Q117 / 120 · VITEEE 2009 · Physics
A bullet of mass $0.02 \mathrm{~kg}$ travelling horizontally with velodity $250 \mathrm{~ms}^{1}$ strikes a block of wood of mass $0.23 \mathrm{~kg}$ which rests on a rough horizontal surface. After the impact, the block and bullet move together and come to rest after travelling a distance of $40 \mathrm{~m}$. The coefficient of sliding friction of the rough surface is $\left(\mathrm{g}=9.8 \mathrm{~ms}^{-2}\right)$
- A$0.75$
- B$0.61$
- Correct answerC$0.51$
- D$0.30$
Solution
After collision the bullet and block move together and comes to rest after covering a distance of $40 \mathrm{~m}$.
By conservation of momentum $$ \begin{array}{l} m_{1} u_{1}+m_{2} u_{2}=m_{1} v_{1}+m_{2} v_{2} \\ \Rightarrow 0.02 \times 250+0.23 \times 0=0.02 v+0.23 v \\ 5+0=v(0.25) \text { or } v=20 \mathrm{~ms}^{-1} \end{array} $$ Now, by conservation of energy $$ \begin{array}{l} \text { or } \frac{1}{2} M v^{2}=\mu R \cdot d \\ \text { or } \frac{1}{2} \times 0.25 \times 400=\mu \times 0.25 \times 9.8 \times 40 \\ \Rightarrow \mu=\frac{200}{9.8 \times 40}=0.51 \end{array} $$
By conservation of momentum $$ \begin{array}{l} m_{1} u_{1}+m_{2} u_{2}=m_{1} v_{1}+m_{2} v_{2} \\ \Rightarrow 0.02 \times 250+0.23 \times 0=0.02 v+0.23 v \\ 5+0=v(0.25) \text { or } v=20 \mathrm{~ms}^{-1} \end{array} $$ Now, by conservation of energy $$ \begin{array}{l} \text { or } \frac{1}{2} M v^{2}=\mu R \cdot d \\ \text { or } \frac{1}{2} \times 0.25 \times 400=\mu \times 0.25 \times 9.8 \times 40 \\ \Rightarrow \mu=\frac{200}{9.8 \times 40}=0.51 \end{array} $$
Hence, the correct answer is option: 'C'.