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A player caught a cricket ball of mass 150 g moving at a rate of 20 m/s. If the catching process is completed in 0.1 s, the force of the blow exerted by the ball on the hand of the player is equal to
 [AIEEE 2006]
  • a)
     150 N
  • b)
    3 N
  • c)
    30 N
  • d)
     300 N
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A player caught a cricket ball of mass 150 g moving at a rate of 20 m/...
F = ma = m(v/t)
F = 150 × 20 / 0.1 × 1000
F = 30N
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A player caught a cricket ball of mass 150 g moving at a rate of 20 m/...
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A player caught a cricket ball of mass 150 g moving at a rate of 20 m/...
To find the force of the blow exerted by the ball on the hand of the player, we can use the equation F = Δp/Δt, where F is the force, Δp is the change in momentum, and Δt is the change in time.

Given values:
Mass of the cricket ball (m) = 150 g = 0.15 kg
Initial velocity of the cricket ball (u) = 0 m/s (since the ball is caught)
Final velocity of the cricket ball (v) = 20 m/s
Time taken to catch the ball (Δt) = 0.1 s

Now, let's calculate the change in momentum (Δp):
Δp = m(v - u)
= 0.15 kg (20 m/s - 0 m/s)
= 3 kg⋅m/s

Now, let's calculate the force (F):
F = Δp/Δt
= (3 kg⋅m/s) / (0.1 s)
= 30 N

Therefore, the force of the blow exerted by the ball on the hand of the player is equal to 30 N.
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