5.12 Exercise Problems

Q1: A girl riding a bicycle along a straight road with a speed of \(5 \mathrm{~m} \mathrm{~s}^{-1}\) throws a stone of mass \(0.5 \mathrm{~kg}\) which has a speed of \(15 \mathrm{~m} \mathrm{~s}^{-1}\) with respect to the ground along her direction of motion. The mass of the girl and bicycle is \(50 \mathrm{~kg}\). Does the speed of the bicycle change after the stone is thrown? What is the change in speed, if so?

Answer: Yes, due to the principle of conservation of momentum. Initial momentum \(=50.5 \times\) \(5 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}\)
Here, the total mass of girl cycle and stone \(m_1=50 \mathrm{~kg}\)
initial speed, \(u_1=5 \mathrm{~m} / \mathrm{s}\), mass of stone \(m_2=0.5 \mathrm{~kg}\) velocity of stone, \(u_2=15 \mathrm{~m} / \mathrm{s}\)
If upsilon is the speed of the girl and cycle (of mass \(m=50 \mathrm{~kg}\) ) after throwing the stone, then applying the principle of conservation of linear momentum, we get
\(
\begin{aligned}
& m_1 u_1=m_2 u_2+m v \\
& m v=m_1 u_2-m_2 u_2 \\
& 50 v=50.5 \times 5-0.5 \times 15=252.5-7.5=245 \\
& v=\frac{245}{50}=4.9 \mathrm{~m} / \mathrm{s} \\
& \text { change in speed }=5-4.9=0.1 \mathrm{~m} / \mathrm{s} .
\end{aligned}
\)

Q2: A person of mass \(50 \mathrm{~kg}\) stands on a weighing scale on a lift. If the lift is descending with a downward acceleration of \(9 \mathrm{~m} \mathrm{~s}^{-2}\), what would be the reading of the weighing scale? \(\left(g=10 \mathrm{~m} \mathrm{~s}^{-2}\right)\)

Answer:

\(
\text { Let } R \text { be the reading of the scale, in newtons. }
\)
When a lift descends with a downward acceleration the apparent weight of a body of mass \(m\) is given by \(w^{\prime}=R=m(g-a)\)
Mass of the person \(\mathrm{m}=50 \mathrm{~kg}\)
Descending acceleration \(\mathrm{a}=9 \mathrm{~m} / \mathrm{s}^2\)
Acceleration due to gravity \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\)
The apparent weight of the person,
\(
\begin{aligned}
& R=m(g-a) \\
& =50(10-9) \\
& =50 \mathrm{~N}
\end{aligned}
\)
\(\therefore\) Reading of the weighing scale \(=\frac{R}{g}=\frac{50}{10}=5 \mathrm{~kg}\).

Q3: The position-time graph of a body of mass \(2 \mathrm{~kg}\) is as given in Fig. 5.4. What is the impulse on the body at \(t=0 \mathrm{~s}\) and \(t=4 \mathrm{~s}\).

Answer: Given, the mass of the body \((\mathrm{m})=2 \mathrm{~kg}\)
From the position-time graph, the body is at \(x=0\) when \(t=0\), i.e., the body is at rest.
\(\therefore\) Impulse at \(t=0, s=0\), is zero
From \(t=0 s\) to \(t=4 s\), the position-time graph is a straight line, which shows that the body moves with uniform velocity.
Beyond \(t=4 \mathrm{~s}\), the graph is a straight line parallel to time ads, i.e., the body is at rest \((v=0)\).
The velocity of the body = slope of a position-time graph
\(
\begin{aligned}
& =\tan \theta=\frac{3}{4} \mathrm{~m s^{-1}} \\
& \text { impulse (at } \mathrm{t}=4 \mathrm{~s})=\text { change in momentum } \\
& =\mathrm{mv}-\mathrm{mu} \\
& =\mathrm{m}(\mathrm{v}-\mathrm{u}) \\
& =2\left(0-\frac{3}{4}\right) \\
& =-\frac{3}{2} \mathrm{~kg}-\mathrm{m} / \mathrm{s}=-1.5 \mathrm{~kg}-\mathrm{m s^{-1}}
\end{aligned}
\)

Q4: A person driving a car suddenly applies the brakes on seeing a child on the road ahead. If he is not wearing a seat belt, he falls forward and hits his head against the steering wheel. Why?

Answer: The only retarding force that acts on him, if he is not using a seat belt comes from the friction exerted by the seat. This is not enough to prevent him from moving forward when the vehicle is brought to a sudden halt.

Q5: The velocity of a body of mass \(2 \mathrm{~kg}\) as a function of \(t\) is given by \(\mathbf{v}(t)=2 t \hat{\mathbf{i}}+t^2 \hat{\mathbf{j}}\). Find the momentum and the force acting on it, at time \(t=2 \mathrm{~s}\).

Answer: Given, the mass of the body \(m=2 \mathrm{~kg}\).
Velocity of the body \(v(t)=2 t \hat{i}+t^2 \hat{j}\)
\(\therefore\) The velocity of the body at \(t=2 \mathrm{~s}\)
\(
\begin{aligned}
& \mathrm{v}=2 \times 2 \hat{i}+(2)^2 \hat{j} \\
& =(4 \hat{i}+4 \hat{j})
\end{aligned}
\)
The momentum of the body \((p)=m v\)
\(
\begin{aligned}
& =2(4 \hat{i}+4 \hat{j}) \\
& =(8 \hat{i}+8 \hat{j}) \mathrm{kg}-\mathrm{m} / \mathrm{s}
\end{aligned}
\)
Acceleration of the body \((\mathrm{a})=\frac{d v}{d t}\)
\(
\begin{aligned}
& =\frac{d}{d t}\left(2 t \hat{i}+t^2 \hat{j}\right) \\
& =(2 \hat{i}+2 t \hat{j})
\end{aligned}
\)
At \(t=2 s\)
\(
\begin{aligned}
& \mathrm{a}=(2 \hat{i}+2 \times 2 \hat{j}) \\
& =(2 \hat{i}+4 \hat{j})
\end{aligned}
\)
Force acting on the body \((\mathrm{F})=\mathrm{ma}\)
\(
\begin{aligned}
& =2(2 \hat{i}+4 \hat{j}) \\
& =(4 \hat{i}+8 \hat{j}) N
\end{aligned}
\)

Q6: A block placed on a rough horizontal surface is pulled by a horizontal force \(F\). Let \(f\) be the force applied by the rough surface on the block. Plot a graph of \(f\) versus \(F\).

Answer: \(\mathrm{F}=\mathrm{F}\) until the block is stationary. \(\mathrm{f}\) remains constant if \(\mathrm{F}\) increases beyond this point and the block starts moving.

Q7: Why are porcelain objects wrapped in paper or straw before packing for transportation?

Answer: In transportation, the vehicle say a truck, may need to halt suddenly. To bring a fragile material, like a porcelain object to a sudden halt means applying a large force and this is likely to damage the object. If it is wrapped up in say, straw, the object can travel some distance as the straw is soft before coming to a halt. The force needed to achieve this is less, thus reducing the possibility of damage.

Q8: Why does a child feel more pain when she falls down on a hard cement floor than when she falls on the soft muddy ground in the garden?

Answer: The body of the child is brought to a sudden halt when she/he falls on a cement floor. The mud floor yields and the body travels some distance before it comes to rest, which takes some time. This means the force which brings the child to rest is less for the fall on a mud floor, as the change in momentum is brought about over a longer period.

Q9: A woman throws an object of mass \(500 \mathrm{~g}\) with a speed of \(25 \mathrm{~m} \mathrm{~s}^1\).
(a) What is the impulse imparted to the object?
(b) If the object hits a wall and rebounds with half the original speed, what is the change in momentum of the object?

Answer:

(a)To solve this question, we look into the theory of Impulse. Impulse is defined as the change in momentum and is written as:
Impulse,\(I=F \Delta t=m \Delta v\)
Where, \(I\) is the impulse, \(\mathrm{F}\) is the force applied on the object, \(\Delta t\) is the time of application of force, \(m\) is the mass of the object and \(v\) is the velocity of the object.
Here, \(m=500 \mathrm{gm}=0.5 \mathrm{~kg}\)
\(
v_{\text {initial }}=0 \mathrm{~m} \mathrm{~s}^{-1}
\)
\(
v_{\text {final }}=25 \mathrm{~m} \mathrm{~s}^{-1}
\)
Therefore, \(
\mathrm{I}=\mathrm{m} \Delta \mathrm{v}=\mathrm{m}\left(\mathrm{v}_{\text {final }}-\mathrm{v}_{\text {initial }}\right)=0.5(25-0)=12.5 \mathrm{~N} \mathrm{-s}
\)
(b) Here, \(m=500 \mathrm{gm}=0.5 \mathrm{~kg}\)
\(
v_{\text {initial }}=-25 \mathrm{~m} \mathrm{~s}^{-1}
\)
\(
v_{\text {final }}=12.5 \mathrm{~m} \mathrm{~s}^{-1}
\)
There, the change in momentum is:
\(
\begin{aligned}
& \Delta p=m \Delta v=0.5[12.5-(-25)] \\
& =18.75 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}
\)

Q10: Why are mountain roads generally made winding upwards rather than going straight up?

Answer:

The mountain roads are generally constructed in a winding fashion so as to increase friction and thereby reduce the skidding of vehicles. This comes from the definition of friction for an object placed at a slope of angle \(\theta\).
\(
f=\mu N=\mu m g \cos \theta
\)
Here, \(\mathrm{f}\) is the frictional force, \(\mu\) is the coefficient of friction, \(\mathrm{m}\) is the mass of the object, \(\mathrm{g}\) is the acceleration due to gravity and \(\theta\) is the angle made by the object with the surface.

Now, winding the road means decreasing the \(\theta\) with respect to the ground. This will increase friction as cosine will increase. Hence the frictional force increases. Going straight up means going at a larger angle so the friction will decrease.
Here, \(\theta\) ‘ is the angle made by the slope of the mountain with the ground whereas, \(\theta\) is the angle made by each turn with the ground. As it is clear that \(\theta^{\prime}>\theta\).

Q11: A mass of \(2 \mathrm{~kg}\) is suspended with thread AB (Fig. 5.5). Thread \(C D\) of the same type is attached to the other end of \(2 \mathrm{~kg}\) mass. The lower thread is pulled gradually, harder and harder in the downward direction so as to apply force on AB. Which of the threads will break and why?

Answer: AB, because the force on the upper thread will be equal to the sum of the weight of the body and the applied force.

To solve such questions, we first draw the free-body diagram to see what forces are acting on the system. Here, we can see ” \(\mathrm{T}\) “. In addition to these forces, we apply a third force via string \(\mathrm{CD}\) in the downward direction given by ” \(\mathrm{F}\) “. So,
\(
T=F+m g
\)
As the tension is the greatest force, given by the sum of the weight and the additional force applied, string AB will break.

Q12: In the above given problem (Q11) if the lower thread is pulled with a jerk, what happens?

Answer: If the force is large and sudden, thread CD breaks because as CD is jerked, the pull is not transmitted to AB instantaneously (transmission depends on the elastic properties of the body). Therefore, before the mass moves, CD breaks.

Q13: Two masses of \(5 \mathrm{~kg}\) and \(3 \mathrm{~kg}\) are suspended with help of massless inextensible strings as shown in Fig. 5.6. Calculate \(T_1[latex] and [latex]T_2\) when whole system is going upwards with acceleration \(=2 \mathrm{~m} \mathrm{~s}^2\) (use \(g=9.8 \mathrm{~m} \mathrm{~s}^{-2}\) ).

Answer:

Given, \(m_1=5 \mathrm{~kg}, m_2=3 \mathrm{~kg}\) \(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2\) and \(\mathrm{a}=2 \mathrm{~m} / \mathrm{s}^2\)
For the upper block
\(
\begin{aligned}
& T_1-T_2-5 g=5 a \\
& \Rightarrow T_1-T_2=5(g+a) \ldots . .(i)
\end{aligned}
\)
For the lower block
\(
\begin{aligned}
& T_2-3 g=3 a \\
& \Rightarrow T_2=3(g+a) \\
& =3(9.8+2) \\
& =35.4 \mathrm{~N}
\end{aligned}
\)
From equation (i)
\(
\begin{aligned}
& T_1=T_2+5(g+a) \\
& =35.4+5(9.8+2) \\
& =94.4 \mathrm{~N}
\end{aligned}
\)

Q14: Block A of weight \(100 \mathrm{~N}\) rests on a frictionless inclined plane of slope angle \(30^{\circ}\) (Fig. 5.7). A flexible cord attached to A passes over a frictionless pulley and is connected to block \(B\) of weight \(W\). Find the weight \(\mathrm{W}\) for which the system is in equilibrium.

Answer: In equilibrium, the force mg \(\sin \theta\) acting on block A parallel to the plane should be balanced by the tension in the string, ie., \(m g \sin \theta=T=F \quad[\because T=F\) given \(] \ldots . .(i)\)
And for block B \(\mathrm{w}=T=F \dots(ii)\)
Where \(\mathrm{w}\) is the weight of block \(B\).
From equations (i) and (ii), we get,
\(
\begin{aligned}
& \mathrm{w}=\mathrm{mg} \sin \theta \\
& =100 \times \sin 30^{\circ} \ldots . .[\because \mathrm{mg}=100 \mathrm{~N}] \\
& =100 \times \frac{1}{2} \mathrm{~N} \\
& =50 \mathrm{~N}
\end{aligned}
\)

Q15: A block of mass \(M\) is held against a rough vertical wall by pressing it with a finger. If the coefficient of friction between the block and the wall is \(\mu\) and the acceleration due to gravity is \(g\), calculate the minimum force required to be applied by the finger to hold the block against the wall?

