Question

A ball of mass $m$ moving with a speed $u$ collides elastically with another identical ball moving with velocity $u_2$. Find the angle between velocities after collision if they collide obliquely and $u_2$ = 0​

• A. $\pi$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{4}$
• D. zero

Answer 1

The correct option is B

$\frac{\pi }{2}$

Let velocity of the first ball after collision be $v_1$ making an angle ${\theta }_1$ with the horizontal anti-clockwise and the velocity of second ball be $v_2$ making an angle ${\theta }_2$ with the horizontal clockwise.

$mu$ = $m{v}_{1}cos{\theta }_1+m{v}_{2}cos{\theta }_2$ ............(i)

0 = $m{v}_{1}sin{\theta }_1-m{v}_{2}sin{\theta }_2$ ............(ii)

$\frac{1}{2}m{u}^2$ = $\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2$ ...........(iii)

Squaring and adding (i) and (ii), we get: $u^2$ = $v_1^2+v_2^2+2v_1{v}_{2}cos({\theta }_1+{\theta }_2)$

Using (iii), we have : $u^2$ = $v_1^2+v_2^2$
$\therefore v_1^2+v_2^2=v_1^2+v_2^2+2v_1{v}_{2}cos({\theta }_1+{\theta }_2)\Rightarrow 2v_1{v}_{2}cos({\theta }_1+{\theta }_2)=0cos({\theta }_1+{\theta }_2)=0{\theta }_1+{\theta }_2={90}^{\circ }=\frac{\pi }{2}$