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Question
A bullet of mass 20 g moving horizontally at a speed of 300 m/s is fired into a wooden block of mass 500 g suspended by a long string. The bullet crosses the block and emerges on the other side. If the centre of mass of the block rises through a height of 20.0 cm, find the speed of the bullet as it emerges from the block.
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Solution
Given:
Mass of bullet, m = 20 gm = 0.02 kg
Horizontal speed of the bullet, u = 300 m/s
Mass of wooden block, M = 500 gm = 0.5 kg
Let the bullet emerges out with velocity v.
Let the velocity of the block be v'.
Using the law of conservation of momentum, we get:
mu = Mv' + mv ...(1)
Now, applying the work-energy principle for the block after the collision, we get:
\[0 - \left( \frac{1}{2} \right)M \times \left( v' \right)^2 = - Mgh\]
\[ \Rightarrow (v' )^2 = 2gh\]
\[ v' = \sqrt{2gh}\]
\[ = \sqrt{20 \times 10 \times 0 . 2} = 2 \text{ m/s}\]
On substituting the value of v' in equation (1), we get:
\[0 . 02 \times 300 = 0 . 5 \times 2 + 0 . 02 \times v\]
\[ \Rightarrow v = \frac{6 - 1}{0 . 02} = \frac{5}{0 . 02}\]
\[ \Rightarrow v = 250 \text{ m/s}\]
\[0 . 02 \times 300 = 0 . 5 \times 2 + 0 . 02 \times v\]
\[ \Rightarrow v = \frac{6 - 1}{0 . 02} = \frac{5}{0 . 02}\]
\[ \Rightarrow v = 250 \text{ m/s}\]
Hence, the speed of the bullet as it emerges out from the block is 250 m/s.
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