Answer:

Here, the force by the finger is given by ” \(\mathrm{F}\) “, the normal force is ” \(\mathrm{N}\) “, the frictional force is ” \(\mathrm{f}\) ” and the weight of the body is acting downwards given by ” \(\mathrm{Mg}\) “.
We first balance the horizontal forces:\(
F=N
\)
Now we balance the horizontal forces. Here, we have to stop the block from falling. So the acceleration should be zero.
\(
f-M g=0—–(1)
\)
The frictional force is given by:
\(
f=\mu N=\mu F—–(2)
\)
Substituting (2) in (1)
\(
\begin{aligned}
& \mu F=M g \\
& \Rightarrow F=\frac{M g}{\mu}
\end{aligned}
\)

Q16: A \(100 \mathrm{~kg}\) gun fires a ball of \(1 \mathrm{~kg}\) horizontally from a cliff of height \(500 \mathrm{~m}\). It falls on the ground at a distance of \(400 \mathrm{~m}\) from the bottom of the cliff. Find the recoil velocity of the gun. (acceleration due to gravity \(=10 \mathrm{~m} \mathrm{~s}^{-2}\) )

Answer: 

\(
\text { Given, the mass of the gun }\left(m_1\right)=100 kg
\)
Mass of the ball \(\left(\mathrm{m}_2\right)=1 \mathrm{~kg}\)
Height of the cliff \(=500 \mathrm{~m}\)
Horizontal distance travelled by the ball \((x)=400 \mathrm{~m}\)
From \(h=\frac{1}{2} \mathrm{gt}^2 \quad \ldots . .(\because\) Initial velocity in the downward direction is zero)
\(500=\frac{1}{2} \times 10 t^2\)
\(t=\sqrt{100}=10 \mathrm{~s}\)
From \(x=u t, u=\frac{x}{t}=\frac{400}{10}=40 \mathrm{~m} / \mathrm{s}\)
It \(v\) is the recoil velocity of the gun, then according to the principle of conservation of linear momentum,
\(
\begin{aligned}
& \mathrm{m}_1 \mathrm{v}=\mathrm{m}_2 \mathrm{u} \\
& \mathrm{v}=\frac{m_2 u}{m_1}=\frac{1}{100} \times 40=0.4 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Q17: Figure 5.8 shows \((x, t),(y, t)\) diagram of a particle moving in 2-dimensions.

If the particle has a mass of \(500 \mathrm{~g}\), find the force (direction and magnitude) acting on the particle.

Answer: Clearly, from the diagram (a), the variation can be related as \(\mathrm{x}=\mathrm{t}\)
\(
\begin{aligned}
& \Rightarrow \frac{d x}{d t}=1 \mathrm{~m} / \mathrm{s} \\
& \mathrm{a}_{\mathrm{x}}=0
\end{aligned}
\)
From diagram (b) \(y=t^2\)
\(
\Rightarrow \frac{d y}{d t}=2 \mathrm{t} \text { or } \mathrm{a}_{\mathrm{y}}=\frac{d^2 y}{d t^2}=2 \mathrm{~m} / \mathrm{s}^2
\)
Hence, \(\mathrm{I}_{\mathrm{y}}=\mathrm{ma}_{\mathrm{y}}=500 \times 10^{-3} \times 2=1 \mathrm{~N} \ldots . .(\because \mathrm{m}=500 \mathrm{~g})\)
\(
\mathrm{F}_{\mathrm{x}}=m \mathrm{a}_{\mathrm{x}}=0
\)
Hence, net force, \(\mathrm{F}=\sqrt{F_x^2+F_y^2}=\mathrm{F}_{\mathrm{y}}=1 \mathrm{~N} \quad\)….(along \(\mathrm{y}\)-axis)

Q18: A person in an elevator accelerating upwards with an acceleration of \(2 \mathrm{~m} \mathrm{~s}^{-2}\), tosses a coin vertically upwards with a speed of \(20 \mathrm{~m}\) \(\mathbf{s}^1\). After how much time will the coin fall back into his hand? \(\left(g=10 \mathrm{~m} \mathrm{~s}^{-2}\right)\)

Answer: Here, the initial velocity of the coin is, \(u=20 \mathrm{~m} \mathrm{~s}^1\)
And the acceleration of the elevator is, \(a=2 \mathrm{~m} \mathrm{~s}^{-2}\)
Here, let’s think about this problem from the elevator reference frame. For the person inside the elevator going up with acceleration \(2 \mathrm{~m} \mathrm{~s}^{-2}\) will experience a net acceleration of \((\mathrm{g}+\mathrm{a})\) which is \(12 \mathrm{~m} \mathrm{~s}^{-2}\).

And as the coin returns back to its original position (because we are in the elevator reference frame), the net displacement (s) is zero.
So we can use the laws of motion to solve the problem.
\(
s=0=u t-\frac{1}{2} a^{\prime} t^2
\)
Here, \(a^{\prime}=a+g=12 \mathrm{~m} \mathrm{~s}^{-2}\)
\(
\begin{aligned}
& \Rightarrow t\left(u-\frac{1}{2} a^{\prime} t\right)=0 \\
& \Rightarrow t=\frac{2 u}{a^{\prime}}=\frac{2 \times 20}{12}=\frac{10}{3}=3.33 \mathrm{~s}
\end{aligned}
\)
So the time of flight of the coin is 3.33 seconds.

Q19: There are three forces \(\mathbf{F}_{\mathbf{1}}, \mathbf{F}_{\mathbf{2}}\) and \(\mathbf{F}_{\mathbf{3}}\) acting on a body, all acting on a point \(\mathrm{P}\) on the body. The body is found to move with uniform speed.
(a) Show that the forces are coplanar.
(b) Show that the torque acting on the body about any point due to these three forces is zero.

Answer:

(a) Since the body is moving with no acceleration, the sum of the forces is zero \(\mathbf{F}_1+\mathbf{F}_2+\mathbf{F}_3=0\). Let \(\mathbf{F}_1, \mathbf{F}_2, \mathbf{F}_3\) be the three forces passing through a point. Let \(\mathbf{F}_1\) and \(\mathbf{F}_2\) be in plane A (one can always draw a plane having two intersecting lines such that the two lines lie on the plane). Then \(\mathbf{F}_1+\mathbf{F}_2\) must be in the plane A. Since \(\mathbf{F}_3=-\left(\mathbf{F}_1+\mathbf{F}_2\right), \mathbf{F}_3\) is also in the plane A.
(b) Consider the torque of the forces about P. Since all the forces pass through \(\mathrm{P}\), the torque is zero. Now consider torque about another point 0. Then torque about 0 is
Torque \(=\mathbf{O P} \times\left(\mathbf{F}_1+\mathbf{F}_2+\mathbf{F}_3\right)\)
Since \(\mathbf{F}_1+\mathbf{F}_2+\mathbf{F}_3=0\), torque \(=0\)

Q20: When a body slides down from rest along a smooth inclined plane making an angle of \(45^{\circ}\) with the horizontal, it takes time \(T\). When the same body slides down from rest along a rough inclined plane making the same angle and through the same distance, it is seen to take time \(p T\), where \(p\) is some number greater than 1 . Calculate the coefficient of friction between the body and the rough plane.

Answer: General Case:
Given that, the body slides down from an inclined plane making an angle of \(45^{\circ}\) with the horizontal, taking time T.
The effective acceleration of the body, in this case, will be \(a=g \sin 45^{\circ}=\frac{g}{\sqrt{2}}\)
Now for this motion, we can write,
\(
\Rightarrow s=u t+\frac{1}{2} a t^2
\)
That gives us,
\(
\begin{aligned}
& \Rightarrow s=0 \cdot T+\frac{1}{2} \frac{g}{\sqrt{2}} T^2 \\
& \text { Or } \Rightarrow s=\frac{g T^2}{2 \sqrt{2}}
\end{aligned}
\)
\(
T=\sqrt{2 s / a}
\)

Smooth Case:
For a smooth inclined plane,
Here, \(M g \sin \theta=M a\)
\(
\Rightarrow a=g \sin \theta
\)
Let \(s=\) Length of inclined plane
\(
\begin{aligned}
& \text { Using, } s=u t+\frac{1}{2} a t^2 \Rightarrow s=0 \times T+\frac{1}{2}(g \sin \theta) T^2(\because t=T) \\
& \Rightarrow \quad s=\frac{1}{2} g \sin \theta T^2 \Rightarrow s=\frac{1}{2 \sqrt{2}} g T^2\left(\because \theta=45^{\circ}\right) \quad \ldots(i)
\end{aligned}
\)
Rough Case:

For rough inclined plane.
\(
\begin{aligned}
& f=\mu N=\mu M g \cos \theta \\
& M g \sin \theta-f=M a^{\prime}
\end{aligned}
\)
\(
\Rightarrow a^{\prime}=(\sin \theta-\mu \cos \theta) g
\)
\(
\begin{aligned}
& \text { Using } s=u t+\frac{1}{2} a^{\prime} t^2 \\
& \Rightarrow \quad s=0 \times(p T)+\frac{1}{2}(\sin \theta-\mu \cos \theta) g \times p^2 T^2(\because t=p T) \\
& \Rightarrow \quad s=\frac{1}{2 \sqrt{2}}(1-\mu) g p^2 T^2 \quad \ldots(i i)
\end{aligned}
\)
From equation (i) and (ii)
\(
\begin{aligned}
& \frac{1}{2 \sqrt{2}} g T^2=\frac{1}{2 \sqrt{2}}(1-\mu) g p^2 T^2 \\
& \Rightarrow \quad 1=(1-\mu) p^2 \Rightarrow \mu=\left(1-\frac{1}{p^2}\right)
\end{aligned}
\)

Q21: Figure 5.9 shows \(\left(v_x, t\right)\), and \(\left(v_y, t\right)\) diagrams for a body of unit mass. Find the force as a function of time.

Answer:

\(
\begin{aligned}
v_x & =2 t; & & 0<t \leq 1 \\
& =2(2-t); & & 1<t<2 \\
& =0; & & 2<\mathrm{t}
\end{aligned}
\)
\(
\begin{aligned}
v_y & =t; \quad 0<t<1 \mathrm{~s} \\
& =1; 1 \mathrm{~s} <t
\end{aligned}
\)
\(
\begin{aligned}
F_x=2 ; & ~0<t<1 \\
= & -2 ; \quad 1 \mathrm{~s}<t<2 \mathrm{~s} \\
= & 0 ; \quad 2 \mathrm{~s}<t
\end{aligned}
\)
\(
\begin{aligned}
F_y & =1; & & 0<t<1 \mathrm{~s} \\
& =0;  & & 1 \mathrm{~s}<t
\end{aligned}
\)
\(
\begin{array}{rlrl}
\mathbf{F}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}; & 0<t<1 \mathrm{~s} \\
& =-2 \hat{\mathbf{i}}; & 1 \mathrm{~s}<t<2 \mathrm{~s} \\
& =0;  & 2 \mathrm{~s}<t
\end{array}
\)

Explanation:

In figure (a), we have \(v_x=2 t\) for \(t<1 s\) and \(v_x=2(2-t)\) for \(1 s<t<2 s\)
Hence the acceleration between 0 s to \(1 \mathrm{~s}\) will be given by,
\(
\Rightarrow a_x=\frac{d v_x}{d t}=\frac{d^2 t}{d t}=2 m s^{-2}
\)
Also \(F_x=m a_x=1 \times 2=2 N\) for \(\mathrm{t}<1 \mathrm{~s}\)
And the acceleration between \(1 \mathrm{~s}\) to \(2 \mathrm{~s}\) will be given by,
\(
\Rightarrow a_x=\frac{d v_x}{d t}=\frac{2 d(2-t)}{d t}=-2 \mathrm{~ms}^{-2}
\)
Also \(F_x=m a_x=1 \times(-2)=-2 N\) for \(1 \mathrm{~s}<\mathrm{t}<2 \mathrm{~s}\)
Now in figure (b), we have \(v_y=t\) for \(t<1 \mathrm{~s}\) and \(v_y=1\) for \(t>1 \mathrm{~s}\)
Hence the acceleration between 0 s to \(1 \mathrm{~s}\) will be given by,
\(
\Rightarrow a_y=\frac{d v_y}{d t}=\frac{d t}{d t}=1 m s^{-2}
\)
Also \(F_y=m a_y=1 \times 1=1 N\) for \(\mathrm{t}<1 \mathrm{~s}\)
And the acceleration between \(1 \mathrm{~s}\) to \(3 \mathrm{~s}\) will be given by,
\(
\Rightarrow a_y=\frac{d v_y}{d t}=\frac{d 1}{d t}=0 \mathrm{~ms}^{-2}
\)
Also \(F_y=m a_y=1 \times 0=0 N\) for \(\mathrm{t}>1 \mathrm{~s}\)
Now the resultant force will be given by,
\(
\Rightarrow \vec{F}=\vec{F}_x \hat{i}+\vec{F}_y \hat{j}
\)
Hence,
\(
\Rightarrow \vec{F}=2 \hat{i}+\hat{j} \text { for } \mathrm{t}<1 \mathrm{~s}
\)
And,
\(
\Rightarrow \vec{F}=-2 \hat{i} \text { for } 1 \mathrm{~s}<\mathrm{t}<2 \mathrm{~s}
\)
And,
\(
\Rightarrow \vec{F}=0 \text { for } t>2 \mathrm{~s}
\)

Q22: A racing car travels on a track (without banking) ABCDEFA (Fig. 5.10). ABC is a circular arc of radius \(2 R . C D\) and FA are straight paths of length \(R\) and DEF is a circular arc of radius \(R=100 \mathrm{~m}\). The coefficient of friction on the road is \(\mu=0.1\). The maximum speed of the car is \(50 \mathrm{~m} \mathrm{~s}^{-1}\). Find the minimum time for completing one round.

