Search any question & find its solution
Question:
Answered & Verified by Expert
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 $\left(g=9.8 \mathrm{~ms}^{-2}\right)$
Options:
Solution:
1350 Upvotes
Verified Answer
The correct answer is:
0.51
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
$\frac{1}{2} M v^2=\mu R \cdot d$
$\begin{array}{ll}\text { or } & \frac{1}{2} \times 0.25 \times 400=\mu \times 0.25 \times 9.8 \times 40 \\ \Rightarrow & \mu=\frac{200}{9.8 \times 40}=0.51\end{array}$
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
$\frac{1}{2} M v^2=\mu R \cdot d$
$\begin{array}{ll}\text { or } & \frac{1}{2} \times 0.25 \times 400=\mu \times 0.25 \times 9.8 \times 40 \\ \Rightarrow & \mu=\frac{200}{9.8 \times 40}=0.51\end{array}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.