A 2 kg ball moving at `24 ms^(-1)` undergoes head on collision with a 4 kg ball moving in the opposite direction at `48 m s^(-1)`. If the coefficient of restitution is `2/3`, their velocities in `m s^(-1)` after collision are    

Options

A. `- 56, - 8`
B. `– 28, -4`
C. `– 14, -2 `
D. `- 7,-1 `

Answer

A

Explanation

Here , `m_1 = 2kg , m_2 = 4kg = 2m_1`
`u_1 = 24 ms^(-1) , u_2 = -48 ms^(-1) = -2u_1`
where `u_1` and `u_2` be the velocities of masses `m_1` and `m_2` before collision.
According to the law of conservation of linear momentum , we get
`m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2v_2 , m_1 u_1 + 2m_1(-2u_1) = m_1 v_1 + 2m_1 v_2`
`= -3 u_1 = v_1 + 2v_2 " ".........(i)`
Coefficient of restitution , `e = (v_2 - v_1)/(u_1 - u_2)`
`v_2 - v_1 = e(u_1 - u_2) = e(u_1 - (-2u_1))`
`v_2 - v_1 = 3u_1 e" " ....(ii)`
Adding (i) and (ii), we get
`3v_2 = 3u_1 e - 3u_1 " or " v_2 = u_1 (e - 1) " " .......(iii)`
Substituting this value of `v_2` in (i), we get
`v_1 = -u_1 (1 + 2e) " " ......(iv)`
Substiting the given values in (iii) and (iv), we get
`v_2 = 24 (2/3 - 1) = -8 ms^(-1)`
and `v_1 = -24(1 + 2 xx 2/3) = -56 ms^(-1)` .