Physics 2021 JAMB Past Questions - Question 20: A ball of mass 500g released from a cliff 20m above the ground, rebounds to a height of 5m. if the ball is in contact with ground for 0.1s, calculate the force acting on the ball on impact with the ground.

Correct Answer

C

Explanation

To calculate the force acting on the ball upon impact with the ground, we can use the impulse-momentum theorem. The change in momentum of the ball when it hits the ground is equal to the impulse imparted to it, which in turn is equal to the force acting on it multiplied by the time of contact.

First, let's find the initial velocity of the ball just before it hits the ground. We can use the equation of motion:

v^2 = u^2 + 2as

Where:
v = final velocity (0 m/s at the highest point)
u = initial velocity
a = acceleration due to gravity (-9.81 m/s^2, taking upward direction as positive)
s = displacement (20 m)

Rearranging the equation, we get:

u^2 = v^2 - 2as
u^2 = 0 - 2*(-9.81)*20
u^2 = 392.4
u ≈ 19.81 m/s

Now, let's find the final velocity of the ball just before it rebounds. We can use the same equation of motion:

v^2 = u^2 + 2as

Where:
v = final velocity
u = initial velocity (19.81 m/s)
a = acceleration due to gravity (-9.81 m/s^2, taking upward direction as positive)
s = displacement (20 m - 5 m = 15 m)

Rearranging the equation, we get:

v^2 = (19.81)^2 + 2*(-9.81)*15
v^2 = 392.04 - 294.3
v^2 = 97.74
v ≈ 9.886 m/s

Now, we can calculate the change in momentum:

Δp = mΔv
Δp = 0.5 * (9.886 - (-19.81))
Δp = 0.5 * 29.696
Δp = 14.848 kg m/s

The impulse imparted to the ball is equal to the change in momentum, so:

Impulse = Δp
Impulse = 14.848 N s

Finally, we can calculate the force:

Force = Impulse / Time
Force = 14.848 / 0.1
Force = 148.48 N

Therefore, the force acting on the ball upon impact with the ground is approximately 148.48 Newtons.