A ball of 0.1 \text{ kg} mass is dropped on a hard floor from a height of 0.45 \text{ m} and rises to a height of 0.20 \text{ m}. If it was in touch with the floor for 0.1 \text{ s}, the net force it applied on the floor while bouncing is : (take the gravitational acceleration g=10 \text{ m s}^{-2})
- A. 1.0 N
- B. 6.0 N ✓
- C. 3.0 N
- D. 5.0 N
Correct Answer: B. 6.0 N
Explanation
Velocity just before impact v_1 = \sqrt{2gh_1} = 3 \text{ m/s}. Velocity after impact v_2 = \sqrt{2gh_2} = 2 \text{ m/s}. Change in momentum \Delta p = 0.1 \times (2 - (-3)) = 0.5 \text{ kg m/s}. The average impact force is \Delta p / \Delta t = 5 \text{ N}. The net force applied on the floor is the impact force plus the weight of the ball (1 \text{ N}), which is 6.0 \text{ N}.
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