JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Conservation of Momentum

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:

Answer: (D)

Solution

For an elastic collision between particles of mass

2 m

and

m

, applying conservation of momentum and energy:

Momentum:

2 m v_{1} = 2 m v_{1 f} cos θ + m v_{2 f} cos ϕ

Energy:

\frac{1}{2} (2 m) v_{1}^{2} = \frac{1}{2} (2 m) v_{1 f}^{2} + \frac{1}{2} m v_{2 f}^{2}

Solving these equations leads to:

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

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

16 v_{1}^{2} cos^{2} θ - 12 v_{1}^{2} \geq 0

cos^{2} θ \geq \frac{3}{4}

cos θ \geq \frac{3}{2}

The maximum deviation occurs when

cos θ = \frac{3}{2}

, giving

θ = \frac{π}{6}

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Single correct option