Illl A 200 g ball is dropped from a height of 2.0 m, bounces on a hard floor,

step 1 Hello everyone and welcome. We have a 200 gram ball dropped from a height of 2 meters and it reb

Illl A 200 g ball is dropped from a...

$\begin{array}{lllll}\text { Illl } \mathrm{A} & 200 & \mathrm{g} & \text { ball } & \text { is }\end{array}$ dropped from a height of $2.0 \mathrm{m},$ bounces on a hard floor, and rebounds to a height of $1.5 \mathrm{m} .$ Figure P9.49 shows the impulse received from the floor. What maximum force does the floor exert on the ball?

$\begin{array}{lllll}\text { Illl } \mathrm{A} & 200 & \mathrm{g} & \text { ball } & \text { is }\end{array}$
dropped from a height of $2.0 \mathrm{m},$ bounces on a hard floor, and rebounds to a height of $1.5 \mathrm{m} .$ Figure P9.49 shows the impulse received from the floor.
What maximum force
does the floor exert on
the ball?

Key Concepts

Gravitational Acceleration

Gravitational acceleration is the constant rate at which objects accelerate towards the ground due to gravity. It is a key concept for calculating the velocities of falling objects and the corresponding kinetic energy at impact, thereby aiding in the analysis of the subsequent collision response.

Collision Dynamics

Collision dynamics involves studying the forces during impacts, particularly how high instantaneous forces (such as maximum force) can be when a moving object collides with a surface and rebounds. It emphasizes understanding the time interval over which the collision occurs to relate the overall impulse to the peak force during the collision.

Energy Conservation and Conversion

In this context, conservation of energy is used to relate the gravitational potential energy at a certain height to the kinetic energy of an object just before impact or just after rebound. This conversion helps determine the velocities associated with the object’s motion, which are critical for calculating momentum changes.

Impulse-Momentum Theorem

This concept states that the impulse applied to an object is equal to its change in momentum. It is fundamental when analyzing collisions or impacts because it links the applied force, the duration of the force, and the resultant momentum change experienced by the object.

Transcript

00:01 Hello everyone and welcome.

00:02 We have a 200 gram ball dropped from a height of 2 meters and it rebounds to a height of 1 .5 meters.

00:09 So we're given the, we've been shown the impulse received from the floor, but we want to find out what the maximum force that is exerted on the ball.

00:21 So the way we're going to do this is by setting up two equations for the impulse.

00:25 So our first equation is the impulse is equal to the base times height.

00:30 Like the area under the curve of the force versus time graph.

00:34 So it's base, which is 5, times height, which is f max, divided by 2.

00:40 So that's one way.

00:42 So the second way that we're going to do this is by calculating the other way, the fact that the impulse is the change in momentum.

00:52 So the p final minus the p initial.

01:04 So the way that we're going to do this is by finding the velocities, but we're not given the velocity.

01:09 So we actually have to find a whiff to solve for it.

01:12 So let's worry about the first one first, and then we can find the second one after that.

01:18 So in the initial drop, the v0 is equal to zero because it's dropped from rest.

01:24 The acceleration is just the force of gravity right now, which is 9 .8 meters per second, and the total displacement is 2 meters.

01:35 So to get, we're going to use the kinematics equation, v final squared is equal to v initial squared minus 2 a d so when we solve for this we're just going to substitute our variables in this should actually be negative 2 and so let's solve we get this is equal to the square root of negative 2 times 9 .8 times negative 2.

02:13 Which is equal to negative 6 .26 meters per second.

02:21 And the reason that it's negative is because it's moving in the negative y direction.

02:25 So we have that, but now we also have to find the second velocity.

02:31 So remember the ball bounces up to 1 .5.

02:34 So we have to do very similar steps.

02:37 We're going to get, but this time, v final is equal to zero...

Please give Ace some feedback