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NDA 1 2024 GAT, question 86

Topic: Physics

Q86

A ball of 0.1 kg0.1\text{ kg} mass is dropped on a hard floor from a height of 0.45 m0.45\text{ m} and rises to a height of 0.20 m0.20\text{ m}. If it was in touch with the floor for 0.1 s0.1\text{ s}, the net force it applied on the floor while bouncing is : (take the gravitational acceleration g=10 m s−2g=10\text{ m s}^{-2})

  1. (a) 1.0 N1.0\text{ N}
  2. (b) 6.0 N6.0\text{ N}Correct answer
  3. (c) 3.0 N3.0\text{ N}
  4. (d) 5.0 N5.0\text{ N}

Solution.

Answer: (b) 6.0 N6.0\text{ N}

  1. The impact speed is √(2 × 10 × 0.45) = 3 m/s downward; the rebound speed is √(2 × 10 × 0.20) = 2 m/s upward.
  2. The upward momentum change is 0.1(2 + 3) = 0.5 kg m/s, giving net force on the ball 5 N.
  3. Add its weight 1 N to obtain the average floor reaction 6 N.
  4. By Newton’s third law the ball exerts 6 N on the floor; 5 N is only the net force on the ball.