Question

A particle of mass $m_{1}$ collides with a particle of $m_{2}$ at rest. After the elastic collision, the two particles moves at an angle of $90^{\circ}$ with respect to each other. The ratio $\frac{m_{2}}{m_{1}}$ is

• A. 1.0
• B. 1.5
• C. 2.0
• D. 2.5

Answer 1

Answer: (A)

According to given situation,

Momentum conservation gives,
$p_{1}'+p_{2}'=p_{1} \ldots \ldots$ (i)
KE conservation gives,
$\frac{p' 2}{m_{1}}+\frac{p'{ }_{2}^{2}}{m_{2}}=\frac{p_{1}^{2}}{m_{1}}$
$\left[\therefore k=\frac{p^{2}}{2 m}\right] \ldots $(ii)
Substituting the value of $p$ from Eq. (i) into Eq. (ii), we have
$\frac{p_{1}' 2}{m_{1}}+\frac{p_{2}' 2}{m_{2}}=\frac{1}{m_{1}}\left(p_{1}'+p_{2}'\right)^{2}$
$\frac{p_{1}' 2}{m_{1}}+\frac{p_{2}^{'2}}{m_{2}}=\frac{p_{1}' 2}{m_{1}}+\frac{p_{2}^{'2}}{m_{1}}+\frac{2 p_{1}' p_{2}'}{m_{1}}$
As particles fly off at $90^{\circ}$ after collision,
$\therefore p_{1} \cdot p_{2}=0$
$p_{1} \cdot p_{2} \cos \theta=0$
$ \Rightarrow p_{1} p_{2}=0$
$\Rightarrow \frac{p' 2}{m_{2}}=\frac{p_{2}' 2}{m_{1}}$
or $\frac{m_{2}}{m_{1}}=1$