Collisions: Definition, Types (Elastic, Inelastic), Examples, Problems

Collisions: We all know about the game called snooker. In snooker, the players hit the ball at different angles to get the ball to move in the desired direction. The physics behind this is the concept of collisions. In this article, we will learn about the Law of conservation of linear and angular momentum, impulse-momentum theorem. We will also see an interesting trick to solve problems related to the momentum by using the concept of centre of mass and relative velocity.

What is Collision?

Collision is a phenomenon in which two or more objects are moving such that at a particular point, they come in contact with each other and apply some force on each other, due to which the momentum of the objects changes. A collision can happen in one dimension, two- dimensions and in three dimensions.

Momentum and Angular Momentum

Momentum can be understood as the measure of how difficult it is to stop a particular particle. A particle with greater momentum will require a greater force to stop. It is a vector quantity, and its direction is in the direction of the velocity of the particle. The SI unit of the momentum is \(\mathrm{kg}\,\mathrm{m}\,{\rm{s}}^{-1}\).
There are two types of momentum.

1. Linear Momentum

The linear momentum is the product of mass and the velocity of the particle. The SI unit of the momentum is \(\mathrm{kg}\,\mathrm{m}\,{\rm{s}}^{-1}\). Linear momentum is given by,

\(p=mv\)

Where,

\(m\) is the mass of the particle

\(v\) is the velocity of the particle

2. Angular Momentum

Just as we have linear momentum for translational motion, we have angular momentum for the rotational or combined motion for the motion of rigid bodies. It is a vector quantity. The SI unit for angular momentum is \(\mathrm{kg}\,\mathrm{m}^{2}\,\mathrm{s}^{-1}\). The angular momentum of a particle about a point ‘\(o\)’ is given by,

\(L_{o}=I_{o} \omega_{o}\)

Where,

\(I_{0}\) is the moment of inertia of the particle about point ‘\(o\)’

\(\omega_{0}\) is the angular velocity of a particle about point ‘\(o\)’

Laws Associated With Momentum

The two laws associated with the momentum are as follows.

1. The law of conservation of linear momentum

If the net force on the system is zero, the net linear momentum of the system remains constant. The individual momentum of the particles of the system may change, but the total momentum will remain constant.

2. The law of conservation of angular momentum

If the net torque on the system about a point is zero, then that point’s angular momentum always remains conserved.

Impulse and Angular Impulse

When a cricket bat hits a ball, then the momentum of the ball changes due to the force applied by the bat. The time interval for which the bat was in contact with the ball is infinitely small, and in that short period, an infinitely large amount of force is applied. Even though the force was very large, tending to infinity, the effect of that force was finite and is known as an impulse.

\(J=\int F d t=m \Delta v\)

Where,

\(m\) is the mass of the particle.

\(\Delta v\) is the change in velocity of the particle.

\(F\) is the force on the particle.

\(dt\) is the duration for which force acts on the particle.

Similarly, we define angular Impulse as the net effect of a very large amount of torque when applied for an infinitesimal amount of time, and it is given by,

\(J_{o}=\int T_{o} d t=I_{o} \Delta \omega\)

Where,

\(I_{o}\) is the moment of inertia of the particle about any point ‘\(o\)’.

\(\Delta \omega\) is the change in angular velocity of the particle about any point ‘\(o\)’.

\(T_{o}\) is the torque on the particle about any point ‘\(o\)’.

\(dt\) is the duration for which torque acts on the particle.

Theorems Associated With Momentum

The two theorems associated with the momentum are as follows:

1. Linear impulse-momentum theorem

According to this theorem, the net in linear momentum is equal to the net Impulse applied to the system.

\(J=m \Delta v=\Delta p\)

2. Angular impulse-momentum theorem

According to this theorem, the net change in angular momentum is equal to the angular Impulse applied.

\(J_{0}=I_{o} \Delta \omega=\Delta L_{o}\)

Elastic and Inelastic Collisions

Coefficient of restitution: The ratio of relative velocity after the collision to the relative velocity before the collision is known as the coefficient of restitution. It is denoted by ’\(e\)’. Its value ranges from zero to one.

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

For a perfectly inelastic collision, the value of the coefficient of restitution is zero. That is, the relative velocity after the collision becomes zero, and both the particles stick to each other after the collision,

For a perfectly elastic collision, the value of the coefficient of restitution is one, and the velocity of approach is equal to the velocity of separation. The kinetic energy remains conserved in case of perfectly elastic collision.

For values other than zero or one for the coefficient of restitution, some amount of the kinetic energy is dissipated.

Role of the Centre of Mass

When the net force on the system is zero, the acceleration of the centre of mass is zero. The velocities of the individual particles of the system can change, but the velocity of the centre of mass remains constant. We can use this fact and the concept of the centre of mass to simplify the problems related to momentum.

If we make the centre of mass as the observer and change the velocities of particles relative to the centre of mass, then the centre of mass will behave as a stationary wall, and the particles of the system will appear as they are hitting a wall and returning back.

