Problem 86 - College Physics

Question

Compare two different situations in which a ball and hard surface exert forces on one another. First, a putty ball is placed gently on the floor and left at rest. Then it is dropped from a height of \(2.00 \mathrm{~m}\) and comes to rest without a bounce, leaving a \(1.15-\mathrm{cm}\) -deep dent in the putty. (a) In which case does the ball exert more force on the floor? In which case is it most likely to dent the floor? Explain. (b) Calculate the force exerted by the ball on the floor (in terms of its weight \(w\) ) in the first case. (c) Determine the average acceleration of the ball and the average force exerted by the ball on the floor (in terms of the ball's weight \(w\) ) in the second case.

Step-by-Step Solution

Answer

The ball exerts more force in the second case, and it's more likely to dent the floor when dropped.

Step 1: Understanding the First Situation

In the first situation, the putty ball is gently placed on the floor and at rest. Here, the only force exerted by the ball is its weight, acting downwards due to gravity. Let's denote the weight of the ball as its force on the floor, which is simply the gravitational force given by: \( F_1 = w \).

Step 2: Understanding the Second Situation

In the second situation, the ball is dropped from a height of 2.00 m and creates a 1.15 cm dent when it hits the floor. The impact force when the ball hits the floor is greater than just its weight because it includes the additional force due to the change in momentum over the time it takes to stop.

Step 3: Finding the Force in the First Case

For the first case, the force exerted by the ball on the floor is just its weight. So, using the weight of the ball as \( w \), the force is simply \( F_1 = w \).

Step 4: Calculating Impact Velocity

To find the force in the second case, we first need the velocity of the ball just before it hits the floor. Using the formula for free-fall: \( v = \sqrt{2gh} \), where \( g = 9.8 \frac{m}{s^2} \) and \( h = 2.00 \ m \), we have: \( v = \sqrt{2 \times 9.8 \times 2} \approx 6.26 \ m/s \).

Step 5: Calculating Average Deceleration

Next, calculate the deceleration within the dent depth. Using: \( a = \frac{v^2}{2d} \), where \( v = 6.26 \, m/s \) and \( d = 0.0115 \, m \), \( a = \frac{6.26^2}{2 \times 0.0115} \approx 1706 \ m/s^2 \).

Step 6: Calculating Force in the Second Case

Now, we calculate the force using \( F = ma \), and transforming using \( F_2 = w \cdot \frac{a}{g} \), where \( a = 1706 \, m/s^2 \) and \( g = 9.8 \, m/s^2 \). Thus, \( F_2 = w \times \frac{1706}{9.8} \approx 174w \).

Step 7: Comparing the Forces

In the first case, the force is simply \( w \). In the second case, the force is approximately \( 174w \). The second case significantly exerts more force and is more likely to dent the floor.

Key Concepts

Force Calculation

When analyzing any physical situation involving motion, understanding how to calculate force is crucial. Here, we have a scenario with a ball interacting with the floor in two different ways.

• In the first scenario, the ball is gently placed on the floor. The only force then is its gravitational weight, a straightforward calculation: - The force is the ball's weight, denoted as \( F_1 = w \).
• In the second scenario, the ball is dropped from a height and hits the floor. Here, the force is more complex. It includes the impact force, which is due to the ball's change in momentum as it comes to rest.

By calculating the force in both scenarios, we can compare them to understand how the force applied changes based on the conditions.

Gravitational Force

Gravitational force is a fundamental force acting on objects due to their mass and Earth's gravity. In our examples, it plays a crucial role in determining the initial downward force on the ball.

• When the putty ball is gently placed on the floor, the gravitational force is its weight: - This is expressed as \( F_1 = w \), where \( w \) is the weight of the ball.
• Gravitational force also influences the ball as it falls from a height of 2 meters. This weight remains constant but becomes a factor in free-fall calculations (e.g., calculating the falling velocity).

Gravitational force acts throughout the ball's descent and when it is resting, establishing a consistent baseline for further force calculations.

Impact Force

Impact force is the extra force experienced when an object collides with a surface. This force results from the object's momentum change over the short time it contacts a surface.

• In the exercise, the putty ball dropped from height creates an impact force way greater than its static weight: - This force is calculated by determining the change in momentum and dividing it by the short time interval as the ball comes to rest.
• The impact force is evidenced by the formation of a dent in the putty ball on collision, highlighting its intensity compared to merely resting.

This force is vital in understanding why the second scenario results in more forceful flooring interaction.

Acceleration

Acceleration is the rate of change of velocity. It's pivotal in understanding how quickly an object's speed is altered. In our scenario, it guides the calculation of impact dynamics.

• After the ball is dropped, it accelerates due to gravity with a value of \( g = 9.8 \, m/s^2 \).
• When the ball hits the floor, it undergoes rapid deceleration due to impact over the dent's depth: - We calculated this as \( a = \frac{v^2}{2d} \), leading to a significant average deceleration of \( 1706 \, m/s^2 \).
This huge reduction in speed supports the high impact force exerted when the ball contacts the floor. By grasping how acceleration changes during impact, we achieve a better understanding of force dynamics in motion.