Question

A ball A moving with a velocity 5 m/s collides elastically with another identical ball at rest such that the velocity of A makes an angle of ${30}^{\circ }$ with the line joining the centres of the balls. Then:

• A. Speed of B after collision is $\frac{5}{2}m/s$
• B. Speed of A after collision is $\frac{5\sqrt{3}}{2}$ m/s
• C. Balls $A$ and $B$ move at right angles after collision
• D. Kinetic energy is not conserved as the collision is not head-on

Answer 1

Answer: (A), (B), (C)

Balls $A$ and $B$ move at right angles after collision
Speed of B after collision is $\frac{5}{2}m/s$
Speed of A after collision is $\frac{5\sqrt{3}}{2}$ m/s

As the collision is elastic so kinetic energy, as well as momentum both, are conserved.
For momentum conservation,
$mv=m{v}_A\cos 30+m{v}_B\cos \theta \Rightarrow v=v_A\cos 30+v_B\cos \theta ....\left(1\right)$

$0=m{v}_A\sin 30-m{v}_B\sin \theta \Rightarrow 0=v_A\sin 30-v_B\sin \theta .....\left(2\right)$

For KE conservation,

$\frac{1}{2}m{v}^2=\frac{1}{2}m(v_A^2{\cos }^{2}30+v_B^2{\cos }^2\theta )$

$v^2=v_A^2{\cos }^{2}30+v_B^2{\cos }^2\theta ....\left(3\right)$ and

$0=v_A^2{\sin }^{2}30+v_B^2{\sin }^2\theta .....\left(4\right)$

$\left(3\right)+\left(4\right),v^2=v_A^2+v_B^2....\left(5\right)$

Now squaring (1) and (2) and then adding, $v^2=v_A^2+v_B^2+v_A{v}_B\cos (30+\theta )$

Using (5), $\cos (30+\theta )=0=\cos 90\Rightarrow \theta =90-30={60}^o$

Thus, A and B will move at the right angle after collision.

Putting $\theta =60$ in (1) and (2), $v=v_A\cos 30+v_B\cos 60\Rightarrow 5=\frac{\sqrt{3}}{2}{v}_A+\frac{v_B}{2}...\left(6\right)$

and $0=\frac{v_A}{2}-\frac{\sqrt{3}}{2}{v}_B...\left(7\right)$

Solving (6) and (7), $v_A=\frac{5\sqrt{3}}{2}m/s$ and $v_B=\frac{5}{2}m/s$

The angle between the ball after collision

$\varphi =\theta +{30}^{\circ }={60}^{\circ }+{30}^{\circ }={90}^{\circ }$

Hence :

Option A,B and C