A bouncing ball-part 2. You lift a ball of radius \(R\) and mass \(m\) vertically up, until its center is a height \(h\) above the ground. Here, you hit the ball, giving it an initial horizontal velocity \(v_{0}\). You may model the collision between the ball and the ground using a spring force acting in the direction normal to the ground. You can assume that all the deformation occurs in the ball and that the spring constant is \(k\). There is no friction between the ball and the ground. (a) What is the (vector) velocity of the ball when it comes in contact with the ground? (b) What is the (vector) velocity of the ball when it reaches its maximum compression? (c) What is the maximum deformation, \(\delta y\), of the ball during the collision with the ground? (d) What is the (vector) velocity of the ball as it loses contact with the ground? (e) Based on your results, can you propose a law for how the velocity of a ball changes during a collision with a (frictionless) wall?

Step by step solution

01

Find the vertical velocity when the ball comes in contact with the ground

The ball is dropped from a height h. The initial vertical velocity is 0. The vertical velocity just before hitting the ground is due to gravitational acceleration. Use the equation of motion to find it: \[ v_y^2 = u_y^2 + 2gh \] Where \( u_y = 0 \) (initial vertical velocity), \( g \) is the acceleration due to gravity, and \( h \) is the height. Substituting the values, we get: \[ v_y = \sqrt{2gh} \] The horizontal velocity, \(v_0\), remains unchanged.

02

Write the vector velocity when the ball touches the ground

The vector velocity of the ball when it touches the ground is the combination of horizontal and vertical components:\[ \mathbf{v} = v_0 \hat{i} + (-\sqrt{2gh}) \hat{j} \]

03

Determine the vertical velocity at maximum compression

At maximum compression, the velocity in the vertical direction is zero because the ball temporarily comes to rest. The ball still has its horizontal velocity component. Hence, the vector velocity at maximum compression is: \[ \mathbf{v}_{compression} = v_0 \hat{i} \]

04

Calculate the maximum deformation of the ball, \( \delta y \)

Apply the conservation of energy principle. At the point of maximum compression, all kinetic energy due to the vertical motion and gravitational potential energy is converted into the elastic potential energy of the ball. Total energy before collision: \[ E_{before} = \frac{1}{2}mv_y^2 + mgh \]Total energy at maximum compression: \[ E_{compression} = \frac{1}{2} k \delta y^2 \]Equating both energies, we get: \[ \frac{1}{2} m (\sqrt{2gh})^2 + mgh = \frac{1}{2} k \delta y^2 \]Simplify to solve for \( \delta y \):\[ \delta y = \sqrt{ \frac{3mgh}{k} } \]

05

Find the vector velocity as the ball loses contact with the ground

As the ball loses contact with the ground, it will have upward vertical velocity due to the restoring force of the spring. At this point, the vertical component of the velocity equals the negative of the impact velocity (elastic collision without energy loss), and the horizontal component remains same. Thus, \[ \mathbf{v}_{leave} = v_0 \hat{i} + (\sqrt{2gh}) \hat{j} \]

06

Propose a law for the velocity change during a collision

The vertical velocity component of the ball reverses direction with the same magnitude (assuming elastic collision and no energy loss), while the horizontal velocity component remains unchanged. Hence, if \( \mathbf{v}_{initial} = v_{0x} \hat{i} + v_{0y} \hat{j} \), then after collision, \[ \mathbf{v}_{final} = v_{0x} \hat{i} - v_{0y} \hat{j} \]

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Key Concepts

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

Vector Velocity

When dealing with a bouncing ball problem, understanding vector velocity is crucial. A vector velocity has two components: horizontal and vertical. To find the overall vector velocity, we combine these two components.

In the context of a bouncing ball, think of the ball's horizontal velocity as a constant value, unaffected by gravity. The vertical velocity, on the other hand, changes due to gravitational acceleration. When the ball is just about to hit the ground, its vertical velocity can be calculated using the equation: \[ v_y = \sqrt{2gh} \] where \( g \) is the acceleration due to gravity, and \( h \) is the height from which the ball falls.

After calculating the vertical velocity, the vector velocity when the ball touches the ground combines the horizontal \( v_0 \) and vertical \( v_y \) components: \[ \mathbf{v} = v_0 \hat{i} + (-\sqrt{2gh}) \hat{j} \] This demonstrates how vector components work together to describe the ball's motion.

Energy Conservation

Energy conservation is a key principle in physics. It states that the total energy in a closed system remains constant. This includes kinetic energy, potential energy, and other forms of energy.

In the bouncing ball scenario, when the ball is at height \( h \), it possesses gravitational potential energy: \[ E_{potential} = mgh \]

As the ball falls, this potential energy is converted into kinetic energy. Upon reaching the ground, the energy has transformed into kinetic energy given by: \[ E_{kinetic} = \frac{1}{2}mv_y^2 \]

When the ball compresses upon impact, all kinetic energy and any remaining gravitational potential energy turn into elastic potential energy within the ball. At maximum compression, the energy equation is: \[ \frac{1}{2} k \delta y^2 = \frac{1}{2} m v_y^2 + mgh \] Here, \( k \) is the spring constant, and \( \delta y \) is the compression. Solving it helps us understand the energy transformation during the process.

Elastic Collision

In physics, an elastic collision is one where there's no loss of kinetic energy in the system. The total kinetic energy before and after collision remains the same.

For a bouncing ball, an elastic collision implies that the ball rebounds with the same speed it had when it hit the ground. If a ball is modeled as perfectly elastic, its vertical velocity just before hitting the ground will be equal in magnitude but opposite in direction to its vertical velocity just after losing contact with the ground.

Therefore, if the ball's vertical velocity before hitting the ground is \( \sqrt{2gh} \), it will bounce back with a velocity \( -\sqrt{2gh} \) immediately after leaving the ground. This conservation principle allows us to predict the ball's rebound behavior accurately.

Gravitational Acceleration

Gravitational acceleration, denoted as \( g \), is the acceleration an object undergoes due to Earth’s gravitational pull. For any object near the Earth's surface, \( g \) approximates to \( 9.8 \: m/s^2 \).

In our exercise, gravitational acceleration affects the ball's vertical motion. When you drop the ball, it accelerates downward due to gravity. The velocity of the ball just before it hits the ground is found using the equation of motion affected by gravitational acceleration: \[ v_y = \sqrt{2gh} \]

This velocity will change direction as the ball hits and then rebounds off the ground. Understanding \( g \) helps us calculate the velocities at different stages of the ball's motion, especially while it is in free fall or rebounding post-collision.