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. $\frac{1-e}{1+e}$
• B. $\frac{1+e}{1-e}$
• C. $\frac{e-1}{e+1}$
• D. $\frac{e+1}{e-1}$

Answer 1

Answer: (A)

$\frac{1-e}{1+e}$

According to the law of conservation of linear momentum, we get
$m\left(2v_0\right)+m\times 0=m{v}_1+m{v}_2$
where $v_1$ and $v_2$ are the velocities of the balls after collision.
$v_1+v_2=2v_0....\left(i\right)$
By definition of coefficient of restitution
$e=\frac{v_2-v_1}{u_1-u_2}=\frac{v_2-v_1}{2v_0}(\because u_1=2v_0,u_2=0)$
Adding (i) and (ii), we get
$2v_2=2v_0+2v_0;v_2=(1+e)v_0$
Subtract (ii) from (i), we get
$2v_1=2v_0-2e{v}_0;v_1=v_0(1-e)$
Their corresponding ratio is
$\frac{v_1}{v_2}=\frac{1-e}{1+e}$.