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A particle (A) of mass $m_{1}$ elastically collides with another stationary particle (B) of mass $m_{2}$ . then:

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If the collision is head on and $\frac{m_{1}}{m_{2}} = \frac{1}{2}$ and the particles fly a part in the opposite direction with equal speed.

If the collision is head on and $\frac{m_{1}}{m_{2}} = \frac{1}{3}$ and the particles fly apart in the opposite direction with equal speed..

If the collision is oblique and $\frac{m_{1}}{m_{2}} = \frac{2}{1}$ and the collision angle between the particles is $60^{°}$ symmetrically.

If the collision is oblique and $\frac{m_{1}}{m_{2}} = \frac{3}{1}$ and the particles fly apart symmetrically at an angle $90^{°}$

answer is B, C.

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Detailed Solution

$m_{1} u + 0 = m_{1} v_{1} + m_{2} v_{2}$ …….(1)

From kinetic energy conservation

$\frac{1}{2} m_{1} u^{2} = \frac{1}{2} m_{1} {v_{1}}^{2} + \frac{1}{2} m_{2} {v_{2}}^{2}$ …..(2)

Putting $v_{1} = v_{2} = v i n e q u a t i o n s$ (1) and (2) we get ; $\frac{m_{1}}{m_{2}} = \frac{1}{3}$

So, choice (2) is correct.

Again for oblique collision , $P^{2} = {P_{1}}^{2} + {P_{2}}^{2} + 2 P_{1} P_{2} cosθ$ ……..(3)

Also from conservation of kinetic energy

$\frac{P^{2}}{2 m_{1}} = \frac{{P_{1}}^{2}}{2 m_{1}} + \frac{{P^{2}}_{2}}{2 m_{2}}$ …….(4)

Since particles fly symmetrically

$P_{1} \sin \frac{θ}{2} = P_{2} \sin \frac{θ}{2}$ …….(5)

Solving eq. (3), (4) and (5) we get

$\frac{m_{1}}{m_{2}} = \frac{2}{1}$

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