After falling from rest from a height of 30 m, a 0.50-kg ball rebounds upward, reaching a height of 20 m. If the contact between ball and ground lasted 2.0 ms, what average force was exerted on the ball?
The correct answer and explanation is:
Let’s solve this step-by-step.
Given data:
- Mass of ball, m=0.50 kgm = 0.50 \, \text{kg}
- Initial drop height, h1=30 mh_1 = 30 \, \text{m}
- Rebound height, h2=20 mh_2 = 20 \, \text{m}
- Contact time with ground, Δt=2.0 ms=2.0×10−3 s\Delta t = 2.0 \, \text{ms} = 2.0 \times 10^{-3} \, \text{s}
- Acceleration due to gravity, g=9.8 m/s2g = 9.8 \, \text{m/s}^2
Step 1: Calculate velocity just before impact
Using conservation of energy or kinematics for free fall from rest: v1=2gh1=2×9.8×30=588≈24.25 m/sv_1 = \sqrt{2gh_1} = \sqrt{2 \times 9.8 \times 30} = \sqrt{588} \approx 24.25 \, \text{m/s}
This is the velocity just before hitting the ground, downward direction.
Step 2: Calculate velocity just after rebound
The ball rebounds and reaches height h2h_2, so initial velocity right after rebound is: v2=2gh2=2×9.8×20=392≈19.80 m/sv_2 = \sqrt{2gh_2} = \sqrt{2 \times 9.8 \times 20} = \sqrt{392} \approx 19.80 \, \text{m/s}
This velocity is upward.
Step 3: Calculate change in velocity (vector)
Choose downward as negative direction. Initial velocity just before impact is v1=−24.25 m/sv_1 = -24.25 \, \text{m/s} (downward), and velocity just after impact is v2=+19.80 m/sv_2 = +19.80 \, \text{m/s} (upward).
Change in velocity Δv\Delta v is: Δv=v2−v1=19.80−(−24.25)=19.80+24.25=44.05 m/s\Delta v = v_2 – v_1 = 19.80 – (-24.25) = 19.80 + 24.25 = 44.05 \, \text{m/s}
Step 4: Calculate average acceleration during contact
a=ΔvΔt=44.052.0×10−3=22025 m/s2a = \frac{\Delta v}{\Delta t} = \frac{44.05}{2.0 \times 10^{-3}} = 22025 \, \text{m/s}^2
This acceleration is upward.
Step 5: Calculate average force exerted by the ground on the ball
Net force on ball during contact includes force from ground FF upward and weight mgmg downward. Using Newton’s second law: Fnet=maF_{\text{net}} = ma
Force from the ground must overcome weight plus provide acceleration, so: F−mg=ma ⟹ F=m(a+g)F – mg = ma \implies F = m(a + g)
Calculate: F=0.50×(22025+9.8)≈0.50×22034.8=11017.4 NF = 0.50 \times (22025 + 9.8) \approx 0.50 \times 22034.8 = 11017.4 \, \text{N}
Final answer:
The average force exerted on the ball by the ground is approximately 11,000 Newtons upward.
Explanation:
When the ball hits the ground, it experiences a very rapid change in velocity from downward to upward motion. The ground applies an impulse force in a very short time (2 milliseconds), causing this abrupt reversal of velocity. This results in a very large acceleration during contact. Since force equals mass times acceleration, the average force from the ground must be very large to cause such a quick velocity change. The weight of the ball adds to the total force the ground must exert. The magnitude of this average force is much greater than the ball’s weight because of the very short contact time and the significant change in velocity.
The calculation assumes constant average force during contact. In reality, the force varies during impact, but this average value is useful to understand the typical magnitude of force involved. This explains why impacts feel strong despite the ball being light.