Sample Exercises

Hint

The geometrical centre of each. No, the CM may lie outside the body, as in the case of a ring, a hollow sphere, a hollow cylinder, a hollow cube, etc.

Hint

\(
r=1.27 ;  r_{c m} = \frac{m_1 r_1+m_2 r_2}{m_1+m_2}
\)
Since centre of mass cannot go beyond bond legth
\(
\begin{aligned}
&r_{\mathrm{cm}} = \frac{0+35.5 \times 1.27}{35.5+1}=\frac{35.5 \times 1.27}{36.5} \\
&=1.24 
\end{aligned}
\)

Hint

The speed of the centre of mass of the system remain unchanged (equal to v) because no external force acts on the system. The force involved such as action and reaction, friction, etc when the child runs on the trolley are internal to the (trolley+child) system.

Hint

Consider two vectors \(\mathrm{OK}=\) vector \(\vec{a}\) and \(\mathrm{OM}=\) \(\vec{b}\), inclined at an angle \(\theta\) as shown in the following figure.
In \(\triangle \mathrm{OMN}\), we can write the relation:
\(
\begin{aligned}
&\sin \theta=\frac{\mathrm{MN}}{\mathrm{OM}}=\frac{\mathrm{MN}}{|\vec{b}|} \\
&\Rightarrow \mathrm{MN}=|\overrightarrow{\mathrm{b}}| \sin \theta
\end{aligned}
\)
\(
\begin{aligned}
|\vec{a} \times \vec{b}| &=|\vec{a}||\vec{b}| \sin \theta \\
&=O K \times M N
\end{aligned}
\)
\(
\begin{aligned}
=2 \times \frac{1}{2} \times \mathrm{OK} \times \mathrm{MN} \\
=& 2 \times \text { Area of } \triangle \mathrm{OMK}
\end{aligned}
\)
\(
\Longrightarrow \text { Area of } \triangle \mathrm{OMK}=\frac{1}{2} \times|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|
\)

Hint

A parallelepiped with origin O and sides a, b, and c is shown in the following figure.

Volume of the given parallelepiped \(=a b c\)
\(
\begin{aligned}
&\overrightarrow{\mathrm{OC}}=\vec{a} \\
&\overrightarrow{\mathrm{OB}}=\vec{b} \\
&\overrightarrow{\mathrm{OC}}=\vec{c}
\end{aligned}
\)
Let cap \(\mathbf{n}\) be a unit vector perpendicular to both b and \(\mathrm{c}\). Hence, cap \(\mathbf{n}\) and a have the same direction
\(
\begin{aligned}
&\therefore \vec{b} \times \vec{c}=b c \sin \theta \hat{\mathbf{n}} \\
&=b c \sin 90^{\circ} \hat{\mathbf{n}} \\
&=b c \hat{n}
\end{aligned}
\)
\(\vec{a} \cdot(\vec{b} \times \vec{c})\)
\(=a \cdot(b c \hat{\mathbf{n}})\)
\(=a b c \cos \theta \hat{\mathbf{n}}\)
\(=a b c \cos 0^{\circ}\)
\(=a b c\)
\(=\) Volume of the parallelepiped

Hint

\(T_1\) and \(T_2\) are the tensions produced in the left and right strings respectively.
At translational equilibrium, we have:
\(
\begin{aligned}
&T_1 \sin 36.9^{\circ}=T_2 \sin 53.1 \\
&\frac{T_1}{T_2}=\frac{\sin 53.1^{\circ}}{\sin 36.9} \\
&=\frac{0.800}{0.600}=\frac{4}{3} \\
&\Rightarrow T_1=\frac{4}{3} T_2
\end{aligned}
\)
For rotational equilibrium, on taking the torque about the centre of gravity, we have:
\(
T_1 \cos 36.9 \times d=T_2 \cos 53.1(2-d)
\)
\(
\begin{aligned}
&T_1 \times 0.800 d=T_2 0.600(2-d) \\
&\frac{4}{3} \times T_2 \times 0.800 d=T_2[0.600 \times 2-0.600 d] \\
&1.067 d+0.6 d=1.2
\end{aligned}
\)
\(
\begin{aligned}
&\therefore d=\frac{1.2}{1.67} \\
&=0.72 \mathrm{~m}
\end{aligned}
\)
Hence, the C.G. (centre of gravity) of the given bar lies \(0.72 \mathrm{~m}\) from its left end.

Hint

Mass of the car, \(m=1800 \mathrm{~kg}\)
Distance between the front and back axles, \(d=1.8 \mathrm{~m}\)
Distance between the centre of gravity and the back axle \(=1.05 \mathrm{~m}\)
The various forces acting on the car are shown in the following figure.
\(R_{\mathrm{f}}\) and \(R_{\mathrm{b}}\) are the forces exerted by the level ground on the front and back wheels respectively.
At translational equilibrium,
\(
\begin{aligned}
R_{\mathrm{f}}+R_{\mathrm{b}} &=\mathrm{mg} \\
&=1800 \times 9.8 \\
&=17640 \mathrm{~N} \dots(i)
\end{aligned}
\)
For rotational equilibrium, on taking the torque about the C.G., we have
\(
\begin{aligned}
R_{\mathrm{f}}(1.05) &=R_{\mathrm{b}}(1.8-1.05) \\
\frac{R_{\mathrm{b}}}{R_{\mathrm{f}}} &=\frac{7}{5} \\
R_{\mathrm{b}} &=1.4 R_{\mathrm{f}} \dots(ii)
\end{aligned}
\)
Solving equations ( \(i\) ) and (ii), we get
\(
1.4 R_{\mathrm{f}}+R_{\mathrm{f}}=17640
\)
\(
R_{\mathrm{f}}=7350 \mathrm{~N}
\)
\(
\begin{aligned}
\therefore \quad R_{\mathrm{b}} &=17640-7350 \\
&=10290 \mathrm{~N}
\end{aligned}
\)
Therefore, the force exerted on each front wheel \(=\frac{7350}{2}=3675 \mathrm{~N}\). The force exerted on each back wheel \(=\frac{10290}{2}=5145 \mathrm{~N}\)

Hint

The moment of inertia for the hollow cylinder \(=\mathrm{I}_1=\mathrm{mr}^2\)
The moment of inertia for the solid sphere \(=\mathrm{I}_2=\frac{2}{5} \mathrm{mr}^2\)
For hollow sphere we have \(\tau=\mathrm{I}_1 \alpha_1\)
For solid sphere we have \(\tau=\mathrm{I}_2 \alpha_2\)
\(
\Rightarrow \frac{\alpha_2}{\alpha_1}=\frac{I_1}{I_2}=\frac{5}{2}>1
\)
Thus \(\alpha_2>\alpha_1\)
\(
\omega(\mathrm{t})=\omega_0+\alpha \mathrm{t}
\)
The angular velocity( \(\omega)\) at a certain time will be greater for solid sphere.

Hint

The moment of inertia of a solid cylinder \(=\mathrm{mr}^2 / 2\)
\(
\begin{aligned}
&=\frac{1}{2} \times 20 \times(0.25)^2 \\
&=0.625 \mathrm{kgm}^2
\end{aligned}
\)
Therefore kinetic energy \(=\frac{1}{2} I \omega^2\) \(=3125 \mathrm{~J}\)
Angular momentum , \(\mathrm{L}=\mathrm{I} \omega\)
\(
=0.625 \times 100
\)
\(
=62.5 \mathrm{~J} \mathrm{~s}
\)

Hint

Since no external force acts on the boy, the angular momentum is constant.
Thus \(I_1 \omega_1=I_2 \omega_2\)
\(
\Longrightarrow \omega_2=\frac{I_1}{I_2} \omega_1=\frac{5}{2} \times 40  \mathrm{rev} / \mathrm{min}=100  \mathrm{rev} / \mathrm{min}
\)

Hint

Here, \(M=3 \mathrm{~kg}, \mathrm{R}=40 \mathrm{~cm}=0.4 \mathrm{~m}\)
Moment of inertia of the hollow cylinder about its axis \(l=M R^2=3(0.4)^2=0.48 \mathrm{kgm}^2\)
Force applied \(\mathrm{F}=30 \mathrm{~N}\)
\(\therefore \quad\) Torque, \(\tau=F \times R=300 \times 0.4=12 Nm\).
If \(\alpha\) is angular acceleration produced, then from \(\tau=I \alpha\), \(\alpha=\frac{\tau}{I}=\frac{12}{0.48}=25\) rads \(^{-2}\)
Linear acceleration, \(a=R \alpha=0.4 \times 25=10 \mathrm{~ms}^{-2}\).

Hint

Given: The angular speed of the rotor is \(200 \mathrm{rad} / \mathrm{s}\).
The torque needed to be transmitted by the engine is \(180 \mathrm{Nm}\).
The power of the rotor required to transmit energy to apply a torque \(\tau\) to rotate a motor with angular speed \(\omega\),
\(
\begin{aligned}
&\mathrm{P}=\tau \omega \\
&=180 \times 200 \mathrm{~W} \\
&=36 \mathrm{~kW}
\end{aligned}
\)

Hint

Given that,
Radius of uniform disc \(=\mathrm{R}\)
Radius of the smaller disc \(=\frac{\mathrm{R}}{2}\)
Let the mass per unit area of the original disc \(=\sigma\)
Therefore mass of the uniform disc \(=\mathrm{M}=\sigma \pi \mathrm{R}^2\)
And the mass of the small disc \(=\sigma \pi\left(\frac{\mathrm{R}}{2}\right)^2=\frac{\sigma \pi \mathrm{R}^2}{4}=\frac{\mathrm{M}}{4}\)
Now as the small disc has been cut from the uniform disc, the remaining portion is considered to be a system of two masses.
The two masses are: \(M\) (concentrated at \(O)\)  & \(-M\) (concentrated at \(O^{\prime}\) )
(negative sign indicating above that the portion is removed from the uniform disc)
Let \(\mathrm{x}\) be the distance through which the centre of mass of the remaining portion shifts from point \(\mathrm{O}\).
The relation between the centre of masses of two masses is give as:
\(
\begin{aligned}
\mathrm{x} &=\frac{\left(\mathrm{m}_1 \mathrm{r}_1+\mathrm{m}_2 \mathrm{r}_2\right)}{\left(\mathrm{m}_1+\mathrm{m}_2\right)} \\
\mathrm{x} &=\frac{\left[\left(\mathrm{M} \times 0-\frac{\mathrm{M}}{4}\right) \times\left(\frac{\mathrm{R}}{2}\right)\right]}{\left(\mathrm{M}-\frac{\mathrm{M}}{4}\right)} \\
&=\frac{\left(\frac{-\mathrm{MR}}{8}\right)}{\left(\frac{3 \mathrm{M}}{4}\right)} \\
=& \frac{-4 \mathrm{R}}{24} \\
\mathrm{x} &=\frac{-\mathrm{R}}{6}
\end{aligned}
\)
Note: The relation between the centre of masses of two masses is calculated by the formula \(\mathrm{x}=\frac{\left(\mathrm{m}_1 \mathrm{r}_1+\mathrm{m}_2 \mathrm{r}_2\right)}{\left(\mathrm{m}_1+\mathrm{m}_2\right)}\) which is found to be \(\frac{-\mathrm{R}}{6}\) here the negative sign indicates that the centre of gravity of the resulting flat body gets shifted towards the left point \(O\).

Hint

Let \(W\) and \(W^{\prime}\) be the respective weights of the metre stick and the coin.
The mass of the metre stick is concentrated at its mid-point, i.e., at the \(50 \mathrm{~cm}\) mark.
Mass of the meter stick \(=\mathrm{m}\) ‘
Mass of each coin, \(m=5 \mathrm{~g}\)
When the coins are placed \(12 \mathrm{~cm}\) away from the end \(P\), the centre of mass gets shifted by \(5 \mathrm{~cm}\) from point \(\mathrm{R}\) toward the end \(\mathrm{P}\). The centre of mass is located at a distance of \(45 \mathrm{~cm}\) from point \(P\).
The net torque will be conserved for rotational equilibrium about point R.
\(
10 \times \mathrm{g}(45-12)-m^{\prime} \mathrm{g}(50-45)=0
\)
\(
\therefore m^{\prime}=\frac{10 \times 33}{5}=66 \mathrm{~g}
\)
Hence, the mass of the metre stick is \(66 \mathrm{~g}\).

Hint

Using law of conservation of energy,
\(
\frac{1}{2} m v^2+\frac{1}{2} I \omega^2=m g h
\)
\(
\text { or } \frac{1}{2} m v^2+\frac{1}{2}\left(\frac{2}{5} m R^2\right) \frac{v^2}{R^2}=m g h
\)
or \(7 / 10 v^{\wedge} 2=\) gh or \(v=\sqrt{\frac{10 g h}{7}}\)

Hence, the velocity of the sphere at the bottom depends only on height \((h)\) and acceleration due to gravity \((g)\). Both these values are constants. Therefore, the velocity at the bottom remains the same from whichever inclined plane the sphere is rolled.

Hint

Given,
Radius of the hoop, \(r=2 \mathrm{~m}\)
Mass of the hoop, \(m=100 \mathrm{~kg}\)
Velocity of the hoop, \(v=20 \mathrm{~cm} / \mathrm{s}=0.2 \mathrm{~m} / \mathrm{s}\)
Total energy of the hoop \(=\) Translational K.E. + Rotational K.E.
\(
E_{\mathrm{T}}=\frac{1}{2} m v^2+\frac{1}{2} I \omega^2
\)
Moment of inertia of the hoop about its centre, \(I=m r^2\)
\(
E_T=\frac{1}{2} m v^2+\frac{1}{2}\left(m r^2\right) \omega^2
\)
Using the relation, \(v=r \omega\)
\(
\begin{aligned}
E_{\mathrm{T}} &=\frac{1}{2} m v^2+\frac{1}{2} m r^2 \omega^2 \\
&=\frac{1}{2} m v^2+\frac{1}{2} m v^2 \\
&=m v^2
\end{aligned}
\)
The work required to be done for stopping the hoop is equal to the total energy of the hoop.
\(\therefore\) Required work to be done, \(W=m v^2\)
\(
\begin{aligned}
&=100 \times(0.2)^2 \\
&=4 \mathrm{~J} .
\end{aligned}
\)

Hint

Given, \(m=5.30 \times 10^{-26} \mathrm{~kg}\),
\(
\begin{aligned}
&\mathrm{I}=1.94 \times 10^{-26} \mathrm{kgm}^2 \text { and } \\
&\mathrm{V}=500 \mathrm{~m} / \mathrm{s} \\
&\mathrm{KE}(\text { translation })=\frac{1}{2} \mathrm{mv}^2 \\
&=\frac{1}{2} \times\left(5.30 \times 10^{-26}\right)(500)^2 \\
&=6.625 \times 10^{-21} \mathrm{~J}
\end{aligned}
\)
given that,
\(\mathrm{KE}(\) rotation \()=\left(\frac{2}{3}\right) \mathrm{KE}\) (translation)
\(\frac{1}{2} l \omega^2=\left(\frac{2}{3}\right) \frac{1}{2} m v^2\)
\(\therefore \omega=\sqrt{2 / 3 \frac{m v^2}{1}}\)
\(=\sqrt{\frac{2 / 3 \times 6.625 \times 10^{-21}}{1.94 \times 10^{-46}}}\)
\(=6.75 \times 10^{12} \mathrm{rad} / \mathrm{sec}\).

