JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Conservation of Momentum

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A spherical body of mass 100 g is dropped from a height of 10 m from the ground. After hitting the ground, the body rebounds to a height of 5 m . The impulse of force imparted by the ground to the body is given by : (given

g = 9.8 m / s^{2}

)

(A) $4.32 kg ms^{- 1}$
(B) $43.2 kg ms^{- 1}$
(C) $23.9 kg ms^{- 1}$
(D) $2.39 kg ms^{- 1}$ ✓ Correct

Answer: (D)

Solution

Speed just before impact:

v_{1} = 2 g h_{1} = 2 (9.8) (10) = 14 m/s

(downward).

Speed just after rebound:

v_{2} = 2 g h_{2} = 2 (9.8) (5) = 72 m/s

(upward).

Impulse

= m (v_{2} + v_{1}) = 0.1 (14 + 72) \approx 0.1 (23.9) = 2.39 kg m/s

JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Single correct option

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