Answer:
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. .