Hint

(a) From the figure \(\mathrm{h}=\mathrm{I} \sin \theta \ldots \ldots \ldots \ldots . . .(1)\)
Using the principle of conservation of energy,
\(
\begin{aligned}
&\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^2+1 \omega^2 \\
&=\frac{1}{2} \mathrm{mv}^2+\frac{1}{2}\left(\frac{\mathrm{mv}^2}{2}\right)\left(\frac{\mathrm{V}^2}{\mathrm{r}^2}\right)
\end{aligned}
\)
[Assuming no slipping of cylinder]
\(
\begin{aligned}
&\mathrm{mgh}=\frac{3}{4} \mathrm{mv}^2 \\
&\Rightarrow \mathrm{h}=\frac{3 \mathrm{~V}^2}{4 \mathrm{~g}}=\frac{3 \times(5)^2}{4 \times 9.8}=1.91 \mathrm{~m} \\
&\mathrm{I}=\frac{\mathrm{h}}{\sin \left(30^{\circ} 0\right.}=3.83 \mathrm{~m}
\end{aligned}
\)
(b) Time taken to reach the bottom \(=2 \mathrm{~T}\). The acceleration of cylinder along plane \(=a\)
\(
\therefore=u \mathrm{~T}+\frac{\mathrm{h}}{\sin \left(30^{\circ}\right)}(\mathrm{a}) \mathrm{T}^2 \ldots \ldots \ldots \ldots(2)
\)
\(
\therefore T=\sqrt{\frac{2 I}{a}}
\)
f.r. \(=1 a=\frac{m^2}{2} \times \frac{a}{r}\)
\(
\therefore \mathrm{f}=\frac{\mathrm{ma}}{2}
\)
\(
\Rightarrow \mathrm{mg} \quad \sin \theta=\frac{3 \mathrm{ma}}{2}
\)
\(
\therefore a=\frac{2}{3} \mathrm{~g} \quad \sin \theta
\)
\(\therefore\) Time taken to reach botton \(=2 \mathrm{~T}\)
\(
\begin{aligned}
&=2 \sqrt{\frac{21}{(2 / 3) g \sin \theta}} \\
&=\sqrt[2]{\frac{(3.83)}{(9.8)^{1 / 2}}} \\
&=3.06 \mathrm{sec}
\end{aligned}
\)

Hint

(a) Moment of inertia of man+platform system \(=7.6 \mathrm{kgm}^2\)
Moment of inertia of the weights \(=2 \times \mathrm{mr}^2=2 \times 5 \times 0.9^2=8.1 \mathrm{kgm}^2\)
total initial \(\mathrm{MOI}=7.6+8.1=15.7 \mathrm{kgm}^2\)
Initial angular momentum of system \(=\mathrm{I}_{\mathrm{i}} \omega_{\mathrm{i}}=15.7 \times 30\)
Moment of inertia of weights when man brings arms close \(=2 \times \mathrm{mr}^2\) \(=2 \times 5 \times(0.2)^2=0.4 \mathrm{kgm}^2\)
Thus the final moment of inertia \(=7.6+0.4=8.0 \mathrm{kgm}^2\)
Let the final angular momentum be \(\omega^{\prime}\)
Then from conservation of angular momentum,
\(
\begin{aligned}
&\mathrm{I}_{\mathrm{i}} \omega_{\mathrm{i}}=\mathrm{I}_{\mathrm{f}} \omega^{\prime} \\
&\Rightarrow \omega^{\prime}=\frac{15.7 \times 30}{8}=58.88 \mathrm{rev} / \mathrm{min}
\end{aligned}
\)

(b) Kinetic energy is not conserved in the given process. In fact, with the decrease in the moment of inertia, kinetic energy increases. The additional kinetic energy comes from the work done by the man to fold his hands toward himself.

Hint

Mass of the bullet, \(m=10 \mathrm{~g}=10 \times 10^{-3} \mathrm{~kg}\)
Velocity of the bullet, \(v=500 \mathrm{~m} / \mathrm{s}\)
Thickness of the door, \(L=1 \mathrm{~m}\)
Radius of the door, \(r=\frac{1}{2} \mathrm{~m}\)
Mass of the door, \(M=12 \mathrm{~kg}\)
Angular momentum imparted by the bullet on the door:
\(
\begin{aligned}
&\alpha=m v r \\
&=\left(10 \times 10^{-3}\right) \times(500) \times \frac{1}{2}=2.5 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1} \ldots(i)
\end{aligned}
\)
Moment of inertia of the door:
\(
\begin{aligned}
&I=\frac{1}{3} M L^2 \\
&=\frac{1}{3} \times 12 \times(1)^2=4 \mathrm{~kg} \mathrm{~m}^2
\end{aligned}
\)
But \(\alpha=I \omega\)
\(
\begin{aligned}
\therefore \omega &=\frac{\alpha}{I} \\
&=\frac{2.5}{4}=0.625 \mathrm{rad} \mathrm{s}^{-1}
\end{aligned}
\)

Hint

\(
v_{\mathrm{A}}=R \omega_{\mathrm{o}} ; v_{\mathrm{B}}=R \omega_{\mathrm{o}} ; v_c=\left(\frac{R}{2}\right) \omega_o ; \text { The disc will not roll }
\)

Angular speed of the disc \(=\omega_0\)
Radius of the disc \(=R\)
Using the relation for linear velocity, \(v=\omega_0 R\)
For point A:
\(v_A=R \omega_0 ;\) in the direction tangential to the right
Forpoint B:
\(v_{\mathrm{B}}=R \omega_0\); in the direction tangential to the left
For point \(C\) :
\(v_c=\left(\frac{R}{2}\right) \omega_o\); in the direction same as that of \(v_{\mathrm{A}}\)
The directions of motion of points \(A, B\), and \(C\) on the disc are shown in the following figure below

Since the disc is placed on a frictionless table, it will not roll. This is because the presence of friction is essential for the rolling of a body.

Hint

Here, \(m=10 \mathrm{~kg}, r=15 \mathrm{~cm}=0.15 \mathrm{~m}\)
\(
\theta=30^{\circ}, \mu_s=0.25
\)
Acceleration of the cylinder down the incline,
\(
a=\frac{2}{3} g \sin \theta=\frac{2}{3} \times 9.8 \sin 30^{\circ}=\frac{9.8}{3} \mathrm{~m} / \mathrm{s}^2
\)
(a) Force of friction,
\(
F=m g \sin \theta-m a=m(g \sin \theta-a)=10\left(9.8 \sin 30^{\circ}-\frac{9.8}{3}\right)=16.4 N
\)
(b) During rolling, the point of contact is at rest. Therefore, work done against friction is zero.
(c) For rolling without slipping/skidding, \(\mu=\frac{1}{3} \tan \theta\) \(\tan \theta=3 \mu=3 \times 0.25=0.75\) \(\theta=37^{\circ}\)

Hint

(a) False
Frictional force acts opposite to the direction of motion of the centre of mass of a body. In the case of rolling, the direction of motion of the centre of mass is backward. Hence, frictional force acts in the forward direction.
(b) True
Rolling can be considered as the rotation of a body about an axis passing through the point of contact of the body with the ground. Hence, its instantaneous speed is zero.
(c) False
This is because when a body is rolling, its instantaneous acceleration is not equal to zero. It has some value.
(d) True
This is because once the perfect rolling begins, force of friction becomes zero. Hence work done against friction is zero.
(e) True
This is because rolling occurs only on account of friction which is a tangential force capable of providing torque. When the inclined plane is perfectly smooth, the wheel will simply slip under the effect of its own weight.

Hint

With the \(x\)-and \(y\)-axes chosen as shown in Fig. above, the coordinates of points \(\mathrm{O}\), A and \(\mathrm{B}\) forming the equilateral triangle are respectively \((0,0)\), \((0.5,0)\), \((0.25,0.25 \sqrt{3})\). Let the masses \(100 \mathrm{~g}\), \(150 \mathrm{~g}\) and \(200 \mathrm{~g}\) be located at \(\mathrm{O}, \mathrm{A}\) and \(\mathrm{B}\) be respectlvely. Then,
\(
\begin{gathered}
X=\frac{m_1 x_1+m_2 x_2+m_3 x_3}{m_1+m_2+m_3} \\
=\frac{100(0)+150(0.5)+200(0.25) \quad g \mathrm{~m}}{(100+150+200) \mathrm{g}} \\
=\frac{75+50}{450} \mathrm{~m}=\frac{125}{450} \mathrm{~m}=\frac{5}{18} \mathrm{~m} \\
Y=\frac{100(0)+150(0)+200(0.25 \sqrt{3}) \quad \mathrm{gm}}{450 \mathrm{~g}} \\
=\frac{50 \sqrt{3}}{450} \mathrm{~m}=\frac{\sqrt{3}}{9} \mathrm{~m}=\frac{1}{3 \sqrt{3}} \mathrm{~m}
\end{gathered}
\)
The centre of mass \(C\) is shown in the figure.

Hint

\(
\begin{aligned}
\mathbf{a} \cdot \mathbf{b} &=(3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}) \cdot(-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}}) \\
&=-6-4-15 \\
&=-25
\end{aligned}
\)
\(
\mathbf{a} \times \mathbf{b}=\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
3 & -4 & 5 \\
-2 & 1 & -3
\end{array}\right|=7 \hat{\mathbf{i}}-\hat{\mathbf{j}}-5 \hat{\mathbf{k}}
\)
Note \(\mathbf{b} \times \mathbf{a}=-7 \hat{\mathbf{i}}+\hat{\mathbf{j}}+5 \hat{\mathbf{k}}\)

Hint

Answer Here \(\mathbf{r}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(\mathbf{F}=7 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}\).
We shall use the determinant rule to find the torque \(\boldsymbol{\tau}=\mathbf{r} \times \mathbf{F}\)
\(
\tau=\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
1 & -1 & 1 \\
7 & 3 & -5
\end{array}\right|=(5-3) \hat{\mathbf{i}}-(-5-7) \hat{\mathbf{j}}+(3-(-7)) \hat{\mathbf{k}}
\)
or \(\tau=2 \hat{\mathbf{i}}+12 \hat{\mathbf{j}}+10 \hat{\mathbf{k}}\)

Hint

The ladder AB is 3 m long, and its foot A is at a distance AC = 1 m from the wall. From Pythagoras theorem,

\(\mathrm{BC}=2 \sqrt{2} \mathrm{~m}\). The forces on the ladder are its weight \(\mathrm{W}\) acting at its centre of gravity D, reaction forces \(F_1\) and \(F_2\) of the wall and the floor respectively. Force \(F_1\) is perpendicular to the wall, since the wall is frictionless. Force \(F_2\) is resolved into two components, the normal reaction \(N\) and the force of friction \(F\). Note that \(F\) prevents the ladder from sliding away from the wall and is therefore directed toward the wall.

For translational equilibrium, taking the forces in the vertical direction,
\(N-W=0 \dots(i)\)
Taking the forces in the horizontal direction,
\(F-F_1=0 \dots(ii)\)
For rotational equilibrium, taking the moments of the forces about A,
\(2 \sqrt{2} F_1-(1 / 2) W=0 \dots(iii)\)
Now \(\quad W=20 \mathrm{~g}=20 \times 9.8 \mathrm{~N}=196.0 \mathrm{~N}\)
From (i) \(N=196.0 \mathrm{~N}\)
From (iii) \(F_1=W / 4 \sqrt{2}=196.0 / 4 \sqrt{2}=34.6 \mathrm{~N}\)
From (ii) \(F=F_1=34.6 \mathrm{~N}\)
\(
F_2=\sqrt{F^2+N^2}=199.0 \mathrm{~N}
\)
The force \(F_2\) makes an angle \(\alpha\) with the horizontal,
\(
\tan \alpha=N / F=4 \sqrt{2}, \quad \alpha=\tan ^{-1}(4 \sqrt{2}) \approx 80^{\circ}
\)

Hint

For the rod of mass \(M\) and length \(l\), \(I=M l^2 / 12\). Using the parallel axes theorem, \(I^{\prime}=I+M a^2\) with \(a=l / 2\) we get,
\(
I^{\prime}=M \frac{l^2}{12}+M\left(\frac{l}{2}\right)^2=\frac{M l^2}{3}
\)
We can check this independently since \(I\) is half the moment of inertia of a rod of mass \(2 M\) and length \(2 l\) about its midpoint,
\(
I^{\prime}=2 M \cdot \frac{4 l^2}{12} \times \frac{1}{2}=\frac{M l^2}{3}
\)

Hint

The tangent to the ring in the plane of the ring is parallel to one of the diameters of the ring. The distance between these two parallel axes is R, the radius of the ring. Using the parallel axes
theorem,

\(
I_{\text {tangent }}=I_{\text {dia }}+M R^2=\frac{M R^2}{2}+M R^2=\frac{3}{2} M R^2 .
\)

Hint

(i) We shall use \(\omega=\omega_0+\alpha t\)
\(\omega_0=\) initial angular speed in rad/s
\(=2 \pi \times\) angular speed in rev \(/ \mathrm{s}\)
\(=\frac{2 \pi \times \text { angular speed in rev } / \mathrm{min}}{60 \mathrm{~s} / \mathrm{min}}\)
\(=\frac{2 \pi \times 1200}{60} \mathrm{rad} / \mathrm{s}\)
\(=40 \pi \mathrm{rad} / \mathrm{s}\)
Similarly \(\omega=\) final angular speed in rad/s
\(
=\frac{2 \pi \times 3120}{60} \mathrm{rad} / \mathrm{s}
\)
\(=2 \pi \times 52 \mathrm{rad} / \mathrm{s}\)
\(=104 \pi \mathrm{rad} / \mathrm{s}\)
\(\therefore \quad\) Angular acceleration
\(
\alpha=\frac{\omega-\omega_0}{t} \quad=4 \pi \mathrm{rad} / \mathrm{s}^2
\)
The angular acceleration of the engine \(=4 \pi \mathrm{rad} / \mathrm{s}^2\)
(ii) The angular displacement in time \(t\) is given by
\(\theta=\omega_0 t+\frac{1}{2} \alpha t^2\)
\(=\left(40 \pi \times 16+\frac{1}{2} \times 4 \pi \times 16^2\right) \mathrm{rad}\)
\(=(640 \pi+512 \pi) \mathrm{rad}\)
\(=1152 \pi \mathrm{rad}\)
Number of revolutions \(=\frac{1152 \pi}{2 \pi}=576\)

