How Is Average Force Calculated for a Rebounding Ball?

[L²Cc]

Homework Statement

After falling from rest a a height of 30 m, a 0.50 kg ball rebounds upward, reaching a height of 20 m. If the contact between ball and ground lasted 2.0 ms, what average force was exerted on the ball?

Homework Equations

The Attempt at a Solution

I figured I had to find the average acceleration (v-u/t) in order to determine the average force acting on the ball. Hence, the initial acceleration is 9.8 m/s. Now, how do I find the final acceleration using the given information; the ball is moving upward so gravity is no longer acting on it.
I know that the final velocity of the ball (traveling 30 m) is 24.25 m/s, based on the formula v²=u²+2as...would this value help me in solving the problem?
I'm clueless about the rest.

 

[Hootenanny]

Now, how do I find the final acceleration using the given information; the ball is moving upward so gravity is no longer acting on it.

 

[L²Cc]

lol as in I mean the acceleration of the ball isn't 9.8 m/s² since it rebounds upward...

 

[Hootenanny]

lol as in I mean the acceleration of the ball isn't 9.8 m/s² since it rebounds upward...

Are you quite sure about that? It is not the initial and final accelerations that are important, it is the velocities just before and after the collision which are vital. You may wish to consider the impulse-momentum theorem.

 

[L²Cc]

Well, we haven't done the impulse-momentum theorem...Would the initial velocity be 24.45 m/s and the final velocity 0 since the ball hits the ground with the first velocity and comes to a rest for 2 ms, and then rebounds upward...Hence, the average velocity would be (0-24.25/0.002s) ?

 

[Hootenanny]

The initial velocity is correct. As for the final velocity you want to find the velocity just after the collision with the floor.

The impulse-momentum theorem simply states;

$$F\cdot dt = m\cdot dv\hspace{1cm}\text{(for constant mass)}$$

 

[L²Cc]

Would the following info help us find the velocity of the ball after the collision:
0.002 s, initial v = 24.45 m/s, and initial s = 30 m ?
The only kinematic equation which does not ask for acceleration is: s= (u+v/s)t ?

 

[Hootenanny]

Okay think of the question this way.

You throw a ball vertically upwards, if the ball reaches a maximum height of 20m above your hand find the intial velocity.

Does that make sense?

 

[L²Cc]

is the initial velocity 19.80 m/s if, say, the acceleration is 9.8 m/s², final v = 0m/s, and s = 20m...?
Although when you square root u² (if we are to use the formula v²=u²+2as), u is a complex number...?!

 

[Hootenanny]

is the initial velocity 19.80 m/s if, say, the acceleration is 9.8 m/s², final v = 0m/s, and s = 20m...?

Correct, so now you know that before the collision the velocity was 24.5m/s and after the collision the velocity was 19.80 m/s. Using the impulse formulae I gave you above can you go from here?

Initial velocity will not be complex if you have defined you coordinate system and manipulated your equation correctly.

 

[L²Cc]

f x dt = 0.50 kg x (24.45 - 19.8)?

 

[Hootenanny]

f x dt = 0.50 kg x (24.45 - 19.8)?

Correct, except your change in velocity should be final - initial i.e. 19.8 - 24.45

Now, you also know the time over which the force acts, so you can calculate the average force.

 

[L²Cc]

F= (0.50 kg x -4.65)/ 0.02 ?

 

[Hootenanny]

Careful, 2 ms $\neq$ 0.02 s

 

[L²Cc]

Yeah, I corrected the error but I guess it didn't save the changes! So the answer is -1.16 x 10^4

 

[Hootenanny]

Yeah, I corrected the error but I guess it didn't save the changes! So the answer is -1.16 x 10^4

No, your an order of magnitude off. It should be -1.16x10³N

 

[L²Cc]

hmmm...The answer key in my textbook claims that the answer is 1.1 x 10^4 upward?

 

[Hootenanny]

In that case, the answer in the book is off by an order of magnitude, the direction is correct (negative implying opposite direction to motion, which in this case is up). One small error I did spot is that your 24.45 should be she 24.25

 

[L²Cc]

Oh ok, Thank you Hootenanny, once again! :)

 

[Hootenanny]

Oh ok, Thank you Hootenanny, once again! :)

Once again a pleasure.

 

[chromium blade]

the answer in the textbook is 100% correct

Impulse is never negative, and the magnitude of the force would be 11012.5 N

remember I = mu - mv in this case