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The (linear) momentum of a particle is defined as \(\overrightarrow{p} = m\overrightarrow{v}\). The momentum of an N-particle-system is the (vector) sum of the momenta of the N particles i.e.,

\[{\overrightarrow{P}}_{sys} = \sum_{i}^{}\mspace{2mu}{\overrightarrow{p}}_{i} = \sum_{i}^{}\mspace{2mu}{\text{ }m}_{i}{\overrightarrow{\text{ }V}}_{i}.\]

But \(\ \sum_{i}^{}\mspace{2mu} m_{i}{\overrightarrow{v}}_{i} = \frac{d}{dt}\sum_{i}^{}\mspace{2mu} m_{i}{\overrightarrow{r}}_{i} = \frac{d}{dt}M{\overrightarrow{R}}_{CM} = M{\overrightarrow{V}}_{CM}\)
Thus, \(\ {\overrightarrow{P}}_{sys} = M{\overrightarrow{V}}_{CM}\)
As we have seen, if the external forces acting on the system add up to zero, the centre of mass moves with constant velocity, which means \(\overrightarrow{P} =\) constant. Thus the linear momentum of a system remains constant (in magnitude and direction), if the external forces acting on the system add up to zero. This is known as the principle of conservation of linear momentum.
Let as see a simple example of a bomb explosion.
Consider a bomb placed on a horizontal surface which suddenly explodes into two parts of masses \(m_{1}\) and \(m_{2}\). The forces that are responsible for the explosion are internal. As there is no external force on the
system, momentum of system remains conserved. The initial momentum of the system is zero, Thus the final momentum of the system must also be zero.
i.e. After explosion, if mass \(m_{1}\) moves with velocity \({\overrightarrow{v}}_{1}\), and mass \(m_{2}\) moves with velocity \({\overrightarrow{v}}_{2}\), then by conservation of linear momentum.

\[m_{1}{\overrightarrow{v}}_{1} + m_{2}{\overrightarrow{v}}_{2} = 0\]

in scalar form,

\[\begin{matrix} & m_{1}v_{1} + m_{2}\left( - v_{2} \right) = 0 \\ & m_{1}v_{1} = m_{2}v_{2}. \end{matrix}\]

Practice Exercise

Q. 1 The ring R in the arrangement shown can slide along a smooth fixed, horizontal rod XY . It is attached to the block \(B\) by a light string. The block is released from rest, with the string horizontal. Then which of the following are true.

(a) One point in the string will have only vertical motion
(b) R and B will always have momentum of same magnitude
(c) When the string becomes vertical, the speeds of \(R\) and \(B\) will be inversely proportional to their masses
(d) R will lose contact with the rod at some point
Q. 2 The figure shows a man of mass \(m\) standing at the end \(A\) to a trolley of mass \(M\) placed at rest on a smooth horizontal surface. The man starts moving towards the end B with a velocity \(u_{rel}\) with respect to the trolley. The length of the trolley is L .

(a) Find the time taken by the man to reach the other end.
(b) As the man walk on the trolley, find the velocity centre of mass of the system (man + trolley).
(c) When the man reaches the end B , find the distance moves by the trolley with respect to ground.
(d) Find the distance moved by the man with respect to ground.
Q. 3 A man of mass 60 kg jumps from a trolley of mass 20 kg standing on smooth surface with absolute velocity \(3\text{ }m/s\). Find the velocity of trolley and total energy produced by man.
Q. 4 Three particles of mass \(20\text{ }g,30\text{ }g\), and 40 g are initially moving along the positive direction of the three coordinate axes respectively with the same velocity of \(20\text{ }cm/s\), when due to their mutual interactional the first particle comes to rest, the second acquires a velocity \(10\widehat{i} + 20\widehat{k}\). What is then the velocity of the third particle?
Q. 5 A bullet of mass \(m\) strikes a block of mass \(M\) connected to a light spring of stiffness \(k\), with a speed \(v_{0}\) and gets into it. Find the loss of K.E. of the bullet.

Answers

Q. 1 (a), (c)
\[\begin{matrix} \text{~}\text{Q.}\text{~}2 & \text{~}\text{(a)}\text{~}\frac{L}{u_{\text{rel}\text{~}}} & \text{~}\text{(b) zero}\text{~} & \text{~}\text{(c)}\text{~}\frac{mL}{m + M} & \text{~}\text{(d)}\text{~}\frac{ML}{m + M} \end{matrix} \]Q. \(3\ 9\text{ }m/s,1.08\text{ }kJ\ \) Q. \(4\ 2.5\widehat{i} + 15\widehat{j} + 5\widehat{k}\ \) Q. \(5\ \frac{{Mmv}_{0}^{2}}{2(M + m)}\)

The total momentum of a system of particles in the \(\mathbf{C}\)-frame of reference is always zero. We can attach a frame of reference to the center of mass of a system, this is called the center of mass or C-frame of reference (figure). Relative to this frame, the center of massis at rest ( \(v_{\text{com}\text{~}} = 0\) ) and according
to equation \(P = M{\overrightarrow{v}}_{\text{com}\text{~}}\), the total momentum of a system of particles in the C-frame of reference is always zero.

\[\overrightarrow{P} = \sum_{i}^{}\mspace{2mu}{\overrightarrow{P}}_{i} = 0\text{~}\text{in the C-frame of reference}\text{~}\]

C-frame of reference

The C -frame is important because many situations can be more simply analyzed in this frame. It is clear that the C -frame moves with a velocity \(v_{\text{com}\text{~}}\) relative to the ground frame. When no external forces act on a system, the C -frame becomes an inertial frame.

Sol. From definition

\[v_{\text{com}\text{~}} = \frac{dr_{\text{com}\text{~}}}{dt} = \frac{1}{M}\sum_{i}^{}\mspace{2mu} m_{i}\frac{dr_{i}}{dt} = \frac{\sum_{i}^{}\mspace{2mu}\mspace{2mu} m_{i}v_{i}}{M}\]

The velocity of the center of mass relative to the observer is

\[v_{\text{com}\text{~}} = \frac{m_{l}v_{l} + m_{2}v_{2}}{m_{l} + m_{2}}\]

The velocity of each particle relative to the center of mass (figure) using the relative motion equations for velocities is

\[v_{lc}^{'} = v_{l} - v_{\text{com}\text{~}} = v_{l} - \frac{m_{l}v_{l} + m_{2}v_{2}}{m_{l} + m_{2}}\]

Answers

Q. 1 (a), (c)
\[\begin{matrix} \text{~}\text{Q.}\text{~}2 & \text{~}\text{(a)}\text{~}\frac{L}{u_{\text{rel}\text{~}}} & \text{~}\text{(b) zero}\text{~} & \text{~}\text{(c)}\text{~}\frac{mL}{m + M} & \text{~}\text{(d)}\text{~}\frac{ML}{m + M} \end{matrix} \]Q. \(3\ 9\text{ }m/s,1.08\text{ }kJ\ \) Q. \(4\ 2.5\widehat{i} + 15\widehat{j} + 5\widehat{k}\ \) Q. \(5\ \frac{{Mmv}_{0}^{2}}{2(M + m)}\)

\(C\)-Frame :

The total momentum of a system of particles in the \(\mathbf{C}\)-frame of reference is always zero. We can attach a frame of reference to the center of mass of a system, this is called the center of mass or C-frame of reference (figure). Relative to this frame, the center of massis at rest ( \(v_{\text{com}\text{~}} = 0\) ) and according
to equation \(P = M{\overrightarrow{v}}_{\text{com}\text{~}}\), the total momentum of a system of particles in the C-frame of reference is always zero.

\[\overrightarrow{P} = \sum_{i}^{}\mspace{2mu}{\overrightarrow{P}}_{i} = 0\text{~}\text{in the C-frame of reference}\text{~}\]

C-frame of reference

The C -frame is important because many situations can be more simply analyzed in this frame. It is clear that the C -frame moves with a velocity \(v_{\text{com}\text{~}}\) relative to the ground frame. When no external forces act on a system, the C -frame becomes an inertial frame.