Answer:
Balancing frictional force for centripetal
force \(\frac{m v^2}{R}=f=\mu N=\mu m g\)
where, \(\mathrm{N}\) is normal reaction
\(\therefore v_{max}=\sqrt{\mu R g}\) (where \(R\) is radius of the circular track)
Segment ABC:
For path \(A B C\) Path length
\(
\begin{aligned}
& =\frac{3}{4}(2 \pi 2 R)=3 \pi R=3 \pi \times 100 \\
& =300 \pi
\end{aligned}
\)
Here, the frictional force is
\(
f=\mu m g
\)
And the centripetal force is
\(
F_C=\frac{m v_{A B C}^2}{2 R}
\)
Therefore,
\(
\frac{m v_{A B C}^2}{2 R}=\mu m g
\)
\(
\Rightarrow v_{A B C}=\sqrt{2 R \mu g}=\sqrt{200}=14.14 \mathrm{~ms}^{-1}
\)
\(
\text { Time for } \mathrm{ABC}=\frac{3 \pi}{2} \frac{200}{14.14}=\frac{300 \pi}{14.14} \mathrm{~s}
\)
Segment DEF:
For path \(D E F\) Path
\(
\begin{aligned}
& \text { length }=\frac{1}{4}(2 \pi R)=\frac{\pi \times 100}{2}=50 \pi \\
& v_{D E F}=\sqrt{\mu R g}=\sqrt{0.1 \times 100 \times 10}=10 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
\(
\text { Time for DEF }=\frac{\pi}{2} \times \frac{100}{10}=5 \pi \mathrm{s}
\)
For paths, \(C D\) and \(F A\) Path length
\(
\begin{aligned}
& =R+R=2 R=200 \mathrm{~m} \\
& \text { Time for } \mathrm{CD ~and ~FA} =\frac{200}{50}=4.0 \mathrm{~s}
\end{aligned}
\)
\(
\text { Total time }=5 \pi+\frac{300 \pi}{14.14}+4=86.3 \mathrm{~s}
\)

Q23: The displacement vector of a particle of mass \(m\) is given by \(\mathbf{r}(t)=\hat{\mathbf{i}} A \cos \omega t+\hat{\mathbf{j}} B \sin \omega t\).
(a) Show that the trajectory is an ellipse.
(b) Show that \(\mathbf{F}=-m \omega^2 \mathbf{r}\).

Answer: (a) The displacement vector of the particle of mass \(\mathrm{m}\) is given by \(r(t)=\hat{i} \mathrm{~A} \cos \omega t+\hat{j} \mathrm{~B} \sin \omega t\)
\(\therefore\) Displacement along \(\mathrm{x}\)-axis is \(x=A \cos \omega t\)
or \(\frac{x}{A}=\cos \omega t \dots(i)\)
Displacement along the \(y\)-axis is,
And \(y=B \sin \omega t\)
or \(\frac{y}{B}=\sin \omega t \dots(ii)\)
Squaring and then adding equations (i) and (ii), we get
\(
\frac{x^2}{A^2}+\frac{y^2}{B^2}=\cos ^2 \omega t+\sin ^2 \omega t=1
\)
This is an equation of the ellipse.
Therefore, the trajectory of the particle is an ellipse.

(b) The velocity of the particle
\(
\begin{aligned}
& v=\frac{d r}{d t}=\hat{i} \frac{d}{d t}(A \cos \omega t)+\hat{j} \frac{d}{d t}(B \sin \omega t) \\
& =\hat{i}[A(-\sin \omega t) \cdot \omega]+\hat{j}[B(\cos \omega t) \cdot \omega] \\
& =-\hat{i} A \omega \sin \omega t+\hat{j} B \omega \cos \omega t
\end{aligned}
\)
Acceleration of the particle \((\mathrm{a})=\frac{d v}{d t}\)
\(
\begin{aligned}
& \text { or } a=-\hat{i} A \omega \frac{d}{d}(\sin \omega t)+\hat{j} B \omega \frac{d}{d t}(\cos \omega t) \\
& =-\hat{i} A \omega[\cos \omega t] \cdot \omega+\hat{j} B \omega[-\sin \omega t] \cdot \omega \\
& =-\hat{i} A \omega^2 \cos \omega t-\hat{j} B \omega^2 \sin \omega t \\
& =-\omega^2[\hat{i} A \cos \omega t+\hat{j} B \sin \omega t] \\
& =-\omega^2 r
\end{aligned}
\)
\(\therefore\) Force acting on the particle, \(\mathrm{F}=\mathrm{ma}=-\mathrm{m} \omega^2 \mathrm{r}\), Hence proved.

Q24: A cricket bowler releases the ball in two different ways
(a) giving it only horizontal velocity, and
(b) giving it horizontal velocity and a small downward velocity. The speed \(v_s\) at the time of release is the same. Both are released at a height \(H\) from the ground. Which one will have greater speed when the ball hits the ground? Neglect air resistance.

Answer:

(a) When a ball is given only horizontal velocity at the time of release \(\left(u_x\right)=v_s\)

During projectile motion. horizontal velocity remains unchanged,
Therefore, \(v_x=u_x=v_s\)
In the vertical direction, \(v_y^2=u_y^2+2 g H\)
\(
v_y=\sqrt{2 g H} \ldots \ldots\left(\because {u}_y=0\right)
\)
\(\therefore\) Resultant speed of the ball at the bottom,
\(
\begin{aligned}
& v=\sqrt{v_x^2+v_y^2} \\
& =\sqrt{v_s^2+2 g H}
\end{aligned}
\)

(b) For (b) also \(\left[\frac{1}{2} m v_{\mathrm{s}}{ }^2+m g H\right]\) is the total energy of the ball when it hits the ground.
So the speed would be the same for both (a) and (b).

Explanation: For case b we also have an additional downward velocity \(\mathrm{v}_{\mathrm{s}}\) in addition to horizontal velocity \(\mathrm{v}_{\mathrm{h}}\). So, the total velocity is
\(
v=\sqrt{v_s^2+v_h^2}
\)
At the point of contact, \(v_h=0\)
So, from the conservation of energy,
\(
\begin{aligned}
& \frac{1}{2} m v_s^2=m g H \\
& \Rightarrow v_s=\sqrt{2 g H}
\end{aligned}
\)
Hence, both techniques will yield the same speed.

Q25: There are four forces acting at a point \(\mathrm{P}\) produced by strings as shown in Fig. 5.11, which is at rest. Find the forces \(\mathbf{F}_1\) and \(\mathbf{F}_{\mathbf{2}}\).

Answer: In this problem, we have to resolve forces horizontally and vertically.
Case: 1 Vertically
\(
\begin{aligned}
& 2 \cos 45^{\circ}+1 \cos 45^{\circ}=F_2 \\
& \Rightarrow F_2=\frac{3}{\sqrt{2}} N
\end{aligned}
\)
Case: 2 Horizontally
\(
\begin{aligned}
& 2 \cos 45^{\circ}=1 \cos 45^{\circ}+F_1 \\
& \Rightarrow F_1=\frac{1}{\sqrt{2}} N
\end{aligned}
\)

Q26: A rectangular box lies on a rough inclined surface. The coefficient of friction between the surface and the box is \(\mu\). Let the mass of the box be \(m\).
(a) At what angle of inclination \(\theta\) of the plane to the horizontal will the box just start to slide down the plane?
(b) What is the force acting on the box down the plane, if the angle of inclination of the plane is increased to \(\alpha>\theta\)?
(c) What is the force needed to be applied upwards along the plane to make the box either remain stationary or just move up with uniform speed?
(d) What is the force needed to be applied upwards along the plane to make the box move up the plane with acceleration \(a\)?

Answer: 

(a) Consider the given diagram, the force of friction(f) on the box will act up the plane.
Therefore,
For the box to just start sliding down,
The horizontal equation is given by,
\(f=m g \sin \theta \dots(i)\)
The vertical equation is given by,
\(
N=m g \cos \theta \dots(ii)
\)
As we know,
\(
\mathrm{f}=\mu \mathrm{N}
\)
Therefore,
From equation (i) & (ii)
\(
\begin{aligned}
& m g \sin \theta=\mu m g \cos \theta \\
& \sin \theta=\mu \cos \theta \\
& \mu=\tan \theta
\end{aligned}
\)
\(
\theta=\tan ^{-1} \mu
\)

(b) 

As given, \(\alpha>\theta\), the angle of inclination of the plane with horizontal it will slide down (f upward) as \(\theta\) is the angle of repose. So the net force downward
\(
F_1=m g \sin \alpha-f
\)
As we know,
\(
f=\mu \mathrm{N}
\)
Therefore,
\(
{F}_1=m g \sin \alpha-\mu \mathrm{N}
\)
\(
F_1=m g \sin \alpha-\mu m g \cos \alpha
\)
\(
F_1=m g[\sin \alpha-\mu \cos \alpha]
\)
Final Answer: \(F_1=m g[\sin \alpha-\mu \cos \alpha]\)

(c)

To move the box upward with uniform velocity or to keep it stationary,
\(
F_2-m g \sin \alpha-f=m a
\)
As we know,
\(
f=\mu \mathrm{N}
\)
Since \(\mathrm{a}=0\), the block has to move with uniform velocity or remain stationary,
Then,
\(
F_2-m g \sin \alpha-\mu N=0
\)
Therefore,
\(
F_2=m g \sin \alpha+\mu N
\)
\(
F_2=m g(\sin \alpha+\mu \cos \alpha)
\)

(d)

As we know,
\(
N=m g \cos \theta
\)
\(
f=\mu N
\)
The force applied \(\mathrm{F}_3\) to move the box upward with acceleration a,
\(
F_3-f-\mu N=m a
\)
Therefore,
\(
\mathrm{F}_3-\mathrm{mg} \sin \theta-\mu \mathrm{mg} \cos \theta=\mathrm{ma}
\)
So,
\(
F_3=m g(\sin \theta+\mu \cos \theta)+m a
\)
Final Answer :
\(
F_3=m g(\sin \theta+\mu \cos \theta)+m a
\)

Q27: A helicopter of mass \(2000 \mathrm{~kg}\) rises with a vertical acceleration of \(15 \mathrm{~m} \mathrm{~s}^{-2}\). The total mass of the crew and passengers is \(500 \mathrm{~kg}\). Give the magnitude and direction of the \(\left(g=10 \mathrm{~m} \mathrm{~s}^{-2}\right)\)

(a) force on the floor of the helicopter by the crew and passengers.
(b) the action of the rotor of the helicopter on the surrounding air.
(c) force on the helicopter due to the surrounding air.

Answer: Given, the mass of the helicopter \(\left(m_1\right)=2000 \mathrm{~kg}\)
Mass of the crew and passengers \(\mathrm{m}_2=500 \mathrm{~kg}\)
Acceleration in vertical direction \(\mathrm{a}=15 \mathrm{~m} / \mathrm{s}^2(\uparrow)\) and \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2(\downarrow)\)
(a) \(F-(500 \times 10)=(500 \times 15)\) or \(F=12.5 \times 10^3 \mathrm{~N}\), where \(F\) is the upward reaction of the floor and is equal to the force downwards on the floor, by Newton’s 3rd law of motion
(b) \(\mathrm{R}-(2500 \times 10)=(2500 \times 15)\) or \(R=6.25 \times 10^4 \mathrm{~N}\), the action of the atr on the system, upwards. The action of the rotor on the surrounding air is \(6.25 \times 10^4 \mathrm{~N}\) downwards.
(c) Force on the helicopter due to the atr \(=6.25 \times 10^4 \mathrm{~N}\) upwards.

Q28: Give the magnitude and direction of the net force acting on
(a) a drop of rain falling down with a constant speed,
(b) a cork of mass \(10 \mathrm{~g}\) floating on water,
(c) a kite skillfully held stationary in the sky,
(d) a car moving with a constant velocity of \(30 \mathrm{~km} / \mathrm{h}\) on a rough road,
(e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields.

For simplicity in numerical calculations, take \(g=10 \mathrm{~m} \mathrm{~s}^{-2}\)

Answer: (a) As the drop of rain is failing with constant speed, in accordance with the first law of motion, the net force on the drop of rain is zero.
(b) As the cork is floating on water, its weight is being balanced by the upthrust (equal to the weight of water dissipated ). Hence net force on the cork is zero.
(c) Net force on a kite skilfully held stationary in the sky is zero because it is at rest.
(d) Since the car is moving with a constant velocity, the net force on the car is zero.
(e) Since the electron is far away from all material agencies producing electromagnetic and gravitational forces, the net force on the electron is zero.

Q29: A pebble of mass \(0.05 \mathrm{~kg}\) is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,
(a) during its upward motion,
(b) during its downward motion,
(c) at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of \(45^{\circ}\) with the horizontal direction?
Ignore air resistance.
For simplicity in numerical calculations, take \(g=10 \mathrm{~m} \mathrm{~s}^{-2}\)

Answer: The only force in each case is the force of gravity, (neglecting effects of air) equal to \(0.5 \mathrm{~N}\) vertically downward. The answers do not change, even if the motion of the pebble is not along the vertical. The pebble is not at rest at the highest point. It has a constant horizontal component of velocity throughout its motion.

