A block of mass 1.9 kg is at rest at the edge of a table, of height 1 m. A bullet of mass 0.1 kg collides with the block and sticks to it. If the velocity of the bullet is \(20 \mathrm{~m} / \mathrm{s}\) in the horizontal direction just before the combined system strikes the floor, is [Take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\). Assume there is no rotational motion and loss of energy after the collision is negligible.]

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Conservation of Momentum

A block of mass 1.9 kg is at rest at the edge of a table, of height 1 m. A bullet of mass 0.1 kg collides with the block and sticks to it. If the velocity of the bullet is

20 m / s

in the horizontal direction just before the combined system strikes the floor, is [Take

g = 10 m / s^{2}

. Assume there is no rotational motion and loss of energy after the collision is negligible.]

(A)21 J✓ Correct
(B)20 J
(C)23 J
(D)19 J

Answer: (A)

Solution

The bullet embeds in the block, so momentum is conserved in the collision:

0.1 \times 20 = (0.1 + 1.9) v_{x}

, giving a horizontal speed

v_{x} = 1 m s^{- 1}

. Falling the

1 m

height adds a vertical component

v_{y} = 2 g h = 2 \times 10 \times 1 = 20 m s^{- 1}

. Just before landing the kinetic energy of the

2 kg

system is

\frac{1}{2} (2) (v_{x}^{2} + v_{y}^{2}) = \frac{1}{2} \times 2 \times (1 + 20) = 21 J

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Single correct option

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JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Conservation of Momentum

Solution

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