A ball whose mass is 0.3 kg hits the floor with a speed of 7 m/s and rebounds upward with a speed of 5 m/s. If the ball was in contact with the floor for 0.5 ms (0.5multiply10-3 s), what was the average magnitude of the force exerted on the ball by the floor?
To find the average magnitude of the force exerted on the ball by the floor, we can use the impulse-momentum principle.
The impulse-momentum principle states that the change in momentum of an object is equal to the force applied to it multiplied by the time over which the force acts. Mathematically, it can be written as:
Impulse = Force * Time
In this case, the impulse is equal to the change in momentum of the ball. The change in momentum is given by the difference between the final momentum and the initial momentum. Mathematically, it can be written as:
Impulse = (final momentum) - (initial momentum)
To find the final momentum, we can use the formula:
Final momentum = Mass * Final Velocity
Similarly, the initial momentum can be calculated using:
Initial momentum = Mass * Initial Velocity
Substituting these values into the impulse-momentum formula, we get:
Impulse = (Mass * Final Velocity) - (Mass * Initial Velocity)
Now, let's calculate the impulse:
Impulse = (0.3 kg * 5 m/s) - (0.3 kg * 7 m/s)
= 1.5 kg·m/s - 2.1 kg·m/s
= -0.6 kg·m/s
Note that the impulse is negative because the ball changes direction.
Finally, we can find the average magnitude of the force by dividing the impulse by the time over which the force acts:
Average Force = Impulse / Time
Substituting the values given in the problem:
Average Force = (-0.6 kg·m/s) / (0.5 * 10^-3 s)
= -1200 N
The average magnitude of the force exerted on the ball by the floor is 1200 Newtons.