A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.600 m. What impulse was given to the ball by the floor?
2.2
To determine the impulse given to the ball by the floor, we can use the principle of conservation of energy. The initial potential energy of the ball is equal to the final potential energy after rebounding.
Step 1: Calculate the initial potential energy (PEi):
Potential energy (PEi) = mass × acceleration due to gravity × height
PEi = 0.120 kg × 9.8 m/s² × 1.25 m
Step 2: Calculate the final potential energy (PEf):
PEf = mass × acceleration due to gravity × height
PEf = 0.120 kg × 9.8 m/s² × 0.600 m
Step 3: Calculate the change in potential energy:
Change in potential energy (ΔPE) = PEf - PEi
Step 4: Calculate the impulse using the relationship between impulse and change in potential energy:
Impulse = ΔPE
Now, let's plug in the values and find the result:
PEi = 0.120 kg × 9.8 m/s² × 1.25 m
PEi = 1.47 J (joules)
PEf = 0.120 kg × 9.8 m/s² × 0.600 m
PEf = 0.706 J (joules)
ΔPE = PEf - PEi
ΔPE = 0.706 J - 1.47 J
ΔPE = -0.764 J (since the potential energy decreases during rebound)
Therefore, the impulse given to the ball by the floor is -0.764 J, indicating a change in momentum in the opposite direction.