Question

A particle of mass $m$ moving with velocity $1\text{m/s}$ collides elastically with another particle of mass $2m$ (initialy at rest). If the incident particle is deflected by ${90}^{\circ }$. The angle $\theta$ made by the heavier mass with the initial direction of motion of mass $m$ will be equal to:
(consider collision to be perfectly elastic)

• A. ${60}^{\circ }$
• B. ${45}^{\circ }$
• C. ${15}^{\circ }$
• D. ${30}^{\circ }$

Answer 1

The correct option is D ${30}^{\circ }$

During collison, $F_{\text{ext}}=0$ on the system of particles.
Applying conservation of momentum along incident direction,
$(m\times 1)+0=0+2m{v}_2\text{cos}\theta$
$1=2v_2\cos \theta ....\left(1\right)$
Conservation of momentum along perpendicular direction, considering upwards as $+ve-$direction.
$(m\times 0)+(2m\times 0)=m{v}_1-2m{v}_2\text{sin}\theta$
$\Rightarrow v_1=2v_2\sin \theta ....\left(2\right)$

From energy conservation during perfectly elastic collision,
$\frac{1}{2}m\times (1{)}^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}2m{v}_2^2$
$\Rightarrow v_1^2+2v_2^2=1....\left(3\right)$
Squaring and adding Eq. $\left(1\right)$ and Eq. $\left(2\right)$,
$\Rightarrow 1+v_1^2=4v_2^2({\sin }^2\theta +{\cos }^2\theta )$
$\Rightarrow 1+v_1^2=4v_2^2....\left(4\right)$
From Eq.$\left(3\right)$ and Eq.$\left(4\right)$,
$v_2^2=\frac{2}{6}=\frac{1}{3}$
$\Rightarrow v_2=\frac{1}{\sqrt{3}}$
and $v_1=\sqrt{\frac{4}{3}-1}=\frac{1}{\sqrt{3}}$
From Eq.$\left(2\right)$,
$\Rightarrow \sin \theta =\frac{v_1}{2v_2}=\frac{1}{2}$
$\therefore \theta ={30}^{\circ }$