Question

Two particles of equal masses moving with same speed collide perfectly in-elastically. After the collision, the combined mass moves with half of the speed of the individual masses. The angle between the initial momenta of individual particle is .

• A. ${60}^{\circ }$
• B. ${90}^{\circ }$
• C. ${120}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

The correct option is C ${120}^{\circ }$

Magnitude of momentum of combined particles after collision is same as that of individual momentum

before collision. So $|\overrightarrow{P_1}|$ = $|\overrightarrow{P_2}|$ = $|\overrightarrow{P}|$ = $P_0$(say)

From conservation of linear momentum

$\overrightarrow{P_1}$ + $\overrightarrow{P_2}$ = $\overrightarrow{P}$

or $P_0$ = $\sqrt{P_1^2+P_2^2+2P_1{P}_{2}cos\theta }$

or $P_0$ = $\sqrt{P_0^2+P_0^2+2P_0^{2}cos\theta }$

or $cos\theta$ = $\frac{-1}{2}$ or $\theta$ = ${120}^{\circ }$