A steel ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.

step 1 So today we're going to choose downwards to be positive. And so the wife final squared equals V

A steel ball is dropped onto a hard floor from a height of...

A steel ball is dropped onto a hard floor from a height of $1.50 \mathrm{m}$ and rebounds to a height of $1.45 \mathrm{m}$. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts $0.0800 \mathrm{ms}\left(8.00 \times 10^{-5} \mathrm{s}\right)$ (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?

A steel ball is dropped onto a hard floor from a height of $1.50 \mathrm{m}$ and rebounds to a height of $1.45 \mathrm{m}$. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up.
(c) Calculate its acceleration during contact with the floor if that contact lasts $0.0800 \mathrm{ms}\left(8.00 \times 10^{-5} \mathrm{s}\right)$ (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?

Key Concepts

Deformation and Contact Time in Collisions

This concept focuses on the deformation and compression that occur during collisions, particularly when objects like balls compress against a rigid surface. By examining the short contact duration and applying kinematic relations, one can estimate the extent of deformation, even when the collision involves nearly rigid bodies.

Impulse-Momentum Theorem and Average Acceleration

This concept relates the change in momentum of an object to the impulse applied over a given time period. During a collision, the large change in velocity over a very short time interval results in a high average acceleration. Understanding this theorem is key to calculating the acceleration experienced by the ball during contact with the floor.

Conservation of Energy

This principle is applied to relate the gravitational potential energy at one height to the kinetic energy just before impact (or just after rebound). It allows one to determine the speed of an object from its height by equating potential and kinetic energies, assuming negligible energy losses in parts of the motion.

Collision Dynamics and Rebound Behavior

This concept deals with the behavior of objects during collisions, particularly in terms of how energy and momentum are transferred or lost. In problems involving rebounding objects, factors like the coefficient of restitution—which quantifies the loss of kinetic energy during impact—are considered to understand the difference between the incident and rebound velocities.

Kinematics of Free Fall

This concept involves using the equations of motion to determine how objects move under the influence of gravity when they are in free-fall. It includes calculating velocities, displacements, and times based on a constant gravitational acceleration, and is crucial for understanding the motion of the ball as it falls towards and rebounds from the floor.

Transcript

00:01 So today we're going to choose downwards to be positive.

00:02 And so the wife final squared equals v y initial squared plus two g times delta y.

00:10 We know that the initial velocity zero and so the final velocity would be equal in the square root of two times 9.8 meters per second squared multiplied by 1.50 meters.

00:25 And the final velocity is than equalling 5.4 two meters per second.

00:32 Of course, the ball a the steel ball is being dropped, so this would be downwards and then for part b, we can say that after rebound, we know that here the we can say doubt tow why sub two is equaling 1.45 meters.

00:58 Hi.

00:59 So the velocity just after hitting the ground here would be equaling.

01:12 We're going again.

01:13 We're choosing upward recusing downwards to be positive.

01:16 So we're actually gonna make this negative and we can say that then v why initial would be equaling negative two times 9.8 meters per second squared multiplied by 1.45 meters.

01:42 And here what happens is that this is equaling 5.33 meters per second this would be upwards.

01:52 So right after it hits the ground, it rebounds and the initial velocity right as it comes back up would be 5.33 meters per second.

02:03 Now here we can say that.

02:06 Then, for part c, it's asking us calculate the acceleration of the ball during contact with the floor.

02:15 So we can say that for part c, we can see the change in velocity would be equaling 5.42 meters per second minus negative 5.33 meters per second.

02:32 Because of course the upwards velocity is in the opposite direction of the downwards velocity.

02:37 This is equaling approximately 10.7 meters per second and then we consider it.

02:43 Then the acceleration by definition would be the change in velocity divided by the change in time.

02:49 This would be equal in 10.7 meters per second.

02:54 We can actually just say 10.75 meters per second and then this would be divided by eight by another point eight because they eight times 10 to the negative fifth seconds and so the acceleration is actually 1.34 times 10 to the fifth meters per second squared...

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