A bullet of mass 0 . 1 kg moving horizontally with speed 400 m s - 1 hits a wooden block of mass 3 . 9 kg kept on a horizontal rough surface. The bullet gets embedded into the block and moves 20 m before coming to rest. The coefficient of friction between the block and the surface is _______.

Step-by-step solution

Let the mass of the bullet be m and mass of the block be M . It is given that V = 400   m   s - 1 . By conservation of momentum, m V = ( m + M ) v ⇒ 0 . 1 × 400 = ( 0 . 1 + 3 . 9 ) v ⇒ v = 10   m   s - 1 The force acting on the block f = μ m ' g ⇒ a = μ g Using v ' 2 - v 2 = - 2 a S ⇒ 0 2 - 10 2 = 2 ( - μ g ) ( S ) ⇒ μ = 100 2 × 10 × 20 = 0 . 25