Question

A bullet of mass 20g 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. The friction coefficient between the block and the surface is $g=10m{s}^{-2}$

• A. 0.16
• B. 0.24
• C. 0.32
• D. 0.50

Answer 1

The correct option is A

0.16

$20\times {10}^{-3}\times 500+10.0\times 0=20\times {10}^{-3}\times 100+10.0V$

$20\times {10}^{-3}\times 400=10.0\Rightarrow V=0.80m{s}^{-1}$ which becomes the initial velocity of the block.

$\mu =\frac{u^2}{2sg}=\frac{(0.80{)}^2}{2\times 0.20\times 10}=0.16$