/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 82 A rubber ball of mass \(m\) is r... [FREE SOLUTION] | 91Ó°ÊÓ

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A rubber ball of mass \(m\) is released from rest at height \(h\) above the floor. After its first bounce, it rises to 90\(\%\) of its original height. What impulse (magnitude and direction) does the floor exert on this ball during its first bounce? Express your answer in terms of the variables \(m\) and \(h\) .

Short Answer

Expert verified
The impulse exerted by the floor is \(1.9m\sqrt{2gh}\) directed upward.

Step by step solution

01

Calculate Initial Momentum Before Bounce

The ball starts from rest, so its initial momentum at the top of the drop is zero. As it falls from height \( h \), its velocity just before hitting the floor can be found using energy conservation. The potential energy at the top is completely converted to kinetic energy at the bottom: \( mgh = \frac{1}{2}mv^2 \). Solving for \( v \), we get \( v = \sqrt{2gh} \). The initial momentum before the bounce is \( mv = m\sqrt{2gh} \) directed downward.
02

Calculate Final Momentum After Bounce

After the first bounce, the ball rises to 90\(\%\) of its initial height. The new height \( h' \) is 0.9\(h\), so the velocity after the bounce can be found similarly: \( mgh' = \frac{1}{2}mv_f^2 \). Solving for \( v_f \), we find \( v_f = \sqrt{2g(0.9h)} = 0.9\sqrt{2gh} \). The final momentum after the bounce is \( mv_f = m(0.9\sqrt{2gh}) \) directed upward.
03

Calculate Change in Momentum

The impulse experienced by the ball is equal to the change in momentum. The initial momentum is downward \(-m\sqrt{2gh}\), and the final momentum is upward \(+m(0.9\sqrt{2gh})\). Therefore, the change in momentum \( \Delta p \) is \( m(0.9\sqrt{2gh}) - (-m\sqrt{2gh}) = m(\sqrt{2gh})(1 + 0.9) = 1.9m\sqrt{2gh} \).
04

Determine Impulse Direction and Magnitude

Since the impulse is equal to the change in momentum, the magnitude of the impulse is \( 1.9m\sqrt{2gh} \) and it is directed upward because it results in the ball moving upwards after the bounce.

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Key Concepts

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

Impulse
Impulse is a fundamental concept in physics that describes the effect of a force applied over a specific time duration. To understand impulse, think of it as the "push" given to an object. This "push" changes the object's momentum, which is the product of its mass and velocity, signifying how difficult it is to stop the object once it's moving.
Impulse is defined by the equation \( J = \Delta p \), where \(J\) represents impulse and \(\Delta p\) is the change in momentum. In our exercise, the impulse caused by the floor's force when the ball bounces is calculated by determining the change from the initial downward momentum to the final upward momentum.
Important points about impulse:
  • Impulse has the same direction as the change in momentum.
  • Since the ball moves in the opposite direction after the bounce, the impulse direction is upward.
  • The magnitude of impulse can be calculated as \(1.9m\sqrt{2gh}\) as derived in the original solution.
This calculation allows us to precisely understand the effect of the floor on the ball's motion during the bounce, indicating the size and direction of the "push" necessary to reverse its velocity.
Momentum
Momentum, usually denoted as \(p\), is an essential concept to analyze the dynamics of moving objects. It is defined as the product of an object's mass and velocity:\( p = mv \). Essentially, momentum measures how much motion an object possesses and how challenging it would be to bring it to rest.
In our scenario, we are interested in how the momentum changes during the ball's bounce. Initial momentum is zero at the top before the ball is released, and as it falls towards the floor, its momentum increases due to gravitational acceleration.
Key aspects of momentum in this context:
  • The direction of momentum matters; it changes from downward before hitting the ground to upward after the bounce.
  • The change in momentum, \(\Delta p\), is crucial for calculating the force exerted by the floor.
  • The momenta before and after the bounce are linked by the factor of the rebound height, 90% of the original.
Understanding momentum helps us predict how interactions affect an object's movement, such as the bounce height in this problem.
Energy Conservation
The principle of energy conservation is a core physics principle stating that energy cannot be created or destroyed, only transformed. This conservation is pivotal in analyzing systems where energy shifts between types.
In the exercise with the rubber ball, gravitational potential energy at the initial height is converted to kinetic energy as the ball falls. At the peak of its bounce, kinetic energy converts back to potential energy.
The fundamental relationship used here is:
  • Gravitational potential energy at height \(h\): \(mgh\).
  • Kinetic energy at the moment of impact with the floor: \(\frac{1}{2}mv^2\).
  • These energies must equal each other, set by the equation \(mgh = \frac{1}{2}mv^2\) to find the velocity before impact.
This can also show why the ball bounces to 90% of the original height, indicating some energy is lost in other forms like sound or heat, common in real-world interactions.
Kinetic Energy
Kinetic energy is the energy of motion, described by the formula \( K = \frac{1}{2}mv^2 \). It represents the energy an object possesses due to its velocity and mass, giving a quantitative measure to its motion.
In the exercise, calculating kinetic energy helps us know the velocity of the ball before and after the bounce. As the ball drops, it converts potential energy into kinetic energy, gaining speed up to the point of impact.
About kinetic energy conversion in our context:
  • At the highest point, all energy is potential, and at impact, it's kinetic.
  • The post-bounce kinetic energy relates to the new height the ball will achieve.
  • Given it reaches 90% of its original height after bouncing, the kinetic energy post-impact is slightly less, shown by a factor \(0.9^2\), reflecting energy losses.
Thus, understanding kinetic energy lets us predict how fast the ball moves and the height achieved after bouncing, central to analyzing motion in energy contexts.

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Most popular questions from this chapter

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