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 arid 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)

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 arid 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) 0.30

Tardigrade Admin · Accepted Answer

After impact the bullet and block move together and come's to rest after covering 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}

or

0.02 \times 250 + 0.23 \times 0 = 0.02 v + 0.23 v

5 + 0 = v (0.25)

\frac{500}{25} v = 20 m s^{- 1}

Now, by conservation of energy

\frac{1}{2} M v^{2} = μ R . d

or

\frac{1}{2} \times 0.25 \times 400 = μ \times 0.25 \times 9.8 \times 40

\Rightarrow μ = \frac{200}{9.8 \times 40} = 0.51