Question

A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of $12 ms^{-1}$. If the mass of the ball is 0.15 kg, the imparted to the ball is

• A. $36 N-s$
• B. $3.6 N-s$
• C. $0.36 N-s$
• D. $0.036 N-s$

Answer 1

The Correct Option is B

To determine the impulse imparted to the ball, we need to understand that impulse is the change in momentum. The formula for impulse (J) is given by:

J = \Delta p = m \times \Delta v

where:

• m is the mass of the ball.
• \Delta v is the change in velocity.

Let's consider the motion of the ball:

• The initial velocity (v_i) of the ball is 12 \, \text{ms}^{-1} towards the bowler.
• When the batsman hits the ball back in the same direction at the same speed, the final velocity (v_f) becomes -12 \, \text{ms}^{-1} because the direction is reversed.

Calculating the change in velocity:

\Delta v = v_f - v_i = -12 \, \text{ms}^{-1} - 12 \, \text{ms}^{-1} = -24 \, \text{ms}^{-1}

Given the mass of the ball m = 0.15 \, \text{kg}, the impulse is calculated as:

J = m \times \Delta v = 0.15 \, \text{kg} \times (-24 \, \text{ms}^{-1}) = -3.6 \, \text{N-s}

The negative sign indicates that the direction of impulse is opposite to the initial direction of motion.

Hence, the magnitude of the impulse imparted to the ball is 3.6 \, \text{N-s}.

Therefore, the correct answer is 3.6 \, \text{N-s}.