Explanation: \(0.5 \mathrm{~N}\), in vertically downward direction, in all cases
Acceleration due to gravity, irrespective of the direction of motion of an object, always acts downward. The gravitational force is the only force that acts on the pebble in all three cases. Its magnitude is given by Newton’s second law of motion as:
\(
F=m \times a
\)
Where,
\(
\begin{aligned}
& F=\text { Net force } \\
& m=\text { Mass of the pebble }=0.05 \mathrm{~kg} \\
& a=g=10 \mathrm{~m} / \mathrm{s}^2 \\
& \therefore F=0.05 \times 10=0.5 \mathrm{~N}
\end{aligned}
\)
The net force on the pebble in all three cases is \(0.5 \mathrm{~N}\) and this force acts in the downward direction.
If the pebble is thrown at an angle of \(45^{\circ}\) with the horizontal, it will have both the horizontal and vertical components of velocity. At the highest point, only the vertical component of velocity becomes zero. However, the pebble will have the horizontal component of velocity throughout its motion. This component of velocity produces no effect on the net force acting on the pebble.

Q30: Give the magnitude and direction of the net force acting on a stone of mass \(0.1 \mathrm{~kg}\),
(a) just after it is dropped from the window of a stationary train,
(b) just after it is dropped from the window of a train running at a constant velocity of \(36 \mathrm{~km} / \mathrm{h}\),
(c) just after it is dropped from the window of a train accelerating with \(1 \mathrm{~m} \mathrm{~s}^{-2}\),
(d) lying on the floor of a train which is accelerating with \(1 \mathrm{~m} \mathrm{~s}^{-2}\), the stone being at rest relative to the train.
Neglect air resistance throughout. For simplicity in numerical calculations, take \(g=10 \mathrm{~m} \mathrm{~s}^{-2}\)

Answer:  (a) \(1 \mathrm{~N}\); vertically downward
Mass of the stone, \(m=0.1 \mathrm{~kg}\)
Acceleration of the stone, \(a=\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\)
As per Newton’s second law of motion, the net force acting on the stone,
\(
\begin{aligned}
& F=m a=m g \\
& =0.1 \times 10=1 \mathrm{~N}
\end{aligned}
\)

(b) When the train is running at a constant velocity, its acceleration is zero. No force acts on the stone due to this motion. Therefore, the force on the stone is the same \((1.0 \mathrm{~N}\).)

(c) \(1 \mathrm{~N}\); vertically downward
It is given that the train is accelerating at the rate of \(1 \mathrm{~m} / \mathrm{s}^2\).
Therefore, the net force acting on the stone, \(F^{\prime}=m a=0.1 \times 1=0.1 \mathrm{~N}\)
This force is acting in the horizontal direction. Now, when the stone is dropped, the horizontal force \(F^{\prime}\), stops acting on the stone. This is because of the fact that the force acting on a body at an instant depends on the situation at that instant and not on earlier situations.
Therefore, the net force acting on the stone is given only by the acceleration due to gravity.
\(
F=m g=1 \mathrm{~N}
\)

(d) \(0.1 \mathrm{~N}\); in the direction of motion of the train
The weight of the stone is balanced by the normal reaction of the floor. The only acceleration is provided by the horizontal motion of the train.
Acceleration of the train, \(a=0.1 \mathrm{~m} / \mathrm{s}^2\)
The net force acting on the stone will be in the direction of motion of the train. Its magnitude is given by:
\(
\begin{aligned}
& F=m a \\
& =0.1 \times 1=0.1 \mathrm{~N}
\end{aligned}
\)

Q31: One end of a string of length \(l\) is connected to a particle of mass \(m\) and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed \(v\) the net force on the particle (directed towards the centre) is :
(i) \(T\),
(ii) \(T-\frac{m v^2}{l}\),
(iii) \(T+\frac{m v^2}{l}\),
(iv) 0
\(T\) is the tension in the string. [Choose the correct alternative].

Answer: (i) \(\mathrm{T}\)
When a particle connected to a string revolves in a circular path around a centre, the centripetal force is provided by the tension produced in the string. Hence, in the given case, the net force on the particle is the tension \(T\), i.e.,
\(
F=T=\frac{m v^2}{l}
\)
Where \(F\) is the net force acting on the particle.

Q32: A body of mass \(0.40 \mathrm{~kg}\) moving initially with a constant speed of \(10 \mathrm{~m} \mathrm{~s}^{-1}\) to the north is subject to a constant force of \(8.0 \mathrm{~N}\) directed towards the south for \(30 \mathrm{~s}\). Take the instant the force is applied to be \(t=0\), the position of the body at that time to be \(x=0\), and predict its position at \(\mathrm{t}=-5 \mathrm{~s}, 25 \mathrm{~s}, 100 \mathrm{~s}\).

Answer: Here, \(m=0.40 \mathrm{~kg}, \mathrm{u}=10 \mathrm{~ms}^{-1}, \mathrm{~F}=-8 \mathrm{~N}\) (retarding force)
As \(\mathrm{a}=\mathrm{F} / \mathrm{m}=8 / 0.4=-20 \mathrm{~ms}^{-2}\)
Also, \(\mathrm{S}=u t+\frac{1}{2} a T^2\)
(i) Position at \(\mathrm{t}=-5 \mathrm{~s}\)
\(
S=10(-5)+1 / 2 \times 0 \times(-5)^2=-50 \mathrm{~m}
\)
(ii) Position at \(\mathrm{t}=25 \mathrm{~s}\)
\(
S_1=10 \times 25+\frac{1}{2} \times(-20) \times(25)^2=-6000 \mathrm{~m}=-6 \mathrm{~km}
\)
(iii) Position at \(\mathrm{t}=100 \mathrm{~s}\)
\(
\begin{aligned}
& S_2=10 \times 30+\frac{1}{2} \times(-20) \times(30)^2=-8700 \mathrm{~m} \\
& \text { At t }=30 \mathrm{~s}, \mathrm{v}=\mathrm{u}+\text { at } \\
& \mathrm{v}=10-20 \times 30=-590 \mathrm{~ms}^{-1}
\end{aligned}
\)
Now , for motion from \(30 \mathrm{~s}\) to \(100 \mathrm{~s}\)
\(
S_3=-590 \times 70+\frac{1}{2}(0) \times(70)^2=-41300 \mathrm{~m}
\)
Total distance \(=S_2+S_3=-8700-41300=-50000 \mathrm{~m}=-50 \mathrm{~km}\).

Q33: A man of mass \(70 \mathrm{~kg}\) stands on a weighing scale in a lift that is moving
(a) upwards with a uniform speed of \(10 \mathrm{~m} \mathrm{~s}^{-1}\),
(b) downwards with a uniform acceleration of \(5 \mathrm{~m} \mathrm{~s}^{-2}\),
(c) upwards with a uniform acceleration of \(5 \mathrm{~m} \mathrm{~s}^{-2}\). What would be the readings on the scale in each case?
(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?

Answer: Here, \(m=70 \mathrm{~kg}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^2\)
The weighing machine in each case measures the reaction \(\mathrm{R}\) i.e., the apparent weight.
a) When the lift moves upwards with a uniform speed, its acceleration is zero.
\(
R=m g=70 \times 10=700 \mathrm{~N}; \text { Reading is } 70 \mathrm{~kg}
\)
(b) When the lift moves downwards with \(\mathrm{a}=5 \mathrm{~ms}^{-2}\)
\(
R=m(g-a)=70(10-5)=350 \mathrm{~N}; \text { Reading is } 35 \mathrm{~kg}
\)
(c) When the lift moves upwards with \(\mathrm{a}=5 \mathrm{~ms}^{-2}\)
\(
R=m(g+a)=70(10+5)=1050 \mathrm{~N}; \text { Reading is } 105 \mathrm{~kg}
\)
(d) If the lift were to come down freely under gravity, downward acceleration \(a=g\)
\(
\therefore R=m(g-a)=m(g-g)=\text { Zero }; \text { Reading would be zero; the scale would read zero. }
\)

Q34: Figure 5.16 shows the position-time graph of a particle of mass \(4 \mathrm{~kg}\). What is the (a) force on the particle for \(t<0, t>4 \mathrm{~s}, 0<t<4 \mathrm{~s}\)? (b) impulse at \(t=0\) and \(t=4 \mathrm{~s}\)? (Consider one-dimensional motion only).

Answer: (a) For \(t<0\)
It can be observed from the given graph that the position of the particle is coincident with the time axis. It indicates that the displacement of the particle in this time interval is zero.
Hence, the force acting on the particle is zero.
For \(t>4 \mathrm{~s}\)
It can be observed from the given graph that the position of the particle is parallel to the time axis. It indicates that the particle is at rest at a distance of \(3 \mathrm{~m}\) from the origin.
Hence, no force is acting on the particle.
For \(0<t<4\)
It can be observed that the given position-time graph has a constant slope. Hence, the acceleration produced in the particle is zero. Therefore, the force acting on the particle is zero.
(b) At \(t=0\)
Impulse \(=\) Change in momentum
\(
=m v-m u
\)
Mass of the particle, \(m=4 \mathrm{~kg}\)
The initial velocity of the particle, \(u=0\)
Final velocity of the particle, \(v=\frac{3}{4} \mathrm{~m} / \mathrm{s}\)
\(
\therefore \text { Impulse }=4\left(\frac{3}{4}-0\right)=3 \mathrm{~kg} \mathrm{~m} / \mathrm{s}
\)
At \(t=4 \mathrm{~s}\)
Initial velocity of the particle, \(u=\frac{3}{4} \mathrm{~m} / \mathrm{s}\)
Final velocity of the particle, \(v=0\)
\(
\therefore \text { Impulse }=4\left(0-\frac{3}{4}\right)=-3 \mathrm{~kg} \mathrm{~m} / \mathrm{s}
\)

Q35: A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.

Answer: Let \(m, m_1\), and \(m_2\) be the respective masses of the parent nucleus and the two daughter nuclei. The parent nucleus is at rest.
The initial momentum of the system (parent nucleus) \(=0\)
Let \(v_1\) and \(v_2\) be the respective velocities of the daughter nuclei having masses \(m_1\) and \(m_2\).
Total linear momentum of the system after disintegration \(=\) \(m_1 v_1+m_2 v_2\)
According to the law of conservation of momentum:
Total initial momentum \(=\) Total final momentum
\(
\begin{aligned}
& 0=m_1 v_1+m_2+v_2 \\
& v_1=\frac{-m_2 v_2}{m_1}
\end{aligned}
\)
Here, the negative sign indicates that the fragments of the parent nucleus move in directions opposite to each other.

Note: By the momentum conservation principle, the total final momentum is zero. Two momentum vectors cannot sum to a null momentum unless they are equal and opposite.

Q36: A stone of mass \(0.25 \mathrm{~kg}\) tied to the end of a string is whirled round in a circle of radius \(1.5 \mathrm{~m}\) with a speed of \(40 \mathrm{rev}. / \mathrm{min}\) in a horizontal plane. If the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks :
(a) the stone moves radially outwards,
(b) the stone flies off tangentially from the instant the string breaks,
(c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle.

Answer: Option (b) is correct. When the string breaks, the stone will move in the direction of the velocity at that instant. According to the first law of motion, the direction of the velocity vector is tangential to the path of the stone at that instant. Hence, the stone will fly off tangentially from the instant the string breaks.

Q37: Explain why
(a) a horse cannot pull a cart and run in empty space,
(b) passengers are thrown forward from their seats when a speeding bus stops suddenly,
(c) it is easier to pull a lawn mower than to push it,
(d) a cricketer moves his hands backward while holding a catch.

Answer: (a) The horse-cart system has no external force in empty space. The mutual forces between the horse and the cart cancel (Third Law). On the ground, the contact force between the system and the ground (friction) causes their motion from rest.

In order to pull a cart, a horse pushes the ground backward with some force. The ground in turn exerts an equal and opposite reaction force upon the feet of the horse. This reaction force causes the horse to move forward.
An empty space is devoid of any such reaction force. Therefore, a horse cannot pull a cart and run in empty space.

(b) Due to inertia of the body not directly in contact with the seat. The passengers in a speeding bus have the inertia of motion. When the bus is suddenly stopped the passengers are thrown forward due to this inertia of motion.

(c) A lawn mower is pulled or pushed by applying force at an angle. When you push, the normal force \((N)\) must be more than its weight, for equilibrium in the vertical direction. This results in greater friction \(f(f \propto N)\) and, therefore, a greater applied force to move. Just the opposite happens while pulling.

(d) To reduce the rate of change of momentum and hence to reduce the force necessary to stop the ball. The ball comes with large momentum after being hit by the batsman. When the player takes a catch it causes a large impulse on his palms which may hurt the cricketer. When he moves his hands backward the time of contact of the ball and hand is increased so the force is reduced.

Q38: Figure 5.17 shows the position-time graph of a body of mass \(0.04 \mathrm{~kg}\). Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the body? What is the magnitude of each impulse?

