Two billiard balls of equal mass 30g strike a rigid wall with same speed of 108 kmph (as shown) but at different angles. If the balls get reflected with the same speed then the ratio of the magnitude of impulses imparted to ball 'a' and ball 'b' by the wall along 'X' direction is :

Solution

Physics sir · 4 steps

The wall only pushes along X, so count only the X-component of velocity that flips.

1. Impulse along X. Impulse equals change in momentum, Jₓ = m(vₓ - uₓ). A smooth rigid wall reverses the X-component of velocity and leaves the Y-component unchanged. So for each ball Jₓ = 2m uₓ, where uₓ is the X-component of the incoming velocity.
2. Ball a. It hits head-on, so uₓ = u and Jₐ = 2mu. With m = 0.03 kg and u = 108 km h⁻¹ = 30 m s⁻¹, this is 1.8 N s.
3. Ball b. Its path makes 45° with the X-axis, so uₓ = u cos 45° = u/√2. The Y-component u sin 45° does not change and gives no impulse along X.

   Jₐ/J_b = 2mu/(2m u/√2) = √2

4. So. The ratio of impulses along X is √2 : 1, which is option (b).

The idea to keep. In a collision with a wall, only the momentum component normal to the wall changes. Impulse along the normal is 2mu cos θ, with θ measured from the normal.

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