Hint

\(
\begin{array}{ll}
\text { (a) We use } & I \alpha=\tau \\
\text { the torque } & \tau=F R \\
& =25 \times 0.20 \mathrm{Nm} \text { (as } R=0.20 \mathrm{~m}) \\
& =5.0 \mathrm{Nm}
\end{array}
\)
\(I=\) Moment of inertia of flywheel about its
\(
\begin{aligned}
&\text { axls }=\frac{M R^2}{2} \\
&=\frac{20.0 \times(0.2)^2}{2}=0.4 \mathrm{~kg} \mathrm{~m}^2 \\
&\alpha=\text { angular acceleration } \\
&\quad=5.0 \mathrm{~N} \mathrm{~m} / 0.4 \mathrm{~kg} \mathrm{~m}^2=12.5 \mathrm{~s}^{-2} \\
&\text { (b) Work done by the pull unwinding } 2 \mathrm{~m} \text { of the } \\
&\text { cord } \\
&=25 \mathrm{~N} \times 2 \mathrm{~m}=50 \mathrm{~J}
\end{aligned}
\)
(c) Let \(\omega\) be the final angular velocity. The kinetic energy galned \(=\frac{1}{2} I \omega^2\), since the wheel starts from rest. Now, \(\omega^2=\omega_0^2+2 \alpha \theta, \quad \omega_0=0\)
The angular displacement \(\theta=\) length of unwound string / radius of wheel \(=2 \mathrm{~m} / 0.2 \mathrm{~m}=10 \mathrm{rad}\)
\(
\omega^2=2 \times 12.5 \times 10.0=250(\mathrm{rad} / \mathrm{s})^2
\)
\(\therefore\) K.E. gained \(=\frac{1}{2} \times 0.4 \times 250=50 \mathrm{~J}\)

Hint

Let us take the mass of Hydrogen atom as 1 unit and Oxygen atom as 16 units. From the symmetry, it is clear that the CoM will lie on the perpendicular bisector of the line joining Hydrogen atoms which will pass through the oxygen atom. Let the distance of CoM from Oxygen atom be \(Y\).
Taking a horizontal line passing through the center of oxygen atom as an axis. \(\mathrm{OH}=\) \(r, m=\) mass of hydrogen atom, \(m^{\prime}=\) mass of Oxygen atom. \(M=2+16=18\) unit.
\(
\begin{aligned}
& Y=(1 / M) \times \left(2 \times \mathrm{~m r} \times \cos 52^{\circ}+16 \times 0\right) \\
& =0.066 \times 10^{-10} \mathrm{~m} . \\
& =6.6 \times 10^{-12} \mathrm{~m}
\end{aligned}
\)
Hence distance of the center of mass of the molecule from the center of oxygen atom is \(6.6 \times 10^{-12} \mathrm{~m}\)

Hint

Let ‘ \(O\) ‘ \((0,0)\) be the origin of the system.
Each brick is mass ‘ \(M\) ‘ \& length ‘ \(L\) ‘.
Each brick is displaced w.r.t. one in contact by ‘ \(L / 10\) ‘
\(\therefore\) The \(\mathrm{X}\) coordinate of the centre of mass
\(
{X}_{c m}=\frac{m\left(\frac{L}{2}\right)+m\left(\frac{L}{2}+\frac{L}{10}\right)+m\left(\frac{L}{2}+\frac{2 L}{10}\right)+m\left(\frac{L}{2}+\frac{3 L}{10}\right)+m\left(\frac{L}{2}+\frac{3 L}{10}-\frac{L}{10}\right)+m\left(\frac{L}{2}+\frac{L}{10}\right)+m\left(\frac{L}{2}\right)}{7 m}
\)
\(
\begin{aligned}
& =\frac{\frac{L}{2}+\frac{L}{2}+\frac{L}{10}+\frac{L}{2}+\frac{L}{5}+\frac{L}{2}+\frac{3 L}{10}+\frac{L}{2}+\frac{L}{5}+\frac{L}{2}+\frac{L}{10}+\frac{L}{2}}{7} \\
& =\frac{\frac{7 \mathrm{~L}}{2}+\frac{5 \mathrm{~L}}{10}+\frac{2 \mathrm{~L}}{5}}{7}=\frac{35 \mathrm{~L}+5 \mathrm{~L}+4 \mathrm{~L}}{10 \times 7}=\frac{44 \mathrm{~L}}{70}=\frac{22}{35} \mathrm{~L}
\end{aligned}
\)

Hint

Two masses \(m_1 \& m_2\) are placed on the \(X\)-axis
\(
\mathrm{m}_1=10 \mathrm{~kg} \text {, } \quad \mathrm{m}_2=20 \mathrm{~kg} \text {. }
\)
The first mass is displaced by a distance of \(2 \mathrm{~cm}\)
\(
\begin{aligned}
& \therefore \overline{\mathrm{x}}_{\mathrm{cm}}=\frac{\mathrm{m}_1 \mathrm{x}_1+\mathrm{m}_2 \mathrm{x}_2}{\mathrm{~m}_1+\mathrm{m}_2}=\frac{10 \times 2+20 \mathrm{x}_2}{30} \\
& \Rightarrow 0=\frac{20+20 \mathrm{x}_2}{30} \Rightarrow 20+20 \mathrm{x}_2=0 \\
& \Rightarrow 20=-20 \mathrm{x}_2 \Rightarrow \mathrm{x}_2=-1 .
\end{aligned}
\)
\(\therefore\) The \(2^{\text {nd }}\) mass should be displaced by a distance \(1 \mathrm{~cm}\) towards left so as to kept the position of centre of mass unchanged.

Hint

Two masses \(m_1 \& m_2\) are kept in a vertical line
\(
\mathrm{m}_1=10 \mathrm{~kg}, \quad \mathrm{~m}_2=30 \mathrm{~kg}
\)
The first block is raised through a height of \(7 \mathrm{~cm}\).
The centre of mass is raised by \(1 \mathrm{~cm}\).
\(
\begin{aligned}
& \therefore 1=\frac{\mathrm{m}_1 \mathrm{y}_1+\mathrm{m}_2 \mathrm{y}_2}{\mathrm{~m}_1+\mathrm{m}_2}=\frac{10 \times 7+30 \mathrm{y}_2}{40} \\
& \Rightarrow 1=\frac{70+30 \mathrm{y}_2}{40} \Rightarrow 70+30 \mathrm{y}_2=40 \Rightarrow 30 \mathrm{y}_2=-30 \Rightarrow \mathrm{y}_2=-1 .
\end{aligned}
\)
The \(30 \mathrm{~kg}\) body should be displaced \(1 \mathrm{~cm}\) downward in order to raise the centre of mass through \(1 \mathrm{~cm}\).

Hint

As the hall is gravity free, after the ice melts, it would tend to acquire a spherical shape. But, there is no external force acting on the system. So, the centre of mass of the system would not move. Zero cm is the answer.

Hint

\(
\mathrm{m}_1=60 \mathrm{~kg}, \quad \mathrm{~m}_2=40 \mathrm{~kg}, \quad \mathrm{~m}_3=50 \mathrm{~kg},
\)
Let \(\mathrm{A}\) be the origin of the system.
Initially Mr. Verma \& Mr. Mathur are at extreme position of the boat.
\(\therefore\) The centre of mass will be at a distance
\(
=\frac{60 \times 0+40 \times 2+50 \times 4}{150}=\frac{280}{150}=1.87 \mathrm{~m} \text { from ‘ } \mathrm{A} \text { ‘ }
\)
When they come to the mid point of the boat the \(\mathrm{CM}\) lies at \(2 \mathrm{~m}\) from ‘ \(\mathrm{A}\) ‘.
\(\therefore\) The shift in \(\mathrm{CM}=2-1.87=0.13 \mathrm{~m}\) towards right.
But as there is no external force in longitudinal direction their \(\mathrm{CM}\) would not shift. So, the boat moves \(0.13 \mathrm{~m}\) or \(13 \mathrm{~cm}\) towards right.

Hint

Let the bob fall at \(A,\). The mass of bob \(=m\).
The mass of cart \(=M\).
Initially their centre of mass will be at
\(
\frac{m \times L+M \times 0}{M+m}=\left(\frac{m}{M+m}\right) L
\)
Distance from \(P\)
When, the bob falls in the slot the \(\mathrm{CM}\) is at a distance ‘ \(\mathrm{O}\) ‘ from \(\mathrm{P}\).
Shift in \(\mathrm{CM}=0-\frac{\mathrm{mL}}{M+\mathrm{m}}=-\frac{\mathrm{mL}}{\mathrm{M}+\mathrm{m}}\) towards left \(=\frac{m L}{M+m}\) towards right.
But there is no external force in horizontal direction.
So the cart displaces a distance \(\frac{\mathrm{mL}}{\mathrm{M}+\mathrm{m}}\) towards right.

Hint

Initially, the monkey & balloon are at rest.
So the CM is at ‘ \(\mathrm{P}\) ‘
When the monkey descends through a distance ‘ \(L\) ‘
The CM will shift
\(t_0=\frac{m \times L+M \times 0}{M+m}=\frac{m L}{M+m}\) from \(P\)
So, the balloon descends through a distance \(\frac{m L}{M+m}\)

Hint

Let the mass of the to particles be \(m_1 \& m_2\) respectively \(\mathrm{m}_1=1 \mathrm{~kg}\),
\(
\mathrm{m}_2=4 \mathrm{~kg}
\)
\(\therefore\) According to question
\(
\begin{aligned}
& 1 / 2 m_1 v_1^2=1 / 2 m_2 v_2^2 \\
& \Rightarrow \frac{m_1}{m_2}=\frac{v_2^2}{v_1^2} \Rightarrow \frac{v_2}{v_1}=\sqrt{\frac{m_1}{m_2}} \Rightarrow \frac{v_1}{v_2}=\sqrt{\frac{m_2}{m_1}} \\
& \text { Now, } \frac{m_1 v_1}{m_2 v_2}=\frac{m_1}{m_2} \times \sqrt{\frac{m_2}{m_1}}=\frac{\sqrt{m_1}}{\sqrt{m_2}}=\frac{\sqrt{1}}{\sqrt{4}}=1 / 2 \\
& \Rightarrow \frac{m_1 v_1}{m_2 v_2}=1: 2
\end{aligned}
\)

Hint

As uranium 238 nucleus emits a \(\alpha\)-particle with a speed of \(1.4 \times 10^7 \mathrm{~m} / \mathrm{sec}\). Let \(v_2\) be the speed of the residual nucleus thorium 234 .
\(
\begin{aligned}
& \therefore m_1 v_1=m_2 v_2 \\
& \Rightarrow 4 \times 1.4 \times 10^7=234 \times v_2 \\
& \Rightarrow v_2=\frac{4 \times 1.4 \times 10^7}{234}=2.4 \times 10^5 \mathrm{~m} / \mathrm{sec}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& m_1 v_1=m_2 v_2 \\
& \Rightarrow 50 \times 1.8=6 \times 10^{24} \times v_2 \\
& \Rightarrow v_2=\frac{50 \times 1.8}{6 \times 10^{24}}=1.5 \times 10^{-23} \mathrm{~m} / \mathrm{sec}
\end{aligned}
\)
so, the earth will recoil at a speed of \(1.5 \times 10^{-23} \mathrm{~m} / \mathrm{sec}\).

Hint

Mass of proton \(=1.67 \times 10^{-27}\)
Let ‘ \(V_p\) ‘ be the velocity of proton
Given momentum of electron \(=1.4 \times 10^{-26} \mathrm{~kg} \mathrm{~m} / \mathrm{sec}\)
Given momentum of antineutrino \(=6.4 \times 10^{-27} \mathrm{~kg} \mathrm{~m} / \mathrm{sec}\)
a) The electron & the antineutrino are ejected in the same direction. As the total momentum is conserved the proton should be ejected in the opposite direction.
\(
1.67 \times 10^{-27} \times V_p=1.4 \times 10^{-26}+6.4 \times 10^{-27}=20.4 \times 10^{-27}
\)
\(\Rightarrow V_p=(20.4 / 1.67)=12.2 \mathrm{~m} / \mathrm{sec}\) in the opposite direction.
b) The electron \& antineutrino are ejected \(\perp^{\mathrm{r}}\) to each other. Total momentum of electron and antineutrino,
\(
\begin{aligned}
& =\sqrt{(14)^2+(6.4)^2} \times 10^{-27} \mathrm{~kg} \mathrm{~m} / \mathrm{s}=15.4 \times 10^{-27} \mathrm{~kg} \mathrm{~m} / \mathrm{s} \\
& \text { Since, } 1.67 \times 10^{-27} \mathrm{~V}_{\mathrm{p}}=15.4 \times 10^{-27} \mathrm{~kg} \mathrm{~m} / \mathrm{s} \\
& \text { So } V_p=9.2 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { Mass }=50 \mathrm{~g}=0.05 \mathrm{~kg} \\
& \mathrm{v}=2 \cos 45^{\circ} \hat{\mathrm{i}}-2 \sin 45^{\circ} \hat{\mathrm{j}} \\
& \mathrm{v}_1=-2 \cos 45^{\circ} \hat{\mathrm{i}}-2 \sin 45^{\circ} \hat{\mathrm{j}}
\end{aligned}
\)
(a) change in momentum \(=m \vec{v}-m \vec{v}_1\)
\(
\begin{aligned}
& =0.05\left(2 \cos 45^{\circ} \hat{i}-2 \sin 45^{\circ} \hat{j}\right)-0.05\left(-2 \cos 45^{\circ} \hat{i}-2 \sin 45^{\circ} \hat{j}\right) \\
& =0.1 \cos 45^{\circ} \hat{i}-0.1 \sin 45^{\circ} \hat{j}+0.1 \cos 45^{\circ} \hat{i}+0.1 \sin 45^{\circ} \hat{j} \\
& =0.2 \cos 45^{\circ} \hat{i} \\
& \therefore \text { magnitude }=\sqrt{\left(\frac{0.2}{\sqrt{2}}\right)^2}=\frac{0.2}{\sqrt{2}}=0.14 \mathrm{~kg} \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
(b) The change in magnitude of the momentum of the ball
\(
=\left|\vec{P}_i\right|-\left|\vec{P}_f\right|=2 \times 0.5-2 \times 0.5=0
\)

i.e. There is no change in magnitude of the momentum of the ball.