Answer: This graph can be of a ball rebounding between two walls situated at position \(0 \mathrm{~cm}\) and \(2 \mathrm{~cm}\). The ball is rebounding from one wall to another, time and again every \(2 \mathrm{~s}\) with uniform velocity.
Impulse, Here velocity \(=\frac{\text { displacement }}{\text { time }}=\frac{2}{100 \times 2}=0.01 \mathrm{~ms}^{-2}\)
Initial momentum \(=\mathrm{mu}=0.04 \times 0.01=4 \times 10^{-4} \mathrm{~kg} \mathrm{~ms}^{-1}\)
Final momentum \(=\mathrm{mv}=0.04 \times (-0.01)=-4 \times 10^{-4} \mathrm{~kg} \mathrm{~ms}^{-1}\)
Magnitude of impulse \(=\) Change in momentum
\(=\left(4 \times 10^{-4}\right)-\left(-4 \times 10^{-4}\right)=8 \times 10^{-4} \mathrm{kgms}^{-1}\)
Time between two consecutive impulses is 2 s.i.e the ball receive an impulse every \(2 \mathrm{~s}\)

Q39: Figure 5.18 shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with \(1 \mathrm{~m} \mathrm{~s}^{-2}\). What is the net force on the man? If the coefficient of static friction between the man’s shoes and the belt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to the belt? (Mass of the \(\operatorname{man}=65 \mathrm{~kg}\).)

Answer:

Here acceleration of conveyor belt \(\mathrm{a}=1 \mathrm{~ms}^{-2}, \mu_{\mathrm{s}}=0.2\) and mass of man \(\mathrm{m}=65 \mathrm{~kg}\). \(\mathrm{t}\) As the man is in an accelerating frame, he experiences a pseudo force \(\mathrm{F}_{\mathrm{s}}=\mathrm{ma}\) as shown in figure

Hence to maintain his equilibrium, he exerts a force \(F=-F_s=m a=65 \times 1=65 \mathrm{~N}\) in the forward direction i.e., the direction of motion of the belt.
\(\therefore\) Net force acting on man \(=65 \mathrm{~N}\) (forward)
As shown in Fig. (b), the man can continue to be stationary with respect to the belt, if the force of friction
\(
\begin{aligned}
& \mu_s N=\mu_s m g=m a_{\max } \\
& a_{\max }=\mu_s g=0.2 \times 10=2 m s^{-2}
\end{aligned}
\)

Q40: A stone of mass \(m\) tied to the end of a string revolves in a vertical circle of radius \(R\). The net forces at the lowest and highest points of the circle directed vertically downwards are: [Choose the correct alternative]

\(\mathrm{T}_1\) and \(v_1\) denote the tension and speed at the lowest point. \(\mathrm{T}_2\) and \(v_2\) denote corresponding values at the highest point.

Answer: (a)The free-body diagram of the stone at the lowest point is shown in the following figure.

According to Newton’s second law of motion, the net force acting on the stone at this point is equal to the centripetal force, i.e.,
\(
F_{\text {net }}=T-m g=\frac{m v_1^2}{R} \ldots(i)
\)
Where, \(v_1=\) Velocity at the lowest point.

The free-body diagram of the stone at the highest point is shown in the following figure.

Using Newton’s second law of motion, we have:
\(T+m g=\frac{m v_2^2}{R} \ldots\) (ii)
Where, \(v_2=\) Velocity at the highest point
It is clear from equations (i) and (ii) that the net force acting at the lowest and the highest points are respectively \((T-m g)\) and \((T+m g)\).

The moral is: do not confuse the actual material forces on a body (tension, gravitational force, etc) with the effects they produce: centripetal acceleration \(v_2^2 / \mathrm{R}\) or \(\boldsymbol{v}_1^2 / \mathrm{R}\) in this example.

Q41: A helicopter of mass \(1000 \mathrm{~kg}\) rises with a vertical acceleration of \(15 \mathrm{~m} \mathrm{~s}^{-2}\). The crew and the passengers weigh \(300 \mathrm{~kg}\). Give the magnitude and direction of the
(a) force on the floor by the crew and passengers,
(b) the action of the rotor of the helicopter on the surrounding air,
(c) force on the helicopter due to the surrounding air.

Answer: (a) Mass of the helicopter, \(m_{\mathrm{h}}=1000 \mathrm{~kg}\)
Mass of the crew and passengers, \(m_{\mathrm{p}}=300 \mathrm{~kg}\)
Total mass of the system, \(m=1300 \mathrm{~kg}\)
Acceleration of the helicopter, \(a=15 \mathrm{~m} / \mathrm{s}^2\)
Using Newton’s second law of motion, the reaction force \(R\), on the system by the floor can be calculated as:
\(
\begin{aligned}
& R-m_{\mathrm{p}} \mathrm{g}=m_p a \\
& =m_{\mathrm{p}}(\mathrm{g}+a) \\
& =300(10+15)=300 \times 25 \\
& =7500 \mathrm{~N}
\end{aligned}
\)
Since the helicopter is accelerating vertically upward, the reaction force will also be directed upward. Therefore, as per Newton’s third law of motion, the force on the floor by the crew and passengers is \(7500 \mathrm{~N}\), directed downward.
(b) Using Newton’s second law of motion, the reaction force \(R^{\prime}\), experienced by the helicopter can be calculated as:
\(
\begin{aligned}
& R^{\prime}-m g=m a \\
& =m(g+a) \\
& =1300(10+15)=1300 \times 25 \\
& =32500 \mathrm{~N}
\end{aligned}
\)
The reaction force experienced by the helicopter from the surrounding air is acting upward. Hence, as per Newton’s third law of motion, the action of the rotor on the surrounding air will be \(32500 \mathrm{~N}\), directed downward.
(c) The force on the helicopter due to the surrounding air is \(32500 \mathrm{~N}\), directed upward.

Q42: Ten one-rupee coins are put on top of each other on a table. Each coin has a mass \(m\). Give the magnitude and direction of
(a) the force on the \(7^{\text {th }}\) coin (counted from the bottom) due to all the coins on its top,
(b) the force on the \(7^{\text {th }}\) coin by the eighth coin,
(c) the reaction of the \(6^{\text {th }}\) coin on the \(7^{\text {th }}\) coin.

Answer: (a) Force on the seventh coin is exerted by the weight of the three coins on its top.
Weight of one coin \(=\mathrm{mg}\)
Weight of three coins \(=3 \mathrm{mg}\)
Hence, the force exerted on the \(7^{\text {th }}\) coin by the three coins on its top is \(3 \mathrm{mg}\).
This force acts vertically downward.
(b) Force on the seventh coin by the eighth coin is because of the weight of the eighth coin and the other two coins (ninth and tenth) on its top.
Weight of the eighth coin \(=m g\)
Weight of the ninth coin \(=\mathrm{mg}\)
Weight of the tenth coin \(=\mathrm{mg}\)
Total weight of these three coins \(=3 \mathrm{mg}\)
Hence, the force exerted on the \(7^{\text {th }}\) coin by the eighth coin is \(3 \mathrm{mg}\). This force acts vertically downward.
(c) The 6th coin experiences a downward force because of the weight of the four coins \(\left(7^{\text {th }}, 8^{\text {th }}, 9^{\text {th }}\right. \), and \(\left.10^{\text {th }}\right)\) on its top. Therefore, the total downward force experienced by the 6th coin is \(4 \mathrm{mg}\). As per Newton’s third law of motion, the 6th coin will produce an equal reaction force on the \(7^{\text {th }}\) coin, but in the opposite direction. Hence, the reaction force of the 6th coin on the \(7^{\text {th }}\) coin is of magnitude \(4 \mathrm{mg}\). This force acts in the upward direction.

Q43: A train runs along an unbanked circular track of radius \(30 \mathrm{~m}\) at a speed of \(54 \mathrm{~km} / \mathrm{h}\). The mass of the train is \(10^6 \mathrm{~kg}\). What provides the centripetal force required for this purpose – The engine or the rails? What is the angle of banking required to prevent wearing out of the rail?

Answer: Radius of the circular track, \(R=30 \mathrm{~m}\)
Speed of the train, \(v=54 \mathrm{~km} / \mathrm{h}=15 \mathrm{~m} / \mathrm{s}\)
Mass of the train, \(m=10^6 \mathrm{~kg}\)
The centripetal force is provided by the lateral thrust by the rail on the flanges of the wheels. By the Third Law, the train exerts an equal and opposite thrust on the rail causing its wear and tear.
Angle of banking \(=\tan ^{-1}\left(\frac{v^2}{R g}\right)=\tan ^{-1}\left(\frac{15 \times 15}{30 \times 10}\right) \simeq 37^{\circ}\)

Q44: A block of mass \(25 \mathrm{~kg}\) is raised by a \(50 \mathrm{~kg}\) man in two different ways as shown in Fig. 5.19. What is the action on the floor by the man in the two cases? If the floor yields to a normal force of \(700 \mathrm{~N}\), which mode should the man adopt to lift the block without the floor yielding?

Answer: \(750 \mathrm{~N}\) and \(250 \mathrm{~N}\) in the respective cases; Method (b)
Mass of the block, \(m=25 \mathrm{~kg}\)
Mass of the man, \(M=50 \mathrm{~kg}\)
Acceleration due to gravity, \(g=10 \mathrm{~m} / \mathrm{s}^2\)
Force applied on the block, \(\mathrm{F}=25 \times 10=250 \mathrm{~N}\)
Weight of the man, \(W=50 \times 10=500 \mathrm{~N}\)
Case (a): When the man lifts the block directly
In this case, the man applies a force in the upward direction. This increases his apparent weight.
\(\therefore\) Action on the floor by the \(\operatorname{man}=250+500=750 \mathrm{~N}\)
Case (b): When the man lifts the block using a pulley
In this case, the man applies a force in the downward direction. This decreases his apparent weight.
\(\therefore\) Action on the floor by the man \(=500-250=250 \mathrm{~N}\)
If the floor can yield to a normal force of \(700 \mathrm{~N}\), then the man should adopt the second method to easily lift the block by applying lesser force.

Q45: A monkey of mass \(40 \mathrm{~kg}\) climbs on a rope (Fig. 5.20) which can stand a maximum tension of \(600 \mathrm{~N}\). In which of the following cases will the rope break: the monkey
(a) climbs up with an acceleration of \(6 \mathrm{~m} \mathrm{~s}^{-2}\)
(b) climbs down with an acceleration of \(4 \mathrm{~m} \mathrm{~s}^{-2}\)
(c) climbs up with a uniform speed of \(5 \mathrm{~m} \mathrm{~s}^{-1}\)
(d) falls down the rope nearly freely under gravity? (Ignore the mass of the rope).

Answer:
Case (a).
Mass of the monkey, \(m=40 \mathrm{~kg}\)
Acceleration due to gravity, \(g=10 \mathrm{~m} / \mathrm{s}\)
The maximum tension that the rope can bear, \(T_{\max }=600 \mathrm{~N}\)
Acceleration of the monkey, \(a=6 \mathrm{~m} / \mathrm{s}^2\) upward
Using Newton’s second law of motion, we can write the equation of motion as:
\(
\begin{aligned}
& T-m g=m a \\
& \therefore T=m(\mathrm{~g}+a) \\
& =40(10+6) \\
& =640 \mathrm{~N}
\end{aligned}
\)
Since \(T>T_{\max }\) the rope will break in this case.
Case (b).
Acceleration of the monkey, \(a=4 \mathrm{~m} / \mathrm{s}^2\) downward
Using Newton’s second law of motion, we can write the equation of motion as:
\(
\begin{aligned}
& m g-T=m a \\
& \therefore T=m(g-a) \\
& =40(10-4) \\
& =240 \mathrm{~N}
\end{aligned}
\)
Since \(T<T_{\max }\) the rope will not break in this case.
Case (c).
The monkey is climbing with a uniform speed of \(5 \mathrm{~m} / \mathrm{s}\). Therefore, its acceleration is zero, i.e., \(a=0\).
Using Newton’s second law of motion, we can write the equation of motion as:
\(
\begin{aligned}
& T-m g=m a \\
& T-m g=0 \\
& \therefore T=m g \\
& =40 \times 10 \\
& =400 \mathrm{~N}
\end{aligned}
\)
Since \(T<T_{\text {max }}\) the rope will not break in this case.
Case (d).
When the monkey falls freely under gravity, its will acceleration becomes equal to the acceleration due to gravity, i.e., \(a=\mathrm{g}\)
Using Newton’s second law of motion, we can write the equation of motion as:
\(
\begin{aligned}
& m g-T=m g \\
& \therefore T=m(\mathrm{~g}-\mathrm{g})=0
\end{aligned}
\)
Since \(T<T_{\max }\) the rope will not break in this case.

Q46: A thin circular loop of radius \(R\) rotates about its vertical diameter with an angular frequency \(\omega\). Show that a small bead on the wire loop remains at its lowermost point for \(\omega \leq \sqrt{g / R}\). What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for \(\omega=\sqrt{2 g / R}\)? Neglect friction.

Answer:

Let the radius vector joining the bead to the centre of the wire make an angle \(\theta\) with the verticle downward direction. if \(\mathrm{N}\) is normal reaction, then from fig.
\(
\begin{aligned}
& m g=N \cos \theta \dots(i)\\
& m r \omega^2=N \sin \theta \dots(ii)
\end{aligned}
\)
or \(m(R \sin \theta) \omega^2=N \sin \theta\)
orm \(R \omega^2=N\)
or \(\cos \theta=\frac{g}{R \omega^2} \dots(iii)\)
\(
\text { As } \mid \cos \theta \mid<=1 \text {, therefore bead will remain at its lowermost point for }
\)
\(\frac{g}{R \omega^2} \leq 1\) or \(\omega \leq \sqrt{\frac{g}{R}}\)
When \(\omega=\sqrt{\frac{2 g}{R}}\) from equation iii
\(
\cos \theta=\frac{g}{R}\left(\frac{R}{2 \mathrm{~g}}\right)=\frac{1}{2}
\)
\(
\theta=60^{\circ}
\)

Q47: Two bodies \(A\) and \(B\) of masses \(5 \mathrm{~kg}\) and \(10 \mathrm{~kg}\) in contact with each other rest on a table against a rigid wall (Fig. 5.21). The coefficient of friction between the bodies and the table is 0.15. A force of \(200 \mathrm{~N}\) is applied horizontally to \(A\). What are (a) the reaction of the partition (b) the action-reaction forces between \(A\) and \(B\)? What happens when the wall is removed? Does the answer to (b) change, when the bodies are in motion? Ignore the difference between \(\mu_s\) and \(\mu_{k}\).

