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 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*}

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.

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

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

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.