6.12 Collisions

COLLISIONS

In this section, we shall apply these laws to a commonly encountered phenomena, namely collisions. Several games such as billiards, marbles, or carrom involve collisions. We shall study the collision of two masses in an idealized form. Consider two masses \(m_{1}\) and \(m_{2}\). The particle \(m_{1}\) is moving with speed \(v_{1 i}\), the subscript ‘ \(i\) ‘ implying initial. We can consider \(m_{2}\) to be at rest. No loss of generality is involved in making such a selection. In this situation, the mass \(m_{1}\) collides with the stationary mass \(m_{2}\) and this is depicted in Figure 6.8.
The masses \(m_{1}\) and \(m_{2}\) fly-off in different directions. We shall see that there are relationships, which connect the masses, the velocities, and the angles.

Elastic and Inelastic Collisions

In all collisions the total linear momentum is conserved; the initial momentum of the system is equal to the final momentum of the system. One can argue this as follows. When two objects collide, the mutual impulsive forces acting over the collision time \(\Delta t\) cause a change in their respective momenta :
\(
\begin{array}{l}
\Delta \vec{p}_{1}=\vec{F}_{12} \Delta t \\
\Delta \vec{p}_{2}=\vec{F}_{21} \Delta t
\end{array}
\)
where \(\vec{F}_{12}\) is the force exerted on the first particle by the second particle. \(\vec{F}_{21}\) is likewise the force exerted on the second particle by the first particle. Now from Newton’s third law, \(\vec{F}_{12}=-\vec{F}_{21}\). This implies
\(\Delta \vec{p}_{1}+\Delta \vec{p}_{2}=\mathbf{0}\)
The above conclusion is true even though the forces vary in a complex fashion during the collision time \(\Delta t\). Since the third law is true at every instant, the total impulse on the first object is equal and opposite to that on the second.

On the other hand, the total kinetic energy of the system is not necessarily conserved. The impact and deformation during the collision may generate heat and sound. Part of the initial kinetic energy is transformed into other forms of energy. A useful way to visualize the deformation during collision is in terms of a ‘compressed spring’. If the ‘spring’ connecting the two masses regains its original shape without loss in energy, then the initial kinetic energy is equal to the final kinetic energy but the kinetic energy during the collision time \(\Delta t\) is not constant. Such a collision is called an elastic collision. On the other hand, the deformation may not be relieved and the two bodies could move together after the collision. A collision in which the two particles move together after the collision is called a completely inelastic collision. The intermediate case where the deformation is partly relieved and some of the initial kinetic energy is lost is more common and is appropriately called an inelastic collision.

ELASTIC COLLISION IN ONE DIMENSION

Consider two elastic bodies \(A\) and \(B\) moving along the same line (Figure 6.9). The body \(A\) has a mass \(m_{1}\) and moves with a velocity \(v_{1}\) towards right and the body \(B\) has a mass \(m_{2}\) and moves with a velocity \(v_{2}\) in the same direction. We assume \(v_{1}>v_{2}\) so that the two bodies may collide. Let \(v_{1}^{\prime}\) and \(v_{2}^{\prime}\) be the final velocities of the bodies after the collision. The total linear momentum of the two bodies remains constant, so that,
\(\begin{aligned}
& m_{1} v_{1}+m_{2} v_{2} &=m_{1} v_{1}^{\prime}+m_{2} v_{2}^{\prime} \dots (i)\\
\text { or, } & m_{1} v_{1}-m_{1} v_{1}^{\prime} =m_{2} v_{2}^{\prime}-m_{2} v_{2} \\
\text { or, } & m_{1}\left(v_{1}-v_{1}^{\prime}\right) &=m_{2}\left(v_{2}^{\prime}-v_{2}\right) \dots (ii)
\end{aligned}\)
Also, since the collision is elastic, the kinetic energy before the collision is equal to the kinetic energy after the collision. Hence,
\(
\begin{aligned}
\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2} &=\frac{1}{2} m_{1} v_{1}^{\prime 2}+\frac{1}{2} m_{2} v_{2}^{\prime 2} \\
\text { or, } \quad m_{1} v_{1}^{2}-m_{1} v_{1}^{\prime 2} &=m_{2} v_{2}^{\prime 2}-m_{2} v_{2}^{2}
\end{aligned}
\)

