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Chapter 12: Problem 11

Stopping a ball. A ball is hitting the floor with a vertical velocity \(v_{0}\). The collision between the ball and floor is elastic. (a) If you were able to move the floor up or down (using a vertical accelerator), is it possible to move the floor in a way so that the ball stops? Explain your answer.

Short Answer

Expert verified
Yes, the ball can stop if the floor moves at the same velocity as the ball's initial velocity (\(v_0\)).

Step by step solution

01

Understand the nature of the collision

Since the collision is elastic, kinetic energy and momentum are conserved during the collision.
02

Write the conservation of momentum equation

Let the mass of the ball be m, the velocity of the ball before collision be \(v_0\), and the velocity of the floor be \(v_f\). Using the conservation of momentum: \[ m \times v_0 = m \times v_b' + m \times v_f' \]where \(v_b'\) is the velocity of the ball after the collision and \(v_f'\) is the velocity of the floor after the collision.
03

Write the conservation of kinetic energy equation

Since the collision is elastic, we also have conservation of kinetic energy:\[ \frac{1}{2} m v_0^2 = \frac{1}{2} m v_b'^2 + \frac{1}{2} m v_f'^2 \]
04

Analyze special cases

For the ball to stop, \(v_b' = 0\). Substituting \(v_b' = 0\) into the equations, we get:\[ m \times v_0 = m \times 0 + m \times v_f' \]\( v_0 = v_f'\)Substitute into kinetic energy conservation: \[ \frac{1}{2} m v_0^2 = \frac{1}{2} m (v_0)^2 + \frac{1}{2} m v_f'^2 \].Since \(v_f' = v_0\), the initial and final kinetic energy remain unchanged, consistent with an elastic collision.
05

Conclude whether the ball can be stopped

From the equations, if the floor moves with a velocity equal to the ball's initial velocity (\(v_0\)), the relative velocity of the ball with respect to the floor would be zero after the collision, hence the ball would stop moving relative to the floor.

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Key Concepts

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

Conservation of Momentum
Momentum, a product of an object's mass and velocity, is a crucial concept in physics. In an isolated system with no external forces, the total momentum remains constant through any interaction.
This principle is called the conservation of momentum.
For an elastic collision between a ball and the floor, the total momentum before and after the collision remains unchanged.
The equation for conservation of momentum is:
speed
Conservation of Kinetic Energy
In an elastic collision, not only is momentum conserved, but so is kinetic energy.
Kinetic energy refers to the energy an object possesses due to its motion, given by the equation:
In the case of the ball and floor, the total kinetic energy before and after the collision must be the same to satisfy the condition of elasticity.
We can express the conservation of kinetic energy in the form of the equation:
In this context, we are accounting for how the ball and the floor's velocities contribute to the system's total kinetic energy.
It's key to remember that this conservation holds true for elastic collisions, meaning no kinetic energy is converted into other forms of energy like heat or sound during the collision.
Relative Velocity
Understanding relative velocity can simplify analyzing a collision event.
Relative velocity is the velocity of one object as observed from the frame of reference of another object.
In our problem, if the ball's velocity initially is Additionally, the velocity of the floor before collision is
Elastic Collisions
In physics, an elastic collision is where both kinetic energy and momentum of the system are conserved.
The interaction between the ball and the floor is considered elastic because there is no loss of total kinetic energy.
Key characteristics of elastic collisions:
  • Kinetic energy remains unchanged before and after the collision
  • Momentum remains unchanged before and after the collision
  • No deformation or generation of heat occurs during the collision

Using these principles helps to understand and solve problems where objects collide without losing their total kinetic energy.
This ensures that all the equations derived from these principles accurately represent the physical situation being analyzed.
In the scenario described, ensuring both kinetic energy and momentum conservation led us to understand that the ball could be stopped if it collides with the floor moving at the same initial velocity as the ball.

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Most popular questions from this chapter

A collision paradox. A riddle: Two cars each with speed \(v_{0}\) collide head on, getting stuck in the collision. If you observe the collision from the side of the road, the change in total kinetic energy is \(\Delta K=2 \cdot(1 / 2) m v_{0}^{2}\). If you instead observe the collision from a system moving with one of the cars, one car has velocity 0 and the other car has velocity \(2 v_{0}\). Then the change in total kinetic energy is \((1 / 2) m\left(2 v_{0}^{2}\right)=4 \cdot(1 / 2) m v_{0}^{2}\), which is double of what you observed from the side of the road. Is this argument correct? Explain.

Kicking a ball. A football is lying at rest on the ground. You kick it. After the kick, it has a horizontal velocity of \(20 \mathrm{~m} / \mathrm{s}\). You are in contact with the ball for \(0.1 \mathrm{~s}\). The mass of the ball is \(0.43 \mathrm{~kg}\). (a) What is the change in the momentum of the ball? (b) What is the impulse on the ball during the collision? (c) What is the average force on the ball during the collision? Assume that you are returning a ball coming toward you at \(20 \mathrm{~m} / \mathrm{s}\). You kick the ball, staying in contact with the ball for \(0.1 \mathrm{~s}\), and return the ball with a velocity of \(20 \mathrm{~m} / \mathrm{s}\) (d) What is the average force on the ball during the collision?

A collision paradox. A riddle: Two cars each with speed \(v_{0}\) collide head on, getting stuck in the collision. If you observe the collision from the side of the road, the change in total kinetic energy is \(\Delta K=2 \cdot(1 / 2) m v_{0}^{2}\). If you instead observe the collision from a system moving with one of the cars, one car has velocity 0 and the other car has velocity \(2 v_{0}\). Then the change in total kinetic energy is \((1 / 2) m\left(2 v_{0}^{2}\right)=4 \cdot(1 / 2) m v_{0}^{2}\), which is double of what you observed from the side of the road. Is this argument correct? Explain.

Bullet and a block. You fire a bullet of mass \(100 \mathrm{~g}\) horizontally into a block of mass \(2 \mathrm{~kg}\) where it gets stuck. The block lies on a frictionless table, but after the collision, the block enters a rough region of the table with a dynamic coefficient of friction, \(\mu=0.5\). After entering the rough region, the block slides a distance of \(10 \mathrm{~cm}\) before stopping. (a) What was the velocity of the bullet? (b) What is the loss of energy during the collision?

A bike and a car. You and your bike has a mass of \(100 \mathrm{~kg}\). (a) How fast would you have to ride in order to have the same momentum as a car of mass \(m=1200 \mathrm{~kg}\) and a velocity of \(50 \mathrm{~km} / \mathrm{h}\) ?
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