6.13 Exercise Problems

Q1: A block of mass \(m\) slides along a frictionless surface as shown in the figure below. If it is released from rest at \(A\), what is its speed at \(B\)?

Answer: Take the block + the earth as the system. Only the block moves, so only the work done on the block will contribute to the gravitational potential energy. As it descends through a height \(h\) between \(A\) and \(B\), the potential energy decreases by \(m g h\). The normal contact force \(\mathcal{N}\) on the block by the surface does no work as it is perpendicular to its velocity. No external force does any work on the system. Hence, increase in kinetic energy \(=\) decrease in potential energy or,
\(
\frac{1}{2} m v^2=m g h \text { or, } v=\sqrt{2 g h} \text {. }
\)

Q2: A pendulum bob has a speed \(3 \mathrm{~m} / \mathrm{s}\) while passing through its lowest position. What is its speed when it makes an angle of \(60^{\circ}\) with the vertical? The length of the pendulum is \(0.5 \mathrm{~m}\). Take \(g=10 \mathrm{~m} / \mathrm{s}^2\).

Answer: Take the bob + earth as the system. The external force acting on the system is that due to the string. But this force is always perpendicular to the velocity of the bob and so the work done by this force is zero. Hence, the total mechanical energy will remain constant. As is clear from Figure (8.11), the height ascended by the bob at an angular displacement \(\theta\) is \(l-l \cos \theta=l(1-\cos \theta)\). The increase in the potential energy is \(m g l(1-\cos \theta)\). This should be equal to the decrease in the kinetic energy of the system. Again, as the earth does not move in the lab frame, this is the decrease in the kinetic energy of the bob. If the speed at an angular displacement \(\theta\) is \(v_1\), the decrease in kinetic energy is

\(
\frac{1}{2} m v_0^2-\frac{1}{2} m v_1^2,
\)
where \(v_0\) is the speed of the block at the lowest position.
Thus,
\(
\frac{1}{2} m v_0^2-\frac{1}{2} m v_1^2=m g l(1-\cos \theta)
\)
or,
\(
\begin{aligned}
v_1 & =\sqrt{v_0{ }^2-2 g l(1-\cos \theta)} \\
& =\sqrt{\left(9 \mathrm{~m}^2 / \mathrm{s}^2\right)-2 \times\left(10 \mathrm{~m} / \mathrm{s}^2\right) \times(0.5 \mathrm{~m})\left(1-\frac{1}{2}\right)} \\
& =2 \mathrm{~m} / \mathrm{s} .
\end{aligned}
\)

Q3: A block of mass \(m\), attached to a spring of spring constant \(k\), oscillates on a smooth horizontal table. The other end of the spring is fixed to a wall. If it has a speed \(v\) when the spring is at its natural length, how far will it move on the table before coming to an instantaneous rest?

Answer: Consider the block + the spring as the system. The external forces acting on the system are (a) the force of gravity, (b) the normal force by the table and (c) the force by the wall. None of these do any work on this system and hence the total mechanical energy is conserved. If the block moves a distance \(x\) before coming to rest, we have,
\(
\begin{aligned}
\frac{1}{2} m v^2 & =\frac{1}{2} k x^2 \\
x & =v \sqrt{m / k} .
\end{aligned}
\)

Q4: A block of mass \(m\) is suspended through a spring of spring constant \(k\) and is in equilibrium. A sharp blow gives the block an initial downward velocity \(v\). How far below the equilibrium position, the block comes to an instantaneous rest?

Answer: Let us consider the block + the spring + the earth as the system. The system has gravitational potential energy corresponding to the force between the block and the earth as well as the elastic potential energy corresponding to the spring force. The total mechanical energy includes kinetic energy, gravitational potential energy and elastic potential energy.

When the block is in equilibrium, it is acted upon by two forces, (a) the force of gravity \(m g\) and (b) the tension in the spring \(T=k x\), where \(x\) is the elongation. For equilibrium, \(m g=k x\), so that the spring is stretched by a length \(x=m g / k\). The potential energy of the spring in this position is
\(
\frac{1}{2} k(m g / k)^2=\frac{m^2 g^2}{2 k} .
\)
Take the gravitational potential energy to be zero in this position. The total mechanical energy of the system just after the blow is
\(
\frac{1}{2} m v^2+\frac{m^2 g^2}{2 k}
\)
The only external force on this system is that due to the ceiling which does no work. Hence, the mechanical energy of this system remains constant. If the block descends through a height \(h\) before coming to an instantaneous rest, the elastic potential energy becomes \(\frac{1}{2} k(m g / k+h)^2\) and the gravitational potential energy – \(m g h\). The kinetic energy is zero in this state. Thus, we have
\(
\frac{1}{2} m v^2+\frac{m^2 g^2}{2 k}=\frac{1}{2} k(m g / k+h)^2-m g h .
\)
Solving this we get,
\(
h=v \sqrt{m / k} .
\)
Compare this with the result obtained in Example (Q3). If we neglect gravity and consider the length of the spring in the equilibrium position as the natural length, the answer is the same. This simplification is often used while dealing with vertical springs.

Q5: A porter lifts a suitcase weighing \(20 \mathrm{~kg}\) from the platform and puts it on his head \(2.0 \mathrm{~m}\) above the platform. Calculate the work done by the porter on the suitcase.

Answer: The kinetic energy of the suitcase was zero when it was at the platform and it again became zero when it was put on the head. The change in kinetic energy is zero and hence the total work done on the suitcase is zero. Two forces act on the suitcase, one due to gravity and the other due to the porter. Thus, the work done by the porter is negative of the work done by gravity. As the suitcase is lifted up, the work done by gravity is
\(
\begin{aligned}
W & =-m g h \\
& =-(20 \mathrm{~kg})\left(9.8 \mathrm{~m} / \mathrm{s}^2\right)(2 \mathrm{~m})=-392 \mathrm{~J}
\end{aligned}
\)
The work done by the porter is \(392 \mathrm{~J} \approx 390 \mathrm{~J}\).

Q6: An elevator weighing \(500 \mathrm{~kg}\) is to be lifted up at a constant velocity of \(0.20 \mathrm{~m} / \mathrm{s}\). What would be the minimum horsepower of the motor to be used?

Answer: As the elevator is going up with a uniform velocity, the total work done on it is zero in any time interval. The work done by the motor is, therefore, equal to the work done by the force of gravity in that interval (in magnitude). The rate of doing work, i.e., the power delivered is
\(
\begin{aligned}
P & =F v=m g v \\
& =(500 \mathrm{~kg})\left(9.8 \mathrm{~m} / \mathrm{s}^2\right)(0.2 \mathrm{~m} / \mathrm{s})=980 \mathrm{~W}
\end{aligned}
\)
Assuming no loss against friction etc., in the motor, the minimum horsepower of the motor is
\(
P=980 \mathrm{~W}=\frac{980}{746} \mathrm{hp}=1 \cdot 3 \mathrm{hp}
\)

Q7: A block of mass \(2.0 \mathrm{~kg}\) is pulled up on a smooth incline of angle \(30^{\circ}\) with the horizontal. If the block moves with an acceleration of \(1.0 \mathrm{~m} / \mathrm{s}^2\), find the power delivered by the pulling force at a time \(4.0 \mathrm{~s}\) after the motion starts. What is the average power delivered during the \(4.0 \mathrm{~s}\) after the motion starts?

Answer: The forces acting on the block are shown in Figure (8-W1). Resolving the forces parallel to the incline, we get
\(
\text { or, } \begin{aligned}
F & -m g \sin \theta=m a \\
& =m g \sin \theta+m a \\
& =(2.0 \mathrm{~kg})\left[\left(9.8 \mathrm{~m} / \mathrm{s}^2\right)(1 / 2)+1.0 \mathrm{~m} / \mathrm{s}^2\right]=11.8 \mathrm{~N} .
\end{aligned}
\)

The velocity at \(t=4.0 \mathrm{~s}\) is
\(
v=a t=\left(1.0 \mathrm{~m} / \mathrm{s}^2\right)(4.0 \mathrm{~s})=4.0 \mathrm{~m} / \mathrm{s} .
\)
The power delivered by the force at \(t=4.0 \mathrm{~s}\) is
\(
P=\vec{F} \cdot \vec{v}=(11.8 \mathrm{~N})(4 \cdot 0 \mathrm{~m} / \mathrm{s}) \approx 47 \mathrm{~W}
\)
The displacement during the first four seconds is
\(
x=\frac{1}{2} a t^2=\frac{1}{2}\left(1.0 \mathrm{~m} / \mathrm{s}^2\right)\left(16 \mathrm{~s}^2\right)=8.0 \mathrm{~m} \text {. }
\)
The work done in these four seconds is, therefore,
\(
W=\vec{F} \cdot \vec{d}=(11.8 \mathrm{~N})(8 \cdot 0 \mathrm{~m})=94.4 \mathrm{~J}
\)
\(
\text { The average power delivered }=\frac{94 \cdot 4 \mathrm{~J}}{4 \cdot 0 \mathrm{~s}}=23.6 \mathrm{~W} \approx 24 \mathrm{~W} \text {. }
\)

Q8: A force \(F=(10+0.50 x)\) acts on a particle in the \(x\) direction, where \(F\) is in newton and \(x\) in meter. Find the work done by this force during a displacement from \(x=0\) to \(x=2.0 \mathrm{~m}\).

Answer: As the force is variable, we shall find the work done in a small displacement \(x\) to \(x+d x\) and then integrate it to find the total work. The work done in this small displacement is
\(
d W=\vec{F} \cdot d \vec{x}=(10+0.5 x) d x
\)
Thus,
\(
\begin{aligned}
W & =\int_0^{20}(10+0.50 x) d x \\
& =\left[10 x+0.50 \frac{x^2}{2}\right]_0^{2.0}=21 \mathrm{~J} .
\end{aligned}
\)

Q9: A body dropped from a height \(H\) reaches the ground with a speed of \(1 \cdot 2 \sqrt{g H}\). Calculate the work done by air friction.

Answer: The forces acting on the body are the force of gravity and the air friction. By the work-energy theorem, the total work done on the body is

\(
W=\frac{1}{2} m\left(1^{\cdot} 2 \sqrt{g H}\right)^2-0=0 \cdot 72 m g H .
\)
The work done by the force of gravity is \(m g H\). Hence, the work done by the air friction is
\(
0.72 m g H-m g H=-0.28 \mathrm{mgH} \text {. }
\)

Q10: A block of mass \(M\) is pulled along a horizontal surface by applying a force at an angle \(\theta\) with the horizontal. The friction coefficient between the block and the surface is \(\mu\). If the block travels at a uniform velocity, find the work done by this applied force during a displacement \(d\) of the block.