Figure 6.9

\(\begin{aligned} \text { or, } \quad m_{1}\left(v_{1}^{2}-v_{1}^{\prime 2}\right) &=m_{2}\left(v_{2}^{\prime 2}-v_{2}^{2}\right) \dots (iii) \\ \text { Dividing (iii) by (ii), } \\ v_{1}+v_{1}^{\prime} &=v_{2}^{\prime}+v_{2} \\ v_{1}-v_{2} &=v_{2}^{\prime}-v_{1}^{\prime} \dots (iv) \end{aligned}\)
Now, \(\left(v_{1}-v_{2}\right)\) is the rate at which the separation between the bodies decreases before the collision. Similarly, \(\left(v_{2}^{\prime}-v_{1}^{\prime}\right)\) is the rate of increase of separation after the collision. So the equation (iv) may be written as

The velocity of separation (after collision) = Velocity of approach (before the collision)…. (6.12)

This result is very useful in solving problems involving elastic collision. The final velocities \(v_{1}^{\prime}\) and \(v_{2}^{\prime}\) may be obtained from equation (i) and (iv). Multiply equation (iv) by \(m_{2}\) and subtract from equation (i).
\(
\begin{array}{ll}
2 m_{2} v_{2}+\left(m_{1}-m_{2}\right) v_{1}=\left(m_{1}+m_{2}\right) v_{1}^{\prime} \\
\text { or, } \quad v_{1}^{\prime}=\frac{\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}} v_{1}+\frac{2 m_{2}}{m_{1}+m_{2}} v_{2}  \quad \ldots \quad(6.13)
\end{array}
\)
Now multiply equation (iv) by \(m_{1}\) and add to equation (i),
\(
\begin{array}{c}
2 m_{1} v_{1}-\left(m_{1}-m_{2}\right) v_{2}=\left(m_{2}+m_{1}\right) v_{2}^{\prime} \\
\text { or, } \quad v_{2}^{\prime}=\frac{2 m_{1} v_{1}}{m_{1}+m_{2}}-\frac{\left(m_{1}-m_{2}\right) v_{2}}{m_{1}+m_{2}}  \quad \ldots \quad(6.14)
\end{array}
\)
Equations (6.13) and (6.14) give the final velocities in terms of the initial velocities and the masses.

Special cases :
Case-1: Elastic collision between a heavy body and a light body:

Let \(m_{1} \gg m_{2}\). A heavy body hits a light body from behind. We have,
\(
\begin{array}{ll}
\text { and } \quad & \frac{m_{1}-m_{2}}{m_{1}+m_{2}} \approx 1, \frac{2 m_{2}}{m_{1}+m_{2}} \approx 0 \\
& \frac{2 m_{1}}{m_{1}+m_{2}} \approx 2 .
\end{array}
\)
With these approximations, the final velocities of the bodies are, from (6.13) and (6.14),
\(v_{1}^{\prime} \approx v_{1} \text { and } v_{2}^{\prime} \approx 2 v_{1}-v_{2} \)
The heavier body continues to move with almost the same velocity. If the lighter body were kept at rest \(v_{2}=0, v_{2}{ }^{\prime}=2 v_{1}\) which means the lighter body, after getting a push from the heavier body will fly away with a velocity double the velocity of the heavier body.

