Question

A particle of mass $m$ moving with velocity $1 \, m \, s^{- 1}$ collides perfectly elastically with another stationary particle of mass $2 \, m$ . If the incident particle is deflected by $90°.$ The heavy mass will make an angle $\theta $ with the initial direction of $m$ equal to

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

Answer 1

Answer: (D)

Conservation of momentum in initial direction of motion of $m$ (let $x-$ direction)
$\left(m \times 1\right)\hat{i}=\left(2 m v\right)\hat{i}$
$v_{x}=\left(\frac{1}{2}\right)\hat{i} \, ms^{- 1}$
$0=\left(m v\right)\hat{j}+\left(2 m v_{y}\right)\hat{j}$
$v_{y}=-\frac{v}{2}$
$KE_{B e f o r e}=KE_{A f t e r}$
$\frac{1}{2}m\left(1\right)^{2}+0=\frac{1}{2}mv^{2}+\frac{1}{2}2m\left(v_{x}^{2} + v_{y}^{2}\right)$
$1=v^{2}+2\left[\left(\frac{1}{2}\right)^{2} + \left(- \frac{v}{2}\right)^{2}\right]$
$1=v^{2}+\frac{1}{2}+\frac{v^{2}}{2}$
$\frac{1}{2}=\frac{3 v^{2}}{2}$
$v=\frac{1}{\sqrt{3}}ms^{- 1}$
Let $v$ be the speed of $m$ after it is deflected by $90^{o}$ .
Conservation of momentum in perpendicular direction of motion of $m$ (let $y-$ direction)
As the collision is elastic therefore,
$tan \theta =\frac{v_{y}}{v_{x}}$
$tan \theta =\frac{\frac{v}{2}}{\frac{1}{2}}=v$
$tan \theta =\frac{1}{\sqrt{3}}$
$\theta =30^{o}$