Problem 18 - College Physics

Question

A ball of mass \(200 \mathrm{~g}\) is released from rest at a height of \(2.00 \mathrm{~m}\) above the floor and it rebounds straight up to a height of \(0.900 \mathrm{~m}\). (a) Determine the ball's change in momentum due to its contact with the floor. (b) If the contact time with the floor was \(0.0950 \mathrm{~s}\), what was the average force the floor exerted on the ball, and in what direction?

Step-by-Step Solution

Answer

(a) The change in momentum is approximately \(2.09 \mathrm{~kg \, m/s}\). (b) The average force is approximately \(22.00 \mathrm{~N}\) directed upwards.

Step 1: Understand the Problem

To solve this exercise, we need to find two things. First, the change in momentum when the ball hits and rebounds off the floor (part a). Second, the average force exerted by the floor on the ball and its direction (part b). We'll tackle these using concepts of physics related to momentum, energy, and force.

Step 2: Calculate Initial Velocity Before Impact

The ball is dropped from a height of 2.00 m. To find its velocity just before hitting the floor, we use the equation for gravitational potential energy conversion to kinetic energy: \[ v^2 = 0 + 2gh \]where \( g = 9.81 \mathrm{~m/s^2} \) and \( h = 2.00 \mathrm{~m} \).Substitute the values to find \( v \):\[ v = \sqrt{2 \times 9.81 \times 2.00} \approx 6.26 \mathrm{~m/s} \]

Step 3: Calculate Velocity After Rebound

After rebounding, the ball reaches a height of 0.900 m. Using the same energy conversion principle, we find the velocity right after rebound:\[ v'^2 = 2gh' \]where \( h' = 0.900 \mathrm{~m} \).Solving for \( v' \):\[ v' = \sqrt{2 \times 9.81 \times 0.900} \approx 4.20 \mathrm{~m/s} \]

Step 4: Calculate Change in Velocity

The change in velocity \( \Delta v \) is the difference between the velocity before impact and after rebound. Since the direction reverses upon the rebound, remember to reverse the sign of velocity after impact:\[ \Delta v = v' - (-v) = v' + v = 4.20 + 6.26 = 10.46 \mathrm{~m/s} \]

Step 5: Calculate Change in Momentum

Change in momentum \( \Delta p \) is found by multiplying the change in velocity by the ball's mass. Mass \( m = 0.200 \mathrm{~kg} \):\[ \Delta p = m \times \Delta v = 0.200 \times 10.46 \approx 2.09 \mathrm{~kg \, m/s} \]

Step 6: Calculate Average Force

Average force during the contact is calculated using the formula:\[ F_{avg} = \frac{\Delta p}{\Delta t} \]where \( \Delta t = 0.0950 \mathrm{~s} \). Substituting the values:\[ F_{avg} = \frac{2.09}{0.0950} \approx 22.00 \mathrm{~N} \]The force direction is upwards, opposite to the direction of gravity at impact.

Key Concepts

Change in Momentum

Momentum is the measure of an object's motion and is the product of an object's mass and velocity. When the ball strikes the floor and rebounds, there is a notable change in momentum.

• Initially, the ball has a downward momentum just before it hits the floor.
• After the rebound, it has upward momentum.

The change in momentum (\( \Delta p \)) is calculated using the formula where the ball's velocity changes direction:\[ \Delta p = m \cdot \Delta v \]Where

• \( m \) is the mass of the ball, which is 0.200 kg.
• \( \Delta v \) is the change in velocity which we can calculate as the sum of the velocities before and after impact due to the reversal of direction: \( 10.46 \ \text{m/s} \).

Thus, the change in momentum is approximately \( 2.09 \ \text{kg m/s} \). This points in the opposite direction of the initial momentum, indicating a complete reversal of motion.

Average Force

After the ball hits the floor, it rebounds, meaning a significant force acted on it during the short time of contact with the surface. Average force describes how much force acts on an object over a given time interval.
When calculating the average force during the contact time of \( 0.0950 \ \text{s} \):

• We use the formula for average force:\[ F_{avg} = \frac{\Delta p}{\Delta t} \]
• Where \( \Delta p = 2.09 \ \text{kg m/s} \)
• \( \Delta t = 0.0950 \ \text{s} \)

This yields an average force of approximately \( 22.00 \ \text{N} \), directed upwards due to the rebound effect. This illustrates how the floor exerts an upward force that overcomes the original downward trajectory of the ball.

Energy Conversion

Energy conversion is a critical factor in understanding the movement and rebound of the ball. When released, the ball possesses gravitational potential energy due to its height. As it falls, this energy converts to kinetic energy.

• At rest and height \( 2.00 \ \text{m} \), the ball has maximum potential energy.
• Before hitting the ground, all potential energy converts to kinetic energy, providing it with velocity.

With rebound:

• Some kinetic energy is lost due to impact and conversions, so it reaches a lesser height of \( 0.900 \ \text{m} \).
• This conversion back into potential energy is less than the initial amount, indicating an inelastic collision.

Thus, part of the mechanical energy is lost, often converted into sound and heat during the impact.

Impact

The moment of impact is crucial as it determines how the ball changes its motion and how energy is dissipated. The force exerted by the floor on the ball during impact is a reaction to its action, according to Newton's third law.
When the ball strikes the floor:

• The floor applies a force that significantly changes the ball's momentum.
• At this point, contact time is brief but forceful, causing the rebound.

The nature of the surface and the ball's material can influence:

• The degree of energy lost during impact.
• The height to which it rebounds, showcasing the elasticity of the collision.

In this scenario, the loss of energy seen through decreased rebound height underscores an inelastic collision, where not all kinetic energy reverts to potential energy.