Hint

It is given that:
Wavelength of light \(=\lambda\)
Momentum of each photon \(=h / \lambda\)
Angle of incidence \(=\theta\)
\(
\begin{aligned}
& \vec{P}_{\text {incidence }}=(h / \lambda) \cos \theta \hat{i}-(h / \lambda) \sin \theta \hat{j} \\
& \vec{P}_{\text {Reflected }}=-(h / \lambda) \cos \theta \hat{i}-(h / \lambda) \sin \theta \hat{j}
\end{aligned}
\)
The change in momentum will be only in the \(x[latex]-axis direction. i.e.
[latex]
\begin{aligned}
& |\Delta P|=\left(\frac{h}{\lambda}\right) \cos \theta-\left(-\frac{h}{\lambda} \cos \theta\right) \\
& =\left(\frac{2 h}{\lambda}\right) \cos \theta
\end{aligned}
\)

Hint

Since the spaceship is removed from any material object \& totally isolated from surrounding, the missions by astronauts couldn’t slip away from the spaceship. So the total mass of the spaceship remain unchanged and also its velocity.
\(
\text { Hence, the spaceship moves with the speed of } 15 \mathrm{~km} / \mathrm{s} \text {. }
\)

Hint

It is given that:
Diameter of hailstone \(=1 \mathrm{~cm}=0.01 \mathrm{~m}\)
\(\Rightarrow\) Radius of hailstone, \(r=0.005 \mathrm{~m}\)
Average speed of hailstone \(=20 \mathrm{~m} / \mathrm{s}\)
Density of hailstone \(=900 \mathrm{~kg} / \mathrm{m} 3=0.9 \mathrm{~g} / \mathrm{cm} 3\)
Volume of the hailstones is given as,
\(
V=\frac{4}{3} \pi r^3
\)
\(\Rightarrow V=\frac{4}{3} \pi(0.005)^3=5.235 \times 10^{-7} \mathrm{~m}^3\)
Mass \(=\) volume \(\times\) density \(=5.235 \times 10^{-7} \times 900\)
\(=4.711 \times 10^{-4} \mathrm{~kg}\)
\(\therefore\) Mass of 2000 hailstone \(=2000 \times 4.711 \times 10^{-4}=0.9422\)
Rate of change of momentum \(=0.9422 \times 20 \approx 19 \mathrm{~N} / \mathrm{m}^2\)
\(\therefore\) The total force exerted on the roof \(=19 \times 100=1900 \mathrm{~N}\)

Hint

It is given that the mass of the ball is \(m\).
Let the ball be dropped from a height \(h\).
The speed of ball before the collision is \(v_1\).
\(
\therefore \mathrm{v}_1=\sqrt{2 \mathrm{gh}}
\)
The speed of ball after the collision is \(v_2\).
\(
v_2=-\sqrt{2 g h}
\)
Rate of change of velocity \(=\) acceleration
\(
\Rightarrow a=\frac{2 \sqrt{2 g h}}{t}
\)
\(\therefore\) Force, \(F=\frac{m \times 2 \sqrt{2 g h}}{t} \dots(1)\)
Using Newton’s laws of motion, we can write:
\(
\begin{aligned}
& v=\sqrt{2 g h}, s=h, u=0 \\
& \Rightarrow \sqrt{2 g h}=g t \\
& \Rightarrow t=\sqrt{\frac{2 h}{g}} \\
& \therefore \text { Total time }=2 \sqrt{\frac{2 h}{g}}
\end{aligned}
\)
\(
\text { Substituting this value of time } t \text { in equation (1), we get: }
\)
\(
\therefore F=\frac{m \times 2 \sqrt{2 g h}}{2 \sqrt{\frac{2 h}{g}}}=m g
\)

Hint

A railroad car of mass \(M\) is at rest on frictionless rails when a man of mass \(m\) starts moving on the car towards the engine. The car recoils with a speed \(v\) backward on the rails.
Let the mass is moving with a velocity \(x\) w.r.t. the engine.
\(\therefore\) The velocity of the mass w.r.t earth is \((x-v)\) towards right
\(\mathrm{V}_{\mathrm{cm}}=0\) (Initially at rest)
\(
\therefore 0=-\mathrm{Mv}+\mathrm{m}(\mathrm{x}-\mathrm{v})
\)
\(\Rightarrow M v=m(x-v) \Rightarrow m x=M v+m v \Rightarrow x=\left(\frac{M+m}{m}\right) v \Rightarrow x=\left(1+\frac{M}{m}\right) v\)

Hint

A gun is mounted on a railroad car. The mass of the car, the gun, the shells and the operator is \(50 \mathrm{~m}\) where \(m\) is the mass of one shell. The muzzle velocity of the shells is \(200 \mathrm{~m} / \mathrm{s}\).
Initial, \(\mathrm{V}_{\mathrm{cm}}=0\).
On applying the law of conservation of linear momentum, we get:
\(
\therefore 0=49 \mathrm{~m} \times \mathrm{V}+\mathrm{m} \times 200 \Rightarrow \mathrm{V}=\frac{-200}{49} \mathrm{~m} / \mathrm{s}
\)
\(\therefore \frac{200}{49} \mathrm{~m} / \mathrm{s}\) towards left.
When another shell is fired, then the velocity of the car, with respect to the platform is, \(\Rightarrow \mathrm{V}^{\prime}=\frac{200}{49} \mathrm{~m} / \mathrm{s}\) towards left.
When another shell is fired, then the velocity of the car, with respect to the platform is, \(\Rightarrow \mathrm{V}^{\prime}=\frac{200}{48} \mathrm{~m} / \mathrm{s}\) towards left
\(\therefore\) Velocity of the car w.r.t the earth is \(\left(\frac{200}{49}+\frac{200}{48}\right) \mathrm{m} / \mathrm{s}\) towards left.

Hint

A small block of mass \(m\) which is started with a velocity \(\vee\) on the horizontal part of the bigger block of mass \(\mathrm{M}\) placed on a horizontal floor.
Since the small body of mass \(m\) is started with a velocity \(\mathrm{V}\) in the horizontal direction, so the total initial momentum at the initial position in the horizontal direction will remain same as the total final momentum at the point \(\mathrm{A}\) on the bigger block in the horizontal direction.
Using the law of conservation of linear momentum, we can write:
Initial momentum = final momentum
\(
\begin{aligned}
& \mathrm{MV}+\mathrm{M} \times \mathrm{O}=(\mathrm{m}+\mathrm{M}) \mathrm{V} \\
& \Rightarrow \mathrm{V}=\frac{m v}{m+M}
\end{aligned}
\)
Therefore, the speed of the bigger block when the smaller block reaches point \(A\) of the surface is \(\frac{m v}{m+M}\).

Hint

It is given that:
Mass of the bugghi, \(\mathrm{m}_{\mathrm{b}}=200 \mathrm{~kg}\)
Velocity of the bugghi, \(V_b=10 \mathrm{~km} / \mathrm{h}\)
Mass of the boy, \(\mathrm{m}_{\text {boy }}=25 \mathrm{~kg}\)
Velocity of the boy, \(\mathrm{V}_{\text {Boy }}=4 \mathrm{~km} / \mathrm{h}\)
Consider the boy and the bugghi as a system.
The total momentum before the process of sitting remains same after the process of sitting.
Using the law of conservation of momentum, we can write:
\(
\begin{aligned}
& m_b V_b+m_{\text {boy }} V_{\text {boy }}=\left(m_b+m_{\text {boy }}\right) V \\
& \Rightarrow 200 \times 10+25 \times 4=(200+25) \times V \\
& \Rightarrow V=\frac{2100}{225}=\frac{28}{3} \mathrm{Km} / \mathrm{h}
\end{aligned}
\)

Hint

Mass of the ball \(=m_1=0.5 \mathrm{~kg}\), velocity of the ball \(=5 \mathrm{~m} / \mathrm{s}\)
Mass of the another ball \(m_2=1 \mathrm{~kg}\)
Let it’s velocity \(=v^{\prime} \mathrm{m} / \mathrm{s}\)
Using law of conservation of momentum,
\(
0.5 \times 5+1 \times v^{\prime}=0 \Rightarrow v^{\prime}=-2.5
\)
\(\therefore\) Velocity of second ball is \(2.5 \mathrm{~m} / \mathrm{s}\) opposite to the direction of motion of \(1^{\text {st }}\) ball.

Hint

It is given that:
Mass of the skater who is skating, \(m_1=60 \mathrm{~kg}\)
Initial speed of this man, \(\mathrm{v}_1=10 \mathrm{~m} / \mathrm{s}\)
Mass of the skater at rest, \(m_2=40 \mathrm{~kg}\)
Initial speedof this man, \(v_2=0\)
Let the velocity of both men after collision be \(v\).
Using the law of conservation of momentum, we can write:
\(
\begin{aligned}
& m_1 v_1+m_2 v_2=\left(m_1+m_2\right) v \\
& \Rightarrow 60 \times 10+0=100 \times v \\
& \Rightarrow v=6 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
The loss in kinetic energy during collision is given as,
\(
\begin{aligned}
& \Delta K E=\frac{1}{2} m_1 v_1^2-\frac{1}{2}\left(m_1+m_2\right) v^2 \\
& \Rightarrow \Delta K E=\frac{1}{2} \times 60 \times(10)^2-\frac{1}{2} \times 100 \times 36 \\
& \Rightarrow \Delta K E=1200 J
\end{aligned}
\)

Hint

It is given that:
Speed of the first particle during collision, \(v(t)=u_1+\frac{t}{\Delta t}\left(v_1-u_1\right) \vee\)
Let \(v^{\prime}\) be the speed of the second particle, during collision.
On applying the law of conservation of linear momentum on both particles, we get:
\(
\begin{aligned}
& \mathrm{m}_1 \mathrm{u}_1+\mathrm{m}_2 \mathrm{u}_2=\mathrm{m}_1 \mathrm{v}(\mathrm{t})+\mathrm{m}_2 \mathrm{v}^{\prime} \\
& \Rightarrow m_1 u_1+m_2 u_2=m_1 u_1+m_1 \times\left(\frac{t}{\Delta t}\right)\left(v_1-u_1\right)+m_2 v^{\prime}
\end{aligned}
\)
On dividing both the sides by \(m_2\), we get:
\(
\begin{aligned}
& u_2=\frac{m_1}{m_2}\left(\frac{t}{\Delta t}\right)\left(v_1-u_1\right)+v^{\prime} \\
& \Rightarrow v^{\prime}=u_2-\frac{m_1}{m_2}\left(\frac{t}{\Delta t}\right)\left(v_1-u_1\right)
\end{aligned}
\)
The speed of the second particle during collision can be written as a function of time and is given by the expression, \(u_2-\frac{m_1}{m_2}\left(\frac{t}{\Delta t}\right)\left(v_1-u_1\right)\).

Hint

Given:
The mass of the both balls is \(m\).
Initial speed of first ball \(=\mathrm{v}\)
Initial speed of second ball \(=0\)
Let the final of balls be \(v_1\) and \(v_2\) respectively.
\(
\begin{aligned}
& e=\frac{\text { velocity of separation }}{\text { velocity of approach }} \\
& \Rightarrow e=\frac{v_1-v_2}{v} \\
& \Rightarrow v_1-v_2=e v \ldots \text { (1) }
\end{aligned}
\)
On applying the law of conservation of linear momentum, we get:
\(
\begin{aligned}
& m\left(v_1+v_2\right)=m v \\
& \Rightarrow v_1+v_2=v \dots(2)
\end{aligned}
\)
According to the given condition,
Final K.E. \(=\frac{3}{4}\) Initial K.E.
\(
\begin{aligned}
& \Rightarrow \frac{1}{2} m v_1^2+\frac{1}{2} m v_2^2=\frac{3}{4} \times \frac{1}{2} m v^2 \\
& \Rightarrow v_1^2+v_2^2=\frac{3}{4} v^2 \\
& \Rightarrow \frac{\left(v_1+v_2\right)^2+\left(v_1-v_2\right)^2}{2}=\frac{3}{4} v^2 \\
& \Rightarrow \frac{\left(1+e^2\right) v^2}{2}=\frac{3}{4} v^2[\text { using the equations (1) and }(2)] \\
& \Rightarrow 1+e^2=\left(\frac{3}{2}\right) \\
& \Rightarrow e^2=\frac{1}{2} \\
& \Rightarrow e=\frac{1}{\sqrt{2}}
\end{aligned}
\)
Hence, the coefficient of restitution is found to be \(\frac{1}{\sqrt{2}}\)

Hint

(a) It is given that:
Mass of first block, \(m_1=2 \mathrm{~kg}\)
Initial speed, \(\mathrm{v}_1=2.0 \mathrm{~m} / \mathrm{s}\)
Mass of second block, \(m_2=2 \mathrm{~kg}\)
Initial speed of this block \(=0\)
For maximum possible loss in kinetic energy, we assume that the collision is elastic and both the blocks move with same final velocity \(\mathrm{v}\) (say).
On applying the law of conservation of linear momentum, we get:
\(
\begin{aligned}
& m_1 v_1+m_2 \times 0=\left(m_1+m_2\right) v \\
& 2 \times 2=(2+2) v \\
& \Rightarrow v=1 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
Loss in K.E. in elastic collision is give by,
\(
\begin{aligned}
& =\frac{1}{2} m_1 v_1^2-\frac{1}{2}\left(m_1+m_2\right) v \\
& =\left(\frac{1}{2}\right) \times 2 \times 2^2-\left(\frac{1}{2}\right)(2+2) \times(1)^2 \\
& =4-2=2 J
\end{aligned}
\)
(b) The actual loss in K.E. \(=\frac{\text { Maximum loss in K.E. }}{2}=1 J\)
Let the final velocities of the blocks be \(v_1\) and \(v_2\) respectively. The coefficient of restitution is e.
\(\therefore\) The loss in K.E. is given by,
\(\therefore\) The loss in K.E. is given by,
\(
\begin{aligned}
& \left(\frac{1}{2}\right) \times 2 \times(2)^2-\left(\frac{1}{2}\right) 2 \times v_1^2-\left(\frac{1}{2}\right) \times 2 v_2^2=1 \\
& \Rightarrow 4-\left(v_1^2+v_2^2\right)=1 \\
& \Rightarrow 4-\frac{\left(1+e^2\right) \times 4}{2}=1 \\
& \Rightarrow 2\left(1+e^2\right)=3 \\
& \Rightarrow 1+e^2=\frac{3}{2} \\
& \Rightarrow e^2=\frac{1}{2} \\
& \Rightarrow e=\frac{1}{\sqrt{2}}
\end{aligned}
\)

