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Center of Mass question

2006 · Shift 0 · Q126
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Center of Mass question

2006 · Shift 0 · Q126

JEE MainPhysicsCenter of MassMCQ+4 / −1
A player caught a cricket ball of mass 150g150g150g moving at a rate of 20m/s.20m/s.20m/s. If the catching process is completed in 0.1s,0.1s,0.1s, the force of the blow exerted by the ball on the hand of the player is equal to
  1. A
    150N150N150N
  2. B
    3N3N3N
  3. C
    30N30N30N
  4. D
    300N300N300N
View written solutionFree

Correct answer: C

  1. Given data
  • Mass of ball: m=150 g=0.150 kgm = 150\text{ g} = 0.150\text{ kg}m=150 g=0.150 kg
  • Initial speed: u=20 m/su = 20\text{ m/s}u=20 m/s
  • Final speed after being caught: v=0v = 0v=0
  • Time taken to stop the ball: Δt=0.1 s\Delta t = 0.1\text{ s}Δt=0.1 s
  1. Use impulse-momentum theorem

Average force is given by:

F=ΔpΔtF = \frac{\Delta p}{\Delta t}F=ΔtΔp​

where change in momentum is

Δp=m(v−u)\Delta p = m(v-u)Δp=m(v−u)

So,

Δp=0.150(0−20)=−3 kg m/s\Delta p = 0.150(0-20) = -3\text{ kg m/s}Δp=0.150(0−20)=−3 kg m/s

Magnitude of change in momentum:

∣Δp∣=3 kg m/s|\Delta p| = 3\text{ kg m/s}∣Δp∣=3 kg m/s
  1. Compute force
F=30.1=30 NF = \frac{3}{0.1} = 30\text{ N}F=0.13​=30 N

Thus, the magnitude of the force exerted by the ball on the player's hand is

30 N\boxed{30\text{ N}}30 N​
  1. Check options
  • A: 150 N150\text{ N}150 N ❌
  • B: 3 N3\text{ N}3 N ❌
  • C: 30 N30\text{ N}30 N ✅
  • D: 300 N300\text{ N}300 N ❌

So the correct option is C.

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