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JEE Advanced 2025 · Paper 1, Physics question 2

In a scattering experiment, a particle of mass 2m2m collides with another particle of mass mm, which is initially at rest. Assuming the collision to be perfectly elastic, the maximum angular deviation θ\theta of the heavier particle, as shown in the figure, in radians is:

Before the collision, particle 2m moves to the right along a horizontal dotted line towards particle m at rest. After the collision, 2m moves off above the line at angle θ to the original direction, and m moves off below the line.
Choose one option
Answer and worked solution

Official answer: D

In an elastic collision of m1m_1 with a stationary m2<m1m_2<m_1, the incoming particle's deflection is limited by
sin⁡θmax=m2m1.\sin\theta_{max}=\frac{m_2}{m_1}.

Why. In the CM frame, m1m_1 keeps its speed u′=m2m1+m2uu'=\frac{m_2}{m_1+m_2}u. The CM itself moves at V=m1m1+m2u>u′V=\frac{m_1}{m_1+m_2}u>u', so the lab velocity can tilt at most sin⁡−1(u′/V)\sin^{-1}(u'/V).

Here sin⁡θmax=m2m=12\sin\theta_{max}=\frac{m}{2m}=\frac12, so θmax=π6\theta_{max}=\frac\pi6.

Topics:Elastic and inelastic collisionsCentre of mass
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