Q. A bullet of mass $0.02\ kg$ travelling horizontally with velocity $250\ ms^{- 1}$ strikes a block of wood of mass $0.23\ kg$ which rests on a rough horizontal surface. After the impact, the block and bullet move together and come to rest after travelling a distance of $40m$. The coefficient of sliding friction of the rough surface is ( $g = 9.8\ ms^{- 2}$)
A) 0.75
B) 0.61
C) 0.51
D) 0.30
Explanation
(c) After impact the mass and block move together and come to rest after a distance of $40\ m$
By conservation of momentum,
$m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}$
$0.02 \times 250 + 0.23 \times 0 = 0.02v + 0.23v$
$5 + 0 = v(0.25)$
[image:6_image92.png]
$\frac{500}{25} = v = 20ms^{- 1}$
Now, by conservation of energy,
$\frac{1}{2}Mv^{2} = \mu R.d$
$\frac{1}{2} \times 0.25 \times 400 = \mu \times 0.25 \times 9.8 \times 40 \Rightarrow \mu = 0.51$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.