A ball of mass $m$ moving at a speed $v$ makes a head-on collision with an identical ball at rest. The kinetic energy of the balls after the collision is three fourths of the original. Find the coefficient of restitution.

Open in App

Solution

$M a s s$ $o f$ $1^{s t}$ $b a l l = m$ $a n d$ $s p e e d$ $= v$ $M a s s$ $o f$ $2^{n d}$ $b a l l$ $= m$
$L e t$ $f i n a l$ $v e l o c i t i e s$ $o f$ $1^{s t}$ $a n d$ $2^{n d}$ $b a l l$ $a r e$ $v_{1}$ $a n d$ $v_{2}$ $r e s p e c t i v e l y$ $U s i n g$ $l a w$ $o f$ $c o n s e r v a t i o n$ $o f$ $m o m e n t u m,$
$m (v_{1} + v_{2})$ $= m v$
$\Rightarrow$ $v_{1} + v_{2}$ $= v$ $. . . (i)$
$A l s o$
$v_{1} - v_{2} = e v$ $. . . . (i i)$
$G i v e n$ $t h a t$
$f i n a l$ $K . E .$ $=$ $\frac{3}{4}$ $I n i t i a l$ $K . E .$
$\frac{1}{2}$ $m v_{1}^{2}$ $+$ $\frac{1}{2}$ $m v_{2}^{2}$ $=$ $\frac{3}{4} \times$ $\frac{1}{2} m v^{2}$
$v_{1}^{2} + v_{2}^{2} =$ $\frac{3}{4} v^{2}$
$\frac{(v_{1} + v_{2})^{2} + (v_{1} - v_{2})^{2}}{2}$ $=$ $\frac{3}{4} v^{2}$
$\frac{(1 + e^{2})}{2} v^{2} = \frac{3}{4} v^{2}$
$1 + e^{2} = \frac{3}{4}$
$e^{2} = \frac{1}{2}$
$e = \frac{1}{\sqrt{2}}$

Suggest Corrections

Similar questions

m

v

\frac{3}{4}

e

v_{1}, v_{2}

m

2 v_{0}

e

5 kg

4 m/s

Join BYJU'S Learning Program