Question

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?

Answer 1

Step-by-step explanation:

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. .