Sol. From definition

\[v_{\text{com}\text{~}} = \frac{dr_{\text{com}\text{~}}}{dt} = \frac{1}{M}\sum_{i}^{}\mspace{2mu} m_{i}\frac{dr_{i}}{dt} = \frac{\sum_{i}^{}\mspace{2mu}\mspace{2mu} m_{i}v_{i}}{M}\]

The velocity of the center of mass relative to the observer is

\[v_{\text{com}\text{~}} = \frac{m_{l}v_{l} + m_{2}v_{2}}{m_{l} + m_{2}}\]

The velocity of each particle relative to the center of mass (figure) using the relative motion equations for velocities is

\[v_{lc}^{'} = v_{l} - v_{\text{com}\text{~}} = v_{l} - \frac{m_{l}v_{l} + m_{2}v_{2}}{m_{l} + m_{2}}\]

\[\begin{matrix} = \frac{m_{2}\left( v_{l} - v_{2} \right)}{m_{l} + m_{2}} = \frac{m_{2}v_{l2}}{m_{l} + m_{2}} \\ v_{2c}\ ^{'} = v_{2} - v_{com} = \frac{m_{l}\left( v_{2} + v_{l} \right)}{m_{l} + m_{2}} = - \frac{m_{l}v_{l2}}{m_{l} + m_{2}} \end{matrix}\]

where \(v_{l2} = v_{1} - v_{2}\) is the relative velocity of the two particles．
Thus，in the C－frame，the two particles appear to be moving in opposite directions with velocities inversely proportional to their masses．

Also relative to the center of mass，the two particles move with equal but opposite momentum since

\[p_{l}^{'} = m_{l}v_{lc}^{'} = \frac{m_{l}m_{2}v_{l2}}{\left( m_{l} + m_{2} \right)} = p_{2}^{'}\]

The expressions for two particle problems are much simpler when they are related to the \(C\)－frame of reference．

Let us find relation between kinetic energy of a system from ground frame and C －frame．We have a system consisting of many particles，let＇s say speed of the \(i^{\text{th}\text{~}}\) particle is \(v_{i}\) ．Then kinetic energy of system， K ，in ground frame will be summation of individual kinetic energies．
where \(v_{i}\) is velocity of the \(i^{\text{it}\text{~}}\) particle in ground frame，\(v_{ic}\) is velocity of the \(i^{\text{th}\text{~}}\) particle in reference to frame attached to the center of mass and \(v_{c}\) is velocity of center of mass in ground frame． zero，as \(\Sigma m_{i}{\overrightarrow{v}}_{i/c} = M{\overrightarrow{v}}_{c/c} = 0\)（total momentum of a system of particles in the C－frame of reference is always zero．）
\(v_{cic}\) is velocity of center of mass in frame of com．Which is zero．Also it represents momentum of system in C －frame which is zero．

\[\left( \frac{1}{2}\Sigma{\text{ }m}_{1}{\overrightarrow{v}}_{i/c}^{2} \right) = K_{sys/c}\]

Thus，we get
now

\[\begin{matrix} & K_{sys} = \Sigma\left( \frac{1}{2}{\text{ }m}_{i}v_{i}^{2} \right) \\ & v_{i} = v_{ic} + v_{c} \end{matrix}\]

attached to the center of mass and \(v_{c}\) is velocity of center of mass in ground frame．

\[\begin{matrix} & K_{sys} = \frac{1}{2}\Sigma{\text{ }m}_{i}\left( {\overrightarrow{v}}_{i/c} + {\overrightarrow{v}}_{c} \right)^{2} \\ & {\text{ }K}_{sys} = \frac{1}{2}\Sigma{\text{ }m}_{i}v_{i/c}^{2} + \frac{1}{2}\Sigma{\text{ }m}_{i}{\overrightarrow{v}}_{c}^{2} + \frac{1}{2} \times 2\left( \Sigma{\text{ }m}_{i} \cdot {\overrightarrow{v}}_{i/c} \cdot {\overrightarrow{v}}_{c} \right) \\ & K_{sys} = \frac{1}{2}\Sigma{\text{ }m}_{i}v_{i/c}^{2} + \frac{1}{2}\left( \Sigma{\text{ }m}_{i} \right){\overrightarrow{v}}_{c}^{2} + {\overrightarrow{v}}_{c}^{2}\left( \Sigma{\text{ }m}_{i} \cdot {\overrightarrow{v}}_{i/c} \right){\overrightarrow{v}}_{c} \\ & \text{~}\text{We can take}\text{~}v_{c}\text{~}\text{out of summ}\text{ation in second and third term as it}\text{~}i \end{matrix}\]

咺埥数 zero，as \(\Sigma m_{i}{\overrightarrow{v}}_{i/c} = M{\overrightarrow{v}}_{c/c} = 0\)（total momentum of a system of particles in the C－frame of reference is
\(v_{cc}\) is velocity of center of mass in frame of com．Which is zero．Also it represents momentum of system in C－frame which is zero．

\[1\]

Where \(K_{\text{sysc}\text{~}}\) means kinetic energy of system in C－frame．This important conclusion will be useful again in rotational dynamics；we can do little manipulation to write the equation as

\[K_{sys} = K_{sys/c} + \frac{p_{c}^{2}}{2M}\]

A system of two Particles

Suppose the masses of the particles are equal to \(m_{1}\) and \(m_{2}\) and their velocities in the K reference frame be \({\overrightarrow{v}}_{1}\) and \({\overrightarrow{v}}_{2}\). respectively. Let us find the expressions defining their moment and the total kinetic energy in the C -frame.
The momentum of the first particle in the C -frame is

\[{\overrightarrow{P}}_{1/c} = m_{1/c} = m_{1}\left( {\overrightarrow{v}}_{1} - {\overrightarrow{v}}_{c} \right)\]

Where \(v_{c}\) is the velocity of the center of mass of the system in the ground frame. Substituting in this formula expression.

\[\begin{matrix} & {\overrightarrow{v}}_{c} = \frac{m_{1}{\overrightarrow{v}}_{1} + m_{2}{\overrightarrow{v}}_{2}}{{\text{ }m}_{1} + m_{2}} \\ & {\overrightarrow{P}}_{1/c} = \mu\left( {\overrightarrow{v}}_{1} - {\overrightarrow{v}}_{2} \right) \end{matrix}\]

Where \(\mu\) is the reduced mass of the system, given by

\[m = \frac{m_{1}{\text{ }m}_{2}}{{\text{ }m}_{1} + m_{2}}\]

Similarly, the momentum of the second particle in the C -frame is

\[{\overrightarrow{P}}_{2/c} = \mu\left( {\overrightarrow{v}}_{2} - {\overrightarrow{v}}_{1} \right)\]

Thus, the momenta of the two particles in the C -frame are equal in magnitude and opposite in direction; the modulus of the momentum of each particle is

\[{\overrightarrow{P}}_{1/c} = \mu v_{rel}\]

Where \(v_{\text{rel}\text{~}} = \left| {\overrightarrow{v}}_{1} - {\overrightarrow{v}}_{2} \right|\) is the velocity of one particle relative to another.
Finally, let us consider kinetic energy. The total kinetic energy of the two particles in the C -frame is

\[K_{\text{sysic}\text{~}} = K_{1} + K_{2} = \frac{{\overrightarrow{P}}^{2}}{2{\text{ }m}_{2}} + \frac{{\overrightarrow{P}}^{2}}{2{\text{ }m}_{2}}\]

we know

\[\mu = \frac{m_{1}m_{2}}{m_{1} + m_{2}}\ \text{~}\text{or}\text{~}\ \frac{1}{m_{2}} + \frac{1}{m_{2}} = \frac{1}{\mu}\]

Then

\[\begin{matrix} & K_{sysic} = \frac{{\overrightarrow{P}}^{2}}{2\mu} = \frac{\mu v_{rel}^{2}}{2} \\ & {\text{ }K}_{sys} = K_{sysic} + K_{c},\text{~}\text{we get}\text{~} \\ & K_{sys} = \frac{\mu v_{rel}^{2}}{2} + \frac{mv_{c}^{2}}{2}\ \left( \text{~}\text{where}\text{~}m = m_{1} + m_{2} \right) \end{matrix}\]

Read this Illustration after collision

Sol. In ground frame :

let v be the final velocity of block & plank when relative motion ceases between block and plank; applying conservation of linear momentum, \(mu = (2m + m)v\)
\[\therefore\ v = u/3 w_{f} =\] change in K.E. \(= K.E_{.f} - K.E_{.i}\)
\[w_{f} = \frac{1}{2}3m\left( \frac{u}{3} \right)^{2} - \frac{1}{2}mu^{2} = - \frac{mu^{2}}{3}\]

In C-frame :

Considering block and plank as a system
Work done by friction is change in kinetic energy
\(w_{f} =\) change in K.E. \(= K.E_{._{f}} - K.E_{._{i}}\)
\[K \cdot E_{\cdot i} = \frac{1}{2}\mu(u - 0)^{2} + \frac{(2m + m)v_{c}^{2}}{2}\ \left( \mu = \frac{2m \times m}{2m + m} \right) K.E_{._{f}} = 0 + \frac{(2m + m)v_{c}^{2}}{2}\ \left( v_{c} \right.\ \] is constant as external force is zero \()\)
\[w_{f} = K.E_{._{f}} - K.E_{._{i}} = - \frac{mu^{2}}{3} \]Note: \(P_{\text{sys}\text{~}} =\) conserved if \(f_{\text{ext.}\text{~}} = 0\) although internal friction are doing work.

