Impact

  • To apply conservation of momentum in simple situations.

  • Describe the concept of coefficient of restitution

  • Apply the definition of the coefficient of restitution to compute the initial upward velocity of an object after impact.

  • To understand the basic ideas of elastic and inelastic collisions.

  • Apply linear momentum conservation to two-dimensional collisions.

\[m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'\] where,
\(m_1 , m_2\) = masses of two bodies
\(v_1, v_2\) = the velocities of two bodies just before impact
\(v_1', v_2'\) = the velocities of two bodies just after impact
The coefficient of restitution is defined as, \[e= \dfrac{\mathrm{Relative\ velocity\ of \ separation}}{\mathrm{Relative\ velocity\ of\ approach}}\]

\[e= \dfrac{(v_2')_n - (v_1')_n}{(v_1)_n - (v_2)_n}\]

momentum of the system is conserved in both elastic and inelastic collisions however; kinetic energy is conserved only in the elastic collisions.
If e = 1, perfectly elastic collision. In such case energy is also conserved and momentum is conserved.
If e = 0, perfectly plastic collision. In such case energy is not conserved, but momentum is conserved.

Solved Examples

Solved Example:

32-3-01

A ball is dropped from a height y above a smooth floor. How high will rebound if the coefficient of restitution between the ball and the floor is 0.60?

Solution:
When a ball is dropped from a height y, its velocity at the time of impact will be $\sqrt{2gy}$.

Since coefficient of restitution is not 1, some energy is lost and final velocity will not be same.

\[\mathrm{Coefficient\ of\ restitution} = \dfrac{\mathrm{Relative\ velocity\ of\ separation}}{\mathrm{Relative\ velocity\ of\ approach}}\] Ground remains stationary all the time.

\[0.6 = \dfrac{v'-0}{\sqrt{2gy}-0}\] \[v' = 0.6 \sqrt{2gy}\]
This becomes 'u' for the return journey.
Use newton's equation of motion. \begin{align*} v^2 &= u^2 + 2as\\ 0 &= 0.36 (2gy) + 2gs\\ s &= 0.36y\\ \end{align*}

Correct Answer: D

Solved Example:

32-3-02

A ball is thrown at an angle of 32.5$^\circ$ from the horizontal towards a smooth floor. At what angle will it rebound if the coefficient of between the ball and the floor is 0.30?

Solution:

$v^{2}=u^{2}+2as, v=4.43\sqrt {y}, e=\dfrac {v_{2}}{v_{1}}, v_{2}=2.656\sqrt {y}$

This becomes 'u' for return journey.

$v^{2}=u^{2}+2as, \mathrm{\ or,\ } s=0.36m$

Correct Answer: D

Solved Example:

32-3-03

A tennis ball is dropped into a cement floor from a height of 2 m. It rebounds to a height of 1.8 m. What fraction of energy did it lose in the process of striking the floor?

Solution:
Initial PE = mgh = mg(2)

Final PE = mgh = mg(1.8)

Energy lost = mg(0.2)

Fraction of Energy lost = $\dfrac{mg(0.2)}{mg(2)}$ = 0.1

Correct Answer: A

Solved Example:

32-3-04

The coefficient of restitution of a perfectly plastic impact is:

Solution:
From the Newton's Law of collision of Elastic bodies. Velocity of separation = e $\times$ Velocity of approach. \[(v_2- v_1) = e(u_1- u_2)\] Where e is a constant of proportionality and it is called the coefficient of restitution and its value lies between 0 to 1. The coefficient of restitution of a perfectly plastic impact is zero, because all the K.E. will be absorbed during perfectly plastic impact.

Correct Answer: A

Solved Example:

32-3-05

Collisions in which objects rebound with the same speed as they had prior to the collision are known as:

Solution:
If e = 1, perfectly elastic collision. In such case energy is also conserved and momentum is conserved. In this situation, the body rebounds with the same speed after hitting the ground.
On the other extreme, if e = 0, perfectly plastic collision. In such case energy is not conserved, but momentum is conserved. The body gets stuck and relative velocity of separation is zero.

Correct Answer: A