Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

A ball of mass 0.15 kg is dropped from a height 10 m, strikes the ground, and rebounds to the same height. The magnitude of impulse imparted to the ball is ( $g=10 \text{ m/s}^2$ ) nearly:

2.1 kg-m/s

1.4 kg-m/s

0 kg-m/s

4.2 kg-m/s

Step-by-Step Solution

Calculate Impact Velocity: The velocity of the ball just before hitting the ground is given by kinematic equation $v = \sqrt{2gh}$ . $v = \sqrt{2 \times 10 \times 10} = \sqrt{200} = 10\sqrt{2} \text{ m/s}$ Taking downward direction as negative, $v_i = -10\sqrt{2} \text{ m/s}$ .
Calculate Rebound Velocity: Since the ball rebounds to the same height, the magnitude of velocity remains the same, but the direction is reversed (upward). $v_f = +10\sqrt{2} \text{ m/s}$
Calculate Impulse: Impulse is defined as the change in momentum ( $I = \Delta p = m(v_f - v_i)$ ). $I = 0.15 \times [10\sqrt{2} - (-10\sqrt{2})]$ $I = 0.15 \times 20\sqrt{2}$ $I = 3\sqrt{2} = 3 \times 1.414 \approx 4.24 \text{ kg-m/s}$ [Source 60, 61]

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