Question

A bullet of mass 0.1 kg moving horizontally with speed 400 ms^-1 hits a wooden block of mass 3.9 kg kept on a horizontal rough surface. The bullet gets embedded into the block and moves 20m before coming to rest. The coefficient of friction between the block and the surface is ______.
(Given g = 10 m/s²)

• A. 0.25
• B. 0.50
• C. 0.65
• D. 0.90

Answer 1

The Correct Option is A

Step 1: Apply the principle of conservation of momentum. 
The total momentum just before and just after the collision is conserved. The initial momentum of the bullet is: \[ p_{\text{initial}} = m_{\text{bullet}} \cdot v_{\text{bullet}} = 0.1 \cdot 400 = 40 \, \text{kg} \cdot \text{m/s}. \] The combined mass of the block and bullet after the collision is: \[ m_{\text{total}} = m_{\text{block}} + m_{\text{bullet}} = 3.9 + 0.1 = 4 \, \text{kg}. \] Let the velocity of the block and bullet system just after the collision be \( u \). From conservation of momentum: \[ p_{\text{initial}} = p_{\text{final}} \implies 40 = 4 \cdot u. \] Solve for \( u \): \[ u = \frac{40}{4} = 10 \, \text{m/s}. \] 
Step 2: Use work-energy principle. 
The block comes to rest after moving \( 20 \, \text{m} \) due to the work done against friction. The work-energy principle states: \[ \Delta KE = W_{\text{friction}}, \] where: \[ \Delta KE = \frac{1}{2} m_{\text{total}} u^2 - 0 = \frac{1}{2} (4) (10)^2 = 200 \, \text{J}. \] The work done by friction is: \[ W_{\text{friction}} = f \cdot d, \] where \( f = \mu m_{\text{total}} g \) is the frictional force and \( d = 20 \, \text{m} \) is the distance. Substitute: \[ 200 = \mu \cdot (4) \cdot (10) \cdot (20). \] Solve for \( \mu \): \[ 200 = \mu \cdot 800 \implies \mu = \frac{200}{800} = 0.25. \] 
Final Answer: The coefficient of friction is: \[ \boxed{0.25}. \]