A ball of mass 1.20kg falls vertically and hits a horizontal floor at 2.50m/s . The ball then rebounds upward at 10.00m/s . If the ball is in contact with the floor for 0.02 seconds, compute the average force exerted on the floor.
m=4Kg , v=6m/s momentum changed in 48 kg m/s
To compute the average force exerted on the floor, we can use the concept of impulse. Impulse is defined as the change in momentum of an object and is equal to the average force applied multiplied by the time interval over which it acts.
We can use the equation for impulse:
Impulse = change in momentum
Impulse = (final momentum) - (initial momentum)
First, let's find the initial momentum of the ball before it hits the floor. We can calculate momentum using the equation:
Momentum = mass × velocity
Given:
Mass of the ball (m) = 1.20 kg
Velocity of the ball before hitting the floor (v₁) = 2.50 m/s
Initial momentum (P₁) = m × v₁
= 1.20 kg × 2.50 m/s
Next, we need to find the final momentum of the ball after it rebounds upward. The velocity of the ball changes direction as it rebounds, so the final velocity will be negative.
Given:
Velocity of the ball after rebounding (v₂) = -10.00 m/s
Final momentum (P₂) = m × v₂
= 1.20 kg × (-10.00 m/s)
Now, we can calculate the impulse by subtracting the initial momentum from the final momentum:
Impulse = P₂ - P₁
Once we have the impulse, we can find the average force exerted on the floor by using the equation:
Average force = Impulse / time interval
Given:
Time interval (t) = 0.02 seconds
Average force = Impulse / t
= (P₂ - P₁) / t
Now, let's plug in the values and calculate the average force exerted on the floor.