Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A ball is dropped on the floor from a height of $10 \text{ m}$ . It rebounds to a height of $2.5 \text{ m}$ . If the ball is in contact with the floor for $0.01 \text{ sec}$ , the average acceleration during contact is:

2100 m/s² downwards

2100 m/s² upwards

1400 m/s²

700 m/s²

Step-by-Step Solution

Velocity just before impact ( $v_i$ ): The ball falls freely from height $h_1 = 10 \text{ m}$ . Using the kinematic equation $v^2 = u^2 + 2gh$ with initial velocity $u=0$ and $g=9.8 \text{ m/s}^2$ : $v_i = \sqrt{2gh_1} = \sqrt{2 \times 9.8 \times 10} = \sqrt{196} = 14 \text{ m/s}$ . Taking the upward direction as positive, $v_i = -14 \text{ m/s}$ (downwards) .
Velocity just after rebound ( $v_f$ ): The ball rises to a height $h_2 = 2.5 \text{ m}$ . Using $v^2 = u^2 - 2gh$ where the final velocity at the peak is 0: $0 = v_f^2 - 2(9.8)(2.5) \implies v_f = \sqrt{49} = 7 \text{ m/s}$ . Since it is rebounding, $v_f = +7 \text{ m/s}$ (upwards) .
Change in Velocity ( $\Delta v$ ): $\Delta v = v_f - v_i = (+7) - (-14) = 21 \text{ m/s}$ .
Average Acceleration: Average acceleration is the change in velocity divided by the time interval ( $\Delta t = 0.01 \text{ s}$ ). $a_{avg} = \frac{\Delta v}{\Delta t} = \frac{21}{0.01} = 2100 \text{ m/s}^2$ .
Direction: Since the change in velocity is positive, the acceleration is directed upwards.

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