Answer: Forces on the block are
(i) its weight \(M g\),
(ii) the normal force \(\mathcal{N}\),
(iii) the applied force \(F\) and
(iv) the kinetic friction \(\mu \mathcal{N}\).

The forces are shown in Figure (8-W2). As the block moves with a uniform velocity, the forces add up to zero. Taking horizontal and vertical components,
\(
F \cos \theta=\mu \mathcal{N}
\)
and \(\quad F \sin \theta+\mathcal{N}=M g\).
Eliminating \(\mathcal{N}\) from these equations,
\(
\begin{aligned}
F \cos \theta & =\mu(M g-F \sin \theta) \\
F & =\frac{\mu M g}{\cos \theta+\mu \sin \theta} .
\end{aligned}
\)
The work done by this force during a displacement \(d\) is
\(
W=F d \cos \theta=\frac{\mu M g d \cos \theta}{\cos \theta+\mu \sin \theta}
\)

Q11: Two cylindrical vessels of equal cross-sectional area A contain water upto heights \(h_1\) and \(h_2\). The vessels are interconnected so that the levels in them become equal. Calculate the work done by the force of gravity during the process. The density of water is \(\rho\).

Answer: Since the total volume of the water is constant, the height in each vessel after interconnection will be \(\left(h_1+h_2\right) / 2\). The level in the left vessel shown in the figure, drops from \(A\) to \(C\) and that in the right vessel rises from \(B\) to \(D\). Effectively, the water in the part \(A C\) has dropped down to \(D B\).

The mass of this volume of water is
\(
m=\rho A\left(h_1-\frac{h_1+h_2}{2}\right)
\)
\(
=\rho A\left(\frac{h_1-h_2}{2}\right) .
\)
The height descended by this water is \(A C=\left(h_1-h_2\right) / 2\). The work done by the force of gravity during this process is, therefore,
\(
=\rho A\left(\frac{h_1-h_2}{2}\right)^2 g \text {. }
\)

Q12: What minimum horizontal speed should be given to the bob of a simple pendulum of length \(l\) so that it describes a complete circle?

Answer: Suppose the bob is given a horizontal speed \(v_0\) at the bottom and it describes a complete vertical circle. Let its speed at the highest point be \(v\). Taking the gravitational potential energy to be zero at the bottom, the conservation of energy gives,
\(
\begin{aligned}
\frac{1}{2} m v_0^2 & =\frac{1}{2} m v^2+2 m g l \\
\text { or, } \quad m v^2 & =m v_0^2-4 m g l \dots(i)
\end{aligned}
\)

The forces acting on the bob at the highest point are \(m g\) due to gravity and \(T\) due to the tension in the string. The resultant force towards the centre is, therefore, \(m g+T\). As the bob is moving in a circle, its acceleration towards the centre is \(v^2 / l\). Applying Newton’s second law and using (i),
\(
\begin{array}{rlrl}
m g+T & =m \frac{v^2}{l}=\frac{1}{l}\left(m v_0^2-4 m g l\right) \\
\text { or, } & m v_0^2 & =5 m g l+T l .
\end{array}
\)
Now, for \(v_0\) to be minimum, \(T\) should be minimum. As the minimum value of \(T\) can be zero, for minimum speed, \(m v_0^2=5 \mathrm{mgl}\) or, \(v_0=\sqrt{5 \mathrm{gl}}\).

Q13: A uniform chain of length \(l\) and mass \(m\) overhangs a smooth table with its two third part lying on the table. Find the kinetic energy of the chain as it completely slips off the table.

Answer: Let us take the zero of potential energy at the table. Consider a part \(d x\) of the chain at a depth \(x\) below the surface of the table. The mass of this part is \(d m=m / l d x\) and hence its potential energy is \(-(m / l d x) g x\).

The potential energy of the \(l / 3\) of the chain that overhangs is \(U_1=\int_0^{l / 3}-\frac{m}{l} g x d x\)
\(
=-\left[\frac{m}{l} g\left(\frac{x^2}{2}\right)\right]_0^{l / 3}=-\frac{1}{18} m g l .
\)
This is also the potential energy of the full chain in the initial position because the part lying on the table has zero potential energy. The potential energy of the chain when it completely slips off the table is
\(
U_2=\int_0^l-\frac{m}{l} g x d x=-\frac{1}{2} m g l .
\)
The loss in potential energy \(=\left(-\frac{1}{18} m g l\right)-\left(-\frac{1}{2} m g l\right)\)
\(
=\frac{4}{9} m g l .
\)
This should be equal to the gain in the kinetic energy. But the initial kinetic energy is zero. Hence, the kinetic energy of the chain as it completely slips off the table is \(\frac{4}{9} m g l\).

Q14: A block of mass \(m\) is pushed against a spring of spring constant \(k\) fixed at one end to a wall. The block can slide on a frictionless table as shown in Figure (8-W6). The natural length of the spring is \(L_0\) and it is compressed to half its natural length when the block is released. Find the velocity of the block as a function of its distance \(x\) from the wall.

Answer: When the block is released, the spring pushes it towards right. The velocity of the block increases till the spring acquires its natural length. Thereafter, the block loses contact with the spring and moves with constant velocity.

Initially, the compression of the spring is \(L_0 / 2\). When the distance of the block from the wall becomes \(x\), where
\(x<L_0\), the compression is \(\left(L_0-x\right)\). Using the principle of conservation of energy,
\(
\frac{1}{2} k\left(\frac{L_0}{2}\right)^2=\frac{1}{2} k\left(L_0-x\right)^2+\frac{1}{2} m v^2 .
\)
Solving this,
\(
v=\sqrt{\frac{k}{m}}\left[\frac{L_0^2}{4}-\left(L_0-x\right)^2\right]^{1 / 2} .
\)
When the spring acquires its natural length, \(x=L_0\) and \(v=\sqrt{\frac{k}{m}} \frac{L_0}{2}\). Thereafter, the block continues with this velocity.

Q15: A particle is placed at the point \(A\) of a frictionless track \(A B C\) as shown in figure (8-W7). It is pushed slightly towards the right. Find its speed when it reaches the point B. Take \(g=10 \mathrm{~m} / \mathrm{s}^2\).

Let us take the gravitational potential energy to be zero at the horizontal surface shown in the figure. The potential energies of the particle at \(A\) and \(B\) are
\(
U_{\mathrm{A}}=\operatorname{Mg}(1 \mathrm{~m})
\)
and
\(
U_B=M g(0.5 \mathrm{~m}) .
\)
The kinetic energy at the point \(A\) is zero. As the track is frictionless, no energy is lost. The normal force on the particle does no work. Applying the principle of conservation of energy,
\(
\begin{aligned}
& U_A+K_A=U_B+K_B \\
& \text { or, } \quad M g(1 \mathrm{~m})=M g(0.5 \mathrm{~m})+\frac{1}{2} M v_B^2 \\
& \text { or, } \quad \frac{1}{2} v_B^2=g(1 \mathrm{~m}-0.5 \mathrm{~m}) \\
& =\left(10 \mathrm{~m} / \mathrm{s}^2\right) \times 0.5 \mathrm{~m} \\
& =5 \mathrm{~m}^2 / \mathrm{s}^2 \\
& \text { or, } \\
& v_B=\sqrt{10} \mathrm{~m} / \mathrm{s} \text {. } \\
&
\end{aligned}
\)

Q16: Figure (8-W8) shows a smooth curved track terminating in a smooth horizontal part. A spring of spring constant \(400 \mathrm{~N} / \mathrm{m}\) is attached at one end to a wedge fixed rigidly with the horizontal part. A \(40 \mathrm{~g}\) mass is released from rest at a height of \(4.9 \mathrm{~m}\) on the curved track. Find the maximum compression of the spring.

At the instant of maximum compression, the speed of the \(40 \mathrm{~g}\) mass reduces to zero. Taking the gravitational potential energy to be zero at the horizontal part, the conservation of energy shows,
\(
m g h=\frac{1}{2} k x^2
\)
where \(m=0.04 \mathrm{~kg}, h=4.9 \mathrm{~m}, k=400 \mathrm{~N} / \mathrm{m}\) and \(x\) is the maximum compression.
Thus,
\(
\begin{aligned}
x & =\sqrt{\frac{2 m g h}{k}} \\
& =\sqrt{\frac{2 \times(0.04 \mathrm{~kg}) \times\left(9.8 \mathrm{~m} / \mathrm{s}^2\right) \times(4.9 \mathrm{~m})}{(400 \mathrm{~N} / \mathrm{m})}} \\
& =9.8 \mathrm{~cm} .
\end{aligned}
\)

Q17: Figure (8-W9) shows a loop-the-loop track of radius \(R\). A car (without an engine) starts from a platform at a distance \(h\) above the top of the loop and goes around the loop without falling off the track. Find the minimum value of h for a successful looping. Neglect friction.

Answer: Suppose the speed of the car at the topmost point of the loop is \(v\). Taking the gravitational potential energy to be zero at the platform and assuming that the car starts with a negligible speed, the conservation of energy shows,
\(
\begin{aligned}
0 & =-m g h+\frac{1}{2} m v^2 \\
\text { or, } \quad m v^2 & =2 m g h \dots(i)
\end{aligned}
\)
where \(m\) is the mass of the car. The car moving in a circle must have radial acceleration \(v^2 / R\) at this instant. The forces on the car are, \(m g\) due to gravity and \(\mathcal{N}\) due to the contact with the track. Both these forces are in the radial direction at the top of the loop. Thus, from Newton’s Law
\(m g+\mathcal{N}=\frac{m v^2}{R}\)
or, \(\quad m g+\mathcal{N}=2 \mathrm{mgh} / R\).
For \(h\) to be minimum, \(\mathcal{N}\) should assume the minimum value which can be zero. Thus,
\(
2 m g \frac{h_{\min }}{R}=m g \quad \text { or, } \quad h_{\min }=R / 2 .
\)

Q18: A heavy particle is suspended by a string of length \(l\). The particle is given a horizontal velocity \(v_0\). The string becomes slack at some angle and the particle proceeds on a parabola. Find the value of \(v_0\) if the particle passes through the point of suspension.

