Question

A bullet of mass \(0.02\,kg\) travelling horizontally with velocity \(250\,m\,s^{-1}\) strikes a block of wood of mass \(0.23\,kg\) 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 \(40\,m\). The coefficient of kinetic friction on the rough surface is \((g = 9.8\,m\,s^{-2})\)

• A. \(0.75\)
• B. \(0.61\)
• C. \(0.51\)
• D. \(0.30\)

Hint:

For embedding collision: use momentum conservation first, then use friction work \(= \mu Mg d\) to stop the body.

Answer 1

The Correct Option is C

Step 1: Apply conservation of momentum during collision.
Bullet embeds in block, so collision is perfectly inelastic.
Initial momentum of bullet:
\[ p_i = m_b u_b = 0.02 \times 250 = 5\,kg\,m\,s^{-1} \]
Total mass after collision:
\[ M = 0.02 + 0.23 = 0.25\,kg \]
Let common velocity after collision be \(v\):
\[ m_b u_b = Mv \Rightarrow v = \frac{5}{0.25} = 20\,m\,s^{-1} \]
Step 2: Apply work-energy theorem for motion on rough surface.
Kinetic energy after collision:
\[ KE = \frac{1}{2}Mv^2 = \frac{1}{2}(0.25)(20^2) = 0.125 \times 400 = 50\,J \]
Step 3: Work done by friction stops the system.
Friction force:
\[ F_f = \mu_k Mg \]
Work done by friction over \(40m\):
\[ W = F_f \cdot d = \mu_k Mg \cdot 40 \]
Since system comes to rest:
\[ \mu_k Mg \cdot 40 = 50 \]
Step 4: Solve for \(\mu_k\).
\[ \mu_k = \frac{50}{(0.25)(9.8)(40)} \]
\[ \mu_k = \frac{50}{98} \approx 0.51 \]
Final Answer:
\[ \boxed{0.51} \]