JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Conservation of Momentum

A bullet of mass 0.1 kg moving horizontally with speed

400 ms^{- 1}

hits a wooden block of mass 3.9 kg kept on a horizontal rough surface. The bullet gets embedded into the block and moves 20 m before coming to rest. The coefficient of friction between the block and the surface is ____.

(Given

g = 10 ms^{2}

)

(A)0.50
(B)0.90
(C)0.65
(D)0.25✓ Correct

Answer: (D)

Solution

While the bullet embeds itself, momentum is conserved, so

m u = (m + M) v

gives

v = \frac{0.1 \times 400}{0.1 + 3.9} = \frac{40}{4} = 10 m s^{- 1}

for the bullet-block combination just after impact. The combination then slides to rest under friction, decelerating at

a = μg

, and

v^{2} = 2 a s

gives

v^{2} = 2 μg s

. Substituting,

100 = 2 μ (10) (20) = 400 μ

. The coefficient of friction between the block and the surface is

0.25

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Single correct option

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JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Conservation of Momentum

Solution

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