A 13 g bullet is stopped in a block of wood
(
m
w
= 3
.
1 kg). The speed of the bullet-
plus-wood combination immediately after the
collision is 0
.
57 m
/
s.
What was the original speed of the bullet?
Answer in units of m
/
s
To find the original speed of the bullet, we can use the principle of conservation of momentum.
The momentum before the collision is equal to the momentum after the collision.
The bullet's momentum before the collision is given by:
momentum_bullet_before = mass_bullet * initial_velocity_bullet
The momentum after the collision is given by:
momentum_after_collision = (mass_bullet + mass_wood) * final_velocity
According to the problem statement, the bullet's mass is 13 g, which is 0.013 kg, and the wood block's mass is 3.1 kg. The final velocity after the collision is 0.57 m/s.
So, the equation for the conservation of momentum is:
mass_bullet * initial_velocity_bullet = (mass_bullet + mass_wood) * final_velocity
Plugging in the known values:
0.013 kg * initial_velocity_bullet = (0.013 kg + 3.1 kg) * 0.57 m/s
Simplifying the equation:
0.013 kg * initial_velocity_bullet = 3.113 kg * 0.57 m/s
Dividing both sides by 0.013 kg:
initial_velocity_bullet = (3.113 kg * 0.57 m/s) / 0.013 kg
Calculating the result:
initial_velocity_bullet = (1.773 kg·m/s) / 0.013 kg
initial_velocity_bullet ≈ 136.38 m/s
Therefore, the original speed of the bullet was approximately 136.38 m/s.
To find the original speed of the bullet, we can use the principle of conservation of momentum. The momentum before the collision is equal to the momentum after the collision.
The momentum of an object is given by the formula:
Momentum = Mass × Velocity
Before the collision, the bullet has a mass of 13 g, which is equal to 0.013 kg. Let's call its original velocity Vb.
The momentum of the bullet before the collision is:
Momentum_bullet_before = Mass_bullet × Velocity_bullet
After the collision, the bullet becomes embedded in the block of wood, so we can consider the bullet and wood as a single system. The mass of the bullet plus the wood (Mw) is 3.1 kg. The velocity of the bullet-plus-wood combination immediately after the collision is given as 0.57 m/s. Let's call it Vbw.
The momentum of the bullet-plus-wood combination after the collision is:
Momentum_bullet_plus_wood_after = Mass_bullet_plus_wood × Velocity_bullet_plus_wood
According to the principle of conservation of momentum, the momentum before the collision is equal to the momentum after the collision. So we can set up the following equation:
Momentum_bullet_before = Momentum_bullet_plus_wood_after
Mass_bullet × Velocity_bullet = Mass_bullet_plus_wood × Velocity_bullet_plus_wood
Plugging in the known values:
0.013 kg × Vb = 3.1 kg × 0.57 m/s
Solving for Vb, we get:
Vb = (3.1 kg × 0.57 m/s) / 0.013 kg
Vb ≈ 13.5 m/s
Therefore, the original speed of the bullet was approximately 13.5 m/s.