Question:
If a ball of mass 0.02 kg bowled by a bowler straight to a batsman is hit back with the same speed with an impulse of 2 Ns, then the speed of the ball bowled is
If a ball of mass 0.02 kg bowled by a bowler straight to a batsman is hit back with the same speed with an impulse of 2 Ns, then the speed of the ball bowled is
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Impulse \(J = \Delta p = m(v_f - v_i)\). When direction reverses, \(v_f = -v_i\).
Updated On: Apr 24, 2026
- \(20 \, \text{ms}^{-1}\)
- \(80 \, \text{ms}^{-1}\)
- \(50 \, \text{ms}^{-1}\)
- \(60 \, \text{ms}^{-1}\)
- \(40 \, \text{ms}^{-1}\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
Impulse = change in momentum. If the ball is hit back with the same speed, the velocity changes from \(+v\) to \(-v\).
Step 2: Detailed Explanation:
Initial momentum = \(mv\)
Final momentum = \(-mv\)
Change in momentum = \(-mv - mv = -2mv\)
Magnitude of impulse = \(2mv = 2\) Ns
\(2 \times 0.02 \times v = 2 \Rightarrow 0.04v = 2 \Rightarrow v = \frac{2}{0.04} = 50 \, \text{ms}^{-1}\)
Step 3: Final Answer:
Speed of the ball bowled = \(50 \, \text{ms}^{-1}\).
Impulse = change in momentum. If the ball is hit back with the same speed, the velocity changes from \(+v\) to \(-v\).
Step 2: Detailed Explanation:
Initial momentum = \(mv\)
Final momentum = \(-mv\)
Change in momentum = \(-mv - mv = -2mv\)
Magnitude of impulse = \(2mv = 2\) Ns
\(2 \times 0.02 \times v = 2 \Rightarrow 0.04v = 2 \Rightarrow v = \frac{2}{0.04} = 50 \, \text{ms}^{-1}\)
Step 3: Final Answer:
Speed of the ball bowled = \(50 \, \text{ms}^{-1}\).
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