The impulse of the net force acting on a particle during a given time interval is equal to the change in momentum of the particle during that interval. The Impulse is a vector quantity.
For any arbitrary force. The impulse \(\overrightarrow{J}\) is defined as

\[\begin{matrix} & \overrightarrow{J} = \int_{i_{1}}^{t_{2}}\mspace{2mu}\mspace{2mu}{\overrightarrow{F}}_{ext}dt \\ & {\overrightarrow{F}}_{ext}dt = d{\overrightarrow{P}}_{sys} \\ & \overrightarrow{J} = \int_{P_{i}}^{P_{f}}\mspace{2mu}\mspace{2mu}{\overrightarrow{P}}_{sys} \\ & \overrightarrow{J} = {\overrightarrow{P}}_{f} - {\overrightarrow{P}}_{i} \end{matrix}\]

The concept of impulse can be better explained by an example shown in figure

A is a block of mass \(m\) moving with a velocity \(v_{1}\), at time \(t = 0\), a constant force \(F\) is applied on it in the direction of velocity for a time \(t\). Due to this force the velocity of the body increases hence momentum increase. If after time \(t\) the velocity of the body becomes \(v_{2}\), then according to momentum conservation we have

\[\begin{matrix} & & \text{~}\text{Initial momentum}\text{~} + \text{~}\text{momentum}\text{ imparted}\text{~} = \text{~}\text{final momentum}\text{~} \\ & \ mv_{1} + Ft = mv_{2} & \text{(i)} \end{matrix}\]

If applied force is opposite to the direction of \(v_{1}\) then we'll have

\[\begin{array}{r} mv_{1} - Ft = mv_{2}\#(ii) \end{array}\]

Equation (i) and (ii) are similar to the equations written for work - energy theorem as work done by the system or on the system are subtracted or added to the initial kinetic energy, gives the final kinetic energy of the system. Similar to that in initial momentum impulse due the forces acting on the system are added or subtracted, gives the final momentum of the system. If force is in the direction of the initial velocity of the particle, impulse is added to the initial momentum and if it is against the velocity, impulse is subtracted from the initial momentum

Impulsive Force

When a force of high magnitude acts for a time that is short compared with the time of observation of the system, it is referred as an impulsive force. An impulsive force can change the momentum of a body by a finite magnitude in a very short time interval.

An impulsive force can only be balanced by another impulsive force.
We defined the impulse in terms of a single force, but the impulse-momentum theorem deals with the change in momentum due to the impulse of the net force-the is, the combined effect of all the forces that act on the particle. In the case of a collision involving two particles, there is often no distinction because each particle is acted upon by only one force, which is due to the other particle. In this case, the change in momentum of one particle is equal to the impulse of the force exerted by the other particle.

Average Force

The impulsive force whose magnitude is represented in figure is assumed to have a constant direction. The magnitude of the impulse of this force is represented by the area under the \(F(t)\) curve. We can represent that same area by the rectangle in figure of width \(\Delta t\) and height \(F_{av}\) where \(F_{av}\) is the magnitude of the average force that acts during the interval \(\Delta t\). Thus

\[J = F_{av}\Delta t\]

Impulsive force is a relative term. There is no clear differentiation between an impulsive and non-impulsive force.

1. Gravitational force and spring force are always non-impulsive.

2. Normal, tension and friction are case dependent.

1. Impulsive Normal :

In case of collision, normal forces at the surface of collision are always impulsive.
eg. (i) \(\ N_{1}\) is Impulsive ;
Normal reaction due to ground \(N_{2}\&{\text{ }N}_{3}\) are Non-impulsive

(a)

(ii) Consider a ball dropped on a large ball.

Both normal forces \(N_{1}\) and \(N_{2}\) are impulsive
\(N_{2}\) is impulsive, as it balance N , for the large ball.

(a)

(iii) Consider a large ball colliding with small ball \(N_{1},{\text{ }N}_{3}\) are impulsive; \(N_{2}\) is non-impulsive \(N_{1}\) can be easily seen to be impulsive, as it is the normal force during collision here \(N_{3}\) balance a component of \(N_{1}\), therefore it is also impulsive.

(b)

2. Impulsive Friction : If the normal between the two objects is impulsive, then the friction between the two will also be impulsive.

3. Impulsive Tensions In a string : When a string is jerked out equal and opposite tension acts suddenly at each end and impulses act on the bodies attached with the string in the direction of the string.

All normal are impulsive but tension T is impulsive only for the ball A

One end of the string is fixed : The impulsive force which acts at the fixed end of the string cannot change the momentum of the fixed object attached at the other end. The object attached to the free end however will undergo a change in momentum in the direction of the string. The momentum remains unchanged in a direction perpendicular to the string. In this direction string cannot exert impulsive forces. Both ends of the string attached to movable objects : In this case equal and opposite impulses act on the two object, producing equal and opposite changes in momentum. The total momentum of the system therefore remains constant, although the momentum of each individual object is changed in the direction of the string. Perpendicular to the string however, no impulse acts and the momentum of each particle in this direction is unchanged.

In case of rod : Tension is always impulsive.
In case of spring: Tensions always non-impulsive.

Introduction

Newton's laws are useful for solving a wide range of problems in dynamics. However, there is one class of problem in which, even though Newton's laws still apply as we have defined them, we may have insufficient knowledge of the forces to permit us to analyze the motion. These problems involve collisions between two or more objects

In this section we will learn how to analyze collisions between two objects. In doing so, we will find that we need a new dynamic variable apart from velocity, acceleration, force and energy, called linear momentum. We will see that the law of conservation of linear momentum, one of the fundamental conservation laws of physics, can be used to study the collisions of objects from the scale of subatomic particles to the scale of galaxies.
In a collision, two object exert forces on each other for an identifiable time interval, so that we can separate the motion into three parts : before, during and after the collision. Before and after the collision, we assume that the objects are far enough apart that they do not exert any force on each other. During the collision, the objects exert forces on each other; these forces are equal in magnitude and opposite in direction, according to Newton' third law. We assume that these forces are much larger than any forces exerted on the two objects by other objects in their environment. The motion of the objects (or at least one of them) changes rather abruptly during the collision, so that we can make a relatively clear separation of the situation before the collision from the situation after the collision.

When a bat strikes a baseball, for example, the bat is in contact with the ball for an interval that is quite short compared with the time during for which we are watching the ball. During the collision the bat exerts a large force on the ball. This force varies with time in a complex way that we can measure only with difficulty.

When an alpha particle ( \(\ ^{4}He\) nucleus) collides with another nucleus, the force exerted on each by the other may be the repulsive electrostatic force associated with the charges on the particles. The particles may not actually come into direct contact with each other, but we still may speak of this interaction as collision because a relatively strong force, acting for a time that is short compared with the time that the alpha particle is under observation, has a substantial effect on the motion of the alpha particle.

Collision

Consider the situation shown in figure. Two block of masses \(m_{1}\) and \(m_{2}\) are moving on the same straight line on a frictionless horizontal table. The block \(m_{2}\), which is ahead of \(m_{1}\) is going with a speed \(u_{2}\) smaller than the speed \(u_{1}\) of \(m_{1}\). A spring is attached to the rear end of \(m_{2}\). Since \(u_{1} > u_{2}\), the block \(m_{1}\) will touch the rear end the spring at some instant.

Since \(m_{1}\) moves faster then \(m_{2}\), the length of the spring will decrease. The spring will compress, it pushes back both the blocks with force kx . This force is in the direction of the velocity of \(m_{2}\), hence \(m_{2}\) will accelerate. However, this is opposite to the velocity of \(m_{1}\) and so \(m_{1}\) will decelerate. The velocity of the front block A (Which was slower initially will gradually increase, and the velocity of the rear block B (which was faster initially) will gradually decrease. The spring will continue to become more and more compressed as long as the rear block B is faster than the front block A . There will be an instant when the
two blocks will have equal velocities.
This corresponds to the maximum compression of the spring. Thus "the spring compression is maximum when the two blocks attain equal velocities"
Now, the spring being already compressed, continues to push back the two blocks. Thus, the front block \(A\) will still be accelerated and the rear block \(B\) will still be decelerated. At the instant of maximum compression velocities were equal and hence after this the front block will move faster than the rear block. And so do the ends of the spring as they are in contact with the blocks. The spring will thus increase its length. This process will continue till the spring acquires its natural length. Once the spring regains its natural length, it stops exerting any force on the blocks. As the two blocks are moving with different velocities by this time, the rear one slower, the rear block will leave contact with the spring and the blocks will move with constant velocities. Their separation will go on increasing.

