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Chapter 6: Problem 87

The coefficient of restitution, \(e\), of a ball hitting the floor is defined as the ratio of the speed of the ball right after it rebounds from the impact to the speed of the ball right before it hits the floor, that is, \(e=v_{\text {after }} / t_{\text {helare }}\). (a) Derive a formula for the coefficient of restitution when a ball is released from an initial height \(H\) and rebounds to a final height \(b\). (b) Calculate the coefficient of restitution for a golf ball that bounces to a height of \(60 \mathrm{~cm}\) after having been released from a height of \(80 \mathrm{~cm}\). SSM

Short Answer

Expert verified
The coefficient of restitution (e) for a ball hitting the floor and bouncing back can be expressed as \(e = \sqrt{\frac{b}{H}}\). For a golf ball bouncing from an initial height of 80cm to a final height of 60cm, the coefficient of restitution is \(\sqrt{0.75}\).

Step by step solution

01

Understand the formula

The coefficient of restitution (e) for a ball hitting the floor is defined as the ratio of the speed of the ball right after the rebound (v_after) to the speed of the ball right before the hit (v_before). This can be expressed as: \(e = \frac{v_{\text{after}}}{v_{\text{before}}}\).
02

Derive formula using height and gravity

We can express the velocity of the ball in terms of height and gravity to better suit our problem. This is because the velocity of an object released from rest under gravity, over a certain height, H, can be expressed as \(v = \sqrt{2gH}\), where g is acceleration due to gravity. Similarly, the velocity after the ball hits the floor can be expressed as \(v' = \sqrt{2gb}\), where b is the final height to which the ball bounces. Substitute these into the formula for e from Step 1 to get:\(e = \frac{\sqrt{2gb}}{\sqrt{2gH}} = \sqrt{\frac{b}{H}}\).
03

Apply the derived formula on a practical example

We can now apply the formula derived in Step 2 to find the coefficient of restitution for a golf ball given the initial and final heights. In this case, the initial height (H) is 80cm and the final height (b) is 60cm. Applying the formula:\(e = \sqrt{\frac{b}{H}} = \sqrt{\frac{60}{80}} = \sqrt{0.75}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Elastic Collisions
When a ball hits the floor and rebounds, it undergoes an elastic collision, a process that involves two core ideas: the transfer of energy and the change in momentum. In purely elastic collisions:
  • Kinetic energy before and after the collision stays the same.
  • The objects involved have no loss or conversion of energy to sound or heat.
For a bouncing ball, the collision isn't perfectly elastic, as it loses some energy. This is why it doesn't bounce back to its original height. Understanding how elastic collisions work can help grasp why the coefficient of restitution, which measures bounciness, is used.
Conservation of Energy
Energy conservation is a crucial principle, informing us that energy in a closed system remains constant. When the ball is released from a height, it has potential energy that changes into kinetic energy as it falls. The equation for potential energy is:
  • \( PE = mgh \)
where \( m \) is mass, \( g \) is acceleration due to gravity, and \( h \) is height.
After the bounce, the kinetic energy converts back into potential energy as the ball rises.
  • The heart of this transformation illustrates why energy is only partly conserved in real-world bounces.
Studying energy shifts helps understand the reduced height to which the ball returns.
Kinematics
Kinematics deals with motion without considering forces. It's about how objects move through space and time. When calculating the coefficient of restitution, kinematics helps us understand:
  • How objects accelerate under gravity (both dropping and bouncing).
  • Using the formula \( v = \sqrt{2gh} \) gives the speed at which the ball impacts and rebounds.
This equation comes from setting the changes in potential energy equal to kinetic energy, thus assisting in finding the speeds required to calculate the coefficient of restitution.
Gravitational Acceleration
Gravitational acceleration, denoted by \( g \), is the acceleration of an object due to Earth's gravity. It's approximately \( 9.8 \, \text{m/s}^2 \). This constant plays a crucial role in:
  • Determining how fast an object accelerates toward the Earth when dropped.
  • Calculating the velocities of the ball using \( v = \sqrt{2gh} \).

The influence of \( g \) ensures that the formulas for velocity are consistent regardless of the object's mass, underlying the principle that all objects fall at the same rate in a vacuum. This principle is essential for solving problems related to bouncing balls and understanding their motion under gravity.

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