Hint

It is given that:
Mass of particles \(=100 \mathrm{~g}\)
Initial speed of the first particle \(=u\)
Final K.E. of the system after collision \(=0.2 \mathrm{~J}\)Initial K.E. of the system, before collision \(=\frac{1}{2} m u^2+0\) i.e. Initial K.E. \(=\frac{1}{2} \times 0.1 \times u^2=0.05 u^2\)
Let \(v_1\) and \(v_2\) be the final velocities of the first and second block respectively.
\(
\begin{aligned}
& m v_1+m v_2=m u \\
& \Rightarrow v_1+v_2=u \ldots(1) \\
& \left(v_1-v_2\right)+e\left(u_1-u_2\right)=0 \\
& \Rightarrow e u=v_2-v_1 \ldots(2)\left[\text { Putting } u_2=0, u_1=u\right]
\end{aligned}
\)
Adding the equations (1) and (2), we get:
\(
\begin{aligned}
& 2 v_2=(1+e) u \\
& \Rightarrow v_2=\left(\frac{u}{2}\right)(1+e) \\
& \therefore v_1=u-\frac{u}{2}(1+e) \\
& v_1=\frac{u}{2}(1-e)
\end{aligned}
\)
According to given condition,
\(
\begin{aligned}
& \frac{1}{2} m v_1^2+\frac{1}{2} m v_2^2=0.2 \\
& \Rightarrow v_1^2+v_2^2=4 \\
& \Rightarrow \frac{u^2}{2}\left(1+e^2\right)=4 \\
& \Rightarrow u^2=\frac{8}{1+e^2}
\end{aligned}
\)
For maximum value of \(u\), denominator should be minimum in the above equation.
\(
\begin{aligned}
& \text { i.e. } \mathrm{e}=0 \\
& \Rightarrow \mathrm{u}^2=8 \\
& \Rightarrow u=2 \sqrt{2} \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
For minimum value of \(u\), denominator should have maximum value.
i.e. \(\mathrm{e}=1\)
\(
\begin{aligned}
& \Rightarrow u^2=4 \\
& \Rightarrow u=2 \mathrm{~m} / \mathrm{s} .
\end{aligned}
\)

Hint

Let the velocity of the ball falling from height \(h_1\) be \(u\) (when it approaches the ground). Velocity on the ground, \(u=\sqrt{2 g h_1}\)
\(
\Rightarrow u=\sqrt{2 \times 9.8 \times 2}
\)
Let the velocity of ball when it separates from the ground be \(v\). (Assuming it goes up to height \(h_2\) )
\(
\begin{aligned}
& \Rightarrow v=\sqrt{2 g h_2} \\
& =\sqrt{2 \times 9.8 \times 1.5}
\end{aligned}
\)
Let the coefficient of restitution be e.
We know, \(v=\) eu
\(
\Rightarrow e=\frac{\sqrt{2 \times 9.8 \times 1.5}}{\sqrt{2 \times 9.8 \times 2}}=\frac{\sqrt{3}}{2}
\)
Hence, the coefficient of restitution is \(\frac{\sqrt{3}}{2}\).

Hint

Given,
Mass of each block, \(M_A=M_B=2 \mathrm{~kg}\)
Initial velocities of block \(A, V_a=1 \mathrm{~m} / \mathrm{s}\)
Initial velocity of block \(B, V_b=0\)
Spring constant of the spring \(=100 \mathrm{~N} / \mathrm{m}\)
Block A strikes the spring with a velocity of \(1 \mathrm{~m} / \mathrm{s}\).
After the collision, it’s velocity decreases continuously. At an instant the whole system (Block A + the compound spring + Block
B) moves together with a common velocity \(V\) (say).
Using the law of conservation of energy, we get:
\(
\begin{aligned}
& \left(\frac{1}{2}\right) M_A V_A^2+\left(\frac{1}{2}\right) M_B V_B^2=\left(\frac{1}{2}\right) M_A V^2+\left(\frac{1}{2}\right) M_B V^2+\left(\frac{1}{2}\right) k x^2 \\
& \left(\frac{1}{2}\right) \times 2(1)^2+0=\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right) \times v^2+\left(\frac{1}{2}\right) x^2 \times 100
\end{aligned}
\)
(where \(\mathrm{x}\) is the maximum compression of the spring) \(\Rightarrow 1-2 v^2=50 x^2 \dots(1)\)
As there is no external force acting in the horizontal direction, the momentum is conserved.
\(
\begin{aligned}
& \Rightarrow M_A V_A+M_B V_B=\left(M_A+M_B\right) V \\
& \Rightarrow 2 \times 1=4 \times V \\
& \Rightarrow V=\left(\frac{1}{2}\right) \mathrm{m} / \mathrm{s} \ldots(2)
\end{aligned}
\)
Substituting this value of \(\mathrm{V}\) in equation (1), we get:
\(
\begin{aligned}
& 1=2 \times\left(\frac{1}{4}\right)+50 x^2 \\
& \Rightarrow \frac{1}{4}=50 x^2 \\
& \Rightarrow x^2=\frac{1}{100} \\
& \Rightarrow x=\frac{1}{10} \mathrm{~m} \\
& \Rightarrow x=10 \mathrm{~cm}
\end{aligned}
\)

Hint

It is given that:
Mass of bullet, \(\mathrm{m}=20 \mathrm{~g}=0.02 \mathrm{~kg}\)
The initial speed, \(v_1=500 \mathrm{~m} / \mathrm{s}\)
Mass of block, \(M=10 \mathrm{~kg}\)
The initial speed of block \(=0\)
Final velocity of bullet, \(\mathrm{v}_2=100 \mathrm{~m} / \mathrm{s}\)
Let the final velocity of block when the bullet emerges out \(=v^{\prime}\)
Applying conservation of linear momentum,
\(
\begin{aligned}
& \mathrm{mv}_1+\mathrm{M} \times 0=\mathrm{mv}_2+\mathrm{Mv}^{\prime} \\
& \Rightarrow 0.02 \times 500=0.02 \times 100+10 \times \mathrm{v}^{\prime} \\
& \Rightarrow \mathrm{v}^{\prime}=0.8 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
Distance covered by the block, \(d=20 \mathrm{~cm}=0.02 \mathrm{~m}\).
Let friction coefficient between the block and the surface \(=\mu\)
Thus, the value of friction force, \(F=\mu m g\)
Change in K.E. of block = Work done by the friction force
\(
\begin{aligned}
& \Rightarrow \frac{1}{2} \times M \times 0-\frac{1}{2} \times M \times\left(v^{\prime}\right)^2=\mu \mathrm{mgd} \\
& \Rightarrow 0-\left(\frac{1}{2}\right) \times 10 \times(0.8)^2=\mu \times 10 \times 10 \times 0.2 \\
& \Rightarrow \mu=0.16
\end{aligned}
\)

Hint

It is given that:
Mass of block, \(M=200 \mathrm{~g}=0.20 \mathrm{~kg}\)
Mass of the particle, \(m=120 \mathrm{gm}=0.12 \mathrm{~kg}\)
Height of the particle, \(\mathrm{h}=45 \mathrm{~cm}=0.45 \mathrm{~m}\)
According to the question, as the block attains equilibrium, the spring is stretched by a distance, \(x=1.00 \mathrm{~cm}=0.01 \mathrm{~m}\).
i.e. \(M \times g=K \times x\)
\(
\begin{aligned}
& \Rightarrow 0.2 \times g=K \times x \\
& \Rightarrow 2=K \times 0.01 \\
& \Rightarrow K=200 \mathrm{~N} / \mathrm{m}
\end{aligned}
\)
The velocity with which the particle \(m\) strikes \(M\) is given by,
\(
\begin{aligned}
& u=\sqrt{2 g h} \\
& u=\sqrt{2 \times 10 \times 0.45} \\
& =\sqrt{9}=3 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
After the collision, let the velocity of the particle and the block be V.
According to law of conservation of momentum, we can write:
\(
\mathrm{mu}=(\mathrm{m}+\mathrm{M}) \mathrm{V}
\)
Solving for \(V\), we get:
\(
V=\frac{0.12 \times 3}{0.32}=\frac{9}{8} \mathrm{~m} / \mathrm{s}
\)
Let the spring be stretched through an extra deflection of \(\delta\).
On applying the law of conservation of energy, we can write:
Initial energy of the system before collision = Final energy of the system
\(
\Rightarrow \frac{1}{2} m u^2+\frac{1}{2} K x^2=\frac{1}{2}(m+M) V^2+\frac{1}{2} K(x+\delta)^2
\)
Substituting appropriate values in the above equation, we get:
\(
\left(\frac{1}{2}\right) \times 0.12 \times 9+\left(\frac{1}{2}\right) \times 200 \times(0.01)^2=\left(\frac{1}{2}\right) 0.32 \times\left(\frac{81}{64}\right)+\left(\frac{1}{2}\right) \times 200 \times(\delta+0.1)^2
\)
On solving the above equation, we get:
\(
\begin{aligned}
& \delta=0.061 \\
& \mathrm{~m}=6.1 \mathrm{~cm}
\end{aligned}
\)

Hint

Given:
Mass of bullet, \(\mathrm{m}=25 \mathrm{~g}=0.025 \mathrm{~kg}\)
Mass of ballistic pendulum, \(M=5 \mathrm{~kg}\)
Vertical displacement, \(\mathrm{h}=10 \mathrm{~cm}=0.1 \mathrm{~m}\)
Let the bullet strikes the pendulum with a velocity \(u\).
Let the final velocity be \(v\).
Using the law of conservation of linear momentum, we can write:
\(
\begin{aligned}
& m u=(M+m) v \\
& \Rightarrow v=\frac{m}{(M+m)} u \\
& \Rightarrow v=\frac{0.25}{5.025} \times u=\frac{u}{201}
\end{aligned}
\)
Applying the law of conservation of energy, we get:
\(
\begin{aligned}
& \left(\frac{1}{2}\right)(M+m) v^2=(M+m) g h \\
& \Rightarrow \frac{u^2}{(201)^2}=2 \times 10 \times 0.1 \\
& \Rightarrow u=201 \times \sqrt{2}=280 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
The bullet strikes the pendulum with a velocity of \(280 \mathrm{~m} / \mathrm{s}\).

Hint

Given:
Mass of bullet, \(m=20 \mathrm{gm}=0.02 \mathrm{~kg}\)
Horizontal speed of the bullet, \(u=300 \mathrm{~m} / \mathrm{s}\)
Mass of wooden block, \(M=500 \mathrm{gm}=0.5 \mathrm{~kg}\)
Let the bullet emerges out with velocity \(v\).
Let the velocity of the block be \(v^{\prime}\).
Using the law of conservation of momentum, we get:
\(
\mathrm{mu}=\mathrm{M} v^{\prime}+\mathrm{mv} \dots(1)
\)
Now, applying the work-energy principle for the block after the collision, we get:
\(
\begin{aligned}
& 0-\left(\frac{1}{2}\right) M \times\left(v^{\prime}\right)^2=-M g h \\
& \Rightarrow\left(v^{\prime}\right)^2=2 g h \\
& v^{\prime}=\sqrt{2 g h} \\
& =\sqrt{20 \times 10 \times 0.2}=2 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
On substituting the value of \(v^{\prime}\) in equation (1), we get:
\(
\begin{aligned}
& 0.02 \times 300=0.5 \times 2+0.02 \times v \\
& \Rightarrow v=\frac{6-1}{0.02}=\frac{5}{0.02} \\
& \Rightarrow v=250 \mathrm{~m} / \mathrm{s} \\
& 0.02 \times 300=0.5 \times 2+0.02 \times v \\
& \Rightarrow v=\frac{6-1}{0.02}=\frac{5}{0.02} \\
& \Rightarrow v=250 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
Hence, the speed of the bullet as it emerges out from the block is \(250 \mathrm{~m} / \mathrm{s}\).

Hint

Mass of the man \(\left(\mathrm{M}_{\mathrm{m}}\right)\) is \(50 \mathrm{~kg}\).
Mass of the pillow \(\left(M_p\right)\) is \(5 \mathrm{~kg}\).
When the pillow is pushed by the man, the pillow will go down while the man goes up. It becomes the external force on the system which is zero.
\(\Rightarrow\) acceleration of centre of mass is zero
\(\Rightarrow\) velocity of centre of mass is constant
\(\therefore\) As the initial velocity of the system is zero.
\(\therefore \mathrm{M}_{\mathrm{m}} \times \mathrm{V}_{\mathrm{m}}=\mathrm{M}_{\mathrm{p}} \times \mathrm{V}_{\mathrm{p}} \dots(1)\)
Given the velocity of pillow is \(80 \mathrm{ft} / \mathrm{s}\).
Which is relative velocity of pillow w.r.t. man.
\(
\vec{V}_{p / m}=\vec{V}_p-\vec{V}_m=V_p-\left(-V_m\right)=V_p+V_m \Rightarrow V_p=V_{p / m}-V_m
\)
Putting in equation (1)
\(
\begin{aligned}
& M_m \times V_m=M_p\left(V_{p / m}-V_m\right) \\
& \Rightarrow 50 \times V_m=5 \times\left(8-V_m\right) \\
& \Rightarrow 10 \times V_m=8-V_m \Rightarrow V_m=\frac{8}{11}=0.727 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
\(\therefore\) Absolute velocity of pillow \(=8-0.727=7.2 \mathrm{ft} / \mathrm{sec}\).
\(\therefore\) Time taken to reach the floor \(=\frac{\mathrm{S}}{\mathrm{V}}=\frac{8}{7.2}=1.1 \mathrm{sec}\).
As the mass of wall \(\gg\) then the pillow
The velocity of the block before the collision = velocity after the collision.
\(\Rightarrow\) Times of ascent \(=1.11 \mathrm{sec}\).
\(\therefore\) Total time taken \(=1.11+1.11=2.22 \mathrm{sec}\).

