# Forces & kinematic equations

1. Jan 13, 2007

### L²Cc

1. The problem statement, all variables and given/known data
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?

2. Relevant equations

3. 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?

2. Jan 13, 2007

### Hootenanny

Staff Emeritus
:surprised

3. Jan 13, 2007

### L²Cc

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

4. Jan 13, 2007

### Hootenanny

Staff Emeritus
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.

5. Jan 13, 2007

### 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) ???

6. Jan 13, 2007

### Hootenanny

Staff Emeritus
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;

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

7. Jan 13, 2007

### 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 ???

8. Jan 13, 2007

### Hootenanny

Staff Emeritus
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?

9. Jan 13, 2007

### 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...?!

10. Jan 13, 2007

### Hootenanny

Staff Emeritus
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.

11. Jan 13, 2007

### L²Cc

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

12. Jan 13, 2007

### Hootenanny

Staff Emeritus
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.

13. Jan 13, 2007

### L²Cc

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

14. Jan 13, 2007

### Hootenanny

Staff Emeritus
Careful, 2 ms $\neq$ 0.02 s

15. Jan 13, 2007

### L²Cc

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

16. Jan 13, 2007

### Hootenanny

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

17. Jan 13, 2007

### L²Cc

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

18. Jan 13, 2007

### Hootenanny

Staff Emeritus
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

19. Jan 13, 2007

### L²Cc

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

20. Jan 13, 2007

### Hootenanny

Staff Emeritus
Once again a pleasure.