Question

A billiard ball of mass ‘M’ moving with velocity ‘v₁’ collides with another ball of the same mass but at rest. If the collision is elastic the angle of divergence after the collision is

• A. \(\frac{V}{2}\)
• B. 0º
• C. 30º
• D. \( \frac{\sqrt{3} V}{2} \)
• E. 90º

Answer 1

The Correct Option is D

Given:

• Ball \( B_1 \) moves with velocity \( V \), collides with stationary \( B_2 \).
• After collision, \( B_1 \) deflects by \( 60^\circ \), and the angle between \( B_1 \) and \( B_2 \) is \( 90^\circ \).

Step 1: Apply Momentum Conservation

Let \( v_1 \) and \( v_2 \) be the speeds of \( B_1 \) and \( B_2 \) after collision.

Horizontal: \( V = v_1 \cos 60^\circ + v_2 \cos 30^\circ \)

Vertical: \( 0 = v_1 \sin 60^\circ - v_2 \sin 30^\circ \)

Step 2: Solve for \( v_2 \)

From vertical momentum: \( v_1 \sin 60^\circ = v_2 \sin 30^\circ \).

\[ v_1 = v_2 \cdot \frac{\sin 30^\circ}{\sin 60^\circ} = v_2 \cdot \frac{0.5}{\sqrt{3}/2} = \frac{v_2}{\sqrt{3}} \]

Substitute into horizontal momentum:

\[ V = \left(\frac{v_2}{\sqrt{3}}\right) \cos 60^\circ + v_2 \cos 30^\circ = \frac{v_2}{2\sqrt{3}} + \frac{\sqrt{3} v_2}{2} \]

\[ V = v_2 \left(\frac{1}{2\sqrt{3}} + \frac{\sqrt{3}}{2}\right) = v_2 \left(\frac{1 + 3}{2\sqrt{3}}\right) = \frac{2v_2}{\sqrt{3}} \]

\[ v_2 = \frac{\sqrt{3} V}{2} \]

Conclusion:

The speed of \( B_2 \) after collision is \( \frac{\sqrt{3} V}{2} \).

Answer 2

1. Analyze the collision:

This is a 2D collision problem. Since it's a billiard ball collision, we can assume it's an elastic collision (kinetic energy is conserved) and the masses of the balls are equal.

2. Define variables:

• \(V\) (Initial velocity of B₁)
• \(V_1\) (Velocity of B₁ after collision)
• \(V_2\) (Velocity of B₂ after collision)

The angle between \(V\) and \(V_1\) is 60°. The angle between \(V_1\) and \(V_2\) is 90°.

3. Apply conservation of momentum:

Since the masses are equal, we can simplify the momentum equations by canceling out the mass term.

In the x-direction:

\[V = V_1 \cos(60^\circ) + V_2 \cos(\theta)\]

In the y-direction (assuming B₁ is initially moving along the x-axis):

\[0 = V_1 \sin(60^\circ) - V_2 \sin(\theta)\]

Where \(\theta\) is the angle \(V_2\) makes with the initial direction of \(V\).

4. Use the given angle between the final velocities:

We know the angle between the final velocities is 90°. Since B₁ is deflected by 60°, the angle \(\theta\) B₂ makes with the initial direction of B₁ must be 30° (since 60°+ 30° gives a 90° angle for B₂ relative to B₁'s final velocity).

5. Solve for \(V_2\):

Now we have:

\[V = \frac{1}{2}V_1 + \frac{\sqrt{3}}{2}V_2\]

\[0 = \frac{\sqrt{3}}{2}V_1 - \frac{1}{2}V_2\]

From the second equation: \(V_2 = \sqrt{3}V_1\)

Substitute this into the first equation:

\[V = \frac{1}{2}V_1 + \frac{\sqrt{3}}{2}(\sqrt{3}V_1)\]

\[V = \frac{1}{2}V_1 + \frac{3}{2}V_1\]

\[V = 2V_1\]

\[V_1 = \frac{V}{2}\]

Now substitute \(V_1\) back into the equation for \(V_2\):

\[V_2 = \sqrt{3}V_1 = \sqrt{3}(\frac{V}{2}) = \frac{\sqrt{3}V}{2}\]