Question

A ball moving with a speed of $9n{s}^{-1}$ strikes an identical stationary ball makes that after the collision the direction of each ball makes an angle of ${30}^0$ with the original line of motion. The speed of two balls after collision is

• A. $3m{s}^{-1},3m{s}^{-1}$
• B. $3\sqrt{3}m{s}^{-1},3\sqrt{3}m{s}^{-1}$
• C. $3\sqrt{3}m{s}^{-1},3m{s}^{-1}$
• D. $3m{s}^{-1},3\sqrt{3}m{s}^{-1}$

Answer 1

The correct option is B $3\sqrt{3}m{s}^{-1},3\sqrt{3}m{s}^{-1}$

As the masses & angles are same for both the balls so their speeds should be same $\left(v\right)$

Before cullision

ball $-1$ ball $-2$

$u_1=9m{s}^{-1}$ $u_2=0$

So, Initial momentum can be given as:
$p_i=m{u}_1=9m$

After cullision,
The momentum along the $x$ direction, $p_x=2mvcos{30}^o$

Momentum along $Y$, $p_y=v\sin {30}^o+(-v\sin {30}^o)=0$

By using conservation of momentum,
Initial Momentum = final momentum after the collision
$Pi=\sqrt{p_x^2+p_y^2}$
$\Rightarrow 9m=2mv\times \frac{\sqrt{3}}{2}$

$\Rightarrow v=3\sqrt{3}m{s}^{-1}$
Hence, the correct option is $\left(B\right)$