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A cricket ball of mass 250 g collides with a bat with velocity 10 m/s and returns with the same velocity within 0.01 second. The force acted on bat is:

A

25 N

B

50 N

C

250 N

D

500 N

Step-by-Step Solution

  1. Convert Units: Mass m=250 g=0.25 kgm = 250 \text{ g} = 0.25 \text{ kg}.
  2. Analyze Momentum Change:
  • Initial velocity vi=10 m/sv_i = 10 \text{ m/s}.
  • Final velocity vf=−10 m/sv_f = -10 \text{ m/s} (since it returns with the same speed in the opposite direction).
  • Change in momentum Δp=m(vf−vi)=0.25×(−10−10)=0.25×(−20)=−5 kg m/s\Delta p = m(v_f - v_i) = 0.25 \times (-10 - 10) = 0.25 \times (-20) = -5 \text{ kg m/s}.
  • Magnitude of change in momentum ∣Δp∣=5 kg m/s|\Delta p| = 5 \text{ kg m/s} [Source 68].
  1. Calculate Force: According to Newton's Second Law, the average force is the rate of change of momentum (or Impulse divided by time). Favg=∣Δp∣ΔtF_{avg} = \frac{|\Delta p|}{\Delta t} Favg=5 kg m/s0.01 s=500 NF_{avg} = \frac{5 \text{ kg m/s}}{0.01 \text{ s}} = 500 \text{ N} [Source 66, 67].
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