Next, if we consider \(m_{2} \gg m_{1}\). A light body hits a heavy body from behind. We have,
The final velocities of the bodies are, from (6.13) and (6.14),
\(v_{1}^{\prime} \approx-v_{1}+2 v_{2} \text { and } v_{2}^{\prime} \approx v_{2} \text {. }\)
The heavier body continues to move with almost the same velocity, the velocity of the lighter body changes. If the heavier body were at rest, \(v_{2}=0\) then \(v_{1}^{\prime}=-v_{1}\), the lighter body returns after collision with almost the same speed. This is the case when a ball collides elastically with a fixed wall and returns with the same speed.

Case-2: Elastic collision of two bodies of equal mass:

Putting \(m_{1}=m_{2}\) in equation (6.13) and (6.14)
\(v_{1}{ }^{\prime}=v_{2} \text { and } v_{2}{ }^{\prime}=v_{1} \)
When two bodies of equal mass collide elastically, their velocities are mutually interchanged.

PERFECTLY INELASTIC COLLISION IN ONE DIMENSION

Final Velocity
When perfectly inelastic bodies moving along the same line collide, they stick to each other. Let \(m_{1}\) and \(m_{2}\) be the masses, \(v_{1}\) and \(v_{2}\) be their velocities before the collision and \(V\) be the common velocity of the bodies after the collision. By the conservation of linear momentum,
\(
\begin{aligned}
m_{1} v_{1}+m_{2} v_{2} &=m_{1} V+m_{2} V \\
\text { or, } \quad V &=\frac{m_{1} v_{1}+m_{2} v_{2}}{m_{1}+m_{2}} \dots (6.15)
\end{aligned}
\)

Loss in Kinetic Energy
The kinetic energy before the collision is
\(
\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}
\)
and that after the collision is \(\frac{1}{2}\left(m_{1}+m_{2}\right) V^{2}\). Using equation (i), the loss in kinetic energy due to the collision is
\(
\begin{aligned}
& \frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}-\frac{1}{2}\left(m_{1}+m_{2}\right) V^{2} \\
=& \frac{1}{2}\left[m_{1} v_{1}^{2}+m_{2} v_{2}^{2}-\frac{\left(m_{1} v_{1}+m_{2} v_{2}\right)^{2}}{m_{1}+m_{2}}\right]
\end{aligned}
\)

\(
\begin{array}{l}
=\frac{1}{2}\left[\frac{m_{1} m_{2}\left(v_{1}^{2}+v_{2}^{2}-2 v_{1} v_{2}\right)}{m_{1}+m_{2}}\right] \\
=\frac{m_{1} m_{2}\left(v_{1}-v_{2}\right)^{2}}{2\left(m_{1}+m_{2}\right)} .
\end{array}
\)
We see that the loss in kinetic energy is positive.

Example 6.18:

A cart A of mass \(50 \mathrm{~kg}\) moving at a speed of \(20 \mathrm{~km} / \mathrm{h}\) hits a lighter cart \(B\) of mass \(20 \mathrm{~kg}\) moving towards it at a speed of \(10 \mathrm{~km} / \mathrm{h}\). The two carts cling to each other. Find the speed of the combined mass after the collision.

Solution: This is an example of an inelastic collision. As the carts move towards each other, their momenta have opposite signs. If the common speed after the collision is \(V\), momentum conservation gives
\(
\begin{array}{c}
(50 \mathrm{~kg})(20 \mathrm{~km} / \mathrm{h})-(20 \mathrm{~kg})(10 \mathrm{~km} / \mathrm{h})=(70 \mathrm{~kg}) V \\
V=\frac{80}{7} \mathrm{~km} / \mathrm{h}
\end{array}
\)

COEFFICIENT OF RESTITUTION

We have seen that for a perfectly elastic collision velocity of separation = velocity of approach and for a perfectly inelastic collision, velocity of separation \(=0\).
In general, the bodies are neither perfectly elastic nor perfectly inelastic. In that case we can write
velocity of separation \(=e\) (velocity of approach),
where \(0<e<1\). The constant \(e\) depends on the materials of the colliding bodies. This constant is known as the coefficient of restitution. If \(e=1\), the collision is perfectly elastic and if \(e=0\), the collision is perfectly inelastic.

