Question

A batsman hits a ball of mass 0.2 kg straight towards the bowler without changing its initial speed of 6 m/s. What is the impulse imparted to the ball?

• A. 2.4 Ns
• B. 1.6 Ns
• C. 4 Ns
• D. 3.2 Ns

Hint:

Impulse is the product of mass and change in velocity. For a ball returning to the bowler with the same speed, the change in velocity is twice the initial speed.

Answer 1

The Correct Option is A

Step 1: Impulse and momentum relation.
Impulse is the change in momentum. The formula for impulse is: \[ I = \Delta p = m \Delta v \] where \( m \) is the mass of the ball and \( \Delta v \) is the change in velocity.
Step 2: Impulse calculation.
The initial velocity of the ball is 6 m/s, and since the direction is reversed (in this case, straight towards the bowler), the final velocity is \( -6 \) m/s. So the change in velocity is: \[ \Delta v = v_{\text{final}} - v_{\text{initial}} = -6 - 6 = -12 \, \text{m/s} \] Now, the impulse is: \[ I = 0.2 \times (-12) = -2.4 \, \text{Ns} \]
Step 3: Conclusion.
The magnitude of the impulse is \( 2.4 \, \text{Ns} \), so the correct answer is (A).