Question

A proton in motion makes head on collision with an unknown particle at rest. If the collision is perfectly elastic and proton rebounds back with $\frac{4}{9}$ of its initial kinetic energy after collision, the mass of unknown particle is

• A. Equal to mass of proton
• B. Twice the mass of proton
• C. $3$ times the mass of proton
• D. $5$ times the mass of proton

Answer 1

The correct option is D $5$ times the mass of proton

For perfectly elastic collision $e=1$

Let the initial velocity of the proton be $u_p$ and mass of the unknown particle be $m$.

Final velocity of the proton after collision $v_p=\frac{(m_p-em)u_p+(1+e)mu}{m_p+m}$

$\therefore$ $v_p=\frac{(m_p-m)u_p+(1+1)m\left(0\right)}{m_p+m}=\frac{(m_p-m)u_p}{m_p+m}$

Final kinetic energy of proton is $\frac{4}{9}$ of its initial kinetic energy.

$\therefore$ $\frac{1}{2}{m}_p{v}_p^2=\frac{4}{9}\times \frac{1}{2}{m}_p{u}_p^2$

Or $\frac{(m_p-m{)}^2{u}_p^2}{(m_p+m{)}^2}=\frac{4}{9}{u}_p^2$

Or $\frac{m_p-m}{m_p+m}=\frac{2}{3}$

Or $3m_p-3m=2m_p+2m$

$⟹$ $m_p=5m$