# Help me with an old test question

1. Nov 13, 2006

### curt2121

So I'm about to prepare for a midterm. We have access to old midterms bu not the answers and I'm havign trouble wiht a problem. I've attached a copy since it contains a diagram. I don't know why, but my initial thinking is that the two side squares will travel in a straight line with half the initial speed of the ball and the middle cube will sit there with its point just touchign the ball (both at rest) anyone care to provide some insight and help solve this for me? Thanks!

2. Nov 13, 2006

### physics girl phd

Have you tried to do the problem with just the top/bottom cubes present and not the side cube? What happens then? That might help you.

3. Nov 13, 2006

### curt2121

well i thought about that and that's also part of why I see the two side cubes speeding off at a 45 degree angle to the horizontal. In my mind, I'm seeing it like a pool ball striking right in the middle of 2 other balls and causing them to speed off at half the speed of the original and at a 45 degree angle

4. Nov 13, 2006

### physics girl phd

That's my GUT instinct too -- but have you actually DONE the math for that simplier problem? Can you show it if you have?

5. Nov 13, 2006

### curt2121

well the math for that isn't difficult it's just a colision in 2d. I'm having reservations now though that the ball will hit the side cubes at a point sqrt((l^2)/2) from their intersection since the ball is going to hit equally between the two cubes, it will hit as if a triangle with the longest side being l and the two other sides being equal so that l^2=2x^2 and x (point of contact)=sqrt((l^2)/2). But at this point I don't know if I'm overcomplicating or working on the right track

6. Nov 13, 2006

### physics girl phd

I didn't get this (but it's getting late here -- I might be wrong). Can you show me the geometry of that?

Last edited: Nov 13, 2006
7. Nov 13, 2006

### physics girl phd

Hint -- I also thought about where they would hit just by picturing a fourth box to the right -- and the circle inside it. Does that help?

I'll be back online in a bit -- I need to go catch some food and a bus home. Some others might take over -- it looks fun. :tongue:

8. Nov 13, 2006

### curt2121

i just don't know anymore. Out of about 40 problems I've gone through, I haven't really had any trouble but this one I've gotten to where i've just overthought about it so I need some solid guidance on this one soon

9. Nov 14, 2006

### Staff: Mentor

I'm no expert in this kind of problem, but it seems like one key is whether the ball touches all 3 squares at once, or does it touch the top and botom squares first. When I draw it, the left square is not touched....

10. Nov 14, 2006

### physics girl phd

If you picture the circle inscribed in an imaginary box to the right, you'll see that it touches only the top and bottom squares... and that the touching point is at the middle of the sides.

11. Nov 15, 2006

### curt2121

ok that's what i'm thinking. so we are agreed on that. did my idea of each of the squares moving off with half the initial velocity of the ball make sense?

12. Nov 15, 2006

### Staff: Mentor

Again, I'm no expert in this kind of question, but what is conserved? KE or momentum?

13. Nov 15, 2006

### George Jones

Staff Emeritus
Even if true, this is something you have to justify.

The two blocks will move in directions that are in the directions of the (impulsive) contact forces. The contact forces are directed perpendicular to the tangents at the points of contact. This justifies your original conclusions about directions.

Consider berkeman's last post. You stated that the collisions are elastic. What does this tell you about things that are conserved?

Take anything that is conserved and conserve it - tally up each conserved quantity before and after the collisions. If a conserved quantity is vectorial in nature, resolve it into components.

One possibility is that, after the inititial collisons, the ball keeps moving to the left and runs into the leftmost block. Does this happen, or does this not happen?

14. Nov 15, 2006

### OlderDan

Not that Geoge's comments need to be expanded, but just to round things out, another possibility is that the ball bounces back in the direction from which it came. If the blocks were connected to one another instead of free to separate, this is what would happen. If you find you need more block speed to prove that conservation holds, you could consider this possibility as well.

15. Nov 24, 2006

### curt2121

well the midterm has come and gone and luckily no problems like this were on it. i've been pretty busy lately but I still want to solve this problem (it bugs me to just leave somethign hanging) so I'm going to try to work on it tomorrow. any further insight is still greatly appreciated

16. Nov 24, 2006

### OlderDan

Good for you! Here is a hint
Why half the speed? Why not some other speed?

17. Nov 26, 2006

### curt2121

ok so i used some numbers since i absolutely hate problems in variables. i treated the two side blocks as one object as suggested above with a mass of 10 kg combines. therefore the ball has mass 5kg. I gave it initial velocity of 2m/s so that after the collision, the ball has a velocity of .66m/s and the squares combined have velocity 1.33m/s combined. As far as i know, the velocities of each cube should be equal since they have the same mass and were in the same collision. therefore each cube has a velocity of .66m/s at 45 degree angles so that the ball and 2 side cubes have the same velocity (.66/2=1/3 of original velocity) so that the two cubes shoot off at 1/3 the original speed of the ball and the ball bounces back with 1/3 its original speed and the middle cube stays sitting there. this seem right? if not i simply give up

18. Nov 27, 2006

### curt2121

dan? anyone?

19. Nov 27, 2006

### OlderDan

Your result does not conserve momentum or energy

You were close way back at the beginning. The only problem was your assumption that each of the two blocks would have half of the initial speed of the ball. If they had only half the speed, and the ball came to rest, momentum and energy would not be conserved. If only the two blocks are moving after the collision, the momentum components of each block in the original direction of the ball must be half the original momentum of the ball. If they go off at 45° and the ball stops as you suspected then

2m*v_f*cos45° = m*v_i

v_f = v_i/(2*cos45°) = 0.707*v_i <<== NOT half v_i

Does this conserve energy?

2*½*m*v²_f = 2*½*m*(0.707*v_i)² = ½*m*v²_i

Sure enough. Energy is conserved. The two blocks go off at 45° with 1/sqrt(2) times the initial velocity, while the ball and the third block remain at rest.