During the whole process, the momentum of the two - blocks system remains constant.
This is because there is no resultant external force acting on the system. Note that the spring being massless, exerts equal and opposite forces on the blocks.
Next consider the energy of the system. As there is no friction anywhere, the sum of the kinetic energy and the elastic potential energy remains constant. The elastic potential energy is \(\frac{1}{2}kx^{2}\) when the spring is compressed by \(x\). If \(v_{1}\) and \(v_{2}\) are the speeds at this time, we have,

\[\frac{1}{2}{\text{ }m}_{1}v_{1}^{2} + \frac{1}{2}{\text{ }m}_{2}v_{2}^{2} + \frac{1}{2}{kx}^{2} = E\]

Where E is the total energy of the system.
Initially the spring is at its natural length so that,

\[\begin{array}{r} \frac{1}{2}{\text{ }m}_{1}u_{1}^{2} + \frac{1}{2}{\text{ }m}_{2}u_{2}^{2} = E\#(i) \end{array}\]

At the time when the compression of the spring is maximum.

\[\begin{matrix} & v_{1} = v_{2} = v \\ & \frac{1}{2}\left( {\text{ }m}_{1} + m_{2} \right)V^{2} + \frac{1}{2}{kx}_{\max}^{2} = E \end{matrix}\]

During maximum compression \(v_{1} = v_{2} = V\)
when the spring acquires its natural length, so that,

\[\begin{array}{r} \frac{1}{2}{\text{ }m}_{1}v_{1}^{2} + \frac{1}{2}{\text{ }m}_{2}v_{2}^{2} = E\#(ii) \end{array}\]

From (i) and (ii),

\[\frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^{2} = \frac{1}{2}m_{1}u_{1}^{2} + \frac{1}{2}m_{2}u_{2}^{2}\]

The kinetic energy before the collision is the same as the kinetic energy after the collision. However, we can not say that the kinetic energy remains constant because it changes as a function of time, during the process.

Now consider a very similar situation as shown figure. The two bodies of masses \(m_{1}\) and \(m_{2}\) are moving along a line with velocities \(u_{1}\) and \(u_{2}\) respectively, \(u_{1}\) should be greater then \(u_{2}\) for two bodies to collide.

Fig. I

After some time the two bodies will come in contact as in the previous situation.

Fig. II

As the velocity of \(m_{1}\left( u_{1} \right)\) is greater than that of \(m_{2}\left( u_{2} \right)\), there will be a deformation in the bodies (as we saw in the previous example of spring mass system). This deformation continues till they acquire same velocity. Now a question arises, that why will they acquire same velocity.
As shown figure II both \(m_{1}\) and \(m_{2}\) exert force on each other equal in magnitude (Newton's III \(\ ^{\text{rd}\text{~}}\) Law) This force is in the direction of the velocity of \(m_{2}\), hence \(m_{2}\) will accelerate. However, this is opposite to the velocity of \(m_{1}\) and so \(m_{1}\) will decelerate. The velocity of the front body \(m_{2}\) (which was slower initially) will gradually increase, and the velocity of the rear body \(m_{1}\) (which was faster initially) will gradually decrease.
When they have attained a common velocity ; if bodies do not possess elastic properties, the two bodies will continue to move together and the process of collision ends here at maximum deformation. This is known as perfectly inelastic collision.

Just before collision

As net external force is zero therefore by conservation momentum, we can find the common velocity.

\[\begin{matrix} & m_{1}u_{1} + m_{2}u_{2} = \left( m_{1} + m_{2} \right)V \\ & \text{ }V = \frac{m_{1}u_{1} + m_{2}u_{2}}{{\text{ }m}_{1} + m_{2}} \end{matrix}\]

If bodies possess elastic property they will try to regain their shape that is the potential energy stored in deformation period will get converted into kinetic energy. If bodies are perfectly elastic then total potential energy will get converted into kinetic energy that is kinetic energy will be same before and after the collision, this type of collision in known as perfectly elastic collision.

Just before collision

Just after collision

By conservation momentum.

\[m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}\]

During the whole process, the momentum of the two -block system remains constant. Before and after collision kinetic energy may be same but not during the whole process.

Force of impact

The total time period of impact \(\Delta t\) is divided into period of deformation and period of restitution (recovery). Impulse of deformation and restitution (recovery) are finite and appreciable although the time interval of impact is extremely small.

Area \(A_{1}\) represents impulse of deformation
Area \(A_{2}\) represent impulse of reformation (restitution).
Deformation is maximum at the end of deformation period or at the start of recovery period.
In collision (impact) we consider the situation just before and just after collision. Just before collision is the moment when deformation period starts and just after collision is the moment when recovery period ends. At these moments one may say that force of impact becomes effectively zero.
Since time interval of impact is very small so impulse due to weight of body or weight dependent force is neglected during collision.

Line of collision (LOC)

When two bodies collide, they exert force on each other through point of contact, perpendicular to the plane of contact. The direction of force of interaction is line of collision. Line of collision is independent of the direction of velocities of colliding bodies.

After collision, only the components of velocity along line of collision changes, the perpendicular components of velocity remain unaffected.
According to initial velocities and line of collision, collision is of two types:
(A) Head on
(B) Oblique

Line of collision is same as the line of motion
When the lines of motion of two bodies are
(velocities) of the two colliding bodies. different, then the collision is oblique.

Head on collision

Oblique collision

Laws of collision

(i) Conservation of momentum

If no external impulsive forces act in a particular direction, the total momentum of system in that direction remains conserved.

Here we talk about external impulsive force because the force of impact is impulsive and changes the momentum of the individual bodies in a very short interval of time. The effect of non-impulsive forces such as gravity, spring force in such a small interval of time is negligible. Thus momentum remains conserved

\[m_{1}{\overrightarrow{v}}_{1} + m_{2}{\overrightarrow{v}}_{2} = m_{1}{\overrightarrow{u}}_{1} + m_{1}{\overrightarrow{u}}_{1}\]

Just before collision

Just after collision

(ii) Newton's Experimental Law.

This is an experimental law which relates the velocity of approach of the bodies before collision and the velocity of separation after the impact.
When the collision is neither perfectly elastic nor perfectly inelastic, for such cases, Newton did certain experiments and gave another law which is now called as Newton's Experimental Law.

Just before collision

Just after collision

\(\frac{v_{2} - v_{1}}{u_{2} - u_{1}} = - e\) (coefficient of restitution) or \(\frac{v_{2} - v_{1}}{u_{1} - u_{2}} = e\)
\(v_{2} - v_{1}\) is velocity of separation and \(u_{1} - u_{2}\) is velocity of approach.
Relative velocity of impact \(= e\) [Relative velocity of approach before impact]
\(\mathbf{v}_{\mathbf{1}}\) and \(\mathbf{v}_{\mathbf{2}}\) : Components of velocities of masses colliding, along the line of contact, after collision (with sign).
\(\mathbf{u}_{1}\) and \(\mathbf{u}_{2}\) : Components of velocities of colliding masses, along the line of contact, before collision (with sign)
Newton's Experimental Law can be stated "velocity of separation is e times velocity of approach" Coefficient of Restitution is the property of colliding bodies which depends on their elastic behaviour.

Note : This Law is valid even when momentum is not conserved i.e. external impulsive forces act.
(iii) Type of collision depends on e

For \(e = 1\), collision is perfectly elastic
for \(0 < e < 1\), collision is inelastic
for \(e = 0\), collision is perfectly inelastic (Bodies will move together)

Two bodies \(m_{1}\) & \(m_{2}\) are moving with velocities \({\overrightarrow{u}}_{1}\) and \({\overrightarrow{u}}_{2}\) along the same line as shown.

1. By conservation of momentum

\[\begin{array}{r} m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}\#(i) \end{array}\]

2. By Newton's experimental law

\[\begin{array}{r} \frac{v_{2} - v_{1}}{u_{1} - u_{2}} = e\#(ii) \end{array}\]

From (i) & (ii)
\[{v_{1} = \frac{\left( m_{1} - {em}_{2} \right)u_{1}}{{\text{ }m}_{1} + m_{2}} + \frac{(1 + e)m_{2}u_{2}}{{\text{ }m}_{1} + m_{2}} }{v_{2} = \frac{(1 + e)m_{1}u_{1}}{{\text{ }m}_{1} + m_{2}} + \frac{\left( m_{2} - {em}_{1} \right)u_{2}}{{\text{ }m}_{1} + m_{2}} }\]for perfectly elastic collision velocities can be obtained by substituting \(e = 1\).

\[v_{1} = \frac{m_{1} - m_{2}}{m_{1} + m_{2}}u_{1} + \frac{2m_{2}}{m_{1} + m_{2}}u_{2}\ \&\ v_{2} = \frac{2m_{1}}{m_{1} + m_{2}}u_{1} + \frac{m_{2} - m_{1}}{m_{1} + m_{2}}u_{2}\]

Special cases:

Case 1: \(m_{1} = m_{2}:\ \) In this case the velocities of particles are exchanged.

