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NEET PHYSICSMedium

A ball of mass 0.15 kg0.15 \text{ kg} is dropped from a height 10 m10 \text{ m}, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (g=10 m/s2g = 10 \text{ m/s}^2) nearly

A

0 kg m/s0 \text{ kg m/s}

B

4.2 kg m/s4.2 \text{ kg m/s}

C

2.1 kg m/s2.1 \text{ kg m/s}

D

1.4 kg m/s1.4 \text{ kg m/s}

Step-by-Step Solution

Given that: Mass of ball = 0.15 kg0.15 \text{ kg}, Height from which ball is dropped = 10 m10 \text{ m}. Impulse, I⃗=Change in linear momentum=ΔP⃗=P⃗f−P⃗i\vec{I} = \text{Change in linear momentum} = \Delta\vec{P} = \vec{P}_f - \vec{P}_i. Velocity of ball at ground (v)=2gh=2×10×10=102 m/s(v) = \sqrt{2gh} = \sqrt{2 \times 10 \times 10} = 10\sqrt{2} \text{ m/s}. I⃗=0.15×102(−j^)−0.15×102(j^)=2×0.15×102(−j^)=4.2(−j^)\vec{I} = 0.15 \times 10\sqrt{2}(-\hat{j}) - 0.15 \times 10\sqrt{2}(\hat{j}) = 2 \times 0.15 \times 10\sqrt{2}(-\hat{j}) = 4.2(-\hat{j}). Magnitude of impulse = 4.2 kg m/s4.2 \text{ kg m/s}.

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