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Impulse & Momentum

Section: Physics

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Question : 1 of 16

Marks: +1, -0

In a scattering experiment, a particle of mass

2 \text{ m}

collides with another particle of mass

m

, which is initially at rest. Assuming the collision to be perfectly elastic, the maximum angular deviation

\theta

of the heavier particle, as shown in the figure, in radians is:

[JEE Adv 2025 P1]

$\pi$
$\tan^{-1} \left( \frac{1}{2} \right)$
$\frac{\pi}{3}$
$\frac{\pi}{6}$

Solution:

In centre of mass frame

Speed does not change only orientation does

For

\theta

to maximize

\overrightarrow{v_{1, C M}} \perp \overrightarrow{v_1}

\Rightarrow \sin \theta = \frac{v_{1, C M}}{v_{C M}}

v_{1, C M} = \frac{\overrightarrow{m_2 v_{12}}}{m_1+m_2} \Rightarrow \overrightarrow{v_1} = \frac{\overrightarrow{m_2 u_1}}{m_1+m_2}

v_{C M} = \frac{\overrightarrow{m_1 v_1 + m_2 \overrightarrow{v_2}}}{m_1+m_2} = \frac{\overrightarrow{m_1 u_1}}{m_1+m_2}

\Rightarrow \sin \theta = \frac{m_2}{m_1} \Rightarrow \sin \theta = \frac{m}{2 m}

\theta = 30^{\circ} = \frac{\pi}{6}

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