Newton's Second Law
A cricket ball of mass 0.16 kg moving horizontally is struck by a bat, and its velocity changes from 30 m/s to 50 m/s in the opposite direction in 0.02 s. What average force does the bat exert on the ball?
Select the correct option:
Solution
640 N
According to NCERT Class 11, Chapter 5 (Laws of Motion), Newton's Second Law in its momentum form states that force equals the rate of change of linear momentum, F = Δp / Δt. Direction matters here because the ball reverses, so we assign the initial direction as positive (+30 m/s) and the rebound as negative (−50 m/s). The change in velocity is Δv = (−50) − (30) = −80 m/s. The magnitude of change of momentum is |Δp| = m|Δv| = 0.16 × 80 = 12.8 kg·m/s. Dividing by the contact time gives F = 12.8 / 0.02 = 640 N. Option 160 N ignores the reversal and uses only the speed difference of 20 m/s. Option 320 N corresponds to a Δv of 40 m/s, again mishandling the sign reversal. Option 480 N corresponds to a Δv of 60 m/s, which double-counts incorrectly. It is important to recognise that the large force here arises mainly from the very short contact time of just 0.02 s, since a smaller time over which the same momentum change occurs forces the average force to be correspondingly larger. This is precisely why follow-through in sports lengthens contact time and reduces peak force on the body. Plausibility check: a hard bat strike over a very short time should produce a large force of several hundred newtons, so 640 N is physically reasonable for a struck cricket ball.
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About This Question
- Subject
- physics
- Chapter
- laws of motion
- Topic
- newton's second law
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
640 N
According to NCERT Class 11, Chapter 5 (Laws of Motion), Newton's Second Law in its momentum form states that force equals the rate of change of linear momentum, F = Δp / Δt. Direction matters here because the ball reverses, so we assign the initial direction as positive (+30 m/s) and the rebound as negative (−50 m/s). The change in velocity is Δv = (−50) − (30) = −80 m/s. The magnitude of change of momentum is |Δp| = m|Δv| = 0.16 × 80 = 12.8 kg·m/s. Dividing by the contact time gives F = 12.8 / 0.02 = 640 N. Option 160 N ignores the reversal and uses only the speed difference of 20 m/s. Option 320 N corresponds to a Δv of 40 m/s, again mishandling the sign reversal. Option 480 N corresponds to a Δv of 60 m/s, which double-counts incorrectly. It is important to recognise that the large force here arises mainly from the very short contact time of just 0.02 s, since a smaller time over which the same momentum change occurs forces the average force to be correspondingly larger. This is precisely why follow-through in sports lengthens contact time and reduces peak force on the body. Plausibility check: a hard bat strike over a very short time should produce a large force of several hundred newtons, so 640 N is physically reasonable for a struck cricket ball.
This medium difficulty physics question is from the chapter laws of motion, covering the topic of newton's second law. It appeared in the 2025 exam.
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