Question

A ball impinges directly upon another ball at rest and is itself reduced to rest by the impact. If half of the K.E. is destroyed in the collision, the coefficient of restitution, is

• A. $\frac{1}{4}$
• B. $\frac{1}{3}$
• C. $\frac{3}{4}$
• D. $\frac{1}{2}$

Answer 1

Answer: (D)

$\frac{1}{2}$

Answer 2

Let masses of the two balls be $m_{1}$ and $m_{2}$ .

By the given question, we have

$m_{1}$ $m_{2}$

$u_{1}$ $u_{2} = 0$

$v_{1} = 0$ $v_{2}$

$\therefore$ $m_{1}u_{1} + m_{2}.0 = m_{1}.0 + m_{2}v_{2}$ or $m_{1}u_{1} = m_{2}v_{2}$ .....(i) and $0 - v_{2} = e(0 - u_{1})$ i.e., $v_{2} = eu_{1}$ .....(ii)

Again for the given condition, K.E. (before impact)

= 2 × K.E. after impact

$\therefore$ $\frac{1}{2}m_{1}u_{1}^{2} + 0 = 2\left( 0 + \frac{1}{2}m_{2}v_{2}^{2} \right)$ .....(iii)

or $\frac{1}{2}m_{1}u_{1}^{2} = 2.\frac{1}{2}m_{2}v_{2}^{2}$

But from (i), $m_{1}u_{1} = m_{2}v_{2}$, $\therefore$ $\frac{1}{2}u_{1} = v_{2}$ i.e., $u_{1} = 2v_{2}$. Putting in (ii), we get $v_{2} = e.2v_{2}$, $\therefore$ $e = \frac{1}{2}$.