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.

Answer 1

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:
$$\begin{gathered} 0-\left(\frac{1}{2}\right)M\times {\left(v\text{'}\right)}^2=-Mgh \\ \Rightarrow (v\text{'}{)}^2=2gh \\ v\text{'}=\sqrt{2gh} \\ =\sqrt{20\times 10\times 0.2}=2\mathrm{m}/\mathrm{s} \end{gathered}$$


On substituting the value of v' in equation (1), we get:

$$\begin{gathered} 0.02\times 300=0.5\times 2+0.02\times v \\ \Rightarrow v=\frac{6-1}{0.02}=\frac{5}{0.02} \\ \Rightarrow v=250\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed of the bullet as it emerges out from the block is 250 m/s.