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 )

Step-by-step solution

After impact the bullet and block move together and comes to rest after covering a distance of 40 ~m . By conservation of momentum gathered m₁ u₁+m₂ u₂=m₁ v₁+m₂ v₂ or 0.02 250+0.23 0=0.02 v+0.23 v gathered array r 5+0=v(0.25) 500 25 =v=20 ~ms ⁻¹ array Now, by conservation of energy 1 2 M v^2= R d array ll or & 1 2 0.25 400= 0.25 9.8 40 & = 200 9.8 40 =0.51 array