Solve the following :


A bullet of mass $20 \mathrm{~g}$ travelling horizontally with a speed of $500 \mathrm{~m} / \mathrm{s}$ passes through a wooden block of mass $10.0 \mathrm{~kg}$ initially at rest on a level surface. The bullet emerges with a speed of $100 \mathrm{~m} / \mathrm{s}$ and the block slides $20 \mathrm{~cm}$ on the surface before coming to rest. Find the friction coefficient between the block and the surface.


$f=m a$

$\mu N=m a$

$\mu m g=m a$

$a=\mu .10$

Use C.O.L.M

$0.02(500)=100(0.02)+10 v$

$v=\frac{0.02 \times 400}{10}=0.8$

Use law of kinematics,

$v^{2}=u^{2}+2 a x$

$a=\frac{v^{2}}{2 x}$

$=\frac{(0.8)^{2}}{2 \times 0.2} \quad\{x=20 \mathrm{~cm}=0.02 \mathrm{~m}\}$

$=\frac{0.64}{2 \times 0.2}=\frac{0.64}{0.4}=1.6$

$a=\mu .10 \Rightarrow \mu=\frac{a}{10}=\frac{1.6}{10}=0.16$


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