A bullet of mass 10 g moving horizontally with a velocity of strikes a wooden block of mass 2 kg which is suspended by a light inextensible string of length 5 m. As a result, the centre of gravity of the block is found to rise a vertical distance of 10 c

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Solution

Q.
A bullet of mass 10 g moving horizontally with a velocity of $400$ ${ms}^{- 1}$ strikes a wooden block of mass 2 kg which is suspended by a light inextensible string of length 5 m. As a result, the centre of gravity of the block is found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges out horizontally from the block will be [2016]

1 $100$ ${ms}^{- 1}$

2 $80$ ${ms}^{- 1}$

3 $120$ ${ms}^{- 1}$

4 $160$ ${ms}^{- 1}$

Ans.
(3)

Mass of bullet, $m = 10$ $g = 0.01$ $kg$

Initial speed of bullet, $u = 400$ ${ms}^{- 1}$

Mass of block, $M = 2$ $kg$

Length of string, $l = 5$ $m$

Speed of the block after collision $= v_{1}$

Speed of the bullet on emerging from block, $v = ?$

Using energy conservation principle for the block,

${(KE + PE)}_{Reference} = {(KE + PE)}_{h}$

$\Rightarrow \frac{1}{2} M v_{1}^{2} = M g h or, v_{1} = \sqrt{2 g h}$

$v_{1} = \sqrt{2 \times 10 \times 0.1} = \sqrt{2}$ ${ms}^{- 1}$

Using momentum conservation principle for block and bullet system,

${(M \times 0 + m u)}_{Before collision} = {(M \times v_{1} + m v)}_{After collision}$

$\Rightarrow 0.01 \times 400 = 2 \sqrt{2} + 0.01 \times v$

$\Rightarrow v = \frac{4 - 2 \sqrt{2}}{0.01} = 117.15 {ms}^{- 1} \approx 120 {ms}^{- 1}$

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