Question

A ball of mass m moving with a constant velocity u strikes against a ball of same mass at rest. If e is the coefficient of restitution, then what will be a ratio of the velocity of two balls after collision?
(A) \[\dfrac{1-e}{1+e}\]
(B) \[\dfrac{e-1}{e+1}\]
(C) \[\dfrac{1+e}{1-e}\]
(D) \[\dfrac{e+1}{e-1}\]

Answer 1

Hint: Here a body is moving while another body is at rest. The masses are the same. We need to find the coefficient of restitution. So, it is clear we have to use the law of conservation of momentum which states that total momentum before the collision must be equal to the total momentum after the collision.
 \[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\].

Complete step by step solution:
Given, \[{{m}_{1}}={{m}_{2}}=m\]
\[{{u}_{1}}=u,\,\,{{u}_{2}}=0\]
Let, \[{{V}_{1}}\] = velocity of ball 1 after collision
\[{{V}_{2}}\] = velocity of ball 2 after collision
The coefficient of restitution,
\[e=\dfrac{{{v}_{2}}-{{v}_{1}}}{{{u}_{2}}-{{u}_{1}}}\]
\[eu={{v}_{2}}-{{v}_{1}}\] … (1)
By the conservation of Linear momentum,
\[{{m}_{1}}{{u}_{1}}+....+{{m}_{2}}{{u}_{2}}\]
\[{{m}_{1}}{{v}_{2}}+{{m}_{2}}{{u}_{2}}\]
\[u={{v}_{1}}+{{v}_{2}}\] … (2)
By solving equation (1) and (2), we get
\[{{v}_{1}}=\dfrac{\left( 1-e \right)u}{2}\]
\[{{v}_{2}}=\dfrac{\left( 1+e \right)u}{2}\]
The ratio of the velocity of two ball
\[\dfrac{{{v}_{1}}}{{{v}_{2}}}=\dfrac{1-e}{1+e}\]

Thus, the correct option is (A)

Additional information: The Super Ball has an almost perfect coefficient of restitution and does things other balls do not. The coefficient of restitution is not a material property, but depends on the severity of the impact.

Note: During such collision problems we also need to keep in mind whether the collision is in one dimension or two-dimension. Also, we have to see whether the collision is elastic or inelastic. In all types of collision, momentum is conserved, while in case of elastic collision kinetic energy is also conserved in addition to conservation of momentum.