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NEET PHYSICSMedium

A ball of mass 0.5 kg moving with a velocity of 2 m/s strikes a wall normally and bounces back with the same speed. If the time of contact between the ball and the wall is one millisecond, the average force exerted by the wall on the ball is:

A

2000 N

B

1000 N

C

5000 N

D

125 N

Step-by-Step Solution

  1. Identify Given Values:
  • Mass (mm) = 0.5 kg
  • Initial velocity (viv_i) = +2+2 m/s (towards the wall)
  • Final velocity (vfv_f) = −2-2 m/s (bounces back with same speed, opposite direction)
  • Time of contact (Δt\Delta t) = 1 millisecond = 1×10−31 \times 10^{-3} s

  1. Calculate Change in Momentum (Impulse): Impulse is defined as the change in momentum (Δp=pf−pi\Delta p = p_f - p_i). Δp=m(vf−vi)\Delta p = m(v_f - v_i) Δp=0.5×(−2−2)=0.5×(−4)=−2 kg m/s\Delta p = 0.5 \times (-2 - 2) = 0.5 \times (-4) = -2 \text{ kg m/s} The magnitude of the impulse is 2 kg m/s2 \text{ kg m/s} [Source 66, 68].

  2. Apply Newton's Second Law: The average force (FavgF_{avg}) is the rate of change of momentum. Favg=∣Δp∣ΔtF_{avg} = \frac{|\Delta p|}{\Delta t} Favg=2 kg m/s1×10−3 s=2000 NF_{avg} = \frac{2 \text{ kg m/s}}{1 \times 10^{-3} \text{ s}} = 2000 \text{ N} [Source 66]

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