Work Energy Power

JEE Main level, single correct, medium

70 of 183

Question 70 of 183

A bullet of mass 0.1kg0.1kg moving horizontally with speed 400ms−1400m{s}^{-1} hits a wooden block of mass 3.9kg3.9kg kept on a horizontal rough surface. The bullet gets embedded into the block and moves 20m20m before coming to rest. The coefficient of friction between the block and the surface is _______.

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Answer: (D) 0.250.25

Let the mass of the bullet be mm and mass of the block be MM. It is given that V=400ms−1V=400m{s}^{-1}. By conservation of momentum,
mV=(m+M)vmV=(m+M)v
⇒0.1×400=(0.1+3.9)v\Rightarrow 0.1\times 400=(0.1+3.9)v
⇒v=10ms−1\Rightarrow v=10m{s}^{-1}
The force acting on the block
f=μm′g⇒a=μgf=\mu m'g \Rightarrow a=\mu g
Using v′2−v2=−2aSv{'}^{2}-{v}^{2}=-2aS
⇒\Rightarrow 02−102=2(−μg)(S){0}^{2}-{10}^{2}=2(-\mu g)(S)
⇒μ=1002×10×20=0.25\Rightarrow \mu =\dfrac{100}{2\times 10\times 20}=0.25

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