Question

A bullet of mass $1\times {10}^{-2}$$kg$ moving horizontally with a velocity of $2\times {10}^{2}m{s}^{-1}$ strikes a block of mass $1.99kg$ and gets embedded into it. If the coefficient of kinetic friction between the block and the horizontal surface is $0.25$, then the distance moved by the block with the bullet before coming to rest is

• A. 0.4m
• B. 0.8m
• C. 0.6m
• D. 0.2m

Answer 1

Answer: (D)

0.2m

From the momentum conservation, we have
$0.01\times 200=(0.01+1.99)\times V$
$V=1m/s$

Also,
$F=\mu (m+M)g$; where $m$ and $M$ is mass of bullet and mass of block respectively.
$(0.01+1.99)a=0.25(0.01+1.99)g$

$a=-0.25g=-2.5m/s^2$ -ve sign indicates deceleration

Now,
$v^2-u^2=2as$
$S=-1^2/(-2\times 2.5)$
$S=0.2m$