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,
$v^{2}=u^{2}+2 a x$
$a=\frac{v^{2}}{2 x}$
$=\frac{(0.8)^{2}}{2 \times 0.2} \quad\{x=20 \mathrm{~cm}=0.02 \mathrm{~m}\}$
$=\frac{0.64}{2 \times 0.2}=\frac{0.64}{0.4}=1.6$
$a=\mu .10 \Rightarrow \mu=\frac{a}{10}=\frac{1.6}{10}=0.16$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.