Just before collision

Just after collision

Case 2:

\[\begin{matrix} & m_{1} \gg {\text{ }m}_{2}\ :\ v_{1} \approx u_{1}\& v_{2} = 2u_{1} - u_{2} \\ & e = 1 \end{matrix}\]

Loss in kinetic energy

The loss in kinetic energy is equal to initial kinetic energy ( \(k_{i}\) ) minus final kinetic energy ( \(k_{f}\) )

\[\begin{matrix} & \Delta KE = k_{i} - k_{f} \\ & k_{i} = \frac{1}{2}{\text{ }m}_{1}u_{1}^{2} + \frac{1}{2}{\text{ }m}_{2}u_{2}^{2} \\ & k_{f} = \frac{1}{2}{\text{ }m}_{1}v_{1}^{2} + \frac{1}{2}{\text{ }m}_{2}v_{2}^{2} \end{matrix}\]

Substituting the values of \(v_{1}\& v_{2}\) we obtain

\[\Delta KE = \frac{1}{2}\frac{{\text{ }m}_{1}{\text{ }m}_{2}}{{\text{ }m}_{1} + m_{2}}\left( u_{1} - u_{2} \right)^{2}\left( 1 - e^{2} \right)\]

This result can also be obtained by using concept of C -frame.
For \(e = 1\), i.e. perfectly elastic collision

\[\Delta KE = 0\]

As we discussed earlier there is no loss of kinetic energy
However, in elastic collision, KE is same before & after the collision,
i.e. \(\frac{1}{2}m_{1}u_{1}^{2} + \frac{1}{2}m_{2}u_{2}^{2} = \frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^{2}\)
This equation is not independent, using Newton's experimental law & momentum conservation we can solve for final velocities more easily.
For \(e = 0\), i.e. perfectly in elastic collision

\[\Delta KE = \frac{1}{2}\frac{{\text{ }m}_{1}{\text{ }m}_{2}}{{\text{ }m}_{1} + m_{2}}\left( u_{1} - u_{2} \right)^{2}\]

in this case, there is maximum loss in kinetic energy.

Practice Exercise

Q. 1 Find the final velocities of the masses after in collision the given situations.
(i)

\[e = 1 \](ii)

(iii)

\[(3\text{ }kg) \longrightarrow 2\text{ }m/s\ e = 5 \](iv)

\[e = 3/5 \](v)

\[e = 1\]

Answers

Q. \(1\ \) (i) \(\ v_{1} = 0,v_{2} = 3\) (Rightwards)
(ii) \(\ v_{1} = .3\) (Rightwards), \(v_{2}1.8\) (Rightwards)
(iii) \(\ v_{1} = 2.1\) (Rightwards), \(v_{2} = 2.6\) (Rightwards)
(iv) \(\ v_{1} = 1.8\) (Leftwards), \(v_{2} = 1.2\) (Rightwards)
(v) \(\ v_{2}\) (Leftwards), \(v_{1}\) (Rightwards)

In, oblique impact, the relative velocity of approach of the bodies doesn't coincide with the line of impact.
Conserving the momentum of the system along and perpendicular to the line of impact (due to absence of any other external impulsive force) we obtain

\[\text{~}\text{and,}\text{~}\begin{array}{r} m_{1}u_{1}cos\theta_{1} + m_{2}u_{2}cos\theta_{2} = m_{1}v_{1}cos\beta_{1} + m_{2}v_{2}cos\beta_{2}\#(1) \\ {\text{ }m}_{1}u_{1}sin\theta_{1} + m2u2sin\theta_{2} = m_{1}v_{1}sin\beta_{1} + m_{2}v_{2}sin\beta_{2}\#(1) \end{array}\]

Since no force is acting on \(m_{1}\) and \(m_{2}\) along the tangent, the individual momentum of \(m_{1}\) and \(m_{2}\) remains conserved.

\[\begin{array}{r} \begin{matrix} \Rightarrow \ m_{1}u_{1}sin\theta_{1} = m_{1}v_{1}sin\beta_{1} \\ \text{~}\text{and}\text{~}\ m_{2}u_{2}sin\theta_{2} = m_{2}v_{2}sin\beta_{1} \end{matrix}\#(3) \end{array}\]

Newton's experimental Law:

(We consider velocities along the line of collision)

\[\begin{array}{r} e = \frac{v_{2}cos\beta_{2} - v_{1}cos\beta_{1}}{u_{1}cos\theta_{1} - u_{2}cos\theta_{2}}\#(4) \end{array}\]

Now we have four equations and four unknown's \(v_{1},v_{2},\beta_{1}\) and \(\beta_{2}\). Solving four equations for four unknown we obtain.

\[\begin{array}{r} v_{1}cos\beta_{1} = \frac{\left( m_{1} - em_{2} \right)u_{1}cos\theta_{1} + m_{2}(1 + e)u_{2}cos\theta_{2}}{m_{1} + m_{2}}\#(5) \\ \text{~}\text{and}\text{~}\ v_{2}cos\beta_{2} = \frac{m_{1}(1 + e)cos\theta_{1} + \left( m_{2} - em_{2} \right)u_{2}cos\theta_{2}}{m_{1} + m_{2}}\#(6) \\ \ \therefore\ v_{1} = \sqrt{\left( v_{1}sin\beta_{1} \right)^{2} + \left( v_{1}cos\beta_{1} \right)^{2}}\#(6) \\ \text{~}\text{and}\text{~}\ tan\beta_{1} = \frac{v_{1}sin\beta_{1}}{v_{1}cos\beta_{1}}\#(6) \\ \ \Rightarrow \ \beta_{1} = \tan^{- 1}\left( \frac{v_{1}sin\beta_{1}}{v_{1}cos\beta_{1}} \right)\#(6) \end{array}\]

[Put \(v_{1}sin\beta_{1}\) from (2) and \(v_{1}cos\beta_{1}\) from (5)]
Similarly, find \(v_{2}\) and \(\beta_{2}\)

\[\text{~}\text{Impulse}\text{~} = \frac{m_{1}m_{2}}{m_{1} + m_{2}}(1 + e)\left( u_{1}cos\theta_{1} - u_{2}cos\theta_{2} \right)\]

Energy loss \(= \frac{m_{1}m_{2}}{2\left( m_{1} + m_{2} \right)}\left( 1 - e^{2} \right)\left( u_{1}cos\theta_{1} - u_{2}cos\theta_{2} \right)^{2}\)

Oblique collision on a Fixed Plane

Let a small ball collides with a smooth horizontal floor with a speed \(u\) at an angle \(\theta\) to the vertical as shown in the figure. Just after the collision, let the ball leaves the floor with a speed v at an angle \(\beta\) to vertical.

It is quite clear that the line of action is perpendicular to the floor. Therefore, the impact takes place along the (normal) vertical. Now we can use Newton's experimental law as

\[e = \frac{\text{~}\text{velocity of }\text{seperation}\text{~}}{\text{~}\text{velocity of approach}\text{~}}\]

\(\Rightarrow \ e\lbrack\) velocity of approach \(\rbrack =\) velocity of seperation

\[\begin{array}{r} \Rightarrow \ e(ucos\theta)( - \widehat{j}) = - (vcos\beta)( + \widehat{j})\#(i) \end{array}\]

or \(vcos\beta = eucos\theta\)

Since impulsive force N acts on the body along the normal, we cannot conserve its momentum. Since along horizontal the component of N is zero, therefore we can conserve the horizontal momentum of the body.

