A rubber ball, with a mass of 40.0 grams is dropped from rest from a height of 1.60 m above the floor. It hits the floor, and then reaches a maximum height of 90.0 cm when it comes back up again. In this problem, use g = 10.0 m/s2.
a) The collision with the floor causes some mechanical energy to be lost (this energy generally ends up as thermal energy). How much mechanical energy is lost in this case?
b) What is the ball's speed, just as it leaves the floor on its way up?
c) What is the magnitude of the impulse experienced by the ball for the entire time it is in contact with the floor?
a. V^2 = Vo^2 + 2g*h
V^2 = 0 + 20*1.6 = 32
V = 5.66 m/s. = Final velocity before striking the gnd.
Vo^2 = V^2 - 2g*h
Vo^2 = 0 - 20*0.9 = 18
Vo = 4.24 m/s = Velocity at which it leaves the gnd.
Energy Lost=0.5m*V^2-0.5m*Vo^2 =
0.02*5.66^2 - 0.02*4.24^2=0.281 Joules
b. Vo = 4.24 m/s.
c. Impulse = m*V = 0.04 * 5.66 = 0,2264
a) To calculate the mechanical energy lost, we need to calculate the initial mechanical energy and the final mechanical energy after the ball bounces back up.
As the ball is dropped from rest, the initial mechanical energy is given by:
Initial Mechanical Energy = mgh
= 0.04 kg * 10 m/s^2 * 1.60 m
= 0.64 J
The final mechanical energy is given by the maximum height reached on the way back up:
Final Mechanical Energy = mgh
= 0.04 kg * 10 m/s^2 * 0.90 m
= 0.36 J
Therefore, the mechanical energy lost is:
Mechanical Energy Lost = Initial Mechanical Energy - Final Mechanical Energy
= 0.64 J - 0.36 J
= 0.28 J
So, 0.28 J of mechanical energy is lost in this case.
b) To find the speed of the ball just as it leaves the floor on its way up, we can use the principle of conservation of mechanical energy. At the maximum height, the ball has only potential energy, as it momentarily stops before changing direction. Therefore, the mechanical energy at the floor level is equal to the mechanical energy at the maximum height.
Initial mechanical energy = Final mechanical energy
mgh = (1/2)mv^2
Here, 'm' is the mass of the ball in kg, 'g' is the acceleration due to gravity, 'h' is the height, and 'v' is the velocity.
Rewriting the equation and solving for velocity 'v':
v^2 = 2gh
v = sqrt(2gh)
Substituting the given values:
v = sqrt(2 * 10 m/s^2 * 0.90 m)
= sqrt(18) m/s
≈ 4.24 m/s
Therefore, the ball's speed just as it leaves the floor on its way up is approximately 4.24 m/s.
c) Impulse is given by the change in momentum. Since the ball is dropped from rest, its momentum is initially zero, and it bounces back with an equal but opposite momentum.
Magnitude of Impulse = Change in Momentum
Change in momentum = 2 * momentum of the ball right after collision with the floor
The momentum of an object is given by:
Momentum = mass * velocity
The velocity just as the ball leaves the floor on its way up is the same magnitude as the velocity just before hitting the floor but in the opposite direction.
Magnitude of Impulse = 2 * (mass * velocity before hitting the floor)
Substituting the given values:
Magnitude of Impulse = 2 * (0.04 kg * 4.24 m/s)
≈ 0.34 kg·m/s
Therefore, the magnitude of the impulse experienced by the ball for the entire time it is in contact with the floor is approximately 0.34 kg·m/s.