After falling from rest from a height of 30m, the 0.50kg ball rebounds upward, reaching a height of 20m. If the contact between ball and ground lasted 2.0 milliseconds, what average force was exerted on the ball? Show all math and include a free body diagram.
The correct answer and explanation is:
Let’s analyze the problem step-by-step.
Given:
- Mass of ball, m=0.50 kgm = 0.50 \, \text{kg}
- Initial 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}
Step 1: Find velocity of the ball just before impact (falling from rest)
Use conservation of energy or kinematic equation.
From rest, velocity just before hitting ground: v1=2gh1v_1 = \sqrt{2 g h_1}
where g=9.8 m/s2g = 9.8 \, m/s^2 v1=2×9.8×30=588≈24.25 m/s(downward)v_1 = \sqrt{2 \times 9.8 \times 30} = \sqrt{588} \approx 24.25 \, m/s \quad \text{(downward)}
Step 2: Find velocity of the ball just after rebound (going upward to 20m)
Velocity just after rebound v2v_2 satisfies: v2=2gh2v_2 = \sqrt{2 g h_2} v2=2×9.8×20=392≈19.8 m/s(upward)v_2 = \sqrt{2 \times 9.8 \times 20} = \sqrt{392} \approx 19.8 \, m/s \quad \text{(upward)}
Step 3: Calculate change in velocity
Since down is positive for v1v_1, then:
- Velocity before impact: v1=+24.25 m/sv_1 = +24.25 \, m/s (down)
- Velocity after rebound: v2=−19.8 m/sv_2 = -19.8 \, m/s (up)
Change in velocity: Δv=v2−v1=−19.8−24.25=−44.05 m/s\Delta v = v_2 – v_1 = -19.8 – 24.25 = -44.05 \, m/s
Step 4: Calculate impulse and average force
Impulse J=mΔvJ = m \Delta v J=0.50×(−44.05)=−22.03 kg\cdotpm/sJ = 0.50 \times (-44.05) = -22.03 \, \text{kg·m/s}
Impulse is equal to average force times contact time: J=Favg×ΔtJ = F_{\text{avg}} \times \Delta t
Solve for FavgF_{\text{avg}}: Favg=JΔt=−22.032.0×10−3=−11015 NF_{\text{avg}} = \frac{J}{\Delta t} = \frac{-22.03}{2.0 \times 10^{-3}} = -11015 \, \text{N}
The negative sign indicates force direction opposite to initial velocity (which is upward).
Step 5: Consider weight of the ball during contact
Net force during contact is Fnet=Favg+WF_{\text{net}} = F_{\text{avg}} + W, where W=mg=0.5×9.8=4.9 NW = mg = 0.5 \times 9.8 = 4.9 \, N downward.
Because gravity acts throughout contact, the total force exerted by the ground on the ball is: Fground=Favg+mg=11015−4.9≈11010 N(upward)F_{\text{ground}} = F_{\text{avg}} + mg = 11015 – 4.9 \approx 11010 \, \text{N} \quad \text{(upward)}
Final answer:
The average force exerted by the ground on the ball is approximately 11,000 N upward during the 2.0 ms contact.
Free Body Diagram Description:
- Before contact: gravity acts downward (mgmg), velocity downward.
- During contact: the ground exerts an upward force FgroundF_{\text{ground}}, and gravity acts downward mgmg.
- After contact: the ball moves upward with velocity v2v_2.
Explanation:
The ball falls from 30 m and gains speed due to gravity. Upon hitting the ground, it reverses direction and rebounds up to 20 m, indicating a loss of kinetic energy but a significant velocity change. The change in velocity happens over a very short time interval (2 milliseconds), producing a large average force according to the impulse-momentum relationship. This force from the ground must not only stop the ball’s downward momentum but also impart enough upward velocity for the rebound. The weight of the ball contributes a constant downward force during contact, but the ground force must be strong enough to overcome this and provide the required acceleration. This explains why the average contact force is very large, on the order of thousands of Newtons, despite the small mass of the ball.