Sample Problems

Q.1. Two bodies of mass \(30\;\rm{kg}\) and \(20\;\rm{kg}\), are moving with the velocity \(10 \mathrm{~m}\,\rm{s}^{-1}\) and \(5\,\rm{m}\,\rm{s}^{-1}\) in the same direction. If the coefficient of restitution is \(0.5\) then,  find the final velocities of the bodies after the collision.
Ans: Given,
Masses of the bodies are, \(m_{1}=30 \mathrm{~kg}\) and \(m_{2}=20 \mathrm{~kg}\)
Velocities of the bodies are, \(v_{1}=10 \mathrm{~m}\,\rm{s}^{-1}\) and \(v_{2}=5\,\rm{m}\,\rm{s}^{-1}\)
The coefficient of restitution is, \(e=0.5\)
Let us solve this using the centre of mass trick.
The velocity of the centre of mass is, \(v_{c m}=\frac{30 \times 10+20 \times 5}{50}=8 \mathrm{~m}\,\rm{s}^{-1}\)
Velocity of the \(m_{1}\) relative to the centre of mass is, \(2\;\rm{m}\,\rm{s}^{-1}\)
Velocity of the \(m_{2}\) relative to the centre of mass is, \(-3\;\rm{m}\,\rm{s}^{-1}\)
Now, the centre of mass will behave as a rigid wall with the coefficient of restitution as \(0.5.\)
For \(m_{1}\)
\(e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}}\)
\(0.5=\frac{v_{-1}-0}{0-2}\)
\(\Rightarrow v_{m 1}=-1\;\rm{m}\,\rm{s}^{-1}\)
For \(m_{2}\),
\(e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}}\)
\(0.5=\frac{v_{t 2}-0}{0-(-3)}\)
\(\Rightarrow v_{m 2}=1.5\;\rm{m}\,\rm{s}^{-1}\)
\(v_{m 2}\) and \({v_{\rm{m1}}}\) are with respect centre of mass, so we will convert it with respect to the ground frame to get the final answer.
\(v_{1}=v_{m 1}+v_{\rm{c m}}\)
\(\Rightarrow v_{1}=-1+8=7\,\rm{m}\,\rm{s}^{-1}\)
\(v_{2}=v_{m 2}+v_{\rm{c m}}\)
\(\Rightarrow v_{2}=1.5+8=9.5 \mathrm{~m}\,\rm{s}^{-1}\)

Q.2. In the above problem, solve without using the centre of mass trick.
Ans: Given,
Masses of the bodies are, \(m_{1}=30 \mathrm{~kg}\) and \(m_{2}=20 \mathrm{~kg}\)
Velocities of the bodies are, \(u_{1}=10 \mathrm{~m}\,\rm{s}^{-1}\) and \(u_{2}=5 \mathrm{~m}\,\rm{s}^{-1}\)
The coefficient of restitution is, \(e=0.5\)

The final velocities is given by,
\(v_{1}=u_{1}\left(\frac{m_{1}-e m_{2}}{m_{1}+m_{2}}\right)+u_{2}\left(\frac{m_{2}(1+e)}{m_{1}+m_{2}}\right)\)
\(v_{2}=u_{1}\left(\frac{m_{1}(1+e)}{m_{1}+m_{2}}\right)+u_{2}\left(\frac{m_{2}-e m_{1}}{m_{1}+m_{2}}\right)\)
Putting in the values and solving we get,
\(v_{1}=10\left(\frac{30-0.5 \times 20}{30+20}\right)+5\left(\frac{20(1+0.5)}{30+20}\right)\)
\(\Rightarrow v_{1}=4+3=7 m s^{-1}\)
\(v_{2}=10\left(\frac{30(1+0.5)}{30+20}\right)+5\left(\frac{20-0.5 \times 30}{30+20}\right)\)
\(\Rightarrow v_{2}=9+0.5=9.5 \mathrm{~m}\,\rm{s}^{-1}\)

Summary

In the given article, we learnt about what is collisions. We discussed linear and angular momentum and the laws associated with the collision, such as Law of conservation of linear momentum and Law of conservation of angular momentum. We also introduced the new quantity called Impulse. Impulse is the net effect of a large quantity of force for an infinitesimal amount of time. We have two types of Impulse, linear and angular Impulse. The impulse-momentum theorem states that the change in momentum is equal to the net Impulse on the system.

When the momentum is conserved, the centre of mass has a constant velocity, and in the frame of the centre of mass, the centre of mass behaves as a stationary wall. This helps us to solve the problems related to momentum easily. The coefficient of restitution is g the ratio of the velocity of separation to the velocity of approach.

FAQs

Q.1.What is momentum?
Ans. Momentum can be understood as the measure of how difficult it is to stop a particular particle. A particle with greater momentum will require a greater force to stop.

Q.2. How is Newton second Law related to momentum?
Ans. The rate of change of momentum with respect to time gives us the magnitude of the force. Therefore, if the net force on the body is zero, then its momentum will remain constant.

Q.3. What is Impulse?
Ans. Impulse is the effect produced by an infinitely large force for an infinitely small time interval. The effect is measured in terms of change in momentum. Thus we can say that the net change in momentum of a body is equal to the Impulse.

Q.4. What is the direction of angular momentum?
Ans. The direction of angular momentum is given by the right-hand curl rule. That is, we curl the fingers of our right hand in the direction of rotation then our thumb points towards the direction of the angular momentum vector.

Q.5. What is the role of the centre of mass?
Ans. The centre of mass remains at rest if the momentum is conserved.

Q.6. Does the kinetic energy remain constant?
Ans. Kinetic energy remains constant only for elastic collision. For collisions in which the coefficient of restitution is more than zero, then some amount of kinetic energy is dissipated.