Example 6.19:

A block of mass \(m\) moving at speed \(v\) collides with another block of mass \(2 \mathrm{~m}\) at rest. The lighter block comes to rest after the collision. Find the coefficient of restitution.

Solution: Suppose the second block moves at speed \(v^{\prime}\) towards right after the collision. From the principle of conservation of momentum,
\(
m v=2 m v^{\prime} \text { or } v^{\prime}=v / 2 .
\)
Hence, the velocity of separation \(=v / 2\) and the velocity of approach \(=v\). By definition,
\(
e=\frac{\text { velocity of the separation }}{\text { velocity of approach }}=\frac{v / 2}{v}=\frac{1}{2} .
\)

ELASTIC COLLISION IN TWO DIMENSIONS

Consider two objects \(A\) and \(B\) of mass \(m_{1}\) and \(m_{2}\) kept on the \(X\)-axis (Figure 6.10). Initially, the object \(B\) is at rest and \(A\) moves towards \(B\) with a speed \(u_{1}\). If the collision is not head-on (the force during the collision is not along the initial velocity), the objects move along different lines. Suppose the object \(A\) moves with a velocity \(\overrightarrow{v_{1}}\) making an angle \(\theta\) with the \(X\)-axis and the object \(B\) moves with a velocity \(\overrightarrow{v_{2}}\) making an angle \(\Phi\) with the same axis. Also, suppose \(\overrightarrow{v_{1}}\) and \(\overrightarrow{v_{2}}\) lie in \(X-Y\) plane. Using conservation of momentum in \(X\) and \(Y\) directions, we get
\(
\text { and } \quad \begin{aligned}
m_{1} u_{1} &=m_{1} v_{1} \cos \theta+m_{2} v_{2} \cos \Phi & \ldots & \text { (i) } \\
0 &=m_{1} v_{1} \sin \theta-m_{2} v_{2} \sin \Phi . & \ldots & \text { (ii) }
\end{aligned}
\)
If the collision is elastic, the final kinetic energy is equal to the initial kinetic energy. Thus,
\(
\frac{1}{2} m_{1} u_{1}^{2}=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2} \dots (iii)
\)
We have four unknowns \(v_{1}, v_{2}, \theta\) and \(\Phi\) to describe the final motion whereas there are only three relations. Thus, the final motion cannot be uniquely determined with this information.

Figure 6.10
In fact, the final motion depends on the angle between the line of force during the collision and the direction of the initial velocity. The momentum of each object must be individually conserved in the direction perpendicular to the force. The motion along the line of force may be treated as a one-dimensional collision.

IMPULSE AND IMPULSIVE FORCE

When two bodies collide, they exert forces on each other while in contact. The momentum of each body is changed due to the force on it exerted by the other. On an ordinary scale, the time duration of this contact is very small and yet the change in momentum is sizeable. This means that the magnitude of the force must be large on an ordinary scale. Such large forces acting for a very short duration are called impulsive forces. The force may not be uniform while the contact lasts.

The change in momentum produced by such an impulsive force is
\(
\vec{P}_{f}-\vec{P}_{i}=\int_{P_{i}}^{P_{f}} d \vec{P}=\int_{t_{i}}^{t_{f}} \frac{d \vec{P}}{d t} d t=\int_{t_{i}}^{t_{f}} \vec{F} d t
\)

Figure 6.11
This quantity \(\int_{t_{i}}^{t_{f}} \vec{F} d t\) is known as the impulse of the force \(\vec{F}\) during the time interval \(t_{i}\) to \(t_{f}\) and is equal to the change in the momentum of the body on which it acts. Obviously, it is the area under the \(F-t\) curve for one-dimensional motion.