Answer: Suppose the string becomes slack when the particle reaches the point \(P\) (figure 8 -W10). Suppose the string \(O P\) makes an angle \(\theta\) with the upward vertical. The only force acting on the particle at \(P\) is its weight \(m g\). The radial component of the force is \(m g \cos \theta\). As the particle moves on the circle upto \(P\),
\(
m g \cos \theta=m\left(\frac{v^2}{l}\right)
\)
\(
\text { or, } \quad v^2=g l \cos \theta \quad \ldots \text { (i) }
\)
where \(v\) is its speed at \(P\). Using conservation of energy,
\(
\frac{1}{2} m v_0^2=\frac{1}{2} m v^2+m g l(1+\cos \theta)
\)
or, \(\quad v^2=v_0^2-2 g l(1+\cos \theta)\) \dots(ii)
\(
\text { From (i) and (ii), } v_0^2-2 g l(1+\cos \theta)=g l \cos \theta
\)
\(
\text { or, } \quad v_0^2=g l(2+3 \cos \theta) . \quad \ldots \text { (iii) }
\)
Now onwards the particle goes in a parabola under the action of gravity. As it passes through the point of suspension \(O\), the equations for horizontal and vertical motions give,
\(
\begin{gathered}
l \sin \theta=(v \cos \theta) t \\
\text { and } \quad-l \cos \theta=(v \sin \theta) t-\frac{1}{2} g t^2
\end{gathered}
\)
\(
\text { or, } \quad-l \cos \theta=(v \sin \theta)\left(\frac{l \sin \theta}{v \cos \theta}\right)-\frac{1}{2} g\left(\frac{l \sin \theta}{v \cos \theta}\right)^2
\)
\(
\text { or, } \quad-\cos ^2 \theta=\sin ^2 \theta-\frac{1}{2} g \frac{l \sin ^2 \theta}{v^2 \cos \theta}
\)
\(
\text { or, } \quad-\cos ^2 \theta=1-\cos ^2 \theta-\frac{1}{2} \frac{g l \sin ^2 \theta}{g l \cos ^2 \theta} \text { [From (i)] }
\)
or, \(\quad 1=\frac{1}{2} \tan ^2 \theta\)
or, \(\quad \tan \theta=\sqrt{2}\).
From (iii), \(v_0=[g l(2+\sqrt{3})]^{1 / 2}\).

Q19: Figure (8-E17) shows a smooth track that consists of a straight inclined part of length \(l\) joining smoothly with the circular part. A particle of mass \(m\) is projected up the incline from its bottom. (a) Find the minimum projection speed \(v_0\) for which the particle reaches the top of the track. (b) Assuming that the projection speed is \(2 v_0\) and that the block does not lose contact with the track before reaching its top, find the force acting on it when it reaches the top. (c) Assuming that the projection speed is only slightly greater than \(v_0\), where will the block lose contact with the track?

Answer: (a)

\(
\text { Net force on the particle between } A \text { & B, } F=m g \sin \theta
\)
\(
\text { work done to reach } \mathrm{B}, \mathrm{W}=\mathrm{FS}=\mathrm{mg} \sin \theta \ell
\)
Again, work done to reach \(B\) to \(C=m g h=m g R(1-\cos \theta)\)
So, Total work done \(=m g[\ell \sin \theta+R(1-\cos \theta)]\)
\(
\text { Now, change in K.E. }=\text { work done }
\)
\(
\Rightarrow 1 / 2 \mathrm{mv}_0{ }^2=\mathrm{mg}[\ell \sin \theta+\mathrm{R}(1-\cos \theta)]
\)
\(
\Rightarrow \mathrm{v}_{\mathrm{o}}=\sqrt{2 g(R(1-\cos \theta)+\ell \sin \theta)}
\)
(b) When the block is projected at a speed \(2 \mathrm{v}_0\). Let the velocity at \({C}\) will be \(\mathrm{V}_{\mathrm{c}}\).
Applying the energy principle,
\(1 / 2 m v_c^2-1 / 2 m\left(2 v_o\right)^2=-m g[\ell \sin \theta+R(1-\cos \theta)]\)
\(
\Rightarrow \mathrm{v}_{\mathrm{c}}^2=4 \mathrm{v}_0-2 \mathrm{~g}[\ell \sin \theta+\mathrm{R}(1-\cos \theta)]
\)
\(
\begin{aligned}
& 4.2 \mathrm{~g}[l \sin \theta+R(1-\cos \theta)]-2 g[l \sin \theta+R(1-\cos \theta)] \\
& =6 g[l \sin \theta+R(1-\cos \theta)]
\end{aligned}
\)
\(
\Rightarrow \mathrm{N}=\frac{\mathrm{mv}_{\mathrm{c}}{ }^2}{\mathrm{R}}=6 \mathrm{mg}[(\ell / \mathrm{R}) \sin \theta+1-\cos \theta]
\)
(c) Let the block loose contact after making an angle \(\theta\)
\(\frac{m v^2}{R}=m g \cos \theta \Rightarrow v^2=R g \cos \theta \dots(i)\)
Again, \(1 / 2 m v^2=m g(R-R \cos \theta) \Rightarrow v^2=2 g R(1-\cos \theta) \dots(ii)\)
From (i) and (ii) \(\cos \theta=2 / 3 \Rightarrow \theta=\cos ^{-1}(2 / 3)\)

Q20: A chain of length \(l\) and mass \(m\) lies on the surface of a smooth sphere of radius \(R>l\) with one end tied to the top of the sphere. (a) Find the gravitational potential energy of the chain with the reference level at the centre of the sphere. (b) Suppose the chain is released and slides down the sphere. Find the kinetic energy of the chain, when it has slid through an angle \(\theta\). (c) Find the tangential acceleration \(\frac{d v}{d t}\) of the chain when the chain starts sliding down.

Answer:

Let us consider a small element which makes angle ‘ \(\mathrm{d} \theta\) ‘ at the centre.
\(
\therefore \mathrm{dm}=(\mathrm{m} / {l}) \mathrm{Rd} \theta
\)
(a)
\(
\text { Gravitational potential energy of } d m \text { with respect to centre of the sphere }
\)
\(
\begin{aligned}
& =(\mathrm{dm}) g R \cos \theta \\
& =(\mathrm{mg} / {l}) R \cos \theta {d} \theta
\end{aligned}
\)
So, Total G.P.E. \(=\int_0^{\ell / r} \frac{\mathrm{mgR}^2}{\ell} \cos \theta \mathrm{d} \theta \quad[\alpha=(\ell / \mathrm{R})]\) (angle subtended by the chain at the centre)
\(
=\frac{m R^2 g}{\ell}[\sin \theta](\ell / R)=\frac{m R g}{\ell} \sin (\ell / R)
\)
(b) When the chain is released from rest and slides down through an angle \(\theta\), the K.E. of the chain is given
K.E. = Change in potential energy.
\(
=\frac{m R^2 g}{\ell} \sin (\ell / R)-m \int \frac{g R^2}{\ell} \cos \theta d \theta
\)
\(
=\frac{m R^2 g}{\ell}[\sin (\ell / R)+\sin \theta-\sin \{\theta+(\ell / R)\}]
\)
(c) Since, K.E. \(=1 / 2 m v^2=\frac{m R^2 g}{\ell}[\sin (\ell / R)+\sin \theta-\sin \{\theta+(\ell / R)\}]\)
Taking the derivative of both sides with respect to ‘ \(\mathrm{t}\) ‘
\((1 / 2) \times 2 \mathrm{v} \times \frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{R}^2 \mathrm{~g}}{\ell}\left[\cos \theta \times \frac{\mathrm{d} \theta}{\mathrm{dt}}-\cos (\theta+\ell / \mathrm{R}) \frac{\mathrm{d} \theta}{\mathrm{dt}}\right]\)
\(\therefore\left(\mathrm{R} \frac{\mathrm{d} \theta}{\mathrm{dt}}\right) \frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{R}^2 \mathrm{~g}}{\ell} \times \frac{\mathrm{d} \theta}{\mathrm{dt}}[\cos \theta-\cos (\theta+(\ell / \mathrm{R}))]\)
When the chain starts sliding down, \(\theta=0\).
So, \(\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{Rg}}{\ell}[1-\cos (\ell / \mathrm{R})]\)

Q21: The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:
(a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
(b) work done by the gravitational force in the above case,
(c) work done by friction on a body sliding down an inclined plane,
(d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity,
(e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest.

Answer: (a)
The formula to calculate the work done is,
\(
\mathrm{W}=\mathrm{Fs} \cos \theta \ldots . . .(i)
\)
Here, the amount of force applied on the body is \(F\), the displacement of the body is \(s\) and the angle between the force applied and the displacement is \(\theta\). As the man lifts the bucket out of a well by means of rope that is tied to the bucket, the angle between the direction of force and the direction of displacement is zero, that is \(\theta=0^{\circ}\).
Substitute the values in the equation (i).
\(
\mathrm{W}=\mathrm{Fscos} 0^{\circ}=\mathrm{Fs}
\)
Thus, the work done in this case is positive.
(b)
As the man lifts the bucket out of a well by means of rope that is tied to the bucket, the angle between the direction of gravitational force and the direction of displacement is opposite, that is \(\theta=180^{\circ}\).
Substitute the values in the equation (i).
\(
\mathrm{W}=\mathrm{Fscos} 180^{\circ}=-\mathrm{Fs}
\)
Thus, the work done in this case is negative.
(c)
The friction is applied on the body sliding down an inclined plane. As the body slides down, it accelerates downward and the force of friction is applied opposite to the force. So, the angle between the direction of frictional force and the direction of displacement is opposite, that is \(\theta=180^{\circ}\).
Substitute the values in the equation (i).
\(
\mathrm{W}=\mathrm{Fscos} 180^{\circ}=-\mathrm{Fs}
\)
Thus, the work done in this case is negative.
(d)
The force is applied to a body that moves on a rough horizontal plane with uniform velocity. So, the angle between the applied force and the direction of displacement is the same, that is \(\theta=0^{\circ}\).
Substitute the values in the equation (i).
\(
\mathrm{W}=\mathrm{Fscos}0^{\circ}=\mathrm{Fs}
\)
Thus, the work done in this case is positive.
(e)
The resistive force acts on a vibrating pendulum in bringing it to rest. As the pendulum vibrates the resistive force acts in the direction opposite to the motion of the pendulum. So, the angle between the direction of resistive force and the direction of displacement is opposite, that is \(\theta=180^{\circ}\).
Substitute the values in the equation (i).
\(
\mathrm{W}=\mathrm{Fs} \cos 180^{\circ}=-\mathrm{Fs}
\)
Thus, the work done in this case is negative.

