JEE Mains · Physics · STD 11 - 4.2 friction
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 \(20\,m\) before coming to rest. The coefficient of friction between the block and the surface is \(........\) \(\left(\text { Given } g=10 \,ms ^2\right. \text { ) }\)
- A \(0.50\)
- B \(0.90\)
- C \(0.65\)
- D \(0.25\)
Answer & Solution
Correct Answer
(D) \(0.25\)
Step-by-step Solution
Detailed explanation
\(P _{ i }= P _{ f }(\text { Collision })\) \(\Rightarrow(0.1)(400)=(0.1+3.9) v\) \(\Rightarrow v =\frac{0.1 \times 400}{4}=10\,m / s\) \(a =\frac{\mu mg }{ m }=\mu g\) Apply equation of motion, \(v ^2=u^2+2 as\) \(\Rightarrow 0=(10)^2-2 \mu g \times 20\)…
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