Question

A ball of mass m moving with a speed `2v_0` collides head-on with an identical ball at rest. If e is the coefficient of restitution, then what will be the ratio of velocity of two balls after collision?
A. `(1-e)/(1+e)`
B. `(1+e)/(1-e)`
C. `(e-1)/(e+1)`
D. `(e+1)/(e-1)`

Answer 1

Correct Answer - A
According to law of conservation of linear momentum , we get
`m(2v_0)+mxx0=mv_1 + mv_2`
where `v_1` and `v_2` are the velocities of the balls after collision.
`v_1+v_2 =2v_0`..(i)
By definition of coefficient of restitution
`e=(v_2-v_1)/(u_1-u_2)=(v_2-v_1)/(2v_0)" " (because u_1=2v_0, u_2=0)`
`v_2-v_1=2ev_0`...(ii)
Adding (i) and (ii), we get
`2v_2-2v_0+2ev_0 , v_2=(1+e)v_0`
Subtract (ii) from (i), we get
`2v_1 =2v_0 - 2ev_0 , v_1=v_0 (1-e)`
Their corresponding ratio is
`v_1/v_2=(1-e)/(1+e)`