A bullet of mass $4\,g$ is fired horizontally with a speed of $300\,m/s$ into $0.8\,kg$ block of wood at rest on a table. If the coefficient of friction between the block and the table is $0.3,$ how far will the block slide approximately?
JEE MAIN 2014, Difficult
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$\text { Given, } m_1=4 g, u_1=300 m / s$
$m_2=0.8 \ kg=800 g, u_2=0 m / s$
From law of conservation of momentum,
$m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2$
Let the velocity of combined system
$=v m / s$
then,
$4 \times 300+800 \times 0=(800+4) \times v$
$v=\frac{1200}{804}=1.49 m / s$
Now, $\mu=0.3$ (given)
$a=\mu g$
$a=0.3 \times 10\left(\text { take } g=10 m / s^2\right)$
$=3 m / s^2$
then, from $v^2=u^2+2 a s$
$(1.49)^2=0+2 \times 3 \times s$
$s=\frac{\left(1.49^2\right)}{6} ; s=\frac{2.22}{6}=0.379 m$
$m_2=0.8 \ kg=800 g, u_2=0 m / s$
From law of conservation of momentum,
$m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2$
Let the velocity of combined system
$=v m / s$
then,
$4 \times 300+800 \times 0=(800+4) \times v$
$v=\frac{1200}{804}=1.49 m / s$
Now, $\mu=0.3$ (given)
$a=\mu g$
$a=0.3 \times 10\left(\text { take } g=10 m / s^2\right)$
$=3 m / s^2$
then, from $v^2=u^2+2 a s$
$(1.49)^2=0+2 \times 3 \times s$
$s=\frac{\left(1.49^2\right)}{6} ; s=\frac{2.22}{6}=0.379 m$
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