Q22: A body of mass \(2 \mathrm{~kg}\) initially at rest moves under the action of an applied horizontal force of \(7 \mathrm{~N}\) on a table with the coefficient of kinetic friction \(=0.1\). Compute the
(a) work done by the applied force in \(10 \mathrm{~s}\),
(b) work done by friction in \(10 \mathrm{~s}\),
(c) work done by the net force on the body in \(10 \mathrm{~s}\),
(d) change in kinetic energy of the body in \(10 \mathrm{~s}\),
and interpret your results.

Answer: Mass of the body, \(m=2 \mathrm{~kg}\)
Applied force, \(F=7 \mathrm{~N}\)
Coefficient of kinetic friction, \(\mu=0.1\)
Initial velocity, \(u=0\)
Time, \(t=10 \mathrm{~s}\)
The acceleration produced in the body by the applied force is given by Newton’s second law of motion as:
\(
a^{\prime}=\frac{F}{m}=\frac{7}{2}=3.5 \mathrm{~m} / \mathrm{s}^2
\)
The frictional force is given as
\(
\begin{aligned}
& f=\mu m g \\
& =0.1 \times 2 \times 9.8=-1.96 \mathrm{~N}
\end{aligned}
\)
The acceleration produced by the frictional force
\(
\mathrm{a}^{\prime \prime}=\frac{1.96}{2}=-0.98 \mathrm{~m} / \mathrm{s}^2
\)
Total acceleration of the body:
\(
\begin{aligned}
& a=a^{\prime}+a^{\prime \prime} \\
& =3.5+(-0.098)=2.52 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
\)
The distance travelled by the body is given by the equation of motion:
\(
\begin{aligned}
& s=u t+\frac{1}{2} a t^2 \\
& =0+1 / 2 \times 2.52 \times (10)^2 \\
& =126 \mathrm{~m}
\end{aligned}
\)
(a) Work done by the applied force, \(W_a=F \times s=7 \times 126=882 \mathrm{~J}\)
(b) Work done by the frictional force, \(W_{\mathrm{f}}=F \times s=-1.96 \times 126=-247 \mathrm{~J}\)
(c) Net force \(=7+(-1.96)=5.04 \mathrm{~N}\)
Work done by the net force, \(W_{\text {net }}=5.04 \times 126=635 \mathrm{~J}\)
(d) From the first equation of motion, the final velocity can be calculated as:
\(
\begin{aligned}
& v=u+a t \\
& =0+2.52 \times 10=25.2 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
Change in kinetic energy \(=\frac{1}{2} m v^2-\frac{1}{2} \mu^2\)
\(
=\frac{1}{2} \times 2\left(v^2-u^2\right)=(25.2)^2-0^2=635 J
\)

Work done by the net force on a body equals change in its kinetic energy.

Q23: Given in Fig. 6.11 are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.

Answer: Total energy of a system is,
Energy \(=\) P.E. + K. E.
\(\therefore \quad\) K.E. \(=\mathrm{E}-\mathrm{P} . \mathrm{E}\)
The Kinetic energy of a body is a positive quantity.
It cannot be negative.
Therefore, the particle will not exist in a region where K.E. becomes negative.
(a) The kinetic energy of a body cannot be negative because it is a positive quantity.
From the above graph (first one) it can be observed that \(V_0>0\) for the region \(x>a\).
The kinetic energy of the system is given as,
\(
\text { K.E }=E-P . E
\)
From the graph, it can be observed that the potential energy is greater than total energy for the given region so the kinetic energy will be negative and it is not possible. The particle will not exist in that region. Thus, the particle will not exist in this region and the minimum total energy of the particle will be zero.
(b) All regions \(-\infty<x<\infty\)
In the given case (second one), the potential energy \(\left(V_0\right)\) is greater than total energy \((E)\) in all regions. Hence, the particle will not exist in this region.
(c) \(
{x}>\mathrm{a} \text { and } \mathrm{x}<\mathrm{b};-\mathrm{V}_1 \text {. }
\)
In the given case (third one), the condition regarding the positivity of K.E. is satisfied only in the region between \(x>a\) and \(x<b\).
The minimum P.E in this case is \(-V_1\).
Therfore, K.E. \(=E-\left(-V_1\right)=E+V_1\).
Therefore, for the positivity of the kinetic energy, the total energy of the particle must be greater than \(V_1\).
So, the minimum total energy the particle must have is \(-V_1\).
(d) \(-b / 2<x<a / 2 ; a / 2<x<b / 2 ; -V_1\)
In the given case (fourth one), the potential energy \(\left(V_0\right)\) of the particle becomes greater than the total energy \((E)\) for \(b / 2<x<b / 2\) and \(-a / 2<x<a / 2\).
Therefore, the particle will not exist in these regions.
The minimum potential energy in this case is \(-V_1\).
Therefore, K.E. \(=E-\left(-V_1\right)=E+V_1\).
Therefore, for the positivity of the kinetic energy, the total energy of the particle must be greater than \(V_1\).
So, the minimum total energy the particle must have is \(-V_1\).

Q24: Answer the following:
(a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere?
(b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity
in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?
(c) An artificial satellite orbiting the earth in a very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth?
(d) In Fig. 6.13(i) the man walks \(2 \mathrm{~m}\) carrying a mass of \(15 \mathrm{~kg}\) on his hands. In Fig. 6.13(ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley and a mass of \(15 \mathrm{~kg}\) hangs at its other end. In which case is the work done greater?

Answer: (a) Heat energy required for burning of the casing of a rocket comes from the rocket itself.
The burning of the casing of a rocket in flight (due to friction) results in the reduction of the mass of the rocket.
According to the conservation of energy: Total Energy \((T, E)=\) Potential energy \((P . E)+\) Kinetic Energy (K.E)
\(
=m g h+\frac{1}{2} m v^2
\)
The reduction in the rocket’s mass causes a drop in the total energy. Therefore, the heat energy required for the burning is obtained from the rocket.

(b) Gravitational force is a conservative force. Since the work done by a conservative force over a closed path is zero, the work done by the gravitational force over every complete orbit of a comet is zero (For a conservative force work done over a path is minus of change in potential energy. Over a complete orbit, there is no change in potential energy).

(c) When an artificial satellite, orbiting around the Earth, moves closer to Earth, its potential energy decreases because of the reduction in height. Since the total energy of the system remains constant, the reduction in P.E. results in an increase in K.E. Hence, the velocity of the satellite increases. However, due to atmospheric friction, the total energy of the satellite decreases by a small amount (K.E. increases, but P.E. decreases, and the sum decreases due to dissipation against friction).

(d)

Case (i):
Mass, \(m=15 \mathrm{~kg}\)
Displacement, \(s=2 \mathrm{~m}\)
Work done \(W=F s \cos \theta\)
Where theta \(=\) Angle between force and displacement.
\(
\begin{aligned}
& =m g s \cos \theta=15 \times 2 \times 9.8 \cos 90^\theta \\
& =0 \quad\left(\because \cos 90^{\circ}=0\right)
\end{aligned}
\)

Case (ii):
Mass, \(m=15 \mathrm{~kg}\)
Displacement, \(s=2 \mathrm{~m}\)
Here, the direction of the force applied on the rope and the direction of the displacement of the rope are the same.
Therefore, the angle between them, \(\theta=0^{\circ}\)
Since \(\cos 0^{\circ}=1\)
Work done, \(W=F_s \cos \theta=m g s\)
\(
=15 \times 9.8 \times 2=294 \mathrm{~J}
\)
Hence, more work is done in the second case.

Q25: Underline the correct alternative :
(a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered.
(b) Work done by a body against friction always results in a loss of its kinetic/potential energy.
(c) The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system.
(d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies.

Answer: (a) Decreases
Explanation: A conservative force does positive work on a body when it displaces the body in the direction of force. As a result, the body advances toward the centre of force. It decreases the separation between the two, thereby decreasing the potential energy of the body.

(b) Kinetic energy
The work done against the direction of friction reduces the velocity of a body. Hence, there is a loss of kinetic energy of the body.

(c) External force
Internal forces, irrespective of their direction, cannot produce any change in the total momentum of a body. Hence, the total momentum of a many-particle system is proportional to the external forces acting on the system.

(d) Total linear momentum, and also total energy (if the system of two bodies is isolated). The total linear momentum always remains conserved whether it is an elastic collision or an inelastic collision.

Q26: Answer carefully, with reasons :
(a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact)?
(b) Is the total linear momentum conserved during the short time of an elastic collision of two balls?
(c) What are the answers to (a) and (b) for an inelastic collision?
(d) If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic? (Note, we are talking here of potential energy corresponding to the force during the collision, not gravitational potential energy).

Answer: (a) No, In an elastic collision, the total initial kinetic energy of the balls will be equal to the total final kinetic energy of the balls. This kinetic energy is not conserved at the instant the two balls are in contact with each other. In fact, at the time of the collision, the kinetic energy of the balls will get converted into potential energy.
(b) Yes, In an elastic collision, the total linear momentum of the system always remains conserved.
(c) Linear momentum is conserved during an inelastic collision, kinetic energy is, of course, not conserved even after the collision is over.
(d) Elastic, In the given case, the forces involved are conservation. This is because they depend on the separation between the centres of the billiard balls. Hence, the collision is elastic.

Q27: Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed \(V\). If the collision is elastic, which of the following (Fig. 6.14) is a possible result after collision?

Answer: Let \(m\) be the mass of each ball bearing. Before the collision, the total K.E. of the system
\(
=1 / 2 m v^2+0=1 / 2 m v^2
\)
After the collision, the K.E. of the system is
Case I: E_1=1 / 2(2 m)(v / 2)^2=1 / 4 m v^2[/latex]
Case II: \(E_2=1 / 2 \mathrm{mv}^2\)
Case III: \(E_3=1 / 2(3 m)(v / 3)^2=1 / 6 m v^2\)
Thus, case II is the only possibility since K.E. is conserved in this case.

Q28: Which of the following potential energy curves in Fig. 6.18 cannot possibly describe the elastic collision of two billiard balls? Here r is the distance between centres of the balls.

