After falling from rest from a height of 30 m, a 0.48-kg ball rebounds upward, reaching a height of 20 m. If the contact between ball and ground lasted 1.8 ms, what average force was exerted on the ball?
I worked it out and got 4.8N
I used the equations of motion for constant acceleration (The object is assumed to fall with the acceleration due to gravity which I took to be 10 m/s^2)
Equations of motion:
V^2=U^2 + 2*a*s
V=U+at
Note: when objects rebounds, acceleration = - 10m/s^2 since the latter is a vector quantity; thus when object falls down and rebounds, sign changes.
Moreover, once the speed with which object rebounds has been found, I used the concept of workdone= force X Distance moved in the direction of the force. Kinetic energy = Workdone
therefore, F X distance = 0.5 X m X V^2
Solve for F and you'll get 4.8 N
To find the average force exerted on the ball, we can use Newton's second law of motion, which states that the force exerted on an object is equal to the mass of the object multiplied by its acceleration.
First, let's find the initial velocity of the ball just before impact by using the equation for free fall:
v = √(2 * g * h)
Where:
v is the initial velocity
g is the acceleration due to gravity (approximately 9.8 m/s^2)
h is the height from which the ball falls (30 m in this case)
Plugging in the values, we have:
v = √(2 * 9.8 * 30)
v = √(588)
v ≈ 24.25 m/s
Next, we can calculate the final velocity of the ball just after impact using the same equation:
v = √(2 * g * h)
Where:
v is the final velocity
g is the acceleration due to gravity (approximately 9.8 m/s^2)
h is the maximum height reached by the ball after rebounding (20 m in this case)
Plugging in the values, we have:
v = √(2 * 9.8 * 20)
v = √(392)
v ≈ 19.8 m/s
To find the change in velocity, we subtract the initial velocity from the final velocity:
Δv = v_final - v_initial
Δv = 19.8 - (-24.25)
Δv = 44.05 m/s
Now, we need to convert the time of contact between the ball and the ground into seconds:
t = 1.8 ms = 1.8 * 10^(-3) s
Finally, we can calculate the average force using Newton's second law:
F = m * (Δv / t)
Where:
F is the average force exerted on the ball
m is the mass of the ball (0.48 kg in this case)
Δv is the change in velocity of the ball
t is the duration of the contact between the ball and the ground
Plugging in the values, we have:
F = 0.48 * (44.05 / 1.8 * 10^(-3))
F ≈ 11,404 N
Therefore, the average force exerted on the ball is approximately 11,404 N.