Hint

Given is a uniform chain of mass \(M\) and length \(L\).
Let us consider a small element of chain at a distance \(x\) from the floor having length ‘ \(d x\) ‘.
Therefore, mass of element,
\(
d m=\frac{M}{L} d x \dots91)
\)
The velocity with which the element strikes the floor is,
\(
v=\sqrt{2 g x} \dots(2)
\)
\(\therefore\) The momentum transferred to the floor is,
\(
P=(d m) v=\frac{M}{L} . d x \sqrt{2 g x}
\)
According to the given condition, the element comes to rest.
Thus, the force \(\left(\right.\) say \(F_1\) ) exerted on the floor = Change in momentum
\(
F_1=\frac{d P}{d t}=\frac{M}{L} \times \frac{d x}{d t} \sqrt{2 g x}
\)
(for the chain element)
\(
\begin{aligned}
& A s \frac{d x}{d t}=v \\
& F_1=\frac{M}{L} \times(\sqrt{2 g x})^2[\text { using equation (2)] } \\
& \Rightarrow F_1=\frac{M}{L} 2 g x=\frac{2 M g x}{L}
\end{aligned}
\)
Again, the force exerted due to length \(\mathrm{x}\) of the chain on the floor due to its own weight is given by,
\(
W=\frac{M}{L}(x) \times g=\frac{M g x}{L}
\)
Thus, the total force exerted is given by.
\(
F=F_1+W
\)
where \(W\) is the weight of chain below the element \(d x\) up to the floor.
\(
\Rightarrow F=\frac{2 M g x}{L}+\frac{M g x}{L}=\frac{3 M g x}{L}
\)

Hint

Given,
Speed of the block \(A=10 \mathrm{~m} / \mathrm{s}\)
The block \(B\) is kept at rest.
Coefficient of friction between floor and block \(B, \mu=0.10\)
Lets \(\mathrm{v}_1\) and \(\mathrm{v}_2\) be the velocities of \(\mathrm{A}\) and \(\mathrm{B}\) after collision respectively.
(a) If the collision is perfectly elastic, linear momentum is conserved.
Using the law of conservation of linear momentum, we can write:
\(
\begin{aligned}
& m u_1+m u_2=m v_1+m v_2 \\
& \Rightarrow 10+0=v_1+v_2 \\
& v_1+v_2=10 \ldots(1)
\end{aligned}
\)
We know,
Velocity of separation ( after collision ) = Velocity of approach (before collision )
\(
\begin{aligned}
& v_1-v_2=-\left(u_1-v_2\right) \\
& \Rightarrow v_1-v_2=-10 \ldots(2)
\end{aligned}
\)
Substracting equation (2) from (1), we get:
\(
\begin{aligned}
& 2 v_2=20 \\
& \Rightarrow v_2=10 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
The deceleration of block \(\mathrm{B}\) is calculated as follows:
Applying the work energy principle, we get:
\(
\begin{aligned}
& \left(\frac{1}{2}\right) \times m \times(0)^2-\left(\frac{1}{2}\right) \times m \times v^2=-m \times a \times s_1 \\
& \Rightarrow-\left(\frac{1}{2}\right) \times(10)^2=-\mu g \times s_1 \\
& \Rightarrow s_1=\frac{100}{2 \times 1 \times 10}=50 \mathrm{~m}
\end{aligned}
\)
(b) If the collision is perfectly inelastic, we can write:
\(
\begin{aligned}
& m \times u_1+m \times u_2=(m+m) \times v \\
& \Rightarrow m \times 10+m \times 0=2 m \times v \\
& \Rightarrow v=\left(\frac{10}{2}\right)=5 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
The two blocks move together, sticking to each other.
\(\therefore\) Applying the work-energy principle again, we get:
\(
\begin{aligned}
& \left(\frac{1}{2}\right) \times 2 m \times(0)^2-\left(\frac{1}{2}\right) \times 2 m \times(v)^2=- m \times \mu g \times s_2 \\
& \Rightarrow \frac{(5)^2}{0.1 \times 10}=s_2 \\
& \Rightarrow s_2=25 \mathrm{~m}
\end{aligned}
\)

Hint

Given:
Initial velocity of \(2 \mathrm{~kg}\) block, \(\mathrm{v}_1=1.0 \mathrm{~m} / \mathrm{s}\)
Initial velocity of the \(4 \mathrm{~kg}\) block, \(\mathrm{v}_2=0\)
Let the velocity of \(2 \mathrm{~kg}\) block, just before the collision be \(\mathrm{u}_1\).
Using the work-energy theorem on the block of \(2 \mathrm{~kg}\) mass:
The separation between two blocks, \(\mathrm{s}=16 \mathrm{~cm}=0.16 \mathrm{~m}\)
\(
\begin{aligned}
& \therefore\left(\frac{1}{2}\right) m \times u_1^2-\left(\frac{1}{2}\right) m \times(1)^2=-\mu \times m g \times s \\
& \Rightarrow u_1=\sqrt{(1)^2-2 \times 0.20 \times 10 \times 0.16} \\
& \Rightarrow u_1=0.6 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
As the collision is perfectly elastic, linear momentum is conserved.
Let \(\mathrm{v}_1, \mathrm{v}_2\) be the velocities of \(2 \mathrm{~kg}\) and \(4 \mathrm{~kg}\) blocks, just after collision.
Using the law of conservation of linear momentum, we can write:
\(
\begin{aligned}
& m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2 \\
& \Rightarrow 2 \times 0.6+4 \times 0=2 v_1+4 v_2 \\
& \Rightarrow 2 v_1+4 v_2=1.2 \ldots \text { (1) }
\end{aligned}
\)
For elastic collision,
Velocity of separation (after collision) = Velocity of approach (before collision)
\(
\begin{aligned}
i . e . v_1-v_2 & =+\left(u_1-u_2\right) \\
& =+(0.6-0) \\
\Rightarrow v_1-v_2 & =-0.6 \ldots(2)
\end{aligned}
\)
Substracting equation (2) from (1), we get:
\(
\begin{aligned}
& 3 v_2=1.2 \\
& \Rightarrow v_2=0.4 \mathrm{~m} / \mathrm{s} \\
& \therefore v_1=-0.6+0.4=-0.2 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
Let the \(2 \mathrm{~kg}\) block covers a distance of \(\mathrm{S}_1\).
\(\therefore\) Applying work-energy theorem for this block, when it comes to rest:
\(
\begin{aligned}
& \left(\frac{1}{2}\right) \times 2 \times(0)^2+\left(\frac{1}{2}\right) \times 2 \times(0.2)^2=-2 \times 0.2 \times 10 \times S_1 \\
& \Rightarrow S_1=1 \mathrm{~cm}
\end{aligned}
\)
Let the \(4 \mathrm{~kg}\) block covers a distance of \(\mathrm{S}_2\).
Applying work energy principle for this block:
\(
\begin{aligned}
& \left(\frac{1}{2}\right) \times 4 \times(0)^2-\left(\frac{1}{2}\right) \times 4 \times(0.4)^2=-4 \times 0.2 \times 10 \times S_2 \\
& \Rightarrow 2 \times 0.4 \times 0.4=4 \times 0.2 \times 10 \times S_2 \\
& \Rightarrow S_2=4 \mathrm{~cm}
\end{aligned}
\)
Therefore, the distance between the \(2 \mathrm{~kg}\) and \(4 \mathrm{~kg}\) block is given as, \(S_1+S_2=1+4=5 \mathrm{~cm}\)

Hint

Given \(\mathrm{h}<<\mathrm{R}\)
\(\mathrm{G}\) mass \(=6 \times 10^{24} \mathrm{~kg}\)
\(\mathrm{mb}=3 \times 10^{24} \mathrm{~kg}\)
Let \(V_e \rightarrow\) Velocity of earth
\(V_b \rightarrow\) Velocity of the block
The two blocks are attracted by gravitational force of attraction.The gravitational potential energy stored will be the K.E. of two blocks.
\(
\begin{aligned}
& G M_e m_b \frac{1}{R+\left(\frac{h}{2}\right)}-\frac{1}{R+h} \\
& =\left(\frac{1}{2}\right) m_e \times v_e^2+\left(\frac{1}{2}\right) M_b \times V_b^2 \dots(i)
\end{aligned}
\)
Again As the internal force acts
\(
M_e V_e=M_b V_b
\)
\(
\Rightarrow \quad V_e=\frac{M_b V_b}{M_e} \ldots(ii)
\)
Putting in equation (i),
\(
\begin{aligned}
& G m_e \times m_b\left[\frac{2}{2 R+h}-\frac{1}{R+h}\right] \\
& =\left(\frac{1}{2}\right) \times M_e \times \frac{M_b^2 v_b^2}{M_e^2}+\left(\frac{1}{2}\right) \times M_b \times V_b^2 \\
& =\left(\frac{1}{2}\right) \times V_b^2 \frac{M_b}{M_e}+\frac{1}{2} \times M_b \times v_b^2 \\
& \Rightarrow G M_e \frac{2 r+2 h-2 R-h}{(2 R+h)(R+h)}
\end{aligned}
\)
\(
\begin{aligned}
& =\left(\frac{1}{2}\right) \times V_b^2 \times\left(\frac{3 \times 10^{24}}{6 \times 10^{24}}+1\right) \\
& \Rightarrow\left[\frac{G M \times h}{2 R^2}\right]=\left(\frac{1}{2}\right) \times V_b^2 \times\left(\frac{3}{2}\right) \\
& \Rightarrow \quad g h=V_b^2 \times\left(\frac{3}{2}\right) \\
& \Rightarrow \quad V_b=\frac{2 g h}{3}
\end{aligned}
\)

Hint

Given
\(
t=4 \mathrm{~s}
\)
Initial angular velocity \(=\omega_0=0\)
Final angular velocity \(=\omega=100 \mathrm{rev} / \mathrm{s}\)
\(
\begin{aligned}
& \omega=\omega_0+\alpha t \\
& \alpha=\frac{\omega}{t} \\
& \alpha=\frac{100}{4} \mathrm{rev} / \mathrm{s}^2=25 \mathrm{rev} / \mathrm{s}^2
\end{aligned}
\)
Now, we have
\(
\begin{aligned}
& \theta=\omega_0 t+\frac{1}{2} \alpha t^2 \\
& \Rightarrow \theta=\frac{1}{2} \times 25 \times 16 \\
& =200^{\circ} \\
& \Rightarrow \theta=200 \times 2 \pi \text { radians } \\
& =400 \pi \text { radians }
\end{aligned}
\)

Hint

Given
Angular displacement of the wheel \(=\theta=50 \times 2 \pi=100 \pi\)
Initial angular velocity of the wheel \(=\omega_0=0\)
After, \(t=5\) seconds
\(
\begin{aligned}
& \theta=\omega_0 t+\frac{1}{2} \alpha t^2 \\
& \Rightarrow 100 \pi=\frac{1}{2} \times \alpha \times(5)^2 \\
& \Rightarrow 100 \pi=\frac{1}{2} \times \alpha \times 25 \\
& \Rightarrow \alpha=8 \pi \mathrm{rad} / \mathrm{s}^2 \text { or } 4 \mathrm{rev} / \mathrm{s} \\
& \omega=\omega_0+2 \alpha t \\
& \Rightarrow \omega=0+8 \pi \times 5=40 \pi \mathrm{rad} / \mathrm{s} \\
& \Rightarrow \omega=20 \mathrm{rev} / \mathrm{s}
\end{aligned}
\)

Hint

It is given that the area under the \(\omega-t\) curve gives the total angular displacement.
\(\therefore\) Maximum angular velocity \(=\omega=\alpha t\)
\(
\omega=4 \times 10=40 \mathrm{rad} / \mathrm{s}
\)

The area under the curve
\(
\begin{aligned}
& =\frac{1}{2} \times 10 \times 40+40 \times 10+\frac{1}{2} \times 40 \times 10 \\
& =800 \mathrm{rad}
\end{aligned}
\)
\(\therefore\) Total angle rotated in \(30 \mathrm{~s}=800 \mathrm{rad}\).

Hint

Given
Angular acceleration of the body \(=\alpha=2 \mathrm{rad} / \mathrm{s}^2\)
Initial angular velocity of the body \(=\omega_0=5 \mathrm{rad} / \mathrm{s}^2\)
Final angular velocity of the body \(=\omega=15 \mathrm{rad} / \mathrm{s}\)
We know that:
\(
\begin{aligned}
& \omega=\omega_0+\alpha t \\
& \Rightarrow t=\frac{\left(\omega-\omega_0\right)}{\alpha}=\frac{(15-5)}{1}=10 \mathrm{~s}
\end{aligned}
\)
Also,
\(
\begin{aligned}
& \theta=\omega_0 t+\frac{1}{2} \alpha t^2 \\
& \Rightarrow \theta=5 \times 10+\frac{1}{2} \times 1 \times 100 \\
& \Rightarrow \theta=50+50=100 \mathrm{rad}
\end{aligned}
\)

Hint

The angular displacement of the body,
\(
\begin{aligned}
& \theta=5 \mathrm{rev}=10 \pi \mathrm{rad} \\
& \alpha=2 \mathrm{rev} / \mathrm{s}^2=4 \pi \mathrm{rad} / \mathrm{s}^2
\end{aligned}
\)
Initial angular velocity, \(\omega_0=0\)
Final angular velocity, \(\omega=\) ?
\(
\begin{aligned}
& \omega^2=(2 \alpha \theta) \\
& \Rightarrow \omega=\sqrt{2 \times 4 \pi \times 10 \pi} \\
& =4 \pi \sqrt{5} \mathrm{rad} / \mathrm{s} \\
& \Rightarrow \omega=2 \sqrt{5} \mathrm{rev} / \mathrm{s}
\end{aligned}
\)

Hint

(a) Radius of disc \(=r=10 \mathrm{~cm}=0.1 \mathrm{~m}\)
Angular velocity of the disc \(=\omega=20 \mathrm{rad} / \mathrm{s}\)
\(\therefore\) Linear velocity of point on the rim \(=v=\omega r\)
\(
\Rightarrow v=20 \times 0.1=2 \mathrm{~m} / \mathrm{s}
\)
(b) Linear velocity of point at the middle of radius
\(
v=\frac{\omega r}{2}=\left(\frac{20 \times 0.1}{2}\right)=1 \mathrm{~m} / \mathrm{s}
\)

Hint

Angular acceleration of the disc,
\(
\alpha=4 \mathrm{rad} / \mathrm{s}^2
\)
Distance of the particle from the axis of rotation, \(r=1 \mathrm{~cm}=0.01 \mathrm{~m}\)
So, \(\omega=\alpha t=4 \mathrm{rad} / \mathrm{s} \ldots \ldots \ldots \ldots(t=1 s)\)
Radial acceleration,
\(
\begin{aligned}
& \alpha_r=\omega^2 r=4^2 \times 0.01 \\
& =0.16 \mathrm{~m} / \mathrm{s}^2=16 \mathrm{~cm} / \mathrm{s}^2
\end{aligned}
\)
Tangential acceleration,
\(
\begin{aligned}
& \alpha_t=\alpha r=0.04 \mathrm{~m} / \mathrm{s}^2 \\
& =4 \mathrm{~cm} / \mathrm{s}^2
\end{aligned}
\)

Hint

It is given that the string is moving on the rim of the disc and the block is connected with the string.
Therefore, the speeds of the block going down and the rim will be the same.

