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# Ball 1 collides with an another identical ball 2 at rest For what values of coefficient of restitution e the velocity of 2nd ball becomes two times of ball 1 after collision?

$(a)\;\frac{1}{3} \quad (b)\;\frac{1}{2} \quad (c)\;\frac{1}{4} \quad (d)\;\frac{1}{6}$

Answer: $\large\frac{1}{3}$
A ball 1 moving with velocity u collides with an identical ball 2 at rest, then after collision
$v_2=\bigg(\large\frac{1+e}{2}\bigg)$$u and v_1=\bigg(\large\frac{1-e}{2}\bigg)$$u$, where $u$ is initial speed of ball 1
Given: $v_2=2 v_1$, solving for $e$, we get:
$v_2=\bigg(\large\frac{1+e}{2}\bigg)$$u = 2v_1=2\bigg(\large\frac{1-e}{2}\bigg)$$u$
$e=\large\frac{1}{3}$
edited Aug 27, 2014