A batsman hits back a ball of mass 0.4 kg straight in the direction of the bowler without changing its initial speed of 15 ms⁻¹. The impulse imparted to the ball is ___________ Ns.

Solution

Physics sir · 3 steps

Same speed, opposite direction: the momentum flips sign, so the change is twice mv.

1. Choose a direction. Take the ball's motion towards the batsman as positive. Before the hit, its velocity is +15 m s⁻¹. After the hit, it goes back towards the bowler at the same speed, so its velocity is -15 m s⁻¹.

   J = Δ p = m v_f - m vᵢ = m(v_f - vᵢ)

2. Put in numbers. J = 0.4 × (-15 - 15) = 0.4 × (-30) = -12 N s. The minus sign only says the impulse points towards the bowler.
3. So. The magnitude of the impulse imparted to the ball is 12 N s.

The common slip: 0. Treating both velocities as +15 m s⁻¹ gives a momentum change of 0, not 12. The speed is unchanged, but the direction has reversed.

The idea to keep. Impulse is the change in momentum, and momentum is a vector. When a body reverses direction, the initial and final velocities have opposite signs, so their magnitudes add.

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