Question

A proton of mass m collides elastically with a particle of mass M (which is originally at rest). The track of proton is deviated by ${60}^o$. The struck particle (of mass M) follows a track at an angle of ${30}^0$ (with the incident proton direction). Find the value $(\frac{m}{M})$.

Answer 1

Before collision, let proton (m) have velocity u.
After scattering, let the velocity of proton be $v_1$ as shown.
similarly, the particle of mass M has velocity $v_2$ after collision.
By conversation of energy:
$\frac{1}{2}m{u}^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}M{v}_2^2$
Momentum is conseved both parallel and perpendicular to the original direction of travel of proton.
$\Rightarrow mu=m{v}_{1}cos{60}^0+M{v}_{2}cos{30}^0$
$m{v}_{1}sin{60}^0=M{v}_{2}sin{30}^0$
$\Rightarrow v_2=\frac{m{v}_{1}sin{60}^0}{Msin{30}^0}=\left(\sqrt{3}\right)\frac{m}{M}{v}_1$
hence,
$\frac{1}{2}m{u}^2=\frac{1}{2}m{v}_1^2+\frac{3}{2}\frac{m^2}{M}{v}_1^2$
$mu=\frac{1}{2}m{v}_1+\frac{3}{2}m{v}_1$
$u^2=v_1^2+\frac{3m}{M}{v}_1^2=v_1^2(1+\frac{3m}{M})$...........(i)
$u=\frac{1}{2}{v}_1+\frac{3}{2}{v}_1=v_1(\frac{1}{2}+\frac{3}{2})$
$\Rightarrow u^2=v_1^2{(\frac{1}{2}+\frac{3}{2})}^2$.......(ii)
Equate (i) and (ii)
$v_1^2{(\frac{1}{2}+\frac{3}{2})}^2=v_1^2(1+\frac{3m}{M})$
$\Rightarrow \frac{1}{4}+\frac{9}{4}+\frac{3}{2}=1+\frac{3m}{M}$
$\Rightarrow \frac{m}{M}=1$