A bullet of mass 6.00g is fired horizontally into a wooden block of mass 1.20kg resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.160. The bullet remains embedded in the block, which is observed to slide a distance 0.200m along the surface before stopping. What is the initial speed of the bullet?

Let the initial sped of the bullet be V.

Momentum will be conserved after bullet hits the block. If v₁ be the common velocity of bullet block system

6.00 x 10⁻³ x V + 0 = (6 x 10⁻³ + 1.2) v₁

= (6 + 1200) x 10⁻³ v₁

v₁ = (6 / 1206) x V

1 / 201 V

Now the moving bullet - block system faces friction force and ultimately they come to rest, so

Kinetic energy of bullet block system = work done by friction

Kinetic energy of bullet block system =

1/2 x (6 + 1200) x 10⁻³ x (1 / 201 V) ²

work done by friction

= frictional force x displacement

μ mg x d (μ is coefficient of friction of surface, m is mass of the bullet block system and d is displacement)

0.16 x (6 + 1200) x 10⁻³ x 9.8 x. 2

= 38.592 x 9.8 x 10⁻³

1/2 x (6 + 1200) x 10⁻³ x (1 / 201) ² V ² =

= 38.592 x 9.8 X 10⁻³

V² = 25382.65

V = 159.32 ms⁻¹

This is the initial speed of bullet ...