Answer: The potential energy of a system of two masses varies inversely as the distance (r) between 1 them i.e., \(V(r) \propto 1 / r\). When the two billiard balls touch each other, P.E. becomes zero i.e., at \(r=R+R=2 R; V(r)=0\). Out of the given graphs, curve (v) only satisfies these two conditions. Therefore, all other curves cannot possibly describe the elastic collision of two billiard balls.

Q29: \(\text { Consider the decay of a free neutron at rest : } n \rightarrow p+e^{-}\)
Show that the two-body decay of this type must necessarily give an electron of fixed energy and, therefore, cannot account for the observed continuous energy distribution in the \(\beta\)-decay of a neutron or a nucleus (Fig. 6.19).

[Note: The simple result of this exercise was one among the several arguments advanced by \(\mathrm{W}\). Pauli to predict the existence of a third particle in the decay products of \(\beta\)-decay. This particle is known as a neutrino. We now know that it is a particle of intrinsic spin \(1 / 2\) (like \(e^{-}, p\) or \(n\) ), but is neutral, and either massless or having an extremely small mass (compared to the mass of an electron) and which interacts very weakly with matter. The correct decay process of the neutron is \(n \rightarrow p+e^{-}+v\) ]

Answer: The decay process of the free neutron at rest is given as:\(n \rightarrow p+e^{-}\)
From Einstein’s mass-energy relation, we have the energy of electron as \(m c^2\) Where,
\(
\begin{aligned}
& m=\text { Mass defect }=\text { Mass of neutron (Mass of proton }+ \text { Mass of electron) } \\
& \mathrm{c}=\text { Speed of light }
\end{aligned}
\)
\(m\) and \(c\) are constants. Hence, the given two-body decay is unable to explain the continuous energy distribution in the \(\beta\)-decay of a neutron or a nucleus. The presence of neutrino on the LHS of the decay correctly explains the continuous energy distribution.

Q30: A block of mass \(1 \mathrm{~kg}\) is pushed up a surface inclined to horizontal at an angle of \(30^{\circ}\) by a force of \(10 \mathrm{~N}\) parallel to the inclined surface (Fig. 6.15). The coefficient of friction between the block and the incline is 0.1. If the block is pushed up by \(10 \mathrm{~m}\) along the incline, calculate
(a) work done against gravity
(b) work done against the force of friction
(c) increase in potential energy
(d) increase in kinetic energy
(e) work done by the applied force.

Answer: 

(a) The various forces acting on the block are as shown in the figure Here, \(m=1 \mathrm{~kg}, \theta=30^{\circ}, \mathrm{F}=10 \mathrm{~N}\) \(\mu=0.1, \mathrm{~d}=10 \mathrm{~m}\)
Work done against gravity is
\(
\begin{aligned}
& W_g=m g d \sin \theta \\
& =(1 \mathrm{~kg}) \times\left(10 \mathrm{~ms}^{-2}\right) \times 10 \times \sin 30^{\circ} \mathrm{m}=50 \mathrm{~J}
\end{aligned}
\)
(b) \(W_{\mathrm{f}}=\mu \mathrm{mg} \cos \theta \quad d=0.1 \times 10 \times 0.866 \times 10=8.66 \mathrm{~J}\).
(c) \(\Delta \mathrm{U}=m g h=1 \times 10 \times 5=50 \mathrm{~J}\)
(d)
\(
\begin{aligned}
& a=\{F-(m g \sin \theta+\mu m g \cos \theta)\}=[10-5.87] \\
& =4.13 \mathrm{~m} / \mathrm{s}^2 \\
& v=u+\text { at or } v^2=u^2+2 a d \\
& \Delta K=\frac{1}{2} m v^2-\frac{1}{2} m u^2=m a d=41.3 \mathrm{~J} \\
&
\end{aligned}
\)
\(
\text { (e) } W=F d=100 \mathrm{~J}
\)

Q31: A curved surface is shown in Fig. 6.16. The portion BCD is free of friction. There are three spherical balls of identical radii and masses. Balls are released from rest one by one from A which is at a slightly greater height than C.

With the surface \(\mathrm{AB}\), ball 1 has large enough friction to cause rolling down without slipping; ball 2 has a small friction, and ball 3 has negligible friction.
(a) For which balls is total mechanical energy conserved?
(b) Which ball (s) can reach D?
(c) For balls which do not reach D, which of the balls can reach back A?

Answer: (a) Energy is conserved for balls 1 and 3. A ball 1 is rolling down without slipping there is no dissipation of energy hence, total mechanical energy is conserved. Bail 3 is having negligible friction hence, there is no loss of energy.
(b) Ball 1 acquires rotational energy, ball 2 loses energy by friction. They cannot cross at c. Ball 3 can cross over.
(c) Ball 1, 2 turn back before reaching C. Because of loss of energy, ball 2 cannot reach back to A. Ball 1 has a rotational motion in the “wrong” sense when it reaches B. It cannot roll back to A, because of kinetic friction.

Q32: A rocket accelerates straight up by ejecting gas downwards. In a small time interval \(\Delta t\), it ejects a gas of mass \(\Delta m\) at a relative speed \(u\). Calculate \(\mathrm{KE}\) of the entire system at \(t+\Delta t\) and \(t\) and show that the device that ejects gas does work \(=(1 / 2) \Delta m u^2\) in this time interval (neglect gravity).

Answer: Let \(M\) be the mass of the rocket at any time \(t\) and \(v_1\) the velocity of the rocket at the same time \(t\).
Let \(\Delta m=\) mass of gas ejected in the time interval \(\Delta t\).
The relative speed of gas ejected \(=u\).
Consider at the time \(t+\Delta t\)
\((K E)_{t+\Delta t}=K E \text { of rocket }+\mathrm{KE} \text { of gas }\)
\(
(K E)_{t+\Delta t}=\frac{1}{2}(M-\Delta m)(v+\Delta v)^2+\frac{1}{2} \Delta m(v-u)^2
\)
\(
=\frac{1}{2} M v^2+M v \Delta v-\Delta m v u+\frac{1}{2} \Delta m u^2
\)
\(
(\mathrm{KE})_{\mathrm{t}}=\mathrm{KE} \text { of the rocket at time } \mathrm{t}=\frac{1}{2} M v^2
\)
\(
(K E)_{t+\Delta t}-(K E)_t=(M \Delta v-\Delta m u) v+\frac{1}{2} \Delta m u^2=\frac{1}{2} \Delta m u^2=W
\)
(By Work – Energy theorem)
Since action-reaction forces are equal.
Hence, \(M \frac{d v}{d t}=\frac{d m}{d t}|u|\)
\(
\begin{aligned}
& \Rightarrow \mathrm{M} \Delta \mathrm{v}=\Delta \mathrm{mu} \\
& \Delta \mathrm{K}=\frac{1}{2} \Delta \mathrm{mu}^2
\end{aligned}
\)
Now, by work-energy theorem,
\(
\begin{aligned}
& \Delta \mathrm{K}=\Delta \mathrm{W} \\
& \Rightarrow \Delta \mathrm{W}=\frac{1}{2} \Delta \mathrm{mu}^2
\end{aligned}
\)

Q33: Two identical steel cubes (masses \(50 \mathrm{~g}\), side \(1 \mathrm{~cm}\) ) collide head-on face to face with a speed of \(10 \mathrm{~cm} / \mathrm{s}\) each. Find the maximum compression of each. Young’s modulus for steel \(=Y=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\).

Answer: \(
\text { Hooke’s law : } \frac{F}{A}=\mathrm{Y} \frac{\Delta L}{L}
\)
where A is the surface area and \(L\) is the length of the side of the cube. If \(k\) is spring or compression constant, then \(F=k \Delta L\)
\(
\therefore k=Y \frac{A}{L}=Y L
\)
\(
\text { Inttial } \mathrm{KE}=2 \times \frac{1}{2} m v^2=5 \times 10^{-4} \mathrm{~J}
\)
\(
\text { Final PE }=2 \times \frac{1}{2} k(\Delta L)^2
\)
\(
\Delta L=\sqrt{\frac{K E}{k}}=\sqrt{\frac{K E}{Y L}}=\sqrt{\frac{5 \times 10^{-4}}{2 \times 10^{11} \times 0.1}}=1.58 \times 10^{-7} \mathrm{~m}
\)

Q34: A balloon filled with helium rises against gravity increasing its potential energy. The speed of the balloon also increases as it rises. How do you reconcile this with the law of conservation of mechanical energy? You can neglect the viscous drag of air and assume that the density of air is constant.

Answer: Let \(m, V, \rho_{H e}\) denote respectively the mass, volume and density of helium balloon and \(\rho_{\text {air }}\) be the density of atr Volume \(V\) of balloon displaces volume \(V\) of air.
\(
\text { So, } \mathrm{V}\left(\rho_{\text {air }}-\rho_{\mathrm{He}}\right) g=m a \dots(1)
\)
Integrating with respect to \(t\),
\(
V\left(\rho_{a i r}-\rho_{H e}\right) g t=m v
\)
\(
\Rightarrow \frac{1}{2} m v^2=\frac{1}{2} m \frac{V^2}{m^2}\left(\rho_{a i r}-\rho_{H e}\right)^2 g^2 t^2=\frac{1}{2 m} V^2\left(\rho_{\text {air }}-\rho_{H e}\right)^2 g^2 t^2 \dots(2)
\)
\(
\text { If the balloon rises to a height } h \text {, from } s=u t+\frac{1}{2} a t^2
\)
\(
\text { we get } h=\frac{1}{2} a t^2=\frac{1}{2} \frac{V\left(\rho_{a t r}-\rho_{h e}\right)}{m} g t^2 \dots(3)
\)
From Eqs. (3) and (2),
\(
\begin{gathered}
\frac{1}{2} m v^2=\left[V\left(\rho_a-\rho_{H e}\right) g\right]\left[\frac{1}{2 m} V\left(\rho_a-\rho_{H e}\right) g t^2\right] \\
=V\left(\rho_a-\rho_{H e}\right) g h
\end{gathered}
\)
Rearranging the terms,
\(
\Rightarrow \frac{1}{2} m v^2+V \rho_{H e} g h=V \rho_{a r} h g
\)
\(
KE_{\text {baloon }} + P E_{\text {baloon }}= \text { change in PE of air. }
\)
So, as the balloon goes up, an equal volume of air comes down, an increase in PE and KE of the balloon is at the cost of PE of air [which comes down].

