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Q.2 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:

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

Answer: D

Solution:

1. Understanding the Problem

A particle of mass $2 m$ collides with a stationary particle of mass $m$ .
The collision is perfectly elastic, meaning:
- Kinetic energy is conserved.
- Momentum is conserved.
We need to find the maximum angular deviation $θ$ of the heavier particle ( $2 m$ ) after the collision.

2. Key Physics Concepts

Conservation of Momentum: Total momentum before and after the collision remains the same.
Conservation of Kinetic Energy: Since the collision is elastic, total kinetic energy before and after the collision is the same.
Scattering Angle: The angle by which the heavier particle is deflected.

3. Step-by-Step Analysis

Step 1: Initial Conditions

Let the initial velocity of the heavier particle ( $2 m$ ) be $u$ .
The lighter particle ( $m$ ) is initially at rest ( $0$ ).

Step 2: Conservation of Momentum

Total momentum before collision:

$P_{initial} = 2 m u$

Total momentum after collision:

$P_{final} = 2 m v_{1} + m v_{2}$

where:

$v_{1}$ = final velocity of the heavier particle.
$v_{2}$ = final velocity of the lighter particle.

By conservation of momentum:

$2 m u = 2 m v_{1} + m v_{2} ⟹ 2 u = 2 v_{1} + v_{2} (1)$

Step 3: Conservation of Kinetic Energy

Total kinetic energy before collision:

$K_{initial} = \frac{1}{2} (2 m) u^{2} = m u^{2}$

Total kinetic energy after collision:

$K_{final} = \frac{1}{2} (2 m) v_{1}^{2} + \frac{1}{2} m v_{2}^{2} = m v_{1}^{2} + \frac{1}{2} m v_{2}^{2}$

By conservation of kinetic energy:

$m u^{2} = m v_{1}^{2} + \frac{1}{2} m v_{2}^{2} ⟹ u^{2} = v_{1}^{2} + \frac{1}{2} v_{2}^{2} (2)$

Step 4: Solving the Equations

From (1):

$v_{2} = 2 (u - v_{1})$

Substitute $v_{2}$ into (2):

$u^{2} = v_{1}^{2} + \frac{1}{2} {[2 (u - v_{1})]}^{2} = v_{1}^{2} + 2 (u - v_{1})^{2}$

Expanding:

$u^{2} = v_{1}^{2} + 2 (u^{2} - 2 u \cdot v_{1} + v_{1}^{2})$ $u^{2} = v_{1}^{2} + 2 u^{2} - 4 u \cdot v_{1} + 2 v_{1}^{2}$ $0 = 3 v_{1}^{2} - 4 u \cdot v_{1} + u^{2}$

This is a quadratic in $v_{1}$ . For real solutions, the discriminant must be non-negative:

$(4 u \cdot v_{1})^{2} - 4 (3) (u^{2}) \geq 0$

Step 5: Maximum Angular Deviation

The maximum deviation occurs when the heavier particle is deflected the most, which happens when the lighter particle carries away the maximum possible momentum.

Using relative velocity and angle constraints for elastic collisions:

The maximum angle $θ$ by which the heavier particle can deviate is given by:

$θ = \sin^{- 1} (\frac{m}{2 m}) = \sin^{- 1} (\frac{1}{2}) = \frac{π}{6}$

6. Final Answer

The maximum angular deviation $θ$ of the heavier particle is:

\boxed{D}