Elastic Collision between two pendulums- max heights reached?

1. Mar 23, 2009

wizzle

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

2. Relevant equations

MaVa+MbVb=MaV'a+MbV'b
PE=mgh
KE=.5mv^2
mgh=.5mv^2=KE of small ball just before collision

3. The attempt at a solution
I first determined the starting height of the pendulum using the pythagorean theorem through sin15=y/.95 y=.2459 m .95-.2495m=.704 m above the x axis (where the pendulum would be vertical).

I tried mgh (with h being .704m)=.5mv^2
v=3.715 m/s

Ke just before collision: .5 mv^2 (small ball) + 0 = .5 mv'^2 (small ball) + .5 mv'^2 (big ball)
I found the KE just before the collision to be 0.138 J, which is the left side of the above equation.

The PE just before the collision is zero, and the PE when the balls reach their maximum height is mgh (small ball) + mgh (big ball) = 0.138 J (KE just before collision)

Does it look like I'm on the right track? I'm not sure how to proceed - once I find out the final velocities of each ball (and I have no idea how to do that right now) will I then use kinematic equations to figure out how long until they reach their peak height and then find out how high that will allow them to go? Any help would be greatly appreciated!

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2. Mar 24, 2009

tiny-tim

Hi wizzle!

Momentum is always conserved in collisions in any direction in which there are no external impulses …

so horizontal momentum will be conserved at the time of the collision …

that equation should give you the extra information you need

3. Mar 24, 2009

wizzle

Hi Tiny-tim,
Thanks a lot for the reply and I hope you can help me a little bit more. Since P=mv, I used the equation for the conservation of momentum: MaVa + 0 = MaVaf + MaVbf. (.02)(3.715) + 0 = (.02)Vaf + (.03)Vbf, with A indicating the small ball and B indicating the large ball. I solved for Vaf and found Vaf = ((7.43 x 10^-2) - 0.03Vbf)/ 0.02.
I then input that into my KE equation and ended up with 0.138 J = .5(0.2)(Vaf from above)^2 +.5(0.3)(Vbf)^2.

The quadratic I ended up with is 2.23 x 10^-2 Vb^2 -9.65 x 10^-2 Vb + 0 = 0 (the fact that my c term was zero certainly weirded me out, but I tried multiple times and got that same result). After solving the quadratic equation, I found V'b to be 4.289 m/s and then input that into Va=-V'a + V'b, finding V'a to equal 5.74 x 10^-1 m/s. I checked to see if it worked out by putting it into my KE equation but it didn't work out, since KE intial, .138, did not equal KE final, .279. I also thought that the smaller ball would have a negative velocity since it would bounce back in the direction it came from, and my result didn't show that. If you can spot any places that I am going wrong I would reallllly appreciate it!!! Thanks again!
-Lauren

4. Mar 26, 2009

tiny-tim

Hi Lauren! Thanks for the PM!

That is the correct technique.

But I suspect you've dropped a factor of 2 somewhere.

It's a lot easier sometimes if you use symbols for everything, and just put the figures in at the end.

In this case, you should have got the equations

gh = v2 + M/m V2

√(2gh) = v + M/m V,

and so the "c" isn't zero …

try again!

5. Mar 26, 2009

wizzle

Hi Tim,
I'm trying to do the problem without inputting numbers, as you suggested, and I think my main problem is that I'm messing up the order of operations for this part:

MaVa + 0 = MaV'a + MbV'b
(MaVa-MbV'b)/Ma = V'a

1/2MaVa^2=1/2Ma((MaVa-MbV'b)/Ma)^2 + 1/2 MbV'b^2

Now that I try to isolate V'b, I think I might be going wrong when I try to expand:
1/2MaVa= 1/2 Ma ((MaVa)^2-2(MaVaMbV'b)+(MbV'b)^2)/Ma^2 + 1/2MbV'b^2

1/2Ma^3Va=1/2((MaVa)^2-2MaVaMbV'b+(MbV'b)^2+1/2MaMbV'B^2

Lo and behold...V'a and V'b, when input into the original EK equation, don't add up to the KE just before the collision.

This question makes me feel like I'm aging more rapidly...and/or forgetting basic algebra.

Last edited: Mar 26, 2009
6. Mar 26, 2009

wizzle

hmm may have spotted an error, checking my calculations again

7. Mar 26, 2009

tiny-tim

Hi wizzle!
You mean MaKE = …
Not following you

8. Mar 26, 2009

wizzle

Oh jeez...been looking at this question for far too long. Now I'm very confused. I was trying to input V'a and V'b into the equation for Total energy (PE+KE initial) = (PE +KE final) since the energy is conserved, but the values I obtained didn't cause the final kinetic energy to be equal to the initial kinetic energy. Is that the wrong strategy?

Last edited: Mar 26, 2009
9. Mar 26, 2009

wizzle

sorry Tim, I don't mean to be wasting your time. my calculations seem to be sloppy, I'm going over them again.

10. Mar 27, 2009

wizzle

Hi Tim,
I think I may have got the answer to the first part. After countless re-calculations, I finally have V'b to equal 3.715 m/s, which was the velocity of the smaller ball just before the collision, and V'a is now equal to 0 m/s, so all the kinetic energy was transferred to the larger ball.

When I input that value into the equation Mgh=.5mv'b^2, I end up with h=.(5v'b^2)/g, and get 0.7041 m as the max height of the larger ball, which means the maximum angle was 75 degrees to the left of the y axis, and the smaller ball stays motionless. I looked at the equations you said I should have ended up with and I'm not sure how they fit in.

Hopefully this question is less annoying than the previous few...

Last edited: Mar 27, 2009
11. Mar 27, 2009

wizzle

I just had a thought...I don't think my answer could be right because in order for the kinetic energy just after the collision to be equal to the kinetic energy pre-collision, the velocities I came up with couldn't be right. My answer would mean that EK final would be greater than EK initial, since the moving ball has a mass of 0.03 kg, and the smaller ball, which I have said is now no longer moving, has a mass of 0.02 kg.

12. Mar 27, 2009

tiny-tim

Yes … the energy is only totally transferred when the balls are of equal mass (just like balls on a pool-table).

Lauren, you must show your full calculations, not just your result, or we won't be able to see where you're going wrong.

Call the masses m and M, and KE of the smaller ball mgh, and the speeds immediately after the collision v and V …

what equations do you get, just for the collision?

13. Mar 27, 2009

wizzle

Thanks for not giving up Tim.

I used m and M as you suggested:

mv+Mv=mV+MV
(mv-MV)/m=V

Now I substituted V, the final velocity of the small ball, into the following equation:
mgh=.5mV^2+.5MV^2
mgh=.5m(mv-MV/m)^2+.5MV^2
mgh=.5m((mv)^2-2mvMV+(MV)^2)/m^2+.5MV^2

I multiplied the .5MV^2 term by (m^2/m^2) so that I could add it to the other term

mgh=.5m((mv)(mv)-2mvMV+(MV)(MV)+.5m^2MV^2)/m^2

Does that make sense so far?

Thanks,
Lauren

14. Apr 27, 2009

wizzle

Luckily my boyfriend's sister, a civil engineering genius, helped me through this. Thanks everyone for your help along the way!