Velocity

\(\Rightarrow \ \left( P_{x} \right)\) body \(=\) Constant
\[{\Rightarrow \ \left( P_{x} \right)_{\text{initial}\text{~}} = \left( P_{x} \right)_{\text{final}\text{~}} }{\Rightarrow musin\theta = mvsin\beta}\]

\[\begin{array}{r} \Rightarrow \ vsin\beta = usin\theta\#(ii) \end{array}\]

Squaring equations (i) and (ii) and adding,

\[\begin{matrix} & v^{2}\cos^{2}\beta + v^{2}\sin^{2}\beta = e^{2}u^{2}\cos^{2}\theta + u^{2}\sin^{2}\theta \\ \Rightarrow & v^{2} = u^{2}\left\lbrack e^{2}\cos^{2}\theta + \sin^{2}\theta \right\rbrack \\ \Rightarrow & v = u\sqrt{\sin^{2}\theta + e^{2}\cos^{2}\theta} \end{matrix}\]

Dividing equation (i) by (ii)

\[= \frac{vcos\beta}{vsin\beta} = \frac{\text{~}\text{eu}\text{~}cos\theta}{usin\theta}\]

\[\Rightarrow \ cot\beta = ecot\theta \Rightarrow \ = \cot^{- 1}(ecot\theta)\].
Impulse
Impulse of the blow \(=\) change of momentum of the body

\[= \{ mvsin\beta\widehat{i} + (mvcos\beta)\}\widehat{j}\} - \left( musin\theta_{\widehat{i}} - mucos\theta\widehat{j} \right)\]

\(\Rightarrow \ \) Impulse \(= m(vsin\beta - usin\theta)\widehat{i} + m(vcos\beta + ucos\theta)\widehat{j}\)
Since v \(sin\beta = usin\theta\)
\(\Rightarrow \ \) Impulse \(= m(vcos\beta + ucos\theta)\widehat{j}\)
Putting v \(cos\beta =\) eu \(cos\theta\) from eq (i), we obtain
Impulse \(= m(1 + e)ucos\theta\widehat{j}\)
\(\therefore\) Magnitude of the impulse \(= m(1 + e)ucos\theta\) (in the direction of line of collision)

Change in kinetic energy

\[\Delta K.E. = \frac{1}{2}mv^{2} - \frac{1}{2}mu^{2} \]Putting the value of v we obtain
\[{= \frac{1}{2}\text{ }m\left\lbrack \left\lbrack u\sqrt{\sin^{2}\theta + e^{2}\sin^{2}\theta} \right\rbrack^{2} - u^{2} \right\rbrack }{= \frac{1}{2}{mu}^{2}\left\lbrack \sin^{2}\theta + e^{2}\cos^{2}\theta - 1 \right\rbrack \Delta}\] K.E. \(= \frac{1}{2}\left( 1 - e^{2} \right){mu}^{2}\cos^{2}\theta\)
Negative sign indicates the loss of K.E. i.e. final K.E. is less than initial K.E.

Practice Exercise

Q. 1 Two spheres are moving towards each other. Both have same radius but their masses are 4 kg and 2 kg . If the velocities are \(4\text{ }m/s\) and \(2\text{ }m/s\) respectively and coefficient of restitution is \(e = 1\). Find, final velocity along line of impact.

Q. 2 Two spheres are moving towards each other. Both have same radius but their masses are 4 kg and 4 kg . If the velocities are \(4\text{ }m/s\) and \(2\text{ }m/s\) respectively and coefficient of restitution is \(e = 1\), find. Final velocity along line of impact.
Q. 3 A ball of mass \(m\) moving with a speed \(u_{1}\) collides elasticity with another identical ball moving with velocity \(u_{2}\).
(a) Find the velocities of the balls after collision if the impact is direct.
(b) Find the angle between velocities after collision if they collide obliquely and \(u_{2} = 0\).
Q. 4 Two equal sphere of mass \(m\) are in contact on a smooth horizontal table. A third identical sphere impinges symmetrically on them and is reduced to rest. Find e and the loss of KE.

Answers

Q. \(1v_{1} = 0,\ v_{2} = 3\sqrt{3}\text{ }m/s\)
Q. \(2\ v_{1} = - \sqrt{3}m/s,\ v_{2} = 2\sqrt{3}m/s\)
Q. 3 (a) \(v_{1} = u_{2},v_{2} = u_{1}\); (b) \(\frac{\pi}{2}\)
Q. \(4\frac{2}{3},\frac{{mu}^{2}6}{6}\)

Q. 1 A square hole is punched out of a circular lamina, the diagonal of the square being the radius of the circle. If ' \(a\) ' be the diameter of the circle, find the distance of centre of mass of the remainder from the centre of the circle.
Sol. Consider the figure shown below. Let AB be the diameter passing through the diagonal OB of the square portion where \(O\) is the centre of the circle.

As mass is proportional to area of a uniform laninar body,
Mass of the portions can be replaced by their respective areas at their centre of mass
Area of circular potion \(= \frac{\pi a^{2}}{4}\)
Area of square portion \(= \frac{a^{2}}{8}\)
If \(G_{1}\) and \(G\) the positions of centre of mass of the cut square portion and remaining portion.

Then

\[\begin{matrix} OG = & \frac{- \frac{\pi a^{2}}{4}(0) - \frac{a^{2}}{8}\left( \frac{a}{4} \right)}{\frac{\pi a^{2}}{4} - \frac{a^{2}}{8}} = \frac{\frac{a}{32}}{\left( \frac{2\pi - 1}{8} \right)} \\ & \ = \frac{a}{4(2\pi - 1)} \end{matrix}\]

\(\therefore\ \) The centre of mass of the remaining parting is at a distance of \(\frac{a}{4(2\pi - 1)}\) from the centre.
Q. 2 Find out the centre of mass of a composite object shown in figure. Object consists of a cone with its base joint with the base of a hemisphere. The dimensions of the object are shown in figure. Assume uniform density of the system.

Sol. The shown object is made up of joining a solid cone and a hemisphere. We already know the location of the centre of mass of a cone and that of a hemisphere. The masses of the two are in proportion of their volume. The masses of cone and hemisphere are

Mass of cone is

\[\begin{matrix} & m_{1} = \rho\frac{1}{3}\pi R^{2}\text{ }h \\ & {\text{ }m}_{2} = \rho\frac{2}{3}\pi R^{3} \end{matrix}\]

and that of hemisphere is
Now we apply the result of two body system to find the centre of mass of the composite body. Let \(l\) be the distance between the independent centre of mass of the bodies cone and hemisphere, then

\[l = \frac{3R}{8} + \frac{h}{4}\]

The position of centre of mass from \(m_{2}\) is

\[\begin{matrix} & x = \frac{m_{1}l}{m_{1} + m_{2}} = \frac{\rho\frac{1}{3}\pi R^{2}h\left( \frac{3R}{8} + \frac{h}{4} \right)}{\rho\frac{1}{3}\pi R^{2}h + \rho\frac{2}{3}\pi R^{3}} \\ & x = \frac{h(3R + 2h)}{8(h + 2R)} \end{matrix}\]

Q. 3 For the figure shown, block of mass \(m\) is released from the rest. Find the distance of the wedge from initial position, when block \(m\) arrives at the bottom of the wedge. All surfaces are frictionless.

Sol. As there is no net external force in the \(x\) direction, thus the momentum of system in \(x\) direction \(\left( P_{\text{sys}\text{~}} \right)_{x}\) is conserved.

\[{Mv}_{1} = {mv}_{2},\text{~}\text{Initially}\text{~}\left( P_{sys} \right)_{x} = 0\]

\(\therefore\) Displacement of centre of mass in x -direction \(= 0\)
i.e. \(\ {Mx}_{1} = {mx}_{2}\)

Let the displacement of wedge M be x backwards
\(\therefore\ \) displacement of block \(= hcot\alpha - x\)
Using equation (i)

\[\begin{matrix} & m(\text{ }hcot\alpha - x) = Mx \\ & x = \frac{m}{\text{ }m + M}\text{ }hcot\alpha \end{matrix}\]

Q. 4 Two bodies A and B of masses m and 2 m respectively are placed on a smooth floor. They are connected by a light spring of stiffness k . A third body C of mass m moves with velocity \(v_{0}\) along the line joining A and B and collides elastically with A . If \(\mathcal{l}_{0}\) be the natural length of the spring then find the minimum separation between the blocks.

Sol. Initially there will be collision between C and A which is elastic, therefore, by using conservation of momentum we obtain,

\[{mv}_{0} = {mv}_{A} + {mv}_{C}\ ;\ v_{0} = v_{A} + v_{C}\]

Since \(e = 1,v_{0} = v_{A} - v_{C}\)
Solving the above two equation, \(v_{A} = v_{0}\) and \(v_{C} = 0\)
Now \(A\) will move and compress the spring which in turn acceleration \(B\) and retard \(A\) and finally both \(A\) and \(B\) will move with same velocity \(v\).
(a) Since net external force is zero, therefore momentum of the system (A and B) is conserved.

