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A block of mass $1.9 k g$ is at rest at the edge of a table, of height $1 m$ . A bullet of mass $0.1 k g$ 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 collision then the kinetic energy 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.]

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$20 J$

$19 J$

$21 J$

$23 J$

answer is B.

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Detailed Solution

Given,
Mass of block, $m_{1} = 1.9 k g$
Mass of bullet, $m_{2} = 0.1 k g$
Velocity of bullet, $v_{2} = 20 m / s$

Let $v$ be the velocity of the combined system. It is an inelastic collision.

Using conservation of linear momentum

$m_{1} \times 0 + m_{2} \times v_{2} = (m_{1} + m_{2}) v$
$\Rightarrow 0.1 \times 20 = (0.1 + 1.9) \times v \Rightarrow v = 1 m / s$

Using work energy theorem

Work done = change in kinetic energy

Let K be the kinetic energy of combined system.

$(m_{1} + m_{2}) g h = K - \frac{1}{2} (m_{1} + m_{2}) V^{2} \Rightarrow 2 \times g \times 1 = K - \frac{1}{2} \times 2 \times 1^{2} \Rightarrow K = 21 J$

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