A 13g bullet traveling 330 m/s penetrates a 2.0 kg Block of wood and emerges going 270 m/s . If the block is stationary on a frictionless surface when hit how fast does it move after the bullet emerges?
momentum before = momentum after
momentum before = .013 * 330
momentum after = .013 * 270 + 2 v
so
2 v = .013 (330 - 270)
it was a gi Jane joke
thank you
To determine how fast the block moves after the bullet emerges, we need to apply the law of conservation of momentum. According to this law, the total momentum before and after the collision should remain constant.
The momentum of an object is calculated by multiplying its mass by its velocity. Let's define the variables:
m₁ = mass of the bullet = 13g = 0.013 kg (converted to kg)
v₁ = velocity of the bullet before collision = 330 m/s
m₂ = mass of the block of wood = 2.0 kg
v₂ = velocity of the block of wood before collision = 0 m/s (since it is stationary)
v₃ = velocity of the block of wood after collision (what we need to calculate)
Using the conservation of momentum equation:
(m₁ * v₁) + (m₂ * v₂) = (m₁ * v₁') + (m₂ * v₂')
where v₁' is the velocity of the bullet after the collision.
Substituting the given values:
(0.013 kg * 330 m/s) + (2.0 kg * 0 m/s) = (0.013 kg * v₁') + (2.0 kg * v₂')
0.013 kg * 330 m/s = 0.013 kg * v₁' + 2.0 kg * v₂'
Now, we need to solve for v₁'. Rearranging the equation:
0.013 kg * 330 m/s - 2.0 kg * v₂' = 0.013 kg * v₁'
Divide both sides of the equation by 0.013 kg:
330 m/s - 153.85 kg/s * v₂' = v₁'
Now, substitute the given value v₂' = 270 m/s:
330 m/s - 153.85 kg/s * 270 m/s = v₁'
Now, solve for v₁':
v₁' = 330 m/s - 270 m/s * 153.85 kg/s
v₁' ≈ 330 m/s - 41,500.5 kg·m/s
v₁' ≈ 330 m/s - 41.5 m/s
v₁' ≈ 288.5 m/s
Therefore, the block of wood moves at approximately 288.5 m/s after the bullet emerges.