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A proton of mass 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 $90^0$ with respect to each other. The mass of unknown particle is :

Option 1)
$\frac{m}{2}$

Option 2)
m

Option 3)
$\frac{m}{\sqrt{3}}$

Option 4)
2m

Answers (2)

best_answer

As we have learnt

Elastic Collision in 2 dimension -

$\fn_jvn \vec{P_{i}}= \vec{P_{f}}$

$m_{1}v_{0}\: \: \hat{i}= \left ( m_{1}v_{1} \cos \Theta + m_{2}v_{2} \cos \beta \right )\hat{i}+\left ( m_{1}v_{1} \sin \Theta - m_{2}v_{2} \sin \beta \right )\hat{j}$

- wherein

From momentum conservation along X

$mu =mv_{1}cos\Theta +Mv_{2}sin\Theta\rightarrow (1)$

& along Y,

$mv_{1}sin\Theta =M\: v_{2}\: cos\Theta\rightarrow (2)$

or

$e = 1 = \frac{velocity \, o\! f\, separation}{velocity\, o\! f\, approach}$

or

$v_{2}=usin\Theta\rightarrow (3)$

$\therefore mv_{1}sin\Theta=M.usin\Theta cos\Theta$

$\therefore v_{1}=\frac{m}{M} u.cos\Theta$

Put v1 &v2 in eqn->1;

$mu=M.u.cos^{2}\Theta + M.u.sin^{2}\Theta$

$m=M.cos^{2}\Theta + M.sin^{2}\Theta$

$m=M\left ( cos^{2}\Theta + sin^{2}\Theta \right )$

$\therefore m=M$

Option 1)

$\frac{m}{2}$

Option 2)

m

Option 3)

$\frac{m}{\sqrt{3}}$

Option 4)

2m

Posted by

Avinash

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