Answer: We assume perfect contact between bodies \(\mathrm{A}\) and \(\mathrm{B}\) and the rigid partition. In that case, the self-adjusting normal force on B by the partition (reaction) equals \(200 \mathrm{~N}\). There is no impending motion and no friction. The action-reaction forces between A and \(\mathrm{B}\) are also \(200 \mathrm{~N}\). When the partition is removed, kinetic friction comes into play.
(a) When the wall exists and blocks A and B are pushing the wall, there can’t be any motion i.e., blocks are at rest. Hence,
(i) reaction of the partition \(=-(\) force applied on \(\mathrm{A})=200 \mathrm{~N}\) towards left.
(ii) action-reaction forces between A and B are \(200 \mathrm{~N}\) each. A presses B towards right with an action force \(200 \mathrm{~N}\) and B exerts a reaction force on A towards left having magnitude \(200 \mathrm{~N}\).
(b) When the wall is removed, motion can take place such that net pushing force provides acceleration to the block system. Hence, taking kinetic friction into account, we have
\(
\begin{aligned}
& 200-\mu\left(m_1+m_2\right) g=\left(m_1+m_2\right) a \\
& \Rightarrow a=\frac{200-\mu\left(m_1+m_2\right) g}{\left(m_1+m_2\right)}=\frac{200-0.15 \times(5+10) \times 10}{(5+10)} \\
& =\frac{200-22.5}{15}=\left(\frac{177.5}{15}\right)=11.8 \mathrm{~ms}^{-2}
\end{aligned}
\)
\(\therefore\) if force exerted by \(\mathrm{A}\) on \(\mathrm{B}\) be \(F_{\mathrm{BA}}\), then considering equilibrium (or free body diagram) of only block \(\mathrm{A}\), we have
\(
\begin{aligned}
& 200-f_{k_1}=m_1 a+F_{\mathrm{BA}} \text { or } 200-\mu m_1 g=m_1 a+F_{\mathrm{BA}} \\
& \Rightarrow F_{\mathrm{BA}}=200-\mu m_1 g-m_1 a=200-(0.15 \times 5 \times 10)-(5 \times 11.8) \\
& =200-7.5-59 \\
& =200-66.5=133.5 \mathrm{~N} \approx 1.3 \times 10^2 \mathrm{~N} \text { toward right }
\end{aligned}
\)
\(
\therefore \text { Force exerted on A by B }  F_{A B}=-F_{B A}=1.3 \times 10^2 \mathrm{~N} \text { Toward left }
\)

Q48: A block of mass \(15 \mathrm{~kg}\) is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with \(0.5 \mathrm{~m} \mathrm{~s}^{-2}\) for \(20 \mathrm{~s}\) and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on the ground, (b) an observer moving with the trolley.

Answer:(a) Mass of the block, \(m=15 \mathrm{~kg}\)
Coefficient of static friction, \(\mu=0.18\)
Acceleration of the trolley, \(a=0.5 \mathrm{~m} / \mathrm{s}^2\)
As per Newton’s second law of motion, the force \((F)\) on the block caused by the motion of the trolley is given by the relation:
\(
F=m a=15 \times 0.5=7.5 \mathrm{~N}
\)
This force is acted in the direction of motion of the trolley.
Force of static friction between the block and the trolley:
\(
\begin{aligned}
& f=\mu m g \\
& =0.18 \times 15 \times 10=27 \mathrm{~N}
\end{aligned}
\)
The force of static friction between the block and the trolley is greater than the applied external force. Hence, for an observer on the ground, the block will appear to be at rest.

When the trolley moves with uniform velocity there will be no applied external force. Only the force of friction will act on the block in this situation.
(b) An observer, moving with the trolley, has some acceleration. This is the case of a non-inertial frame of reference. The frictional force, acting on the trolley backward, is opposed by a pseudo force of the same magnitude. keeping the box at rest relative to the observer. When the trolley moves with uniform velocity there is no pseudo-force for the moving (inertial) observer and no friction. However, this force acts in the opposite direction. Thus, the trolley will appear to be at rest for the observer moving with the trolley.

Q49: The rear side of a truck is open and a box of \(40 \mathrm{~kg}\) mass is placed \(5 \mathrm{~m}\) away from the open end as shown in Fig. 5.22. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with \(2 \mathrm{~m} \mathrm{~s}^{-2}\). At what distance from the starting point does the box fall off the truck? (Ignore the size of the box).

Answer: Force experienced by box, \(\mathrm{F}=\mathrm{ma}=40 \times 2=80 \mathrm{~N}\) Frictional force \(F_f=\mu m g=0.15 \times 40 \times 10=60 \mathrm{~N}\).
Acceleration of the box due to friction \(=\mu g=0.15 \times 10=1.5 \mathrm{~m} \mathrm{~s}^{-2}\). But the acceleration of the truck is greater. The acceleration of the box relative to the truck is \(0.5 \mathrm{~m} \mathrm{~s}^{-2}\) towards the rear end.
The time taken for the box to fall off the truck \(=\sqrt{\frac{2 \times 5}{0.5}}=\sqrt{20} \mathrm{~s}\).
\(
\text { During this time, the truck covers a distance } S=u t+\frac{1}{2} a t^2 = 1 / 2 \times 2 \times 20=20 \mathrm{~m} \text {. }
\)

Q50: A disc revolves with a speed of \(33 \frac{1}{3} \mathrm{rev} / \mathrm{min}\), and has a radius of \(15 \mathrm{~cm}\). Two coins are placed at \(4 \mathrm{~cm}\) and \(14 \mathrm{~cm}\) away from the centre of the record. If the coefficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record?

Answer: Given
angular frequency \(\omega=33 \frac{1}{3} \mathrm{rev} / \mathrm{min}=\frac{100}{3} \times \frac{1}{60} \mathrm{rev} / \mathrm{s}\)
\(
\omega=2 \times \frac{22}{7} \times \frac{100}{3} \times \frac{1}{60}=3.5 \mathrm{rad} / \mathrm{s}
\)
For a coin to revolve with the record, the force of friction must be enough to provide the necessary centripetal force,
i.e.,
\(
\begin{aligned}
& \frac{m v^2}{R} \leq \mu m g \Rightarrow m \omega^2 R \leq \mu m g \\
& \Rightarrow R \leq \frac{\mu g}{\omega^2} \Rightarrow \frac{0.15 \times 10}{(3.5)^2}=0.12 \mathrm{~m} \text { or } 12 \mathrm{~cm}
\end{aligned}
\)
The coin should be maximum \(12 \mathrm{~cm}\) from the centre to revolve with the disc.
Hence, the coin placed at \(4 \mathrm{~cm}\) from centre will revolve with the disc.

Q51: You may have seen in a circus a motorcyclist driving in vertical loops inside a ‘death well’ (a hollow spherical chamber with holes, so the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed required at the uppermost position to perform a vertical loop if the radius of the chamber is \(25 \mathrm{~m}\)?

Answer:

In a death well, a motorcyclist does not fall at the top point of a vertical loop because both the force of normal reaction and the weight of the motorcyclist act downward and are balanced by the centripetal force. This situation is shown in the following figure.
The net force acting on the motorcyclist is the sum of the normal force \(\left(F_{\mathrm{N}}\right)\) and the force due to gravity \(\left(F_{\mathrm{g}}=m \mathrm{~g}\right)\).

The equation of motion for the centripetal acceleration \(a_C\), can be written as:
\(
\begin{aligned}
& F_{\text {net }}=m a_C \\
& F_{\mathrm{N}}+F_{\mathrm{g}}=m a_c \\
& F_{\mathrm{N}}+m \mathrm{~g}=\frac{m v^2}{r}
\end{aligned}
\)
where \(N\) is the normal force (downwards) on the motorcyclist by the ceiling of the chamber.
The minimum possible speed at the uppermost point corresponds to \(N=0\).
\(
\begin{aligned}
& m \mathrm{~g}=\frac{m v_{\min }{ }^2}{r} \\
& \begin{aligned}
\therefore v_{\min } & =\sqrt{r \mathrm{~g}} \\
& =\sqrt{25 \times 10}=15.8 \mathrm{~m} / \mathrm{s}
\end{aligned}
\end{aligned}
\)

Q52: Suppose the ceiling in the previous problem is that of an elevator which is going up with an acceleration of \(2 \cdot 0 \mathrm{~m} / \mathrm{s}^2\). Find the elongations.

Answer:

When the ceiling of the elevator is going up with an acceleration a, then a pseudo-force acts on the block in the downward direction.
\(
\mathrm{a}=2 \mathrm{~m} / \mathrm{s}^2
\)
From the free-body diagram of the block,
\(
\begin{aligned}
& k x=m g+m a \\
& \Rightarrow \mathrm{kx}=2 \mathrm{~g}+2 \mathrm{a} \\
& =2 \times 9.8+2 \times 2 \\
& =19.6+4 \\
& \Rightarrow x=\frac{23.6}{100}=0.236 \approx 0.24 \mathrm{~m} \\
&
\end{aligned}
\)
When \(1 \mathrm{~kg}\) body is added,
\(
\text { total mass }=(2+1) \mathrm{kg}=3 \mathrm{~kg}
\)
Let elongation be \(x^{\prime}\).
\(
\begin{aligned}
& \therefore \mathrm{kx}^{\prime}=3 \mathrm{~g}+3 \mathrm{a}=3 \times 9.8+6 \\
& \Rightarrow x^{\prime}=\frac{35.4}{100} \\
& =0.354 \approx 0.36 \mathrm{~m}
\end{aligned}
\)
So, further elongation \(=x^{\prime}-x\)
\(
=0.36-0.24=0.12 \mathrm{~m}
\)

Q53: Find the acceleration of the blocks \(A\) and \(B\) in the three situations shown in Figure (5-E17).

Answer:

(a) \(5 a+T-5 g=0 \Rightarrow T=5 g-5 a\) …(i) (From FBD-1)
Again \((1 / 2)-4 g-8 a=0 \Rightarrow T=8 g-16 a\) ..(ii) (from FBD-2)
From equation (i) and (ii), we get
\(
5 \mathrm{~g}-5 \mathrm{a}=8 \mathrm{~g}+16 \mathrm{a} \Rightarrow 21 \mathrm{a}=-3 \mathrm{~g} \Rightarrow \mathrm{a}=-1 / 7 \mathrm{~g}
\)
So, acceleration of \(5 \mathrm{~kg}\) mass is \(\mathrm{g} / 7\) upward and that of \(4 \mathrm{~kg}\) mass is \(2 \mathrm{a}=2 \mathrm{~g} / 7\) (downward).

(b)

\(
\begin{aligned}
& 4 a-t / 2=0 \Rightarrow 8 a-T=0 \Rightarrow T=8 a \ldots \text { (ii) [From FBD -4] } \\
& \text { Again, } T+5 a-5 g=0 \Rightarrow 8 a+5 a-5 g=0 \\
& \Rightarrow 13 a-5 g=0 \Rightarrow a=5 g / 13 \text { downward. (from FBD -3) }
\end{aligned}
\)
\(
\text { Acceleration of mass }(A) \mathrm{kg} \text { is } 2 a=10 / 13(\mathrm{~g}) \& 5 \mathrm{~kg}(B) \text { is } 5 \mathrm{~g} / 13 \text {. }
\)

(c)

\(
T+1 a-1 g=0 \Rightarrow T=1 g-1 a \ldots \text { (i) [From FBD -5] }
\)
Again, \(\frac{T}{2}-2 g-4 a=0 \Rightarrow T-4 g-8 a=0 \ldots \text { (ii) [From FBD -6] }\)
\(
\Rightarrow 1 g-1 a-4 g-8 a=0[\text { From (i)] }
\)
\(\Rightarrow \mathrm{a}=-(\mathrm{g} / 3)\) downward.
Acceleration of mass \(1 \mathrm{~kg}(\mathrm{~b})\) is \(\mathrm{g} / 3\) (up)
Acceleration of mass \(2 \mathrm{~kg}(A)\) is \(2 \mathrm{~g} / 3\) (downward).

Q54: A monkey is climbing on a rope that goes over a smooth light pulley and supports a block of equal mass at the other end (figure 5-E19). Show that whatever force the monkey exerts on the rope, the monkey and the block move in the same direction with equal acceleration. If initially, both were at rest, their separation will not change as time passes.

Answer: 

Suppose the monkey accelerates upward with acceleration ‘a’ & the block, accelerates downward with acceleration \(a_1\). Let Force exerted by a monkey is equal to ‘ \(T\) ‘
From the free-body diagram of the monkey
\(
\therefore \mathrm{T}-\mathrm{mg}-\mathrm{ma}=0 \dots(i)
\)
\(
\Rightarrow \mathrm{T}=\mathrm{mg}+\mathrm{ma} \text {. }
\)
Again, from the FBD of the block,
\(
\mathrm{T}=\mathrm{ma}_1-\mathrm{mg}=0 .
\)
Again, from the FBD of the block,
\(
\begin{aligned}
& T=m a_1-m g=0 . \\
& \Rightarrow m g+m a+m a_1-m g=0[\text { From (i) }] \Rightarrow m a=-m a_1 \Rightarrow a=a_1 .
\end{aligned}
\)
Acceleration ‘-a’ downward i.e. ‘a’ upward.
The block & the monkey move in the same direction with equal acceleration.
If initially, they are rest (no force is exerted by the monkey) no motion of the monkey of block occurs as they have the same weight (same mass). Their separation will not change as time passes.