Angular speed of the disc, \(\omega=10 \mathrm{rad} / \mathrm{s}\)
Radius of the pulley, \(r=20 \mathrm{~cm}\)
Linear velocity of the rim, \(v=\) Tangential velocity \(=r \omega\) \(v=10 \times 20=200 \mathrm{~cm} / \mathrm{s}=2 \mathrm{~m} / \mathrm{s}\)
Therefore, velocity of the block \(=2 \mathrm{~m} / \mathrm{s}\)

Hint

(a)

The distance of mass at \(\mathrm{A}\) from the axis passing through side \(\mathrm{BC}\),
\(
(A D)=\frac{\sqrt{3}}{2} \times 10=5 \sqrt{3} \mathrm{~cm}
\)
Therefore, we have
Moment of inertia of mass about the axis \(B C\),
\(
\begin{aligned}
& l=m r^2=200 \times(5 \sqrt{3})^2 \\
& =200 \times 25 \times 3 \\
& =15000 \mathrm{gm}-\mathrm{cm}^2 \\
& =1.5 \times 10^{-3} \mathrm{~kg}-\mathrm{m}^2
\end{aligned}
\)

(b) let the axis of rotation passes through \(\mathrm{A}\) and is perpendicular to the plane of triangle.
Therefore, we have
Net moment of inertia,
\(
\begin{aligned}
& l=m r^2+m r^2 \\
& =2 m r^2 \\
& =2 \times 200 \times 10^2 \\
& =400 \times 100
=40000 \mathrm{gm}-\mathrm{cm}^2=4 \times 10^{-3} \mathrm{~kg}-\mathrm{m}^2
\end{aligned}
\)

Hint

Looking into the figure

Given in the question that a perpendicular line passes through 50th particle.
moment of inertia along BC
\(
I=m \times r^2
\)
Where,
\(\mathrm{m}=\) mass of the particle
\(r=\) distance or radius
Now let us consider those two particles at position \(49 \mathrm{~cm}\) and \(51 \mathrm{~cm}\).
Moment inertial due to these two particles will be
\(
\begin{aligned}
& =(49 \times 1)^2+(51 \times 1)^2 \\
& =100 \mathrm{gm} / \mathrm{cm}^2
\end{aligned}
\)
Thus, we will get 49 such sets and one particle at \(100 \mathrm{~cm}\). Therefore, total moment of inertia,
\(
I=\left(I_1+I_2+I_3 \ldots \ldots+I_{49}\right)+I^{\prime}
\)
Here, \(I^{\prime}\) is the moment of inertia of particle at \(100 \mathrm{~cm}\).
\(
=100\left\{1^2+2^2+3^2+\ldots+49^2\right\}+100(50)^2
\)
Using Arithmetic Progression, the equation for sum of square of \(n\) natural numbers is given by
\(
\sum n^2=\left(\frac{n(n+1)}{2}\right)^2
\)
Solving for moment of Inertia, with \(=50\), we will get-
\(
100 \times\left(\frac{50(51)}{2}\right)^2
\)
\(
\begin{aligned}
&=0.429 \mathrm{~kg}-\mathrm{m}^2\\
&=0.43 \mathrm{~kg}-\mathrm{m}^2 \text {. }
\end{aligned}
\)

Hint

The two bodies of mass \(\mathrm{m}\) and radius \(\mathrm{r}\) are moving along the common tangent. Therefore moment of inertia of the first body about \(X Y\) tangent. \(I_1=m r^2+2 / 5 \mathrm{mr}^2=\frac{7}{5} m r^2[latex]
[latex]
\text { Moment of inertia of the second body } X Y \text { tangent } I_2=\mathrm{mr}^2+2 / 5 \mathrm{mr}^2=7 / 5 \mathrm{mr}^2
\)
\(
\text { Therefore, net moment of inertia } I=I_1+I_2=7 / 5 \mathrm{mr}^2+7 / 5 \mathrm{mr}^2=14 / 5 \mathrm{mr}^2 \text { units. }
\)

Hint

Given Length of the rod, \(l=1 \mathrm{~m}\)
Mass of the rod, \(m=0.5 \mathrm{~kg}\)
Let the rod moves at a distance \(d\) from the centre.
On applying parallel axis theorem, we get
Moment of inertia about that axis,
\(
I=\left(\frac{m l^2}{12}\right)+m d^2=0.10
\)
\(
\begin{aligned}
& I=\frac{0.5 \times l^2}{12}+0.5 \times d^2=0.10 \\
& \Rightarrow \frac{1}{12}+d^2=0.2 \\
& \Rightarrow d^2=0.118 \\
& \Rightarrow d=0.342 \mathrm{~m} \text { from the centre }
\end{aligned}
\)

Hint

Moment of inertia of the ring about a point on the rim of the ring and the axis perpendicular to the plane of the ring \(=m r^2+m r^2=2 m r^2\) (from parallel axis theorem)
We know that
\(
m K^2=2 m r^2
\)
\({K}=\) Radius of the gyration
\(
\Rightarrow K=\sqrt{2 r^2}=\sqrt{2} r
\)

Hint

Moment of inertia of the disc about the centre and perpendicular to the plane of the disc \(=\frac{1}{2} m r^2\)
Radius of gyration of the disc about a point \(=\) Radius of the disc
Therefore, \(m k^2=\frac{1}{2} m r^2+m d^2\)
( \(k\) = Radius of gyration about acceleration point; \(d\) = Distance of that point from the centre)
\(
\begin{aligned}
& \Rightarrow k^2=\frac{r^2}{2}+d^2 \\
& \Rightarrow r^2=\frac{r^2}{2}+d^2 \\
& \Rightarrow \frac{r^2}{2}=d^2 \\
& \Rightarrow d=\frac{r}{\sqrt{2}}
\end{aligned}
\)

Hint

Let a small cross-sectional area is at a distance \(x\) from \(AB\) axis. Therefore mass of that small section \(=\mathrm{m} / \mathrm{a}^2 \times \mathrm{a} \times \mathrm{~dx}\) Therefore moment of inertia about \(AB\) axis
\(
\begin{aligned}
& I=2 \int_0^{\frac{a}{2}} \frac{m}{a^2} \times\left(a \times x^2 d x\right) \\
& =m \frac{a^2}{12}
\end{aligned}
\)
Therefore, for two lines, the
\(
I=2 \times m \frac{a^2}{12}=m \frac{a^2}{6}
\)
Now, considering diagonals, the moment of inertia for pair of perpendicular diagonals becomes
\(
I^{\prime}=2 \times m \frac{a^2}{12}=m \frac{a^2}{6}
\)
Therefore, by perpendicular axis theorem, moment of inertia becomes-
\(
\begin{aligned}
&\mathrm{I}+\mathrm{I}=\mathrm{I}^{\prime}\\
&\Rightarrow 2 \mathrm{I}=\mathrm{m} \frac{\mathrm{a}^2}{6}\\
&\Rightarrow \mathrm{I}=\mathrm{m} \frac{\mathrm{a}^2}{12}
\end{aligned}
\)

Hint

The surface density of a circular disc of radius a depends upon the distance from the centre as \(P(r)=A+B r\)
Therefore the mass of the ring of radius \(r\) will be =(A+B r) \times 2 \pi r d r \times r^2[/latex]
Therefore moment of inertia about the centre will be \(=\int_0^a(A+B r) 2 \pi r \times r^2 \times d r=\int_0^a 2 \pi A r^3 d r+\int_0^a 2 \pi B r^4 d r\)
\(
I=2 \pi\left(\frac{A a^4}{4}+\frac{B a^5}{5}\right)
\)

Hint

At the highest point, the total force acting on the particle is its weight acting downward.
\(
\text { Range of the particle }=\left(\frac{u^2 \sin 2 \theta}{g}\right)
\)
At the highest point, we have
Total force acting on the particle \(=m g(\) downward \()\)
Distance between the line of force and the point of projection
\(
\frac{\text { (total range) }}{2}=\frac{u^2 \sin 2 \theta}{2 g}
\)
\(
\begin{aligned}
& \text { So, } \vec{\tau}=\vec{F} \times d_{\perp}=m g \times u^2 \frac{\sin 2 \theta}{2 g} \\
& \vec{\tau}=m u^2 \frac{\sin 2 \theta}{2} \\
& =m u^2 \sin \theta \cos \theta
\end{aligned}
\)
Therefore, the direction of the torque is perpendicular to the plane of the motion.

Hint

Distance between the line of force and point of suspension, \(r=l \sin \theta\)
Torque, \(\vec{\tau}=\vec{F} \times \vec{r}\)
\(
\Rightarrow \tau=w r \sin \theta=w l \sin \theta
\)
Here, \(w\) is the weight of the bob.
The torque will be zero when the force acting on the body passes through the point of suspension, i.e., at the lowest point of suspension.

Hint

The force exerted into the wrench is given as \(F=6 \mathrm{~N}\), the angle of motion is \(30^{\circ}\). The distance from the nut to the wrench end is \(16 \mathrm{~cm}\).
The formula used is that of a torque which tells us the mechanics of force which helps the object to rotate. The formula is
\(
\text { Torque }=\text { F.rsin } \theta
\)
where \(F\) is the force applied on the object; \(r\) is the radius of the object turning. Turning angle is given by \(\theta\)
A force of \(6 \mathrm{~N}\) acting at an angle of \(30^{\circ}\) is just able to loosen the wrench at a distance \(8 \mathrm{~cm}\) from it. Therefore total torque in this case \(=6 \sin 30^{\circ} \times(8 / 100)\)
In this case, the total Torque is
\(
\begin{aligned}
& =\mathrm{F} \times 16 / 100
\end{aligned}
\)
To loosen the nut, torque in both cases should be the same. Thus, we have
\(
\begin{aligned}
& F \times \frac{16}{100}=6 \sin 30^{\circ} \times \frac{8}{100} \\
& F=\frac{(8 \times 3)}{16}=1.5 \mathrm{~N}
\end{aligned}
\)

Hint

The torque about a point \(=\) Total force \(\times\) perpendicular distance from the point to that force Let anticlockwise torque \(=+\) ve
And clockwise acting torque \(=-\) ve
Force acting at the point \(B\) is \(15 \mathrm{~N}\)
Therefore torque at \(\mathrm{O}\) due to this force
\(
\begin{aligned}
& =15 \times 6 \times 10^{-2} \times \sin 37^{\circ} \\
& =15 \times 6 \times 10^{-2} \times 3 / 5=0.54 \mathrm{~N}-\mathrm{m} \text { (anticlock wise) }
\end{aligned}
\)
Force acting at the point \(\mathrm{C}\) is \(10 \mathrm{~N}\) Therefore, torque at \(O\) due to this force \(=10 \times 4 \times 10^{-2}=0.4 \mathrm{~N}-\mathrm{m}\) (clockwise)
Force acting at the point \(\mathrm{A}\) is \(20 \mathrm{~N}\)
Therefore, Torque at \(O\) due to this force \(=20 \times 4 \times 10^{-2} \times \sin 30^{\circ}\) \(=20 \times 4 \times 10^{-2} \times 1 / 2=0.4 \mathrm{~N}\)-m (anticlockwise)
Therefore resultant torque acting at ‘ \(O\) ‘ \(=0.54-0.4+0.4=0.54 \mathrm{~N}-\mathrm{m}\).

Hint

Let \(\mathrm{N}\) be the normal reaction on the block.
From the free body diagram of the block, it is clear that forces \(N\) and mgcos \(\theta\) pass through the same line. So, there will be no torque due to \(\mathrm{N}\) and \(\mathrm{mg} \cos \theta\). The only torque will be produced by \(\mathrm{mg} \sin \theta\).
\(
\therefore \vec{\tau}=\vec{F} \times \vec{r}
\)
a is the edge of the cube.Therefore, we have
\(
\begin{aligned}
& r=\frac{a}{2} \\
& \therefore \tau=m g \sin \theta \times \frac{a}{2} \\
& =\frac{1}{2} m g a \sin \theta
\end{aligned}
\)

Hint

A rod of mass \(m\) and length \(L\), lying horizontally, is free to rotate about a vertical axis passing through its centre.
A force \(F\) is acting perpendicular to the rod at a distance \(L / 4\) from the centre.
Therefore torque about the centre due to this force
\(
\tau=\mathrm{F} \times \mathrm{r}=\mathrm{FL} / 4 .
\)
Let the torque produces an angular acceleration \(\alpha\).
\(
\begin{aligned}
& \tau=I \alpha \\
& \Rightarrow \tau=\frac{m L^2}{12} \times \alpha \ldots \ldots\left(I \text { of a rod }=\frac{m L^2}{12}\right) \\
& \Rightarrow F \frac{L}{4}=\frac{m L^2}{12} \times \alpha \\
& \Rightarrow \alpha=\frac{3 F}{m L}
\end{aligned}
\)
Now, \(\theta=\frac{1}{2} \alpha t^2 \ldots \ldots \ldots\) (initially at rest)
\(
\Rightarrow \theta=\frac{3 F t^2}{2 m L}
\)

Hint

A square plate of mass \(120 \mathrm{gm}\) and edge \(5 \mathrm{~cm}\) rotates about one of the edge.
Let take a small area of the square of width \(\mathrm{dx}\) and length a which is at a distance \(\mathrm{x}\) from the axis of rotation.
Therefore mass of that small area
\(\mathrm{m} / \mathrm{a}^2 \times \mathrm{a} \times \mathrm{dx}(\mathrm{m}=\) mass of the square \(; \mathrm{a}=\) side of the plate \()\)
\(
\begin{aligned}
& l=\int_0^a \frac{m}{a^2} \times a x^2 d x \\
& =\frac{m}{a}\left[\frac{x^3}{3}\right]_0^a=\frac{m a^2}{3}
\end{aligned}
\)
Now, torque produced,
\(
\begin{aligned}
& \tau=\left(\frac{m a^2}{3}\right) \times \alpha \\
& =\left\{\frac{120 \times 10^{-3} \times 5^2 \times 10^{-4}}{3}\right\} \times .02 \\
& =40 \times 25 \times 10^{-7} \times 0.2 \\
& =2 \times 10^{-5} N-m .
\end{aligned}
\)

Hint

Moment of inertia of a square plate about its diagonals,
\(
I=\frac{m a^2}{12}
\)
Therefore, we have
Torque produced,
\(
\begin{aligned}
& \tau=I \alpha \\
& \Rightarrow \tau=\left(\frac{m a^2}{12}\right) \times \alpha \\
& =\frac{120 \times 10^{-3} \times 5^2 \times 10^{-4}}{12} \times 0.2 \\
& =10 \times 25 \times 10^{-4} \times 0.2 \\
& =0.5 \times 10^{-5} \mathrm{~N}-\mathrm{m}
\end{aligned}
\)

Hint

A flywheel of moment of inertia \(5 \mathrm{~kg} \mathrm{~m}\) is rotated at a speed of \(60 \mathrm{rad} / \mathrm{s}\). The flywheel comes to rest due to the friction at the axle after 5 minutes.
Therefore, the angular deceleration produced due to frictional force \(=\omega=\omega_0+\alpha t\)
\(
\begin{aligned}
\alpha & =-\frac{\omega_0}{t} \\
\alpha & =-\left(\frac{60}{300}\right)=-\frac{1}{5} \mathrm{rad} / \mathrm{s}^2
\end{aligned}
\)
(a)
Therefore torque produced by the frictional force \((R)\) is \(\tau=\mathrm{I} \times \alpha=5 \times(-1 / 5)=\mathrm{1 N}-\mathrm{m}\) opposite to the rotation of wheel.
(b) By conservation of energy,
Total work done in stopping the wheel by frictional force = Change in energy
\(
\begin{aligned}
& W=\frac{1}{2} I \omega^2 \\
& =\frac{1}{2} \times 5 \times(60 \times 60) \\
& =9000 \text { joule }=9 \mathrm{~kJ}
\end{aligned}
\)
(c) Angular velocity after 4 minutes
\(
\Rightarrow \omega=\omega_0+\alpha \mathrm{t}=60-240 / 5=12 \mathrm{rad} / \mathrm{s}
\)
Therefore angular momentum about the centre \(L=I \times \omega=5 \times 12=60 \mathrm{~kg}-\mathrm{m}^2 / \mathrm{s}\).

