A bullet of mass $4 \mathrm{~g}$ is fired horizontally with a speed of $300 \mathrm{~m} / \mathrm{s}$ into…
A bullet of mass $4 \mathrm{~g}$ is fired horizontally with a speed of $300 \mathrm{~m} / \mathrm{s}$ into $0.8 \mathrm{~kg}$ 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?
$0.19 \mathrm{~m}$
$0.379 \mathrm{~m}$
$0.569 \mathrm{~m}$
$0.758 \mathrm{~m}$
Solution
Given, $\mathrm{m}_1=4 \mathrm{~g}, \mathrm{u}_1=300 \mathrm{~m} / \mathrm{s}$ $\mathrm{m}_2=0.8 \mathrm{~kg}=800 \mathrm{~g}, \mathrm{u}_2=0 \mathrm{~m} / \mathrm{s}$ From law of conservation of momentum, $\mathrm{m}_1 \mathrm{u}_1+\mathrm{m}_2 \mathrm{u}_2=\mathrm{m}_1 \mathrm{v}_1+\mathrm{m}_2 \mathrm{v}_2$ Let the velocity of combined system $=\mathrm{v} \mathrm{m} / \mathrm{s}$ then, $4 \times 300+800 \times 0=(800+4) \times v$ $\mathrm{v}=\frac{1200}{804}=1.49 \mathrm{~m} / \mathrm{s}$
Now, $\mu=0.3$ (given)
$
\begin{aligned}
&\mathrm{a}=\mu \mathrm{g} \\
&\mathrm{a}=0.3 \times 10\left(\text { take } \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right) \\
&=3 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$
then, from $\mathrm{v}^2=\mathrm{u}^2+2 \mathrm{as}$
$
(1.49)^2=0+2 \times 3 \times s
$
$\mathrm{s}=\frac{\left(1.49^2\right)}{6} ; \mathrm{s}=\frac{2.22}{6}=0.379 \mathrm{~m}$