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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 (See figure).
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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,\[F = \mu mg\]
Change in K.E. of block = Work done by the friction force
\[\Rightarrow \frac{1}{2} \times M \times 0 - \frac{1}{2} \times M \times \left( v' \right)^2 = \mu \text{mgd}\]
\[ \Rightarrow 0 - \left( \frac{1}{2} \right) \times 10 \times (0 . 8 )^2 = \mu \times 10 \times 10 \times 0 . 2\]
\[ \Rightarrow \mu = 0 . 16\]
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