ height below the
Earth’s surface.
Potential energy, \(U=-m g h\).
Kinetic energy, \(K=\frac{1}{2} m v^2\).
Total energy, \(E=U+K\).
Let us suppose for some value of height \(h\) and velocity \(v\), the total energy becomes zero. In that case, we have zero total mechanical energy but non-zero kinetic energy.
Now, non-zero kinetic energy implies that the particle has non-zero momentum.

Q12: If the linear momentum of a particle is known, can you find its kinetic energy? If the kinetic energy of a particle is known can you find its linear momentum?

Answer: The momentum of a particle of mass ‘ \(m\) ‘ and velocity ‘ \(v\) ‘ (magnitude) is given by
\(
p=m v \cdots(1)
\)
and its kinetic energy is given by
\(
K=\frac{1}{2} m v^2 \dots(2)
\)
From equation (1) and (2), we can express the kinetic energy of the particle as
\(
K=\frac{p^2}{2 m}
\)
Similarly, momentum can be expressed as,
\(
p=\sqrt{2 m K}
\)
So, we can find the kinetic energy of the particle from its momentum and vice-versa.
*Note: We can only find the magnitude of momentum from kinetic energy.

Q13: What can be said about the centre of mass of a uniform hemisphere without making any calculation? Will its distance from the centre be more than \(r / 2\) or less than \(r / 2\)?

Answer:

The location of the center of mass of any system is close to the part where most of the mass of the system is present. In this case, more mass of the hemisphere is present in the lower portion which is present at a distance less than \(\mathrm{r} / 2 \) from the center.
So, the distance of the center of mass will be less than \(r / 2\) from the center.

Q14: You are holding a cage containing a bird. Do you have to make less effort if the bird flies from its position in the cage and manages to stay in the middle without touching the walls of the cage? Does it make a difference whether the cage is completely closed or it has rods to let air pass?

Answer: Case 1-Completely closed cage:
In this case, our hand has to support the weight of the cage, the bird and the air inside the cage. When the bird is flying inside the cage, it flaps its wings up and down. With every down stroke of the wings, it pushes the air which further exerts a force against the wall of the cage. This force is nearly equal to the weight of the bird. So effectively, there is no reduction in our effort to hold the cage.
Case 2- Open cage with rods:
In this case, the air pushed downward by the wings of the bird is somewhat dispersed due to the incoming air from outside the cage. So effectively, we have to apply a little less effort in this case to hold the cage.

Q15: A fat person is standing on a light plank floating on a calm lake. The person walks from one end to the other on the plank. His friend sitting on the shore watches him and finds that the person hardly moves any distance because the plank moves backward about the same distance as the person moves on the plank. Explain.

Answer: Since the person is heavier than the plank, the center of mass of the person and the plank system effectively lies on the person. Since there is no external force on the system, the center of mass remains at the same position and so the observer on the ground feels that the person on the plank has not moved much.

Q16: A high-jumper successfully clears the bar. Is it possible that his centre of mass crossed the bar from below it? Try it with appropriate figures.

Answer:

At the instant when the jumper is just above the bar and about to cross it, his arms, head, and legs are below the crossbar. So it is possible that his center of mass may have crossed the bar from below it.

Q17: Which of the two persons shown in Figure (9-Q1) is more likely to fall down? Which external force is responsible for his falling down?

Answer: The person on the right side in the above picture is more likely to fall. The force responsible for his falling is the same force which is responsible for the acceleration of the vehicle that they are sitting in.
Explanation:
From the frame of reference of the vehicle, the two persons experience a pseudo force in the backward direction. While the person on the left does not fall as he gets support from the wall of the vehicle behind his back, the person on the right does not have any support and therefore, is more likely to fall from his seat.

Q18: Suppose we define a quantity ‘Linear Momentum’ as linear momentum \(=\) mass \(\times\) speed.
The linear momentum of a system of particles is the sum of the linear momenta of the individual particles. Can we state a principle of conservation of linear momentum as “linear momentum of a system remains constant if no external force acts on it”?

Answer: Since we are using speed in the definition of linear momentum, so the linear momentum of a system of particles in this case is just the algebraic sum of the linear momentum of the individual particles. So, the principle of conservation of linear momentum does not hold true for every scenario here because we are not using the direction of the velocity of the particles.

Q19: Use the definition of linear momentum from the previous question. Can we state the principle of conservation of linear momentum for a single particle?

Answer: If the external force on the particle is zero, then the acceleration of the particle is also zero and so its speed remains constant and so momentum is conserved.

Q20: To accelerate a car we ignite petrol in the engine of the car. Since only an external force can accelerate the centre of mass, is it proper to say that “the force generated by the engine accelerates the car”?

Answer: No, it is not proper to say this. The force generated by the engine rotates the wheels of the car. Now, there is a rolling frictional force on the wheel of the car from the ground. This friction force acts as an external force and accelerates the center of mass of the car system.

Q21: A ball is moved on a horizontal table with some velocity. The ball stops after moving some distance. Which external force is responsible for the change in the momentum of the ball?

Answer: Friction is responsible for the change in the momentum of the ball. Friction opposes the motion of the ball and so the ball stops after moving some distance.

Q22: Consider the situation of the previous problem. Take “the table plus the ball” as the system. Friction between the table and the ball is then an internal force. As the ball slows down, the momentum of the system decreases. Which external force is responsible for this change in momentum?

Answer: There is friction between the ground and the legs of the table, and this force prevents the table from sliding in the opposite direction of the motion of the ball. Due to this friction, the kinetic energy of the ball is converted into heat and so the ball slows down and eventually stops. So, friction between the ground and the table acts as an external force to change the momentum.

Q23: When a nucleus at rest emits a beta particle, it is found that the velocities of the recoiling nucleus and the beta particle are not along the same straight line. How can this be possible in view of the principle of conservation of momentum?

Answer: In view of the principle of conservation of momentum, an antineutrino is also emitted along with the beta particle. The momentum of the recoiling nucleus and beta particle along with the antineutrino particle vectorially adds up to make the total momentum zero as was before the decay. So, the momentum is conserved in this way.

Q24: A van is standing on a frictionless portion of a horizontal road. To start the engine, the vehicle must be set in motion in the forward direction. How can the persons sitting inside the van do it without coming out and pushing from behind?

Answer: Consider the persons and the van as a system. The center of mass of the system is at rest. The center of mass of this system should remain at rest as there is no external force to change its position. If the persons inside the van move in the backward direction, then the van has to move in the forward direction to keep the center of mass at the same position as it was before.
In this way, the van can be set in motion in the forward direction.

Q25: In a one-dimensional elastic collision of equal masses, the velocities are interchanged. Can velocities in a onedimensional collision be interchanged if the masses are not equal?

Answer: No, if the masses are unequal then the velocities in a one-dimensional collision cannot be interchanged as it would violate the principle of conservation of momentum.