A ball moving with velocity `2 ms^(-1)` collides head on with another stationary ball of double the mass. If the coefficient of restitution is `0.5`, then their velocities (in `ms^(-1)`) after collision will be

Options

A. `0,1`
B. `1,1`
C. `1, 0.5`
D. `0, 2`

Answer

`(4)`

Explanation

Here, `m_(1) = m, m_(2)= 2m`
`u_(1) = 2m /s, u_(2) = 0`
Coefficient of restitution,` e = 0.5`.
Let `v_(1)` and `v_(2)` be their repective velocities after collision.
Applying the law of conservation of linear momentum, we get
`m_(1)u_(1) + m_(2)u_(2) = m_(1)v_(1) + m_(2)v_(2)`
`therefore m xx 2 + 2m xx 0 = mxx v_(1) + 2m xx v_(2)`
or `2m = mv_(1) + 2mv_(2)`
`or 2 = (v_(1) + 2v_(2)) ---(i)`
By definition of coefficient of restitution,
`e = (v_(2)-v_(1))/(u_(1) - u_(2))`
`or e(u_(1) - u_(2)) = v_(2) - v_(1)`
`0.5 (2-0) = v_(2) - v_(1) ----(ii)`
`1 = v_(2) - v_(1)`
Solving equations (i) and (ii), we get `v_(1) = 0m/s, v_(2) = 1 m/s`.