Final answer:
The third piece of the exploded shell has a downward velocity of 6.25 m/s, found using the conservation of momentum principle.
Step-by-step explanation:
To find the velocity of the third piece after the explosion, we use the conservation of momentum. Before the explosion, the total momentum of the shell is the product of its mass and its velocity. The conservation of momentum principle states that, in the absence of external forces, the total momentum of a system remains constant before and after an event. Therefore, the initial momentum of the shell must be equal to the sum of the momenta of all three pieces after the explosion.
Before the explosion:
Total initial momentum = mass of shell × velocity = 8.0 kg × 5.0 m/s = 40.0 kg·m/s upwards
After the explosion (in the vertical direction):
Momentum of the first piece = 0 kg·m/s (since it moves horizontally)
Momentum of the second piece = mass × velocity = 1.0 kg × 15 m/s = 15.0 kg·m/s upwards
Momentum of the third piece (upwards or downwards) + 15.0 kg·m/s (second piece's momentum) = 40.0 kg·m/s (total momentum)
Let's consider the momentum of the third piece is 'p' in a downward direction (negative). We can establish an equation:
- p + 15.0 kg·m/s = 40.0 kg·m/s
p = -25.0 kg·m/s (since it's in the opposite direction)
Finally, to find the velocity of the third piece, we divide its momentum by its mass (8.0 kg - 3.0 kg - 1.0 kg = 4.0 kg):
Velocity of the third piece = -25.0 kg·m/s ÷ 4.0 kg = -6.25 m/s
Therefore, the third piece is moving downwards with a velocity of 6.25 m/s.