Impulse And Momentum
A 0.15 kg ball moving at 20 m/s is stopped by a catch in 0.05 s. What average force is exerted by the hand?
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About This Question
- Subject
- physics
- Chapter
- laws of motion
- Topic
- impulse and momentum
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
60 N
- Impulse-Momentum Theorem: Impulse is defined as the change in momentum: J=Δp=m(vf−vi).
- Step 1: Calculate Delta p:
- m=0.15 kg, u=20 m/s, v=0 m/s.
- Δp=0.15×(−20)=−3 kg·m/s.
- Step 2: Relate to Force: Force is the rate of change of momentum (Newton's 2nd Law in derivative form).
- Favg=ΔtΔp
- Calculation:
- ∣Favg∣=0.05 s3 kg\cdotpm/s=5300=60 N.
- Result: The hand exerts an average force of 60 N.
This easy difficulty physics question is from the chapter laws of motion, covering the topic of impulse and momentum. It appeared in the 2025 exam.
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