Question

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 40 m. 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. 0.3

Answer 1

Answer: (C)

0.51

By conservation of momentum,
$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$
or $0.02\times 250+0.23\times 0=0.02v+0.23v$
$5+0=v\left(0.25\right)$
$v=\frac{500}{25}$

$v=20m{s}^{-1}$
Now, by conservation of energyor
$\frac{1}{2}M{v}^2=\mu Mg.d$
or $\frac{1}{2}\times 0.25\times 400=\mu \times 0.25\times 9.8\times 40$
$\Rightarrow \mu =\frac{200}{9.8\times 40}=0.51$