Q131 / 160 · TS EAMCET 2020 (10 Sep Shift 1) · Physics
A bullet of mass $25 \mathrm{~g}$ moves horizontally at a speed of $250 \mathrm{~m} / \mathrm{s}$ is fired into a wooden block of mass $1 \mathrm{~kg}$ suspended by a long string. The bullet crosses the block and emerges on the other side. If the centre of the mass of the block rises through a height of $20 \mathrm{~cm}$. The speed of the bullet as it emerges from the block is (take, $g=10 \mathrm{~m} / \mathrm{s}^2$ )
- A300 m/s
- B220 m/s
- C150 m/s
- Correct answerD170 m/s
Solution
The given situation is shown in the following figure
Let $v_1$ and $v_2$ are final velocities of bullet and block, respectively, as block rises upto height $h$. Hence, according to conservation of energy, $\frac{1}{2} M v_2^2=M g h \Rightarrow v_2=\sqrt{2 g h}$ Given, $h=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}$ $\Rightarrow \quad v_2=\sqrt{2 \times 10 \times 20 \times 10^{-2}}=2 \mathrm{~m} / \mathrm{s}$ Now, conservation of momentum gives $\begin{aligned} m u & =m v_1+M v_2 \\ \frac{25}{1000} \times 250 & =\frac{25}{1000} \times v_1+1 \times 2 \\ \frac{25}{4} & =\frac{25}{1000} \times v_1+2 \Rightarrow v_1=170 \mathrm{~ms}^{-1} \end{aligned}$
Let $v_1$ and $v_2$ are final velocities of bullet and block, respectively, as block rises upto height $h$. Hence, according to conservation of energy, $\frac{1}{2} M v_2^2=M g h \Rightarrow v_2=\sqrt{2 g h}$ Given, $h=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}$ $\Rightarrow \quad v_2=\sqrt{2 \times 10 \times 20 \times 10^{-2}}=2 \mathrm{~m} / \mathrm{s}$ Now, conservation of momentum gives $\begin{aligned} m u & =m v_1+M v_2 \\ \frac{25}{1000} \times 250 & =\frac{25}{1000} \times v_1+1 \times 2 \\ \frac{25}{4} & =\frac{25}{1000} \times v_1+2 \Rightarrow v_1=170 \mathrm{~ms}^{-1} \end{aligned}$
Hence, the correct answer is option: 'D'.