JEE Advanced 2025 — Paper 1 — PHY-I solution

JEE Advanced2025Paper 1PHY-I

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

A
$π$
B
$tan^{- 1} (\frac{1}{2})$
C
$\frac{π}{3}$
D
$\frac{π}{6}$

Solution

Let the initial speed of the $2 m$ particle be $v_{1}$ , and let $v_{1}^{'}$ and $v_{2}^{'}$ be the final speeds of the $2 m$ and $m$ particles at angles $θ$ and $ϕ$ to the original direction. Conservation of momentum gives:

$2 m v_{1} = 2 m v_{1}^{'} cos θ + m v_{2}^{'} cos ϕ (i)$

$0 = 2 m v_{1}^{'} sin θ - m v_{2}^{'} sin ϕ (ii)$

Elastic collision: $2 v_{1}^{2} = 2 v_{1}^{'2} + v_{2}^{'2} (iii)$

Eliminating $ϕ$ from (i) and (ii) and substituting into (iii) yields a quadratic in $v_{1}^{'}$ :

$3 v_{1}^{'2} - 4 v_{1} v_{1}^{'} cos θ + v_{1}^{2} = 0$

For real solutions, the discriminant must be non-negative:

$(4 v_{1} cos θ)^{2} - 12 v_{1}^{2} \geq 0 \Rightarrow cos^{2} θ \geq \frac{3}{4}$

The maximum $θ$ corresponds to equality, $cos θ = \frac{3}{2}$ , giving $θ_{m a x} = \frac{π}{6}$ . Hence option (D).

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