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).

Answer 1

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

\[\Rightarrow \frac{1}{2} \times M \times 0 - \frac{1}{2} \times M \times \left( v' \right)^2 = \mu \text{mgd}\]

\[ \Rightarrow 0 - \left( \frac{1}{2} \right) \times 10 \times (0 . 8 )^2 = \mu \times 10 \times 10 \times 0 . 2\]

\[ \Rightarrow \mu = 0 . 16\]