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: [2025]

Question

Smart Achievers · Accepted Answer

(4)

Collision is perfectly elastic so both momentum and kinetic energy will be conserved.

$m$ ...(i)

$2 m v_{1 f} \sin θ = m v_{1 f} \sin ϕ$ ...(ii)

$\frac{1}{2} (2 m) v_{1}^{2} + \frac{1}{2} m {(0)}^{2} = \frac{1}{2} (2 m) v_{1 f}^{2} + \frac{1}{2} m v_{2 f}^{2}$

$2 v_{1}^{2} = 2 v_{1 f}^{2} + v_{2 f}^{2}$ ...(iii)

$From eq. (i), (ii) and (iii),$

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

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

$\cos^{2} θ \geq \frac{3}{4} or, \cos θ \geq \frac{\sqrt{3}}{2}$

$∴$ $θ = \frac{π}{6}$

Solution

Q.
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: [2025]

1 $π$

2 $\tan^{- 1} (\frac{1}{2})$

3 $\frac{π}{3}$

4 $\frac{π}{6}$

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Solution

Q. In a scattering experiment, a particle of mass 2m 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: [2025]

1 π

2 tan-1(12)

3 π3