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 configuration of pieces after collision is shown in the figure below. The value of x to the nearest integer is ____________.

Solution

Physics sir · 4 steps

No outside force acts, so momentum is conserved along x and along y separately; the x-direction alone gives v.

1. Before collision. Only the 10 kg ball moves, along x. Its momentum is 10 × 10√3 = 100√3 kg m s⁻¹ along x, and zero along y.
2. After collision. The first ball stops. The 20 kg ball splits into two 10 kg pieces. Piece 1 moves along y at 10 m s⁻¹, so it carries no x-momentum. Piece 2 moves with speed v at 30° below the x-axis.

   100√3 = 10 v cos 30° = 10 v × √3/2 = 5√3 v

3. Check along y. Solving gives v = 20 m s⁻¹. Along y, piece 1 has 10 × 10 = 100 upward and piece 2 has 10 × 20 × sin 30° = 100 downward. They cancel, as they must, because the initial y-momentum was zero.
4. So. The piece moving at 30° to the x-axis has speed 20 m s⁻¹, so x = 20.

The common slip: 10. Using the full 20 kg instead of 10 kg for the piece gives v = 10. Each piece is half the ball.

The idea to keep. In any explosion or collision with no external force, write momentum conservation separately along x and along y. A piece moving along one axis drops out of the other equation.

Answer from the NTA final answer key. Solution by PrepWiser.