A 16g bullet moving horizontally at 400 m/s penetrates a 3.2kg wood block resting on a horizontal surface.(Figure 1)
Part A
If the bullet slows down to 300 m/s after emerging from the block, what is the speed of the block immediately after the bullet emerges?
To determine the speed of the wood block immediately after the bullet emerges, we can use the principle of conservation of momentum.
The momentum of an object is calculated by multiplying its mass by its velocity. In this case, we have two objects: the bullet and the wood block.
Given information:
- Mass of the bullet (m1): 16 g = 0.016 kg
- Velocity of the bullet before entering the block (v1i): 400 m/s
- Velocity of the bullet after exiting the block (v1f): 300 m/s
- Mass of the wood block (m2): 3.2 kg
- Velocity of the wood block (v2f): unknown
Using the conservation of momentum, we can write the equation:
(m1 * v1i) + (m2 * v2i) = (m1 * v1f) + (m2 * v2f)
Since the bullet is the only object in motion initially, the velocity of the wood block before the collision (v2i) can be considered zero.
Plugging in the values:
(0.016 kg * 400 m/s) + (3.2 kg * 0 m/s) = (0.016 kg * 300 m/s) + (3.2 kg * v2f)
Simplifying the equation:
(0.016 kg * 400 m/s) = (0.016 kg * 300 m/s) + (3.2 kg * v2f)
Rearranging the equation to solve for v2f:
3.2 kg * v2f = (0.016 kg * 400 m/s) - (0.016 kg * 300 m/s)
v2f = [(0.016 kg * 400 m/s) - (0.016 kg * 300 m/s)] / 3.2 kg
Calculating the value:
v2f = [6.4 kg m/s - 4.8 kg m/s] / 3.2 kg
v2f = 1.6 kg m/s / 3.2 kg
v2f = 0.5 m/s
Therefore, the speed of the wood block immediately after the bullet emerges is 0.5 m/s.