Q35: A graph of potential energy \(V(x)\) verses \(x\) is shown in Fig. 6.12. A particle of energy \(E_0\) is executing motion in it. Draw the graph of velocity and kinetic energy versus \(x\) for one complete cycle AFA.

Answer: \(\text { (i) KE versus } x \text { graph : }\)
Step 1: Write the total energy of the particle.
By the law of conservation of energy, total mechanical energy is given by
\(
T E=K E+P E
\)
Here, \(T E=E_0\)
Therefore, \(\mathrm{E}_0=\mathrm{KE}+\mathrm{V}(\mathrm{x})\)
\(
\Rightarrow \mathrm{KE}=\mathrm{E}_0-\mathrm{V}(\mathrm{x})
\)
Step 2: Find \(K E\) and \(P E\) of the particle at \(A, B, C, D\) and at \(F\).
At A: Potential energy is maximum.
\(
T E=P E
\)
\(
K E=0
\)
At B : Potential energy will be less than the maximum value of it.
\(
\therefore \mathrm{PE}<\mathrm{E}_0
\)
\(
\mathrm{KE}>\mathrm{0}
\)
At \(C\) & \(D: P E=0\)
\(
K E=E_0
\)
\(
\text { At } F: P E=E_0, K E=0
\)
(ii) Velocity versus \(x\) graph:
Step 3: Find the velocity of the particle at \(A, B, C, D\) and at \(F\).
At \(A:\) As \(K E=0\)
\(\Rightarrow \operatorname{velocity}(\mathrm{v})=0\)
At B: Particle has some value of KE i.e., it has some value of velocity but less than the maximum value.
Since \(K E=E_0-V(x) \Rightarrow v^2=\frac{2}{m}\left(E_0-V(x)\right)\)
Therefore, the variation will be parabolic and symmetric about \(\mathrm{x}\)-axis.
From \(C\) to \(D: K E\) = maximum
\(
\Rightarrow v= \pm v_{\max }
\)
At \(F: K E=0\)
\(
\Rightarrow v=0
\)
Step 4: Draw a plot of velocity versus \(x\).

Q36: A ball of mass \(m\), moving with a speed \(2 v_0\), collides inelastically \((\mathrm{e}>0)\) with an identical ball at rest. Show that
(a) For head-on collision, both the balls move forward.
(b) For a general collision, the angle between the two velocities of scattered balls is less than \(90^{\circ}\).

Answer: (a) For head-on collision:
Conservation of momentum \(\Rightarrow 2 m v_0=m v_1+m v_2\)
Or \(2 v_0=v_1+v_2\)
and \(e=\frac{v_2-v_1}{2 v_0} \Rightarrow v_2=v_1+2 v_0 e\)
\(\therefore 2 v_1=2 v_0^0-2 e v_0\)
\(\therefore \quad v_1=v_0(1-e)\)
Since \(e<1 \Rightarrow v_1\) has the same sign as \(v_0\), therefore the ball moves on after collision.
(b) Conservation of momentum \(\Rightarrow \mathbf{p}=\mathbf{p}_1+\mathbf{p}_2\) But KE is lost \(\Rightarrow \frac{p^2}{2 m}>\frac{p_2{ }^2}{2 m}+\frac{p_2{ }^2}{2 m}\)

\(
\therefore p^2>p_1^2+p_2^2
\)
Thus \(\mathbf{p}, \mathbf{p}_1\) and \(\mathbf{p}_2\) are related as shown in the figure.
\(\theta\) is acute (less than \(\left.90^{\circ}\right)\left(p^2=p_1{ }^2+p_2{ }^2\right.\) would give \(\left.\theta=90^{\circ}\right)\)

Q37: The bob A of a pendulum released from horizontal to vertical hits another bob B of the same mass at rest on a table as shown in Fig. 6.13.
If the length of the pendulum is \(1 \mathrm{~m}\), calculate
(a) the height to which bob A will rise after the collision.
(b) the speed with which Bob B starts moving.
Neglect the size of the bobs and assume the collision to be elastic.

Answer: (a) Two balls have the same mass and the collision between them is elastic, therefore, ball A transfers its entire momentum to ball on the table. Hence, ball A will come to rest after the collision and does not rise at all.
(b) Speed with which bob B starts moving
\(=\) Speed with which bob \(\mathrm{A}\) hits bob \(\mathrm{B}\)
\(
\begin{aligned}
& =\sqrt{2 g h} \\
& =\sqrt{2 \times 9.8 \times 1} \\
& =\sqrt{19.6} \\
& =4.42 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Q38: Consider a one-dimensional motion of a particle with total energy E. There are four regions A, B, C, and D in which the relation between potential energy \(V\), kinetic energy \((K)\), and total energy \(E\) is as given below:
Region A : \(V>E\)
Region B : \(V<E\)
Region \(\mathrm{C}: K>E\)
Region D : \(V>K\)
State with reason in each case whether a particle can be found in the given region or not.

Answer: We know that
Total energy \(E=P E+K E\) \(\Rightarrow \mathrm{E}=\mathrm{V}+\mathrm{K} \ldots \ldots .(\mathrm{i})\)
Region A: No, as KE will become negative.
\(
\text { Given, V > E, From equation (i) }
\)
\(
K=E-V
\)
As \(\mathrm{V}>\mathrm{E} \Rightarrow \mathrm{E}-\mathrm{V}<0\)
Hence, \(K<0\), this is not possible.
Region B: Yes, total energy can be greater than PE for non-zero K.E.
\(
\text { Given, } V<E \Rightarrow E-V>0
\)
\(
\text { This is possible because total energy can be greater than } \mathrm{PE}(\mathrm{V})
\)
Region C: Yes, KE can be greater than total energy if its PE is negative.
\(
\text { Given, } K>E \Rightarrow K-E>0
\)
From equation (i) \(\mathrm{PE}=\mathrm{V}=\mathrm{E}-\mathrm{K}<0\)
This is possible because PE can be negative.
Region D: Yes, as PE can be greater than KE.
\(
\text { Given, } V>K
\)
\(
\text { This is possible because for a system } P E(V) \text { may be greater than } K E(K) \text {. }
\)

Q39: A raindrop of mass \(1.00 \mathrm{~g}\) falling from a height of \(1 \mathrm{~km}\) hits the ground with a speed of \(50 \mathrm{~m} \mathrm{~s}^{-1}\). Calculate
(a) the loss of P.E. of the drop.
(b) the gain in K.E. of the drop.
(c) Is the gain in K.E. equal to the loss of P.E.? If not why.
Take \(g=10 \mathrm{~m} \mathrm{~s}^{-2}\)

Answer: (a) Loss of PE \(=m g h=1 \times 10^{-3} \times 10 \times 10^{-3}=10 \mathrm{~J}\)
(b) Gain in \(\mathrm{KE}=\frac{1}{2} m v^2=\frac{1}{2} \times 10^{-3} \times 2500=1.25 \mathrm{~J}\)
(c) No, because a part of PE is used up in doing work against the viscous drag of air.

Q40: Two pendulums with identical bobs and lengths are suspended from a common support such that in the rest position the two bobs are in contact (Fig. 6.14). One of the bobs is released after being displaced by \(10^{\circ}\) so that it collides elastically head-on with the other bob.
(a) Describe the motion of two bobs.
(b) Draw a graph showing the variation in energy of either pendulum with time, for \(0 \leq t \leq 2 T\), where \(T\) is the period of each pendulum.

Answer: (a) Consider the adjacent diagram in which the bob B is displaced through an angle \(\theta\) and released.
At \(t=0\), suppose bob \(B\) is displaced by \(\theta=10^{\circ}\) to the right. It is given potential energy \(E_1=E\). Energy of \(A\), \(E_2=0\)
When \(B\) is released, it strikes \(A\) at \(t=T / 4\). In the head-on elastic collision between \(B\) and \(A\) comes to rest and \(A\) gets the velocity of \(B\). Therefore. \(E_1=0\) and \(E_2=E\). At \(t=27 T / 4, B\) reaches its extreme right position when \(K E\) of \(A\) is converted into \(P E\) \(=\mathrm{E}_2=\mathrm{E}\). Energy of \(\mathrm{B}, \mathrm{E}_1=0\).
At \(t=3 T / 4\), A reaches its mean position. when its \(P E\) is converted into \(K E=E_2=E\). It collides elastically with \(B\) and transfers the whole of its energy to \(B\). Thus, \(E_2=0\) and \(E_1=E\). The entire process is repeated.
(b) The values of energies of \(B\) and \(A\) at different time intervals are tabulated here. The plot of energy with time \(0 \leq t \leq 2 T\) is shown separately for 8 and \(A\) in the figure below.
\(
\begin{array}{|c|c|c|}
\hline \text { Time (t) } & \begin{array}{c}
\text { Energy of A } \\
\left(\mathrm{E}_1\right)
\end{array} & \begin{array}{c}
\text { Energy of B } \\
\left(\mathrm{E}_2\right)
\end{array} \\
\hline 0 & \mathrm{E} & 0 \\
\hline \mathrm{T} / 4 & 0 & \mathrm{E} \\
\hline 2 \mathrm{~T} / 4 & 0 & \mathrm{E} \\
\hline 3 \mathrm{~T} / 4 & \mathrm{E} & 0 \\
\hline 4 \mathrm{~T} / 4 & \mathrm{E} & 0 \\
\hline 5 \mathrm{~T} / 4 & 0 & \mathrm{E} \\
\hline 6 \mathrm{~T} / 4 & 0 & \mathrm{E} \\
\hline 7 \mathrm{~T} / 4 & \mathrm{E} & 0 \\
\hline 8 \mathrm{~T} / 4 & \mathrm{E} & 0 \\
\hline
\end{array}
\)

Q41: Suppose the average mass of raindrops is \(3.0 \times 10^{-5} \mathrm{~kg}\) and their average terminal velocity \(9 \mathrm{~m} \mathrm{~s}^{-1}\). Calculate the energy transferred by rain to each square metre of the surface at a place that receives \(100 \mathrm{~cm}\) of rain in a year.

