Q 11-05-194JEE MainJEE Main 2018 (15 Apr, Shift 2)Easy
A proton of mass $m$ collides elastically with a particle of unknown mass at rest. After the collision, the proton and the unknown particle are seen moving at an angle of $90^\circ$ with respect to each other. The mass of unknown particle is:
Answer: (D) $m$
Let the unknown mass be $M$. Momentum: $m\vec u = m\vec v_1 + M\vec v_2$. Kinetic energy: $\frac12mu^2 = \frac12mv_1^2 + \frac12Mv_2^2$.
Squaring the momentum equation, with $\vec v_1\perp\vec v_2$:
$$m^2u^2 = m^2v_1^2 + M^2v_2^2$$
The energy equation times $m$ gives $m^2u^2 = m^2v_1^2 + mMv_2^2$. Comparing, $M^2 = mM$, so
$$M = m$$
Solution by Sreeraj P, M.Sc Physics