Hence \({mv}_{0} = (m + 2\text{ }m)v\)
\[\Rightarrow \ v = v_{0}/3 \](b) If \(x_{0}\) is the maximum compression, then using energy conservation

\[\begin{matrix} \frac{1}{2}mv_{0}^{2} & \ = \frac{1}{2}(m + 2m)v^{2} + \frac{1}{2}kx_{0}^{2} \\ \Rightarrow \ \frac{1}{2}mv_{0}^{2} & \ = \frac{1}{2}(3m)\frac{v_{0}^{2}}{9} + \frac{1}{2}kx_{0}^{2}\ \Rightarrow \ x_{0} = v_{0}\sqrt{\frac{2m}{3k}} \end{matrix}\]

Hence minimum distance \(D = \mathcal{l}_{0} - x_{0} = \mathcal{l}_{0} - v_{0}\sqrt{\frac{2\text{ }m}{3k}}\)
Q. 5 A 20 g bullet pierces through a plate of mass \(M_{1} = 1\text{ }kg\) and then comes to rest inside a second plate of mass \(M_{2} = 2.98\text{ }kg\) (refer figure). It is found that the two plates, initially at rest, now move with equal velocity v . Find the velocity of the bullet ( \(inm/s\) ) when it is between \(M_{1}\) and \(M_{2}\). Given that it entered \(M_{1}\) with \(100\text{ }m/s\).

Sol. From the principle of conservation of linear momentum we have
and

\[\begin{matrix} & mu = M_{1}v + mv^{'} \\ & mv^{'} = \left( m + M_{2} \right)v \end{matrix}\]

or
and

\[20u = 1000v + 20v^{'}\]

or

\[20v^{'} = (20 + 2980)v\]

and

\[\begin{array}{r} u = 50v + v^{'}\#(i) \end{array}\]

From (i) and (ii), we get

\[\begin{array}{r} v^{'} = 150v\#(ii) \end{array}\]

\[3u = v^{'} + 3v^{'} = 4v^{'}\ \text{~}\text{or}\text{~}\ v^{'} = 3u/4 = 75\]

Q. 6 A block of mass 4 kg is moving with a velocity of \(7\text{ }m/s\) on a surface. It collides with another block of mass 3 kg elastically. The surface is smooth for 4 kg block but rough for 3 kg block \((\mu = 0.4)\). Find the time (in sec) after which next collision will occur.

Sol. \(\ 4\text{ }kg \longrightarrow 7\text{ }m/s\ 3\text{ }kg\ \square \longrightarrow v_{1}\ \square v_{2}\)
Let after collision their velocities are \(v_{1}\& v_{2}\) applying conservation of momentum

\[\begin{matrix} & & 4 \times 7 + 0 = 4v_{1} + 3v_{2} \\ & 4v_{1} + 3v_{2} = 28 & \text{(1)} \end{matrix}\]

also \(e = - \left( \frac{v_{2} - v_{1}}{u_{2} - u_{1}} \right)\)

\[\begin{matrix} & & 1 = - \left( \frac{v_{2} - v_{1}}{0 - 7} \right) \\ & v_{2} - v_{1} = 7 & \text{(2)} \end{matrix}\]

solving (1) & (2) \(\Rightarrow v_{2} = 8\text{ }m/s\)

\[v_{1} = 1\text{ }m/s\]

time after \(2^{\text{nd}\text{~}}\) block stops \(= \frac{u}{a} = \frac{8}{4} = 2sec\)
distance travelled by \(2^{\text{nd}\text{~}}\) block till this moment \(s = 8t + \frac{1}{2}at^{2}\)

\[s = 8 \times 2 - \frac{1}{2}4 \times 2^{2} = 8m\]

so time elasped till \(2^{\text{nd}\text{~}}\) collision \(= \frac{8}{1} = 8sec\).
Q. 7 A particle of mass ' \(m\) ' is projected with velocity \(v_{0}\) at an angle ' \(\alpha\) ' with the horizontal. The coefficient of restitution for any of its impact with the smooth ground is e .

Find total time taken by the particle before it stops moving vertically?
Sol. Total time taken by the particle to stop

\[\begin{matrix} T & \ = \frac{2v_{0}sin\alpha}{\text{ }g} + \frac{2{ev}_{0}sin\alpha}{\text{ }g} + \frac{2e^{2}v_{0}sin\alpha}{\text{ }g} + \ldots \\ & \ = \frac{2v_{0}sin\alpha}{\text{ }g}\left( 1 + e + e^{2} + \ldots \right)\ = \frac{2v_{0}sin\alpha}{\text{ }g(1 - e)} \end{matrix}\]

Q. 8 After falling from rest through a height \(h\) a body of mass \(m\) begins to raise a body of mass \(M(M > m)\) connected to it through a pulley.
(a) Determine the time it will take for the body of mass M to return to its original position.
(b) Find the fraction of kinetic energy lost when the body of mass M is jerked into motion.

Sol. (a) The speed of the body \(B\) just before the string becomes taut is \(v = \sqrt{2gh}\). When the string is jerked, large impulsive reactions are generated in the string. At this moment effect of gravity is negligible. So momentum of the system is conserved at this instant. Let v' be the common speed of the two bodies after they are jerked into motion. From conservation of momentum, we have

\[mv = (M + m)v^{'}\ \text{~}\text{or}\text{~}\ v^{'} = \frac{m}{M + m}v\]

The acceleration of the system is

\[\Sigma F = Mg - mg = (M + m)a\ \text{~}\text{or}\text{~}\ a = - \frac{M - m}{M + m}\text{ }g\]

The acceleration is negative, (opposite to \(v^{'}\) )
Let the system return to original position at time \(t\).

\[\begin{matrix} 0 & \ = v^{'}t + \frac{1}{2}at^{2} \\ \text{~}\text{or}\text{~}\ t & \ = - \frac{2v^{'}}{a} = \frac{2m}{M - m}\sqrt{\frac{2h}{g}} \end{matrix}\]

(b) The fractional loss of kinetic energy is

\[\frac{\frac{1}{2}mv^{2} - \frac{1}{2}(M + m)v^{'2}}{\frac{1}{2}mv^{2}} = \frac{M}{M + m}\]

Q. 9 A block of mass \(m\) is connected to another block of mass \(M\) by massless spring constant \(k\) The blocks are kept on a smooth horizontal plane. Initially, the blocks are at rest and the spring is unstretched when a constant force F starts acting on the block of mass M to pull it. Find the maximum extension of the spring.

Sol. Let us take the two blocks plus the spring as the system. The centre of mass of the system moves with an acceleration \(a = \frac{F}{m + M}\). Let us work from a reference frame with its origin at the centre of mass. As this frame is acceleration with respect to the ground we have to apply a pseudo force ma towards left on the block of mass \(m\) and Ma towards left on the block of mass M. The net external force on m is

\[F_{1} = ma = \frac{mF}{\text{ }m + M}\text{~}\text{towards left}\text{~}\]

and the net external force on M is

\[F_{2} = F - Ma = F = \frac{MF}{\text{ }m + M} = \frac{mF}{\text{ }m + F}\text{~}\text{towards right}\text{~}\]

The situation from this frame is shown in figure. As the centre of mass is at rest in this frame, the block move in opposite direction and come to instantaneous rest at some instant. The extension of the spring will be maximum of this instant. Suppose the left block through a distance \(x_{2}\) from the initial positions. The total work done by the external forces \(F_{1}\) and \(F_{2}\) in this period are

\[W = F_{1}x_{1} + F_{2}x_{2} = \frac{mF}{\text{ }m + F}\left( x_{1} + x_{2} \right)\]

This should be equal to the increase in the potential energy of the spring as there is no change in the kinetic energy. Thus,

\[\begin{matrix} & \frac{mF}{\text{ }m + F}\left( x_{1} + x_{2} \right) = \frac{1}{2}k\left( x_{1} + x_{2} \right)^{2} \\ \text{~}\text{or,}\text{~} & x_{1} + x_{2} = \frac{2mF}{k(\text{ }m + M)} \end{matrix}\]

This is the maximum extension of the spring.
Q. 10 A glass ball collides with a smooth horizontal surface with a velocity \(\overrightarrow{v} = a\widehat{i} - b\widehat{j}\). If the coefficient of restitution of collision be e, find the velocity of the ball just after the collision. (Take y -axis along vertical)
Sol. Collision takes place along the normal. Therefore the magnitude normal component \(\left( v_{y} \right)\) of the velocity of the glass ball is changed to \(v_{y^{'}} = ev_{y}\) just after the collision whereas the horizontal component \(\left( v_{x} \right)\) of its velocity remains constant due to the absence of any horizontal force.
\(\Rightarrow \ \) The velocity of the ball just after the impact

\[\begin{matrix} & = \overrightarrow{v} = {\overrightarrow{v}}_{x} + {\overrightarrow{v}}_{y} \\ \Rightarrow \ & {\overrightarrow{v}}^{'} = v_{x}^{'}\widehat{i} + v_{y}^{'}\widehat{j} \end{matrix}\]

where, \(v_{x}^{'} = a\& v_{y}^{'} = eb\)

\[\Rightarrow \ {\overrightarrow{v}}^{'} = a\widehat{i} + eb\widehat{j}\]

