# AP Problem

1. Dec 2, 2004

### aber

Hey im in 12 grade but this is advanced placement so its college level... Anyways i could use some help on this problem i have

A 5kg ball initally rest at the edfe of a 2-meter long, 1.2 meter high frictionless tavle, as show above. A hard plastic cude of mass .5kg slides across the table at a speed of 26 m/s and strikes the ball, causing the ball to leave the table in the direction in which the cude was moving. The figure below shows a graph of the force exerted on the ball by the cude as a function of time

graph is:

as time goes from 0-4 seconds, the force rises o-2.0(x10^3 Newtons).
from 4.0-6.0 seconds the F stays at 2.0 (x10^3 Newtons)
6-10 seconds force falls to 0. so it looks like a table /**\ type thing...

Anyways, it wants the total impulse given to the ball?

2. Dec 2, 2004

### aber

i know impulse=change in potentional, or the intengral of dt? i havent taken calculas yet so i have no idea about derivatives and integrations

3. Dec 2, 2004

### Skomatth

Impulse is $$F \Delta t$$

You can easily determine this from the graph, just the same way you get displacement from a velocity vs time graph, or work from a Force vs displacement graph.

4. Dec 2, 2004

### aber

Whats that symbol infront of time.... we havent learned displacement yet...

5. Dec 2, 2004

### aber

nevermind change in time.... ok thanks

6. Dec 2, 2004

### Skomatth

$$F \Delta t$$ is read F delta t and the delta means change in (time for this case). I'm amazed you have learned anything about displacement (maybe you refer to it simply as distance which is a scalar quantity unlike displacement, a vector). But to concentrate on this problem how would you find impulse looking at the graph?

7. Dec 2, 2004

### aber

so i did J=2000(6-4), J=4,000.. so basically area under the FxT graph is the impulse... sasme as area in VxT=distance... next part is to find the horizontal velocity of the ball, which so if i know mv of block, i can use J=change in momentum... so 4000=(26x.5)-(5xV). so Velocity= hm.... nevermind that doesnt seem right... i guess we dont know change in momentum

8. Dec 2, 2004

### aber

o wait, wouldnt it be 4000=(.5x26)-(.5v), V=7974... that defiantly isnt right

9. Dec 3, 2004

### NateTG

Do you know any calculus? (oops, nevermind, just saw the inital post.)

The impulse is going to be equal to the area under the force curve in the force-time graph. You can check to make sure that force times time gives units equal to momentum, and be sure to check the scales.

Last edited: Dec 3, 2004