Homework Help: Physics AP problem (mechanics)

1. Jan 8, 2005

physicsgirl101

The followin is an AP problem that I have a question about:

A 5-kilogram ball intially rests at the edge of a 2-meter-long, 1.2-meter-high frictionless table, as shown above. A hard plastic cube of mass 0.5 kilogram slides across the table at a speed of 26 meters per second and strakes the ball, causing the ball to leave the table in the direction in which the cube was moving. The figure below shows a graph of the force exerted on the ball by the cube as a function of time.

(I did the part of the problem dealing with the graph finding the total impulse.)

I need help doing b and c

b.) Determine the horizontal velocity of the ball immediately after the collision.

c.) Determine the following for the cube immediately after the collision.
i. its speed
ii. its direction of travel (right or left), if moving

I was confused how to set up these two problems, is the collision elastic? How should I find the velocities immediately after the hit? Thank you for any help you can give!!!!
: )

2. Jan 8, 2005

physicsgirl101

Thanks so much to anyone who can help me!!!!

3. Jan 8, 2005

apchemstudent

we should see the graph.... to see if it is in fact an elastic collision..

4. Jan 8, 2005

physicsgirl101

here is a description of the graph:

the y axis is force (x 10^3 N) and the x axis is time (x 10^-3 seconds)

starting from the orgin, it is a straight line with a slope of 1/2 and goes to the point (4,2) from there it is horizontal until the point (6,2) from there the line has a slope of -1/2 and ends at the point (10,0)

i have a scanner, but I don't know how to post the graph on the site, I don't understand what the graph has to do with these questions, it was related to a question involving impulse, which i understood but didn't post. Thanks for your help!!!

5. Jan 8, 2005

cdhotfire

with the information given, I have, using momentum:

$(26)(.5) + (5)(0) = .5V_1 + 5V_2$
momentum before = momentum after

dont think one can do it another way, maybe if we see the graph it would change.

Last edited: Jan 8, 2005
6. Jan 8, 2005

cdhotfire

^
|
I changed my post

7. Jan 8, 2005

physicsgirl101

Thanks so much for your response!..... I actually set it up the same way you did....setting up that equation for momentum.
but to find V1 and V2 I had to make another equation setting the energy before and after equal..... but i thought energy isn't conserved in a collision(???) so that doesn't really make sense. So how else do you use the equation you provided to find V1 and V2?

(By the way, I got V1 = 4.727 m/s (to the right) and V2 = 21.27 m/s (to the left) ).....
Thanks....

8. Jan 8, 2005

physicsgirl101

Last edited: Jan 8, 2005
9. Jan 8, 2005

UrbanXrisis

You can attach a file/picture to your posts, however, it takes time (not just a couple of min) for the mentors to approve your picture.

10. Jan 8, 2005

UrbanXrisis

you see "Mange Attachments"? Click that and add the file you want.

Last edited by a moderator: May 1, 2017
11. Jan 8, 2005

physicsgirl101

I clicked it and nothing happened.... my computer must not allow it or something... can I email it to you?

12. Jan 8, 2005

UrbanXrisis

OKay, when you hit post reply, go past the text box and to where it says "Additional Options"

Here, hit "Mange Attachments" to add an attachment (in your case a picture)

13. Jan 8, 2005

apchemstudent

well to determine if the collision was elastic:

the 5 kg balls should be moving at :

5v = 12

v = 2.4m/s

However though the actual speed that the balls should be travelling at, if it was an elastic collision would be 4.727m/s so no the collision was not elastic.

I(initial) = I(final)
.5v(initial) = .5(v1) + 5(v2)

.5v(initial)^2/2 = .5(v1)^2/2 + 5(v2)^2/2

If you solve for v2, you'll find the answer of 4.727m/s

Edit: i noticed that you've already determined that it wasn't elastic...

C) well you can still solve it since momentum is always conserved...

m1(v1)initial = m1(v1) + m2(v2)

Lets say m1 = .5 kg and m2=5kg
you know v1 initial and v2, so you can solve for v1...

B) You are given the impulse from the graph. You can determine the change in velocity of the ball, since you already know the mass...

Last edited: Jan 8, 2005
14. Jan 8, 2005

cdhotfire

Srry, I was gone.

I think it was an elastic collision. Because if it was not, then the awnser could not be determined.