Q55: A body of mass \(m\) is suspended by two strings making angles \(\alpha\) and \(\beta\) with the horizontal. Find the tensions in the strings.

Answer: Take the body of mass \(m\) as the system. The forces acting on the system are
(i) \(m g\) downwards (by the earth),
(ii) \(T_1\) along the first string (by the first string) and
(iii) \(T_2\) along the second string (by the second string).

These forces are shown in Figure (5-W1). As the body is in equilibrium, these forces must add to zero. Taking horizontal components,
\(
T_1 \cos \alpha-T_2 \cos \beta+m g \cos \frac{\pi}{2}=0
\)
or,
\(
T_1 \cos \alpha=T_2 \cos \beta \dots(i)
\)
Taking vertical components,
\(
T_1 \sin \alpha+T_2 \sin \beta-m g=0 \dots(ii)
\)
Eliminating \(T_2\) from (i) and (ii),
\(
\begin{aligned}
& \qquad T_1 \sin \alpha+T_1 \frac{\cos \alpha}{\cos \beta} \sin \beta=m g \\
& \text { or, } T_1=\frac{m g}{\sin \alpha+\frac{\cos \alpha}{\cos \beta} \sin \beta}=\frac{m g \cos \beta}{\sin (\alpha+\beta)} . \\
& \text { From (i), }
\end{aligned}
\)

Q56: Two bodies of masses \(m_1\) and \(m_2\) are connected by a light string going over a smooth light pulley at the end of an incline. The mass \(m_1\) lies on the incline and \(m_2\) hangs vertically. The system is at rest. Find the angle of the incline and the force exerted by the incline on the body of mass \(m_1\) (figure 5-W2).

Answer: Figure (5-W3) shows the situation with the forces on \(m_1\) and \(m_2\) shown. Take the body of mass \(m_2\) as the system. The forces acting on it are

(i) \(m_2 g\) vertically downward (by the earth),
(ii) \(T\) vertically upward (by the string).
As the system is at rest, these forces should add to zero.
This gives
\(T=m_2 g \dots(i)\).
Next, consider the body of mass \(m_1\) as the system. The forces acting on this system are
(i) \(m_1 g\) vertically downward (by the earth),
(ii) \(T\) along the string up the incline (by the string),
(iii) \(\mathcal{N}\) normal to the incline (by the incline).
As the string and the pulley are all light and smooth, the tension in the string is uniform everywhere. Hence, same \(T\) is used for the equations of \(m_1\) and \(m_2\). As the system is in equilibrium, these forces should add to zero. Taking components parallel to the incline,
\(
T=m_1 g \cos \left(\frac{\pi}{2}-\theta\right)=m_1 g \sin \theta \dots(ii)
\)
Taking components along the normal to the incline,
\(
\mathcal{N}=m_1 g \cos \theta \text {. } \dots(iii)
\)
Eliminating \(T\) from (i) and (ii),
or,
\(
\begin{aligned}
m_2 g & =m_1 g \sin \theta \\
\sin \theta & =m_2 / m_1 \\
\theta & =\sin ^{-1}\left(m_2 / m_1\right) .
\end{aligned}
\)
giving
From (iii) \(\quad \mathcal{N}=m_1 g \sqrt{1-\left(m_2 / m_1\right)^2}\).

Q57: A bullet moving at \(250 \mathrm{~m} / \mathrm{s}\) penetrates \(5 \mathrm{~cm}\) into a tree limb before coming to rest. Assuming that the force exerted by the tree limb is uniform, find its magnitude. Mass of the bullet is \(10 \mathrm{~g}\).

Answer: The tree limb exerts a force on the bullet in the direction opposite to its velocity. This force causes deceleration and hence the velocity decreases from \(250 \mathrm{~m} / \mathrm{s}\) to zero in \(5 \mathrm{~cm}\). We have to find the force exerted by the tree limb on the bullet. If \(a\) be the deceleration of the bullet, we have,
\(
u=250 \mathrm{~m} / \mathrm{s}, v=0, x=5 \mathrm{~cm}=0^{\circ} 05 \mathrm{~m}
\)
giving, \(\quad a=\frac{(250 \mathrm{~m} / \mathrm{s})^2-0^2}{2 \times 0.05 \mathrm{~m}}=625000 \mathrm{~m} / \mathrm{s}^2\).
The force on the bullet is \(F=m a=6250 \mathrm{~N}\).

Q58: The force on a particle of mass \(10 \mathrm{~g}\) is \((\vec{i} 10+\overrightarrow{j 5}) \mathrm{N}\). If it starts from rest what would be its position at time \(t=5 \mathrm{~s}?\)

Answer: We have \(F_x=10 \mathrm{~N}\) giving
\(
a_x=\frac{F_x}{m}=\frac{10 \mathrm{~N}}{0.01 \mathrm{~kg}}=1000 \mathrm{~m} / \mathrm{s}^2 \text {. }
\)
As this is a case of constant acceleration in \(x\)-direction,
\(
\begin{aligned}
x & =u_x t+\frac{1}{2} a_x t^2=\frac{1}{2} \times 1000 \mathrm{~m} / \mathrm{s}^2 \times(5 \mathrm{~s})^2 \\
& =12500 \mathrm{~m}
\end{aligned}
\)
Similarly, \(a_y=\frac{F_y}{m}=\frac{5 \mathrm{~N}}{0.01 \mathrm{~kg}}=500 \mathrm{~m} / \mathrm{s}^2\) and \(y=6250 \mathrm{~m}\).
Thus, the position of the particle at \(t=5 \mathrm{~s}\) is, \(\vec{r}=(\vec{i} 12500+\vec{j} 6250) \mathrm{m}\)

Q59: With what acceleration ‘ \(a\) ‘ should the box of the figure (5-W4) descend so that the block of mass \(M\) exerts a force \(\mathrm{Mg} / 4\) on the floor of the box?

Answer: The block is at rest with respect to the box which is accelerated with respect to the ground. Hence, the acceleration of the block with respect to the ground is ‘ \(a\) ‘ downward. The forces on the block are
(i) \(M g\) downward (by the earth) and
(ii) \(\mathcal{N}\) upward (by the floor).
The equation of motion of the block is, therefore
\(M g-\mathcal{N}=M a\)
If \(\mathcal{N}=M g / 4\), the above equation gives \(a=3 g / 4\). The block and hence the box should descend with an acceleration \(3 g / 4\).

Q60: A block ‘A’ of mass \(m\) is tied to a fixed point \(C\) on a horizontal table through a string passing round a massless smooth pulley B (figure 5-W5). A force \(F\) is applied by the experimenter to the pulley. Show that if the pulley is displaced by a distance \(x\), the block will be displaced by \(2 x\). Find the acceleration of the block and the pulley.

Answer: Suppose the pulley is displaced to \(B^{\prime}\) and the block to \(A^{\prime}\) (figure 5-W6). The length of the string is \(C B+B A\) and is also equal to \(C B+B B^{\prime}+B^{\prime} B+B A^{\prime}\). Hence, \(C B+B A^{\prime}+A^{\prime} A=C B+B B^{\prime}+B^{\prime} B+B A^{\prime}\) or, \(A^{\prime} A=2 B B^{\prime}\).

The displacement of \(A\) is, therefore, twice the displacement of \(B\) in any given time interval. Differentiating twice, we find that the acceleration of \(A\) is twice the acceleration of \(B\).

To find the acceleration of the block we will need the tension in the string. That can be obtained by considering the pulley as the system.
The forces acting on the pulley are
(i) \(F\) towards right by the experimenter,
(ii) \(T\) towards left by the portion \(B C\) of the string and
(iii) \(T\) towards left by the portion \(B A\) of the string.
The vertical forces, if any, add to zero as there is no vertical motion.
As the mass of the pulley is zero, the equation of motion is
\(F-2 T=0\) giving \(T=F / 2\).
Now consider the block as the system. The only horizontal force acting on the block is the tension \(T\) towards the right. The acceleration of the block is, therefore, \(a=T / m=\frac{F}{2 m}\). The acceleration of the pulley is \(a / 2=\frac{F}{4 m}\).

Q61: A smooth ring \(A\) of mass \(m\) can slide on a fixed horizontal rod. A string tied to the ring passes over a fixed pulley \(B\) and carries a block \(C\) of mass \(M(=2 \mathrm{~m})\) as shown in figure (5-W7). At an instant, the string between the ring and the pulley makes an angle \(\theta\) with the rod. (a) Show that, if the ring slides with a speed \(v\), the block descends with speed \(v \cos \theta\). (b) With what acceleration will the ring start moving if the system is released from rest with \(\theta=30^{\circ}\)?

Answer: (a) Suppose in a small time interval \(\Delta t\) the ring is displaced from \(A\) to \(A^{\prime}\) (figure 5-W8) and the block from \(C\) to \(C^{\prime}\). Drop a perpendicular \(A^{\prime} P\) from \(A^{\prime}\) to \(A B\). For small displacement \(A^{\prime} B \approx P B\). Since the length of the string is constant, we have

\(
\begin{aligned}
A B+B C & =A^{\prime} B+B C^{\prime} \\
\text { or, } \quad A P+P B+B C & =A^{\prime} B+B C^{\prime}
\end{aligned}
\)
\(
\begin{array}{lc}
\text { or, } & A P=B C^{\prime}-B C=C C^{\prime} \quad\left(\text { as } A^{\prime} B \approx P B\right) \\
\text { or, } & A A^{\prime} \cos \theta=C C^{\prime}
\end{array}
\)
\(
\text { or, } \quad \frac{A A^{\prime} \cos \theta}{\Delta t}=\frac{C C^{\prime}}{\Delta t}
\)
or, (velocity of the ring) \(\cos \theta=\) (velocity of the block).
(b) If the initial acceleration of the ring is \(a\), that of the block will be \(a \cos \theta\). Let \(T\) be the tension in the string at this instant. Consider the block as the system. The forces acting on the block are
(i) \(M g\) downward due to the earth, and
(ii) \(T\) upward due to the string.
The equation of motion of the block is
\(
M g-T=M a \cos \theta \dots(i)
\)
Now consider the ring as the system. The forces on the ring are
(i) \(M g\) downward due to gravity,
(ii) \(\mathcal{N}\) upward due to the rod,
(iii) \(T\) along the string due to the string.
Taking components along the rod, the equation of motion of the ring is
\(
T \cos \theta=m a \dots(ii)
\)
From (i) and (ii)
\(
\begin{aligned}
M g-\frac{m a}{\cos \theta} & =M a \cos \theta \\
a & =\frac{M g \cos \theta}{m+M \cos ^2 \theta} .
\end{aligned}
\)
Putting \(\theta=30^{\circ}, M=2 \mathrm{~m}\) and \(g=9.8 \mathrm{~m} / \mathrm{s}^2\); therefore \(a=6.78 \mathrm{~m} / \mathrm{s}^2\).

Q62: A light rope fixed at one end of a wooden clamp on the ground passes over a tree branch and hangs on the other side (figure 5-W9). It makes an angle of \(30^{\circ}\) with the ground. A man weighing \((60 \mathrm{~kg}\) ) wants to climb up the rope. The wooden clamp can come out of the ground if an upward force greater than \(360 \mathrm{~N}\) is applied to it. Find the maximum acceleration in the upward direction with which the man can climb safely. Neglect friction at the tree branch. Take \(g=10 \mathrm{~m} / \mathrm{s}^2\).

Answer: Let \(T\) be the tension in the rope. The upward force on the clamp is \(T \sin 30^{\circ}=T / 2\). The maximum tension that will not detach the clamp from the ground is, therefore, given by
\(
\begin{aligned}
& \frac{T}{2}=360 \mathrm{~N} \\
& T=720 \mathrm{~N} .
\end{aligned}
\)
If the acceleration of the man in the upward direction is \(a\), the equation of motion of the man is
\(
T-600 \mathrm{~N}=(60 \mathrm{~kg}) a
\)
The maximum acceleration of the man for safe climbing is, therefore
\(
a=\frac{720 \mathrm{~N}-600 \mathrm{~N}}{60 \mathrm{~kg}}=2 \mathrm{~m} / \mathrm{s}^2 .
\)

Q63: Three blocks of masses \(m_1, m_2\) and \(m_3\) are connected as shown in the figure (5-W10). All the surfaces are frictionless and the string and the pulleys are light. Find the acceleration of \(m_1\).

Answer:Suppose the acceleration of \(m_1\) is \(a_0\) towards the right. That will also be the downward acceleration of the pulley \(B\) because the string connecting \(m_1\) and \(B\) is constant in length. Also, the string connecting \(m_2\) and \(m_3\) has a constant length. This implies that the decrease in the separation between \(m_2\) and \(B\) equals the increase in the separation between \(m_3\) and \(B\). So, the upward acceleration of \(m_2\) with respect to \(B\) equals the downward acceleration of \(m_3\) with respect to \(B\). Let this acceleration be \(a\). The acceleration of \(m_2\) with respect to the ground \(=a_0-a\) (downward) and the acceleration of \(m_3\) with respect to the ground \(=a_0+a\) (downward).

