Center of Mass, Momentum & Collision

JEE Main level, single correct, medium

73 of 130

Question 73 of 130

A spherical body of mass 100g100g is dropped from a height of 10m10m from the ground. After hitting the ground, the body rebounds to a height of 5m5m. The impulse of force imparted by the ground to the body is given by: (given g=9.8ms−2g=9.8m{s}^{-2})

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Answer: (D) 2.39kgms−12.39kgm{s}^{-1}

Using equation of motion with constant acceleration, we can write:
v1=2gh1{v}_{1}=\sqrt{2g{h}_{1}} and v2=2gh2{v}_{2}=\sqrt{2g{h}_{2}}
Now, required impulse
I⃗=Δp⃗=p⃗f−p⃗i\vec{I}=\Delta \vec{p}={\vec{p}}_{f}-{\vec{p}}_{i}
Given, M=0.1kgM=0.1kg
Therefore,
I=Δp=0.1(2×9.8×5−(−2×9.8×10))I=\Delta p=0.1(\sqrt{2\times 9.8\times 5}-(-\sqrt{2\times 9.8\times 10}))
=0.1(72+14)≈2.39kgms−1=0.1(7\sqrt{2}+14)\approx 2.39kgm{s}^{-1}

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