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A proton of mass $\mathrm{m}$ collides elastically with a particle of unknown mass at rest. After the collision, the proton and the unknown particle are seen moving at an angle of $90^{\circ}$ with respect to each other. The mass of unknown particle is:
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Verified Answer
The correct answer is:
$\mathrm{m}$
$\mathrm{m}$
Apply principle of conservation of momentum along $x$-direction,
$$
\begin{aligned}
m u &=m v_1 \cos 45^{\circ}+M v_2 \cos 45^{\circ} \\
m u &=\frac{1}{\sqrt{2}}\left(m v_1+M v_2\right)
\end{aligned}
$$
Along $y$-direction,
$$
\begin{aligned}
&o=m v_1 \sin 45^{\circ}-M v_2 \sin 45^{\circ} \\
&o=\left(m v_1-M v_2\right) \frac{1}{\sqrt{2}}
\end{aligned}
$$
Coefficient of restution $e=1$
$$
=\frac{v_2-v_1 \cos 90}{u \cos 45}
$$
( $\because$ Collision is elastic)
$$
\begin{aligned}
&\Rightarrow \frac{v_2}{\frac{u}{\sqrt{2}}}=1 \\
&\Rightarrow u=\sqrt{2} v_2
\end{aligned}
$$
Solving eqs (i), (ii), \& (iii), we get mass of unknown particle, $M=m$
$$
\begin{aligned}
m u &=m v_1 \cos 45^{\circ}+M v_2 \cos 45^{\circ} \\
m u &=\frac{1}{\sqrt{2}}\left(m v_1+M v_2\right)
\end{aligned}
$$
Along $y$-direction,
$$
\begin{aligned}
&o=m v_1 \sin 45^{\circ}-M v_2 \sin 45^{\circ} \\
&o=\left(m v_1-M v_2\right) \frac{1}{\sqrt{2}}
\end{aligned}
$$
Coefficient of restution $e=1$
$$
=\frac{v_2-v_1 \cos 90}{u \cos 45}
$$
( $\because$ Collision is elastic)
$$
\begin{aligned}
&\Rightarrow \frac{v_2}{\frac{u}{\sqrt{2}}}=1 \\
&\Rightarrow u=\sqrt{2} v_2
\end{aligned}
$$
Solving eqs (i), (ii), \& (iii), we get mass of unknown particle, $M=m$
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