Answer: \(
m=3.0 \times 10^{-5} \mathrm{~kg} \rho=10^{-3} \mathrm{~kg} / \mathrm{m}^2 v=9 \mathrm{~m} / \mathrm{s}
\)
\(
A=1 \mathrm{~m}^2 \quad \mathrm{~h}=100 \mathrm{~cm} \Rightarrow n=1 \mathrm{~m}^3
\)
\(
M=\rho v=10^{-3} \mathrm{~kg}, E=\frac{1}{2} M v^2=\frac{1}{2} \times 10^3 \times(9)^2=4.05 \times 10^4 \mathrm{~J}
\)

Q42: An engine is attached to a wagon through a shock absorber of length \(1.5 \mathrm{~m}\). The system with a total mass of \(50,000 \mathrm{~kg}\) is moving with a speed of \(36 \mathrm{~km} \mathrm{~h}^{-1}\) when the brakes are applied to bring it to rest. In the process of the system being brought to rest, the spring of the shock absorber gets compressed by \(1.0 \mathrm{~m}\). If \(90 \%\) of the energy of the wagon is lost due to friction, calculate the spring constant.

Answer:
\(
\begin{aligned}
K E & =\frac{1}{2} m v^2 \cong \frac{1}{2} \times 5 \times 10^4 \times 10^2 \\
& =2.5 \times 10^5 \mathrm{~J}
\end{aligned}
\)
\(10 \%\) of this is stored in the spring.
\(
\begin{array}{r}
\frac{1}{2} k x^2=2.5 \times 10^4 \\
x=1 \mathrm{~m} \\
k=5 \times 10^4 \mathrm{~N} / \mathrm{m} .
\end{array}
\)

Q43: An adult weighing \(600 \mathrm{~N}\) raises the centre of gravity of his body by \(0.25 \mathrm{~m}\) while taking each step of \(1 \mathrm{~m}\) length in jogging. If he jogs for \(6 \mathrm{~km}\), calculate the energy utilized by him in jogging assuming that there is no energy loss due to friction of ground and air. Assuming that the body of the adult is capable of converting \(10 \%\) of energy intake in the form of food, calculate the energy equivalents of food that would be required to compensate for energy utilized for jogging.

Answer: In \(6 \mathrm{~km}\) there are 6000 steps.
\(
\begin{aligned}
\therefore E & =6000(\mathrm{mg}) \mathrm{h} \\
& =6000 \times 600 \times 0.25 \\
& =9 \times 10^5 \mathrm{~J} .
\end{aligned}
\)
This is \(10 \%\) of intake.
\(
\therefore \quad \text { Intake energy }=10 \mathrm{E}=9 \times 10^6 \mathrm{~J} .
\)

Q44: On complete combustion a litre of petrol gives off heat equivalent to \(3 \times 10^7 \mathrm{~J}\). In a test drive a car weighing \(1200 \mathrm{~kg}\). including the mass of driver runs \(15 \mathrm{~km}\) per litre while moving with a uniform speed on a straight track. Assuming that friction offered by the road surface and air to be uniform, calculate the force of friction acting on the car during the test drive if the efficiency of the car engine were 0.5.

Answer: With 0.5 efficiency, 1 litre generates \(1.5 \times 10^7 \mathrm{~J}\), which is used for 15 km drive.
\(
\begin{aligned}
& \therefore F d=1.5 \times 10^7 \mathrm{~J} . \text { with } d=15000 \mathrm{~m} \\
& \therefore F=1000 \mathrm{~N} \text { : force of friction. }
\end{aligned}
\)

Short Question Answers

Q1: A rough inclined plane is placed on a cart moving with a constant velocity \(u\) on horizontal ground. A block of mass \(M\) rests on the incline. Is any work done by force of friction between the block and incline? Is there then a dissipation of energy?

Answer: Let us consider the diagram according to the situation. Just imagine the situation that wedge is moving with velocity u. As the block Mis at rest with respect to the inclined plane. There is no pseudo force acting on the block because the wedge is moving with constant velocity. So, \(f=\) frictional force \(=M g \sin \theta\) (upward as shown in the figure)

As the block rests on the incline, the force of friction acting between the block and the incline opposes the tendency of sliding of the block. Since block is not in motion with respect to the incline, therefore, work done by the force of friction between the block and the inclined plane is zero. Also due to this reason, there is no dissipation of energy.

Q2: Why is electrical power required at all when the elevator is descending? Why should there be a limit on the number of passengers in this case?

Answer: When the elevator is descending, then electric motor (mainly the induction motor are used) provides some force to overcome the weight of the elevator to prevent it from falling freely under gravity, then electric power is required to run the electric motor which is holding the elevator via cable to prevent it from falling freely. This cable is able to sustain some limiting value of tension developed in it. Due to this reason, it is needed to limit the number of passengers in the elevator.

Q3: A body is being raised to a height \(h\) from the surface of the earth. What is the sign of work done by
(a) applied force
(b) gravitational force?

Answer: (a) External force is applied on the body to lift it in an upward direction against its weight, therefore, the angle between the applied force and displacement is \(=0^{\circ}\)
Work done by the applied force
\(
W=F \cdot S=F s \cos =F s \cos 0^{\circ}=F S\left(\cos 0^{\circ}=1\right)
\)
i.e., the sign of work done by applied force is positive.
(b) As shown in the figure the gravitational force acts in the downward direction and displacement in the upward direction, therefore, the angle between them is \(=180^{\circ}\).
Work done by the gravitational force
\(
W=F s \cos 180^{\circ}=-F s\left(\cos 180^{\circ}=-1\right)
\)

Q4: Calculate the work done by a car against gravity in moving along a straight horizontal road. The mass of the car is \(400 \mathrm{~kg}\) and the distance moved is \(2 \mathrm{~m}\).

Answer: The weight of the car \((\mathrm{mg})\) vertically downward and the car is moving along the horizontal road, so the displacement of the car is in the along horizontal, i.e. angle between them is \(90^{\circ}\).
Work done by weight of the car
\(W=F s \cos 90^{\circ}=0\)
\(
\left(\cos 90^{\circ}=0\right)
\)
Hence, the work done by the car against gravity will also be zero.

Q5: A body falls towards earth in the air. Will its total mechanical energy be conserved during the fall? Justify.

Answer: If we neglect the air resistance then the total mechanical energy of the body is conserved but if there is some air resistance then a small part of its energy is utilized against the resistive force of air, which is nonconservative force. So, the total mechanical energy of the body falling freely under gravity in this case is not conserved. In this condition, gain in \(\mathrm{KE}<\) loss in \(\mathrm{PE}\).
Let \(\mathrm{E}\) be the total mechanical energy.
The initial mechanical energy of the body

Q6: A body is moved along a closed loop. Is the work done in moving the body necessarily zero? If not, state the condition under which work done over a closed path is always zero.

Answer: Key concept: If the work done by a force on a body depends upon the initial and final positions only of that body, then the force is conservative e.g., gravitational, electrostatic, and magnetic forces. If the work done by a force on a body which has moved in closed path and has come back to its initial position is zero, the force is conservative.

Work done in moving along a closed loop is not always zero. Work done in moving a closed path is zero when forces acting on the body are conservative but for non-conservative forces like friction force, viscous force, etc. work done in a closed path is not zero.

Q7: In an elastic collision of two billiard balls, which of the following quantities remain conserved during the short time of collision of the balls (i.e., when they are in contact)?
(a) Kinetic energy.
(b) Total linear momentum?
Give a reason for your answer in each case.

Answer: During the collision, no external force is acting on the balls, therefore total linear momentum of the system of two balls is always conserved.
During collision when the balls are in contact, there may be deformation, hence some kinetic energy of the system will be transformed to the potential energy of the system and consequently kinetic energy will not be conserved.
Important point: Though the kinetic energy of the system will not be conserved but the total energy of the system will be conserved.

Q8: Calculate the power of a crane in watts, which lifts a mass of \(100 \mathrm{~kg}\) to a height of \(10 \mathrm{~m}\) in \(20 \mathrm{~s}\).

Answer: According to the problem, mass \(=m=100 \mathrm{~kg}\) height \(=h-10 \mathrm{~m}\), time interval, \(t=20 \mathrm{~s}\) Power is the rate of doing work with respect to time.
\(
\text { We know from the formula that, the power }=\text { work done/ time }=\frac{F d s \cos \theta}{t i m e}
\)
\(
\begin{aligned}
& \mathrm{F}=\mathrm{mgh} \\
& m g=100 \times 10=1000 \\
& \mathrm{~h}=10 \mathrm{~m} \\
& \mathrm{t}=20 \mathrm{~s}
\end{aligned}
\)
\(\theta=0\), as the force and the displacement are in the same direction.
Hence, power \(=\frac{1000 \times 10 \cos 0}{20}=500\) watts.

Q9: The average work done by a human heart while it beats once is 0.5 J. Calculate the power used by the heart if it beats 72 times in a minute.

Answer: According to the problem, average work done by a human heart per beat \(=0.5 \mathrm{~J}\) Total work done during 72 beats in 1 minute
\(
=72 \times 0.5 \mathrm{~J}=36 \mathrm{~J}
\)
Power \(=\) Work done \(=36 \mathrm{~J} / 60 \mathrm{~s}=0.6 \mathrm{~W}\)

Q10: Give an example of a situation in which an applied force does not result in a change in kinetic energy.

Answer: Assume a ball is tied to a string and is moving in a vertical circle. Work done by tension force will be zero and hence tension force will not cause any change in KE of the ball. Because at any instant of time, the displacement is tangential and the force is central in nature, i.e., the tension in the string and the small displacement at any instant are perpendicular to each other.

Q11: Two bodies of unequal mass are moving in the same direction with equal kinetic energy. The two bodies are brought to rest by applying retarding force of the same magnitude. How would the distance moved by them before coming to rest compare?

Answer: According to the work-energy theorem,
Change in KE is equal to work done by all the forces acting on the body. Let us assume that only one force (retarding force) is acting on the body, therefore,
KE of the body \(=\) Work done by retarding force KE of the body \(=\) Retarding force \(x\) Displacement
As KE of the bodies and retarding forces applied on them are same, therefore, both bodies will travel equal distances before coming to rest.

Q12: A bob of mass \(m\) suspended by a light string of length \(L\) is whirled into a vertical circle as shown in Fig. 6.11. What will be the trajectory of the particle if the string is cut at
(a) Point B?
(b) Point C?
(c) Point X?

