Question

A ball of mass 10 kg moving with a velocity 10√3 m/s along the x-axis, hits another ball of mass 20 kg which is at rest. After the collision, first ball comes to rest while the second ball disintegrates into two equal pieces. One piece starts moving along y-axis with a speed of 10 m/s. The second piece starts moving at an angle of 30° with respect to the x-axis. The velocity of the ball moving at 30° with x-axis is x m/s. The value of x to the nearest integer is __________. 

Hint:

In 2D collisions, solve momentum equations independently for the x and y directions.

Answer 1

Correct Answer: 20

To solve this problem, we apply the law of conservation of momentum. Initially, only the 10 kg ball is moving with velocity \(v = 10\sqrt{3} \, \text{m/s}\) along the x-axis. After collision:

• The first ball comes to rest.
• The 20 kg ball disintegrates into two 10 kg pieces moving in different directions.

Let's denote:

• Piece 1 velocity along the y-axis, \(v_1 = 10 \, \text{m/s}\).
• Piece 2 velocity \(v_2\) at an angle \(30^\circ\) with the x-axis.

Step 1: Conservation of Momentum in the x-direction
The initial momentum in the x-direction is:

\(p_{\text{initial,x}} = 10 \, \text{kg} \times 10\sqrt{3} \, \text{m/s}\).
The final momentum in the x-direction is:

\(p_{\text{final,x}} = 10\, \text{kg} \times v_2\cos(30^\circ)\).

Setting them equal:

\(10 \times 10\sqrt{3} = 10 \times v_2 \times \frac{\sqrt{3}}{2}\)

\(100\sqrt{3} = 5\sqrt{3}v_2\)

\(v_2 = 20 \, \text{m/s}\)

Step 2: Verify Range
The calculated velocity \(v_2 = 20 \, \text{m/s}\) matches the given range of 20,20.

Conclusion: The velocity \(x\) of the ball moving at \(30^\circ\) with the x-axis is 20 m/s.