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Question

A bullet of mass 20 g travelling horizontally with a speed of 500 m/s passes through a wooden block of mass 10.0 kg initially at rest on a level surface. The bullet emerges with a speed of 100 m/s and the block slides 20 cm on the surface before coming to rest. Find the friction coefficient between the block and the surface.
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Solution

It is given that:
Mass of bullet, m = 20 g =0.02 kg
The initial speed, v1 = 500 m/s
Mass of block, M = 10 kg
The initial speed of block = 0
Final velocity of bullet, v2= 100 m/s
Let the final velocity of block when the bullet emerges out = v'

Applying conservation of linear momentum,
mv1 + M × 0 = mv2 + Mv'
⇒ 0.02 × 500 = 0.02 × 100 + 10 × v'
⇒ v' = 0.8 m/s

Distance covered by the block, d = 20 cm = 0.02 m .
Let friction coefficient between the block and the surface = μ
Thus, the value of friction force, F=μmgF=μmg

Change in K.E. of block = Work done by the friction force
12×M×0-12×M×v'2=μmgd 0-12×10×(0.8)2=μ×10×10×0.2 μ=0.16

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