Hint

The earth’s angular speed decreases by \(0.0016 \mathrm{rad} /\) day in 100 years.
Rate of change of angular velocity, i.e., angular acceleration,
\(
\begin{aligned}
& \alpha=\left(\frac{0.0016}{100}\right) \mathrm{rad} / \text { day } \\
& \Rightarrow \alpha=\left\{\frac{0.0016}{(86400)^2 \times 100 \times 365}\right\} \ldots \ldots \ldots \cdot[1 \text { year }=365 \text { days }=365 \times 86400 \mathrm{sec}]
\end{aligned}
\)
Torque produced by the ocean water in decreasing the Earth’s angular velocity,
\(
\begin{aligned}
& \tau=I \alpha=\frac{2}{5} m r^2 \alpha \\
& =\frac{2}{5} \times 6 \times 10^{24} \times\left(64 \times 10^5\right)^2 \times\left\{\frac{0.0016}{86400^2 \times 100 \times 365}\right\} \\
& =5.8 \times 10^{20} N-m
\end{aligned}
\)

Hint

Initial angular velocity of the wheel,
\(\omega_0=600 \mathrm{rpm}\)
\(=\frac{600}{60}=10\) revolutions per second
After 10 seconds,
Final angular velocity of the wheel,
\(\omega=0\)
So, \(\omega_0=-a t\)
\(
\Rightarrow \alpha=-\frac{10}{10}=-1 \mathrm{rev} / \mathrm{s}^2
\)
Now, \(t=5 s\)
We know that
\(
\begin{aligned}
& \omega^{\prime}=\omega_0+a t \\
& \Rightarrow \omega^{\prime}=10-1 \times 5=5 \mathrm{rev} / \mathrm{s}
\end{aligned}
\)

Hint

Initial angular velocity of the wheel,
\(\omega=100 \mathrm{rev} / \mathrm{min}\)
\(=\frac{100}{60}=\frac{5}{3} \mathrm{rev} / \mathrm{s}=\frac{10 \pi}{3} \mathrm{rad} / \mathrm{s}\)
\(\theta=10 \mathrm{rev}=20 \pi \mathrm{rad}\)
\(r=0.2 m\)
Angular deceleration produced by the tangential force in order to stop the wheel after 10 revolutions,
\(
\alpha=\frac{\omega^2}{2 \theta}
\)
Torque by which the wheel will come to rest,
\(
\tau=I_{c m} \alpha
\)
Or, \(\Rightarrow \tau=F \times r=I_{c m} \alpha\)
Putting the values of \(I_{c m}\) and \(\alpha\), we get
\(
\begin{aligned}
& F \times r=\frac{1}{2} m r^2 \times\left(\frac{10 \pi}{3}\right)^2 \times \frac{1}{2 \times 20 \pi} \\
& \Rightarrow F=\frac{1}{2} \times 10 \times 0.2 \times \frac{100 \pi^2}{(9 \times 2 \times 20 \pi)} \\
& \Rightarrow F=\frac{5 \pi}{18}=\frac{15.7}{18}=0.87 \mathrm{~N}
\end{aligned}
\)

Hint

A cylinder is moving with an angular velocity \(50 \mathrm{rev} / \mathrm{s}\) brought in contact with another identical cylinder in rest. The first and second cylinder has common acceleration and deacceleration as \(1 \mathrm{rad} / \mathrm{s}^2\) respectively.
After \(t\) seconds, let the angular velocity of two cylinders be \(\omega\).
For the first cylinder,
\(
\begin{aligned}
& \omega_0=50 \mathrm{rev} / \mathrm{s}, \alpha=1 \mathrm{rev} / \mathrm{s}^2 \\
& \therefore \omega=50-\alpha t \\
& \Rightarrow t=\frac{(\omega-50)}{-1}
\end{aligned}
\)
For the second cylinder,
\(
\begin{aligned}
& \omega_0=0, \alpha=1 \mathrm{rev} / \mathrm{s}^2 \\
& \therefore \omega=\alpha t \\
& \Rightarrow t=\frac{\omega}{1}
\end{aligned}
\)
On equating the value of \(t\), we get
\(
\begin{aligned}
& \omega=\left(\frac{\omega-50}{-1}\right) \\
& \Rightarrow 2 \omega=50 \\
& \Rightarrow \omega=25 \mathrm{rev} / \mathrm{s} \\
& \therefore t=\frac{25}{1} s=25 \mathrm{~s}
\end{aligned}
\)

Hint

Initial angular velocity \(=20 \mathrm{rad} / \mathrm{s}\)
Therefore \(\alpha=2 \mathrm{rad} / \mathrm{s}^2\)
\(\Rightarrow \mathrm{t}_1=\omega / \alpha=20 / 2=10 \mathrm{sec}\)
Therefore \(10 \mathrm{sec}\) it will come to rest.
Since the same torque is continues to act on the body it will produce same angular acceleration and since the initial kinetic energy = the kinetic energy at a instant.
So initial angular velocity \(=\) angular velocity at that instant
Therefore time require to come to that angular velocity,
\(\mathrm{t}_2=\omega / \alpha=20 / 2=10 \mathrm{sec}\)
therefore time required \(=t_1+t_2=20 \mathrm{sec}\).

Hint

Total moment of inertia of the system about the axis of rotation,
\(
\begin{aligned}
& I_{\text {net }}=\left(m_1 r_1^2+m_2 r_2^2\right) \\
& \tau_{\text {net }}=F_1 r_1-F_2 r_2 \\
& \text { Also, } \tau_{n e t}=I_{\text {net }} \times \alpha
\end{aligned}
\)
On equating the value of \(\tau_{\text {net }}\) and putting the value of \(I_{\text {net }}\), we get
\(
\begin{aligned}
& F_1 r_1-F_2 r_2=\left(m_1 r_1^2+m_2 r_2^2\right) \times \alpha \\
& (-2 \times 10 \times 0.5)+(5 \times 10 \times 0.5)=\left[5\left(\frac{1}{2}\right)^2+2\left(\frac{1}{2}\right)^2\right] \alpha \\
& \Rightarrow 15=\frac{7}{4} \alpha \\
& \Rightarrow \alpha=\frac{60}{7}=8.57 \mathrm{rad} / \mathrm{s}^2
\end{aligned}
\)

Hint

(a) Total moment of inertia of the system about the axis of rotation,
\(
I_{n e t}=\left(m_1 r_1^2+m_2 r_2^2+\frac{m l^2}{12}\right)
\)
\(\mathrm{m}\) and \(\mathrm{I}\) are the mass and length of the rod, respectively.
\(
\tau_{\text {net }}=F_1 r_1-F_2 r_2
\)
Also, \(\tau_{n e t}=I_{n e t} \times \alpha\)
On equating the value of \(\tau_{\text {net }}\) and putting the value of \(l_{\text {net }}\), we get \(F_1 r_1-F_2 r_2=\left(m_1 r_1^2+m_2 r_2^2+\frac{m l^2}{12}\right) \times \alpha\)
\(
(-2 \times 10 \times 0.5)+(5 \times 10 \times 0.5)=\left[5\left(\frac{1}{2}\right)^2+2\left(\frac{1}{2}\right)^2+\frac{(1)^2}{12}\right] \alpha
\)
\(
\begin{aligned}
& \Rightarrow 15=(1.75+0.084) \alpha \\
& \Rightarrow \alpha=\frac{1500}{(175+8.4)}=\frac{1500}{183.4} \\
& =8.1 \mathrm{rad} / \mathrm{s}^2 \ldots \ldots \ldots(g=10) \\
& \left.=8.01 \mathrm{rad} / \mathrm{s}^2 \ldots \ldots \text { (if } g=9.8\right)
\end{aligned}
\)

(b) From the free body diagram of the block of mass \(2 \mathrm{~kg}\),
\(
\begin{aligned}
& T_1-m_1 g=m_1 a \\
& \Rightarrow T_1=2(a+g) \\
& =2(\alpha r+g) \ldots \ldots \text { (using, } a=\alpha r) \\
& =2(8 \times 0.5+9.8) \\
& \Rightarrow T_1=27.6 \mathrm{~N}
\end{aligned}
\)
From the free body diagram of the block of mass \(5 \mathrm{~kg}\),
\(
\begin{aligned}
& m_2 g-T_2=m_2 a \\
& \Rightarrow T_2=m_2(g-a) \\
& =5(g-a)=5(9.8-8 \times 0.5) \ldots \ldots \ldots \ldots(a=\alpha r) \\
& =5 \times 5.8=29 N
\end{aligned}
\)

Hint

According to the equation, we have
\(
M g-T_1=M a \dots(1)
\)
\(
T_2=\text { ma………(2) }
\)
\(
\Rightarrow\left(T_1-T_2\right)=\frac{I a}{r^2} \ldots \ldots \ldots(3)[\text { Because, } a=r a]
\)
If we add the equations (1) and (2), we get
\(
\begin{aligned}
& M g+T_2-T_1=M a+m a . \\
& \Rightarrow M g-I \frac{a}{r^2}=M a+m a \\
& \Rightarrow\left(M+m+\frac{I}{r^2}\right) a=M g \\
& \Rightarrow a=\frac{M g}{M+m+\frac{I}{r^2}}
\end{aligned}
\)

Hint

Moment of inertia of the bigger pulley, \(\mathrm{I}=0.20 \mathrm{~kg}-\mathrm{m}^2\), \(\mathrm{r}=10 \mathrm{~cm}=0.1 \mathrm{~m}\),
The smaller pulley is light. Therefore, on neglecting its moment of inertia, we have Mass of the block, \(m=2 \mathrm{~kg}\)
From the free body diagram, we get
\(
\begin{aligned}
& m g-T=m a \ldots \dots(1) \\
& T r=I \alpha \text { And } \\
& a=\alpha r \\
& \Rightarrow T=\frac{I a}{r^2} \dots(2)
\end{aligned}
\)
Using equations (1) and (2), we get
\(
\begin{aligned}
& m g=\left(m+\frac{I}{r^2}\right) a \\
& \Rightarrow a=\frac{m g}{m+\frac{I}{r^2}} \\
& =\frac{2 \times 9.8}{2+\left(\frac{0.2}{0.01}\right)} \\
& =\frac{19.6}{22}=0.89 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
\)

Hint

Given
\(
\begin{aligned}
& m=2 \mathrm{~kg}, \mathrm{I}_1=0.10 \mathrm{~kg}-\mathrm{m}^2 \\
& \mathrm{r}_1=5 \mathrm{~cm}=0.05 \mathrm{~m} \\
& \mathrm{I}_2=2.20 \mathrm{~kg}-\mathrm{m}^2 \\
& \mathrm{r}_2=10 \mathrm{~cm}=0.1 \mathrm{~m}
\end{aligned}
\)
From the free body diagram, we have
\(
\begin{aligned}
& m g-T_1=m a \ldots(1) \\
& \left(T_1-T_2\right) r_1=I_1 \alpha \ldots \ldots(2) \\
& T_2 r_2=I_2 \alpha \ldots \ldots(3)
\end{aligned}
\)
Substituting the value of \(T_2\) in the equation (2), we get
\(
\begin{aligned}
& \Rightarrow\left(T_1-I_2 \frac{\alpha}{r_2}\right) r_1=I_1 \alpha \\
& \Rightarrow T_1-I_2 \frac{a}{r_2^2}=I_1 \frac{a}{r_1^2} \\
& \Rightarrow T_1=\left\{\left(\frac{I_1}{r_1^2}\right)+\left(\frac{I_2}{r_2^2}\right)\right\} a
\end{aligned}
\)
Substituting the value of \(T_1\) in the equation (1), we get
\(
\begin{aligned}
& m g-\left\{\left(\frac{I_1}{r_1^2}\right)+\left(\frac{I_2}{r_2^2}\right)\right\} a=m a \\
& \Rightarrow \frac{m g}{\left\{\left(\frac{I_1}{r_1^2}\right)+\left(\frac{I_2}{r_2^2}\right)\right\}+m}=a \\
& \Rightarrow a=\frac{2 \times 9.8}{\frac{0.1}{0.0025}+\frac{0.2}{0.01}+2} \\
& =0.316 m / s^2 \\
& \Rightarrow T_2=I_2 \frac{a}{r_2^2} \\
& =\frac{0.20 \times 0.316}{0.01}=6.32 \mathrm{~N}
\end{aligned}
\)

Hint

Free the body diagram of the system is given below.

For block of mass \(\mathrm{M}\),
\(
M g-T_1=M a \dots(1)
\)
\(\left(T_1-T_2\right) R=I \alpha\) using,\(a=\alpha r\)
\(\Rightarrow\left(T_1-T_2\right)=I \frac{a}{R^2} \ldots \ldots(2)(\) For pully 1\()\)
\(
\text { Similarly, }\left(T_2-T_3\right)=I \frac{a}{R^2} \ldots \ldots \ldots \text { (3) (For pully 2) }
\)
For block of mass \(\mathrm{m}\),
\(T_3-m g=m a \ldots \ldots \ldots(4)(\) For block \(m)\)
Adding equations (2) and (3), we get
\(
\left(T_1-T_3\right)=\frac{2 I a}{R^2} \dots(5)
\)
Adding equations ( 1 ) and (4), we get
\(
-m g+M g+\left(T_3-T_1\right)=M a+m a \dots(6)
\)
Using equations (5) and (6), we get
\(
\begin{aligned}
& M g-m g=M a+m a+\frac{2 I a}{R^2} \\
& \Rightarrow a=\frac{(M-m) g}{\left(M+m+\frac{2 I}{R^2}\right)}
\end{aligned}
\)