Answer: Key concept: According to the situation shown above that a bob of mass \(m\) is whirled into a vertical circle, the required centripetal force is obtained from the net force towards center at any point of time in the string. Tension in the string is variable and it is always towards center. But the gravitational force on the bob is always towards center. The speed of the body will be different at different points. So the equations of dynamical equilibrium \(\left(F_{\mathrm{c}}=m \mathrm{a}_{\mathrm{c}}, \mathrm{F}_{\mathrm{t}}=m \mathrm{a}_{\mathrm{t}}\right)\) must be satisfied at all the points. Let when the string makes an angle \(90^{\circ}\) with vertical, the speed of mass is \(v\).
Apply Newton’s law perpendicular to the string:
\(
m g \sin \theta =m a_t \Rightarrow a_t=g \sin \theta
\)
The above equation gives tangential acceleration as a function of angle. At lowest point \(=0^{\circ}\) and at highest point \(=180^{\circ}\). So at both points \(\sin \theta=0\). Hence \(a_t=0\) at both points \(L\) and \(H\).
\(
\text { At point } m, \theta=90^{\circ} \text {, then } a_t=g \text {. It is the maximum value of } a_t
\)
\(
\text { Apply Newton’s law along the string: } T-m g \cos =m a_c
\)
\(
\text { or } \quad T=m g \cos +m v^2 / r \ldots \text { (i) }
\)
As the body goes up, its velocity will go on decreasing and angle \(\theta\) will go on increasing. The maximum speed of the body will be at the lowest point \(\mathrm{L}\) and the minimum at the highest point \(\mathrm{H}\). Then from the above relation, we can find that tension will be maximum at the lowest point and minimum at the highest point.
Tension at lowest point \(\left(\theta=0^{\circ}, v=v_L\right): \quad T_L=m g+m \frac{v_L^2}{r} \dots(ii)\)
Tension at highest point \(\left(\theta=180^{\circ}, v=v_H\right): T_H=-m g+m \frac{v_H^2}{r} \dots(iii)\)

When the string is cut, the tension in the string becomes zero and centripetal force is not provided. Hence, bob tends to move in along the direction of its velocity.
(a) If the string is cut at any point, then the velocity of the body of mass \(\mathrm{m}\) is along the tangent to the circle. Tangent at point \(B\) is vertically downward so the trajectory of the particle is a straight line.
(b) Tangent at point \(\mathrm{C}\) is horizontally towards right.
So the trajectory of the particle is the parabola.
(c) Tangent at point X makes some angle with the horizontal. Again bob will follow a parabolic path with vertex higher than \(\mathrm{C}\).

Q13: When you lift a box from the floor and put it on an almirah the potential energy of the box increases, but there is no change in its kinetic energy. Is it a violation of the conservation of energy?

Answer: No. Work done in lifting the box increases the potential energy of the box. During lifting at every point, the force applied by us on the box in the upward direction is equal to the gravitational force acting on the box in the downward direction. Therefore, there is no change in the velocity of the box. As a result, the kinetic energy of the box will not change.

Q14: A particle is released from the top of an incline of height \(h\). Does the kinetic energy of the particle at the bottom of the incline depend on the angle of the incline? Do you need any more information to answer this question in Yes or No?

Answer: No, the kinetic energy of the particle at the bottom of the inclined plane does not depend on the angle of inclination. When the particle reaches the ground, all its potential energy, while at the top of the inclined plane, is converted into kinetic energy. As we know that kinetic energy depends only on the height of the particle, it will be the same for different angles of inclination.
No, we do not need any other information to answer this question.

Q15: Can the work by kinetic friction on an object be positive? Zero?

Answer: Yes. Let us consider a block A which is resting on another block B. Block B is resting on a smooth horizontal surface. Let the coefficient of friction between the blocks be \(\mu\).

When a force \(\mathrm{F}\) is applied on the block \(\mathrm{B}\) in the forward direction as shown in the above figure, block \(\mathrm{A}\) moves with block B in the direction of the applied force. The frictional force on the block \(A\) and the displacement will be in the forward direction. Therefore, the work done by the frictional force is positive.

If we consider the reference frame of block B, then the displacement of block A will be zero. Therefore, the work done by the frictional force is zero.

Q16: Can static friction do nonzero work on an object? If yes, give an example. If no, give a reason.

Answer: Yes. Let us consider a block A which is resting on another block B. Block B is resting on a smooth horizontal surface. Let the coefficient of kinetic friction between the blocks be \(\mu_k\).

When a force \(\mathrm{F}\) is applied on the block \(\mathrm{B}\) in the forward direction as shown in the above figure, block \(\mathrm{A}\) moves with block \(\mathrm{B}\) in the direction of the applied force. The friction force on block A and the displacement will be in the forward direction. Therefore, the work done by the friction force is positive. In this case, block A will remain in contact with block B. This shows that static friction is doing a nonzero work on an object.

Q17: Can normal force do a nonzero work on an object? If yes, give an example. If no, give a reason.

Answer: Yes. Let us consider an elevator accelerating upward with a body placed in it. In this case, the normal reaction offered by the floor of the elevator on the body is greater than the weight of the body acting in the downward direction. If a person is observing this from the ground, then, for him, the normal reaction is doing a positive work, as the elevator is moving upward.

Q18: Can the kinetic energy of a system be increased without applying any external force on the system?

Answer: Yes. Let us consider an isolated system of two particles falling towards each other under their mutual gravitational force of attraction. Here, the net force on the system is zero, but the velocities of the particles keep on increasing. Also, the kinetic energy of the system is increased without applying any external force on it.

Q19: Is the work-energy theorem valid in noninertial frames?

Answer: In a non-inertial frame, the pseudo force also comes into account. As we know that pseudo force does not exist, the work-energy theorem is not valid in non-inertial frames.

Q20: A heavy box is kept on a smooth inclined plane and is pushed up by a force \(F\) acting parallel to the plane. Does the work done by the force \(F\) as the box goes from \(A\) to \(B\) depend on how fast the box was moving at \(A\) and \(B\)? Does the work by the force of gravity depend on this?

Answer: (i) No. As the surface is smooth and the friction is zero, work done by the force will only depend on the force and the displacement.
(ii) No, because gravitational force is a conservative force, and work done by a conservative force will depend only on the force and the displacement

Q21: One person says that the potential energy of a particular book kept in an almirah is \(20 \mathrm{~J}\) and the other says it is \(30 \mathrm{~J}\). Is one of them necessarily wrong?

Answer: No, both are correct. We measure potential energy from a reference level chosen by the observer. Therefore, in this case, both observers are measuring the potential energy from different reference levels.

Q22: A book is lifted from the floor and is kept in an almirah. One person says that the potential energy of the book is increased by \(20 \mathrm{~J}\) and the other says it is increased by \(30 \mathrm{~J}\). Is one of them necessarily wrong?

Answer: Yes, one of them is necessarily wrong. We measure potential energy from a reference level chosen by the observer. However, the change in the potential energy of a body does not depend on the level of reference.

Q23: In one of the exercises to strengthen the wrist and fingers, a person squeezes and releases a soft rubber ball. Is the work done on the ball positive, negative or zero during compression? During expansion?

Answer: (i) During compression, the work done on the ball is positive as the direction of the force applied by the fingers is along the compression of the ball.
(ii) During expansion, the work done is negative as expansion takes place against the force applied by the fingers on the ball.

Q24: In the tug of war, the team that exerts a larger tangential force on the ground wins. Consider the period in which a team is dragging the opposite team by applying a larger tangential force on the ground. List which of the following works are positive, which are negative and which are zero?
(a) work by the winning team on the losing team
(b) work by the losing team on the winning team
(c) work by the ground on the winning team
(d) work by the ground on the losing team
(e) total external work on the two teams.

Answer: (a) Work by the winning team on the losing team is positive, as the displacement of the losing team is along the force applied by the winning team.
(b) Work by the losing team on the winning team is negative, as the displacement of the winning team is opposite to the force applied by losing team.
(c) Work by the ground on the winning team is positive.
(d) Work by the ground on the losing team is negative.
(e) Total external work on the two teams is positive.

Q25: When an apple falls from a tree what happens to its gravitational potential energy just as it reaches the ground? After it strikes the ground?

Answer: When an apple falls from a tree, its gravitational potential energy decreases as it reaches the ground. After it strikes the ground, its potential energy will remain unchanged.

Q26: When you push your bicycle up on an incline the potential energy of the bicycle and yourself increases. Where does this energy come from?

Answer: When a person pushes his bicycle up on an inclined plane, the potential energies of the bicycle and the person increase because moving up on the inclined plane the kinetic energy decreases. and as mechanical energy is the sum of kinetic energy and potential energy, and remains constant for a conservative system. Therefore, potential energy must increase in this case.

Q27: The magnetic force on a charged particle is always perpendicular to its velocity. Can the magnetic force change the velocity of the particle? Speed of the particle?

Answer: The magnetic force on a charged particle is always perpendicular to its velocity. Therefore, the work done by the magnetic force on the charged particle is zero. Here, the kinetic energy and speed of the particle remain unaffected, while the velocity changes due to the change in the direction of its motion.

Q28: A ball is given a speed \(v\) on a rough horizontal surface. The ball travels through a distance \(l[latex] on the surface and stops. (a) What are the initial and final kinetic energies of the ball? (b) What is the work done by the kinetic friction?

Answer: (a) Initial kinetic energy of the ball, [latex]K_{\mathrm{i}}=\frac{1}{2} \mathrm{~m} \mathrm{v}^2\)
Here, \(m\) is the mass of the ball. The final kinetic of the ball is zero.

(b) Work done by the kinetic friction is equal to the change in kinetic energy of the ball.
\(
\begin{aligned}
\therefore \text { Work done by the kinetic friction }=K_f-K_i & =0-\frac{1}{2} \mathrm{mv}^2 \\
& =-\frac{1}{2} \mathrm{mv}^2
\end{aligned}
\)

Q29: Consider the situation of the previous question from a frame moving with a speed \(v_0\) parallel to the initial velocity of the block. (a) What are the initial and final kinetic energies? (b) What is the work done by the kinetic friction?

Answer: \(
\text { The relative velocity of the ball w.r.t. the moving frame is given by } v_r=v-v_0
\)
\(
\text { (a) Initial kinetic energy of the ball }=\frac{1}{2} m v_r^2=\frac{1}{2} m\left(v-v_0\right)^2
\)
\(
\text { Also, final kinetic energy of the ball }=\frac{1}{2} m\left(0-v_0\right)^2=\frac{1}{2} m v_0^2
\)
(b) Work done by the kinetic friction = final kinetic energy – initial kinetic energy
\(
\begin{aligned}
& =\frac{1}{2} m\left(v_0\right)^2-\frac{1}{2} m\left(v-v_0\right)^2 \\
& =-\frac{1}{2} m v^2+m v v_0
\end{aligned}
\)