These accelerations will be used in Newton’s laws. Let the tension be \(T\) in the upper string and \(T^{\prime}\) in the lower string. Consider the motion of the pulley \(B\).
The forces on this light pulley are
(a) \(T\) upwards by the upper string and
(b) \(2 T^{\prime}\) downwards by the lower string.
As the mass of the pulley is negligible,
\(
2 T^{\prime}-T=0
\)
giving
\(
T^{\prime}=T / 2 \dots(i)
\)
Motion of \(m_1\) :
The acceleration is \(a_0\) in the horizontal direction. The forces on \(m_1\) are
(a) \(T\) by the string (horizontal).
(b) \(m_1 g\) by the earth (vertically downwards) and
(c) \(\mathcal{N}\) by the table (vertically upwards).
In the horizontal direction, the equation is
\(
T=m_1 a_0 \dots(ii)
\)
Motion of \(m_2\) : acceleration is \(a_0-a\) in the downward direction. The forces on \(m_2\) are
(a) \(m_2 g\) downward by the earth and
(b) \(T^{\prime}=T / 2\) upward by the string.
Thus,
\(
m_2 g-T / 2=m_2\left(a_0-a\right) \dots(iii)
\)
Motion of \(m_3\) : The acceleration is \(\left(a_0+a\right)\) downward. The forces on \(m_3\) are
(a) \(m_3 g\) downward by the earth and
(b) \(T^{\prime}=T / 2\) upward by the string. Thus,
\(
m_3 g-T / 2=m_3\left(a_0+a\right) \text {. } \dots(iv)
\)
We want to calculate \(a_0\), so we shall eliminate \(T\) and \(a\) from (ii), (iii) and (iv).
Putting \(T\) from (ii) in (iii) and (iv),
\(
\begin{aligned}
a_0-a & =\frac{m_2 g-m_1 a_0 / 2}{m_2}=g-\frac{m_1 a_0}{2 m_2} \\
\text { and } \quad a_0+a & =\frac{m_3 g-m_1 a_0 / 2}{m_3}=g-\frac{m_1 a_0}{2 m_3} .
\end{aligned}
\)
\(
\text { Adding, } \quad 2 a_0=2 g-\frac{m_1 a_0}{2}\left(\frac{1}{m_2}+\frac{1}{m_3}\right)
\)
\(
\text { or, } \quad a_0=g-\frac{m_1 a_0}{4}\left(\frac{1}{m_2}+\frac{1}{m_3}\right)
\)
\(
\text { or, } \quad a_0\left[1+\frac{m_1}{4}\left(\frac{1}{m_2}+\frac{1}{m_3}\right)\right]=g
\)
\(
a_0=\frac{g}{1+\frac{m_1}{4}\left(\frac{1}{m_2}+\frac{1}{m_3}\right)} .
\)

Q64: A particle slides down a smooth inclined plane of elevation \(\theta\), fixed in an elevator going up with an acceleration \(a_0\) (figure 5-W12). The base of the incline has a length \(L\). Find the time taken by the particle to reach the bottom.

Answer: Let us work in the elevator frame. A pseudo force \(m a_0\) in the downward direction is to be applied on the particle of mass \(m\) together with the real forces. Thus, the forces on \(m\) are (figure 5-W13)
(i) \(\mathcal{N}\) normal force,
(ii) \(m g\) downward (by the earth),
(iii) \(m a_0\) downward (pseudo).

Let \(a\) be the acceleration of the particle with respect to the incline. Taking components of the forces parallel to the incline and applying Newton’s law,
\(
\begin{aligned}
m g \sin \theta+m a_0 \sin \theta & =m a \\
a & =\left(g+a_0\right) \sin \theta .
\end{aligned}
\)
or,
th respect to the elevator. In This is the acceleration with this frame, the distance travelled by the particle is \(L / \cos \theta\). Hence,
\(
\frac{L}{\cos \theta}=\frac{1}{2}\left(g+a_0\right) \sin \theta \cdot t^2
\)
or,
\(
t=\left[\frac{2 L}{\left(g+a_0\right) \sin \theta \cos \theta}\right]^{1 / 2} .
\)

Q65: All the surfaces shown in Figure (5-W14) are assumed to be frictionless. The block of mass \(m\) slides on the prism which in turn slides backward on the horizontal surface. Find the acceleration of the smaller block with respect to the prism.

Answer: Let the acceleration of the prism be \(a_0\) in the backward direction. Consider the motion of the smaller block from the frame of the prism.
The forces on the block are (figure 5-W15a)
(i) \(\mathcal{N}\) normal force,
(ii) \(m g\) downward (gravity),
(iii) \(m a_0\) forward (pseudo).

The block slides down the plane. Components of the forces parallel to the incline give
\(
m a_0 \cos \theta+m g \sin \theta=m a
\)
or,
\(
a=a_0 \cos \theta+g \sin \theta \dots(i)
\)
Components of the force perpendicular to the incline give
\(
\mathcal{N}+m a_0 \sin \theta=m g \cos \theta \dots(ii)
\)
Now consider the motion of the prism from the lab frame. No pseudo force is needed as the frame used is inertial. The forces are (figure 5-W15b)
(i) \(M g\) downward,
(ii) \(\mathcal{N}\) normal to the incline (by the block),
(iii) \(\mathcal{N}^{\prime}\) upward (by the horizontal surface).
Horizontal components give,
\(
\mathcal{N} \sin \theta=M a_0 \quad \text { or, } \mathcal{N}=M a_0 / \sin \theta \dots(iii)
\)
Putting in (ii)
\(
\frac{M a_0}{\sin \theta}+m a_0 \sin \theta=m g \cos \theta
\)
or,
\(
a_0=\frac{m g \sin \theta \cos \theta}{M+m \sin ^2 \theta} .
\)
From (i),
\(
\begin{aligned}
a & =\frac{m g \sin \theta \cos ^2 \theta}{M+m \sin ^2 \theta}+g \sin \theta \\
& =\frac{(M+m) g \sin \theta}{M+m \sin ^2 \theta} .
\end{aligned}
\)

Short Question Answers

Q1: The apparent weight of an object increases in an elevator while accelerating upward. A moongphaliwala sells his moongphali using a beam balance in an elevator. Will he gain more if the elevator is accelerating up?

Answer: No, the accelerating elevator will affect the weight of both sides of the beam balance. So, the net effect of the accelerating elevator cancels out, and we get the actual mass.

Q2: A boy puts a heavy box of mass \(M\) on his head and jumps down from the top of a multistoried building to the ground. How much is the force exerted by the box on his head during his free fall? Does the force greatly increase during the period he balances himself after striking the ground?

Answer: During free fall:
Acceleration of the boy \(=\) Acceleration of mass \(M=g\)
Acceleration of mass \(M\) w.r.t. boy, \(a=0\)
So, the force exerted by the box on the boy’s head \(=M \times a=0\)
Yes, the force greatly increases during the period he balances himself after striking the ground because of the weight of the box.

Q3: A person drops a coin. Describe the path of the coin as seen by the person if he is in (a) a car moving at constant velocity and (b) in a freely falling elevator.

Answer: (a) a car moving at a constant velocity- In the car, the path of the coin will be vertically downward because the only force acting on the coin is gravity in the downward direction.
(b) in a free-falling elevator.- the coin as well as the person will be in a condition of weightlessness. So, the coin will remain stationary with respect to the person.
Because all fee falls objects weightlessness.

Q4: Is it possible for a particle to describe a curved path if no force acts on it? Does your answer depend on the frame of reference chosen to view the particle?

Answer: If no force acts on the particle it cannot change its direction. So, it is not possible for a particle to describe a curved path if no force acts on it. Yes, the answer depends on the frame of reference chosen to view the particle if the frame of reference describes a curved path.

Q5: You are riding in a car. The driver suddenly applies the brakes and you are pushed forward. Who pushed you forward?

Answer: We are pushed forward because of the inertia of motion, as our body opposes the sudden change.

Q6: It is sometimes heard that an inertial frame of reference is only an ideal concept and no such inertial frame actually exists. Comment.

Answer: We can’t find a body whose acceleration is zero with respect to all other bodies in the universe because everybody in the universe is moving with respect to other bodies. As we live on earth which itself is accelerates due to its revolution around the sun and spinning about its own axis, so whatever observations and measurements, we make, are w.r.t to earth which itself is not an inertial frame. Similarly, all other planets are also in motion around the sun so ideally no inertial frame is possible.

Q7: An object is placed far away from all the objects that can exert force on it. A frame of reference is constructed by taking the origin and axes fixed in this object. Will the frame be necessarily inertial?

Answer: Yes, if the force on the object is zero, its acceleration w.r.t. all the other objects will be zero. So, the frame will necessarily be an inertial frame.

Q8: Figure (5-Q1) shows a light spring balance connected to two blocks of mass \(20 \mathrm{~kg}\) each. The graduations in the balance measure the tension in the spring. (a) What is the reading of the balance? (b) Will the reading change if the balance is heavy, say \(2 \cdot 0 \mathrm{~kg}\)? (c) What will happen if the spring is light but the blocks have unequal masses?

Answer: The reading of the balance \(=\) Tension in the string
And tension in the string \(=20 \mathrm{~g}\)
So, the reading of the balance \(=20 \mathrm{~g}=200 \mathrm{~N}\)
(b) If the balance is heavy, the reading will not change because the weight of the spring balance does not affect the tension in the string.
(c) If the blocks have unequal masses, the spring balance will accelerate towards the heavy block with an acceleration a. Then the reading will be equal to the tension in the string.

Suppose \(m_1>m_2\).
Then tension in the string,
\(
T=\frac{2 m_1 m_2 g}{m_1+m_2}
\)

Q9: The acceleration of a particle is zero as measured from an inertial frame of reference. Can we conclude that no force acts on the particle?

Answer: No. The acceleration of the particle can also be zero if the vector sum of all the forces is zero, i.e. no net force acts on the particle.

Q10: Suppose you are running fast in a field when you suddenly find a snake in front of you. You stop quickly. Which force is responsible for your deceleration?

Answer: The force of friction acting between my feet and the ground is responsible for my deceleration.

Q11: If you jump barefooted on a hard surface, your legs get injured. But they are not injured if you jump on a soft surface like sand or pillow. Explain.

Answer: In both cases, the change in momentum is the same but the time interval during which momentum changes to zero is less in the first case. So, by
\(
\mathrm{F}=\frac{\mathrm{dp}}{\mathrm{dt}} \text {, }
\)
force in the first case will be more. That’s why we are injured when we jump barefoot on a hard surface.

Q12: According to Newton’s third law each team pulls the opposite team with equal force in a tug of war. Why then one team wins and the other loses?

Answer: The forces on the rope must be equal and opposite, according to Newton’s third law. But not all the forces acting on each team are equal. The friction between one team and the ground does not depend on the other team and can be larger on one side than on the other. In addition, the grips on the rope need not be equal and opposite. Thus, the net force acting on each team from all sources need not be equal.

Q13: A spy jumps from an airplane with his parachute. The spy accelerates downward for some time when the parachute opens. The acceleration is suddenly checked and the spy slowly falls on the ground. Explain the action of the parachute in checking the acceleration.

Answer: Air applies a velocity-dependent force on the parachute in an upward direction when the parachute opens. This force opposes the gravitational force acting on the spy. Hence, the net force in the downward direction decreases and the spy decelerates.

Q14: Consider a book lying on a table. The weight of the book and the normal force by the table on the book are equal in magnitude and opposite in direction. Is this an example of Newton’s third law?

Answer: No, this is not an example of Newton’s third law. According to Newton’s law, if a body A exerts a force on body B, then B exerts a force on A equal in magnitude and opposite in direction.
The forces act on different bodies. So, the normal by table on the book is action and the reaction pair is the force on a table by the book. Weight is due to the force of the earth on the book, not due to the table. Hence this is not an action-reaction pair.

Q15: Two blocks of unequal masses are tied by a spring. The blocks are pulled stretching the spring slightly and the system is released on a frictionless horizontal platform. Are the forces due to the spring on the two blocks equal and opposite? If yes, is lit an example of Newton’s third law?

Answer: Yes, the forces due to the spring on the two blocks are equal and opposite.
But it’s not an example of Newton’s third law because there are three objects ( 2 blocks +1 spring). Spring force on one block and force by the same block on the spring is an action-reaction pair.

Q16: When a train starts, the head of a standing passenger seems to be pushed backward. Analyse the situation from the ground frame. Does it really go backward? Coming back to the train frame, how do you explain the backward movement of the head on the basis of Newton’s laws?

Answer: No, w.r.t. the ground frame, the person’s head is not really pushed backward.
As the train moves, the lower portion of the passenger’s body starts moving with the train, but the upper portion tries to be in rest according to Newton’s first law and hence, the passenger seems to be pushed backward.

Q17: A plumb bob is hung from the ceiling of a train compartment. If the train moves with an acceleration ‘ \(a\) ‘ along a straight horizontal track, the string supporting the bob makes an angle \(\tan ^{-1}(a / g)\) with the normal to the ceiling. Suppose the train moves on an inclined straight track with uniform velocity. If the angle of incline is \(\tan ^{-1}(a / g)\), the string again makes the same angle with the normal to the ceiling. Can a person sitting inside the compartment tell by looking at the plumb line whether the train is accelerated on a horizontal straight track or it is going on an incline? If yes, how? If no, suggest a method to do so.

Answer: No, a person sitting inside the compartment can’t tell just by looking at the plumb line whether the train is accelerating on a horizontal straight track or moving on an incline.
case-1:
the tension in the string is ( When the train is accelerating along the horizontal)
\(
m \sqrt{g^2+a^2}
\)
case-2:
the tension in the string is (when it is moving on the inclined plane ) – mg
Therefore, we can differentiate between the two cases by measuring the tension in the string