Therefore the magnitude of the velocity \({\overrightarrow{v}}^{'} = \left| {\overrightarrow{v}}^{'} \right| = \sqrt{a^{2} + e^{2}b^{2}}\) and the direction is given as \(\theta = \tan^{- 1}\left( \frac{v_{x^{'}}}{v_{y^{'}}} \right) = \tan^{- 1}\left( \frac{a}{eb} \right)\) to the normal (vertical)
Q. 11 A stationary body explodes into four identical fragments such that three of them fly off mutually perpendicular to each other each with same kinetic energy \(\left( E_{0} \right)\). Find the energy of explosion.
Sol. Let three of the fragments move along \(X,Y\& Z\) axes. Therefore their velocities can be given as

\[{\overrightarrow{v}}_{1} = vi,{\overrightarrow{v}}_{2} = vj\&{\overrightarrow{v}}_{3} = v\widehat{k}\text{~}\text{where}\text{~}v = \text{~}\text{speed of each of the three fragments. Let the velocity of}\text{~}\]

the fourth fragment be \(\overrightarrow{v}\). Since, in explosion no net external force is involved, the net momentum of the system remains conserved just before and after the explosion.

\[\Rightarrow \ (\overrightarrow{p})_{f} = (\overrightarrow{p})_{i}\]

\(\Rightarrow \ m{\overrightarrow{v}}_{1} + m{\overrightarrow{v}}_{2} + m{\overrightarrow{v}}_{3} + m{\overrightarrow{v}}_{4} = 0\) ( \(P_{i} = 0\) because the body was stationary), putting the values of

\[{\overrightarrow{v}}_{1},{\overrightarrow{v}}_{2}\&{\overrightarrow{v}}_{3},\text{~}\text{we obtai}\text{n,}\text{~}\]

\[{\overrightarrow{v}}_{4} = - v(\widehat{i} + \widehat{j} + \widehat{k})\]

Therefore, \(v_{4} = \sqrt{3}v\)
The energy of explosion ( \(\bigtriangleup KE\) ) system

\[\begin{matrix} \Rightarrow \ & E = KE_{f} - KE_{i} \\ & \ = \left( \frac{1}{2}mv_{1}^{2} + \frac{1}{2}mv_{2}^{2} + \frac{1}{2}mv_{3}^{2} + \frac{1}{2}mv_{4}^{2} \right) - (0) \end{matrix}\]

Putting \(v_{1} = v_{2} = v_{3} = v,v_{4} = \sqrt{3}v\)

\[\begin{matrix} & & E = \left( \frac{1}{2}{mv}^{2} + \frac{1}{2}{mv}^{2} + \frac{1}{2}{mv}^{2} + \frac{1}{2}\text{ }m(\sqrt{3}v)^{2} \right) - (0) \\ & E = 3{mv}^{2} & \text{(i)} \end{matrix}\]

As we know from question that kinetic energy of three are equal and equal to \(E_{0}\)
\(\therefore\ \frac{1}{2}{mv}^{2} = E_{0}\), Putting this value in equation (i)
we obtain, \(E = 6E_{0}\).
Q. 12 A man of mass m climbs a rope of length L suspended below a balloon of mass M . The balloon is stationary with respect to ground. If the man begins to climb up the rope at a speed \(v\) (relative to rope) in what direction and with speed (relative to ground) will the balloon move?
Sol. Balloon is stationary
\(\Rightarrow \ \) No net external force acts on it.
\(\Rightarrow\) The conservation of linear momentum of the system (balloon + man) is valid
\[\Rightarrow \ M{\overrightarrow{v}}_{b} + m{\overrightarrow{v}}_{mb} = 0 \]where \({\overrightarrow{v}}_{m} = {\overrightarrow{v}}_{mb} + {\overrightarrow{v}}_{b}\)
\[\Rightarrow \ M{\overrightarrow{v}}_{b} + m\left\lbrack {\overrightarrow{v}}_{mb} + {\overrightarrow{v}}_{b} \right\rbrack = 0 \]where \(v_{mb} =\) velocity of man relative to the balloon (rope)
\[\Rightarrow \ {\overrightarrow{v}}_{b} = - \frac{m{\overrightarrow{v}}_{mb}}{M + m} \]Where \(v_{mb} = v \Rightarrow v_{b} = \frac{mv}{M + m}\) and directed opposite to that of the man.

Q. 13 A particle of mass \(m\) strikes elastically with a disc of radius \(R\), with a velocity \(\overrightarrow{v}\) as shown in the figure. If the mass of the disc is equal to that of the particle and the surface of the contact is smooth, find the velocity of the disc just after the collision.

Sol. We see that impact takes place along the normal. Therefore, the particle and the disc change their momentum along that line. However, no external force acts on the system along the normal line. Hence we can conserve the linear momentum of the system (disc + particle) along the normal. Since the masses of the disc and particle are equal, so the exchange of momentum takes place along the normal. That means, the particle completely delivers the part (component) of its momentum \((mvcos\theta)\) along the normal

\(\Rightarrow \ \) Velocity of the disc, \({\overrightarrow{v}}_{1} = (vcos\theta)\widehat{j}\) where, \(cos\theta = cos30^{\circ} = \frac{\sqrt{3}}{2}\)
\[\Rightarrow \ {\overrightarrow{v}}_{1} = \frac{\sqrt{3}v}{2}\widehat{j} \]Q. 14 A wedge of mass \(m\) rest on horizontal surface. The inclination of the wedge is \(\alpha\). A ball of mass m moving horizontal with speed \(u\) hits the inclined face of the wedge inelastically and after hitting slides up the inclined face of the wedge. Find the velocity of the wedge just after collision. Neglect any friction.

Sol. Let velocity of the ball after collision is \({\overrightarrow{v}}_{2}\) (w.r. to wedge) in directions as shown in the figure. Conserving momentum along horizontal, we get
\[mu = m\left\lbrack v_{2}cos\alpha + v_{1} \right\rbrack + {Mv}_{1}\]

\[\begin{array}{r} \Rightarrow mu = {mv}_{2}cos\alpha + (M + m)v_{1}\#(i) \end{array}\]

Since common normal is along \(y^{'}\), therefore momentum of ball remains constant along the incline (along \(x\) ' \(\because{\overrightarrow{F}}_{x^{'}} = 0\) )

\[\begin{array}{r} \Rightarrow \ ucos\alpha = v_{2} + v_{1}cos\alpha\#(ii) \end{array}\]

from equation (i) and (ii), we get

\[mu = mu\cos^{2}a - mv_{1}\cos^{2}a + (M + m)v_{1}\]

\[\Rightarrow \ v_{1} = \frac{mu\sin^{2}\alpha}{M + m\sin^{2}\alpha}\]

Q. 15 A missile of mass M moving with velocity \(v = 200\text{ }m/s\) explodes in midair breaking into two parts of mass \(M/4\& 3M/4\). If the smaller piece flies off at an angle of \(60^{\circ}\) with respect to the original direction of motion with an initial speed of \(400\text{ }m/s\), what is the magnitude and direction of the initial velocity of the other piece.

Sol.

by COLM along \(x\) axis :

\[\begin{matrix} & & 200M = 400 \times \frac{M}{4}cos60^{\circ} + \frac{3M}{4}\text{ }Vcos\theta \\ & \frac{3}{4}vcos\theta = 150\ \Rightarrow vcos\theta = 200 & \text{(i)} \end{matrix}\]

by COLM along \(y\) axis :

\[\begin{matrix} & & 0 = 400\frac{M}{4}sin60^{\circ} - \frac{3M}{4}\text{ }Vsin\theta \\ & & 3vsin\theta = \frac{400\sqrt{3}}{2} \\ & vsin\theta = \frac{200}{\sqrt{3}}\ \ldots\text{~}\text{(ii)}\text{~} & \text{(ii)} \end{matrix}\]

(ii) / (i) gives

\[\begin{matrix} & tan\theta = \frac{1}{\sqrt{3}} \\ \Rightarrow & \theta = 30^{\circ} \\ \Rightarrow & v = \frac{400}{\sqrt{3}}\text{ }m/s \end{matrix}\]