Question

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.

Answer 1

Correct Answer: 12

The impulse imparted to the ball can be calculated using the formula for impulse, which is the change in momentum of the object. Momentum is defined as the product of mass and velocity. Given the initial speed \(v_i\) and the final speed \(v_f\) (in the opposite direction), the impulse \(J\) is calculated as follows:

Step 1: Identify initial parameters.

Mass of the ball, \(m = 0.4\) kg.

Initial speed of the ball, \(v_i = 15\) m/s (initially towards the bowler).

Final speed of the ball, \(v_f = -15\) m/s (since it rebounds back towards the bowler, opposite to the initial direction).

Step 2: Calculate initial and final momentum.

Initial momentum, \(p_i = m \times v_i = 0.4 \, \text{kg} \times 15 \, \text{m/s} = 6 \, \text{kg·m/s}\).

Final momentum, \(p_f = m \times v_f = 0.4 \, \text{kg} \times (-15) \, \text{m/s} = -6 \, \text{kg·m/s}\).

Step 3: Calculate the impulse.

The impulse is the change in momentum:

\(J = p_f - p_i = -6 \, \text{kg·m/s} - 6 \, \text{kg·m/s} = -12 \, \text{kg·m/s}\).

Since impulse is a vector quantity, and we typically speak about its magnitude, the magnitude of \(J\) is \(12\) Ns.

Step 4: Validate the result against the range.

The calculated impulse magnitude \(12\) Ns falls within the provided range [12, 12], confirming it's the correct value.

Final Answer: The impulse imparted to the ball is 12 Ns.