A bullet of mass $0.02 \mathrm{~kg}$ travelling horizontally with velocity $250 \mathrm{~ms}^{-1}$ 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…

Question

A bullet of mass $0.02 \mathrm{~kg}$ travelling horizontally with velocity $250 \mathrm{~ms}^{-1}$ 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 $\left(g=9.8 \mathrm{~ms}^{-2}\right)$

Aicharya · Accepted Answer

After impact the bullet and block move together and comes to rest after covering a distance of $40 \mathrm{~m}$.

By conservation of momentum $\begin{gathered} m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2 \\ \text { or } \quad 0.02 \times 250+0.23 \times 0=0.02 v+0.23 v \end{gathered}$ $\begin{array}{r} 5+0=v(0.25) \\ \frac{500}{25}=v=20 \mathrm{~ms}^{-1} \end{array}$ Now, by conservation of energy $\frac{1}{2} M v^2=\mu R \cdot d$ $\begin{array}{ll}\text { or } & \frac{1}{2} \times 0.25 \times 400=\mu \times 0.25 \times 9.8 \times 40 \\ \Rightarrow & \mu=\frac{200}{9.8 \times 40}=0.51\end{array}$

A bullet of mass $0.02 \mathrm{~kg}$ travelling horizontally with velocity $250 \mathrm{~ms}^{-1}$ strikes a…

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