Respuesta :
The initial speed of the bullet should be 159.32 m/s.
What was the initial speed of the bullet?
Let the initial speed 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
What kind of collision took place and calculate the impulse of the block?
The impulse of the friction force produced by the surface during the impact is disregarded when we apply conservation of momentum to the collision. This makes sense given that the force is far less than the forces that the bullet and block will exert on one another when they collide. The block moves after the collision and is affected by this force, which removes all of the kinetic energy.
Know more about bullet:
brainly.com/question/7773692
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