A 0.005kg bullet is fired at 400m/s straight into a stationary 1.0kg block of wood. The bullet exits the far side of the block and continues on the same straight line. The block also moves off in this same direction with a velocity of 1.2m/s. Calculate the velocity of the bullet after it exits the block.
conservation of momentum
initial momentum=final momentum
.005*400=blockmass*1.2+ bulletmass*Vf
solve for Vf
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
1. Calculate the total momentum before the collision:
Total initial momentum = (mass of the bullet x velocity of the bullet) + (mass of the block x velocity of the block)
= (0.005 kg x 400 m/s) + (1.0 kg x 0 m/s)
= 2 kgm/s
The total initial momentum is 2 kgm/s.
2. Calculate the total momentum after the collision:
Total final momentum = (mass of the bullet x velocity of the bullet after exiting the block) + (mass of the block x velocity of the block after the collision)
= (0.005 kg x velocity of the bullet after exiting the block) + (1.0 kg x 1.2 m/s)
Since the bullet exits the block and continues on the same straight line, the velocity of the bullet after exiting the block and the velocity of the block after the collision are in the same direction and can be added algebraically.
Final momentum = (0.005 kg x velocity of the bullet after exiting the block) + (1.0 kg x 1.2 m/s)
Total final momentum = 2 kgm/s
Therefore, we have the equation:
2 kgm/s = (0.005 kg x velocity of the bullet after exiting the block) + (1.0 kg x 1.2 m/s)
3. Solve for the velocity of the bullet after it exits the block:
(0.005 kg x velocity of the bullet after exiting the block) = 2 kgm/s - (1.2 kgm/s)
(0.005 kg x velocity of the bullet after exiting the block) = 0.8 kgm/s
velocity of the bullet after exiting the block = 0.8 kgm/s / 0.005 kg
velocity of the bullet after exiting the block ≈ 160 m/s
Therefore, the velocity of the bullet after it exits the block is approximately 160 m/s.
To solve this problem, we can apply the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
The momentum before the collision is given by the product of mass and velocity. Let's denote the velocity of the bullet after it exits the block as 'v', and the velocity of the block as 'V'.
The momentum before the collision:
m_bullet * v_bullet + m_block * v_block
The momentum after the collision:
m_bullet * v
Since the bullet is fired straight into the block, it completely stops upon collision. Therefore, the velocity of the bullet after the collision is zero.
Now, let's calculate the total momentum before the collision:
(0.005 kg * 400 m/s) + (1.0 kg * 0 m/s)
= (2 kg * m/s)
Since momentum is conserved, it remains the same after the collision. So we have:
(0.005 kg * v) + (1.0 kg * 1.2 m/s) = 2 kg * m/s
Now, solving for v:
0.005 kg * v = 2 kg * m/s - (1.0 kg * 1.2 m/s)
0.005 kg * v = 2 kg * m/s - 1.2 kg * m/s
0.005 kg * v = 0.8 kg * m/s
v = (0.8 kg * m/s) / 0.005 kg
v ≈ 160 m/s
Therefore, the velocity of the bullet after it exits the block is approximately 160 m/s.