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,
$p_1^{\prime }+p_2^{\prime }=p_1\dots \dots$ (i)
KE conservation gives,
$\frac{p^{\prime }2}{m_1}+\frac{p^{\prime }{}_2^2}{m_2}=\frac{p_1^2}{m_1}$
$\left[\therefore k=\frac{p^2}{2m}\right]\dots$(ii)
Substituting the value of $p$ from Eq. (i) into Eq. (ii), we have
$\frac{p_1^{\prime }2}{m_1}+\frac{p_2^{\prime }2}{m_2}=\frac{1}{m_1}{\left(p_1^{\prime }+p_2^{\prime }\right)}^2$
$\frac{p_1^{\prime }2}{m_1}+\frac{p_2^{{}^{\prime }2}}{m_2}=\frac{p_1^{\prime }2}{m_1}+\frac{p_2^{{}^{\prime }2}}{m_1}+\frac{2p_1^{\prime }{p}_2^{\prime }}{m_1}$
As particles fly off at $90^{\circ }$ after collision,
$\therefore p_1\cdot p_2=0$
$p_1\cdot p_2\cos \theta =0$
$\Rightarrow p_1{p}_2=0$
$\Rightarrow \frac{p^{\prime }2}{m_2}=\frac{p_2^{\prime }2}{m_1}$
or $\frac{m_2}{m_1}=1$