"Most collisions between objects involve the loss of some kinetic energy and are said to be inelastic. In the general case, the final velocities are not determinable from just the initial velocities. If you know the velocity of one object after the collision, you can determine the other "

http://hyperphysics.phy-astr.gsu.edu/hbase/inecol2.html#c1

oh, and I though for putting pictures in forums one had to put [-IMG]www.blahblah.[-/IMG]

withouth the -.

Last edited: Jan 8, 2005
15. Jan 8, 2005

apchemstudent

No, if it was not elastic, then it can still be determined as in your quote"...If you know the velocity of one object after the collision, you can determine the other " which you do know from the graph...

16. Jan 8, 2005

cdhotfire

Oops, I didnt even look at the graph, or read what she posted after your comment . Point taken.

Well seems she didnt even need our help, she had the awnser. Right after my post.

"By the way, I got V1 = 4.727 m/s (to the right) and V2 = 21.27 m/s (to the left)."

Last edited: Jan 8, 2005
17. Jan 8, 2005

apchemstudent

18. Jan 9, 2005

Andrew Mason

The area under the graph gives you the impulse received by the ball. So that is the momentum transferred to the ball and the momentum lost from the cube.

AM

19. Jan 9, 2005

physicsgirl101

Thanks for your help, but I'm not sure I completely understand what you're saying. First, why did you say the 5kg ball shold be moving at 5v=12 (I really don't know anything about impulse.... other than that it's the area of the graph, so i'm assuming this 12 is impulse..... ) So when you say it should be moving at 2.4m/s is that correct?? Then why did you say 4.727m/s also? (I'm confused about which one you're saying is incorrect/correct)---but I did actually get 4.727m/s for that one....

So is the collision inelastic (because they touch for a fraction of a second??)but the objects do end up going in opposite directions(??) Basically, if it's inelastic, what is happening---are they sticking together, or just touching for a short while before separating which makes it inelastic?

for B) I did gather that the total impulse was 12---but how do I calculate the change in velocity---I basically don't really know any equations involving impulse.

(Btw, thanks so much for helping!)

20. Jan 9, 2005

physicsgirl101

Then how do I get the correct answer...../whyis it wrong?

21. Jan 9, 2005

physicsgirl101

Thats what I thought---the area under the graph would be impulse (that's pretty much all I know about impulse) so I got 12. What do you mean by "that is the momentum transferred to the ball and the momentum lost from the cube"? Do you mean that impulse is equal to the momentum transferred to the ball which is equal to the momentum lost from the cube?
Thanks...

22. Jan 9, 2005

apchemstudent

Yes it is exactly what he's saying...

Law of conservation of momentum:

I(initial) = I(final)

m1 = .5 kg m2 = 5kg

m1v(initial) = m1(v1) + m2(v2)

m2v2 = m1v(initial) - m1(v1) <-------------- momentum lost from the cube....

23. Jan 9, 2005

Andrew Mason

Yes. Newton's third law: the force speeding up the ball is identical in magnitude and duration and opposite in direction to the force slowing down the cube.

So $\Delta p_{ball} + \Delta p_{cube} = 0$

The impulse (area) is 12 Nsec (= 12 kg m/sec) so:

$\Delta p_{ball} = - \Delta p_{cube} = 12$

Increase in ball speed = 12/5.
Decrease in cube speed = 12/.5

Figure out their speeds from that.

AM

24. Jan 9, 2005

physicsgirl101

APchemstudent said:
"However though the actual speed that the balls should be travelling at, if it was an elastic collision would be 4.727m/s so no the collision was not elastic.

I(initial) = I(final)
.5v(initial) = .5(v1) + 5(v2)

.5v(initial)^2/2 = .5(v1)^2/2 + 5(v2)^2/2

If you solve for v2, you'll find the answer of 4.727m/s"

I thought you said 4.727m/s was INCORRECT because you said that if the collision was inelastic then it would be 4.727m/s and that the collision was, in fact, inelastic. (???)

***Also, I had used those same two equations to solve for V2, but I thought that was wrong because the second equation ((1/2)mv^2 = (1/2)mv^2) is energy and energy is NOT conserved in an inelastic collision---and we determined this was an inelastic collision, right?****

25. Jan 9, 2005

apchemstudent

energy was not conserved. You cannot use KE initial = KE final... so the only equation you can use is I(initial) = I(final)