How Do You Solve for Final Velocities in Elastic Collisions?

  • #1
pineapplebanana
8
0
Homework Statement
A rubber ball (m, = 0.0500 kg) hangs at the end of a 0.500 m pendulum string. The ball is held to the side at an angle of 10.0°, as shown in the diagram. When the ball is released from rest, it swings down to hit a larger rubber ball (m₂ = 0.0800 kg) sitting on a frictionless surface. The two balls collide elastically. Determine the speed of the second ball after the collision.
Relevant Equations
PEi=KEf, Pi=Pf, mvi= mvf + MVf
So i started off breaking up the problem into two sequences, right before the collision and after the collision has happened. I need to find the first ball's speed immediately before the collision which is no problem. PEi = KEf > mghi = (1/2)mvf (vf being the velocity right before the collision.
solving for vf, vf=(2ghi)^0.5, hi=0.5cos(10) I ended up getting 3.1082m/s for the speed before the collision. Now comes the part where I am struggling, I just can't think of a way to solve for Vf without having vf (first balls speed after collision). I know since there is two unknown variables I have to use two different equations and make a system, but I cant figure out how or which equation to use. Ive tried using the head on collision equation vi + Vi = vf + Vf, but that ends up canceling out Vf in my first equation. Here is what i have been able to do if that makes more sense than what I have typed out.

1699581590398.png
 
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  • #2
I feel like the only way I can get this to work is if the speed of the ball on the string after the collision is zero
 
  • #3
Conservation of Energy is another equation you are to work with, i.e. the collision is perfectly elastic.
 
  • #4
erobz said:
Conservation of Energy is another equation you are to work with, i.e. the collision is perfectly elastic.
I used conservation of Energy to find the first balls speed right before it collides with the second ball. I asked my prof and said I should only have to use collision equations to find the speeds after they hit. I dont understand how F=ma is going to work in this case as its for force? and not energy?
 
  • #5
pineapplebanana said:
I used conservation of Energy to find the first balls speed right before it collides with the second ball. I asked my prof and said I should only have to use collision equations to find the speeds after they hit. I dont understand how F=ma is going to work in this case as its for force? and not energy?
Yes you used conservation of energy before the collision. Thats good... But the collision itself is elastic meaning energy is not converted to heat, deformation, sound, etc... it is conserved immediately after the collision.

Write ##KE_{before} = KE_{after}##
 
  • #6
erobz said:
Yes you used conservation of energy before the collision. Thats good... But the collision itself is elastic meaning energy is not converted to heat, deformation, sound, etc... it is conserved immediately after the collision.

Write ##KE_{before} = KE_{after}##
1699584203797.png

yes I've done that but i think my brain has gone to mush and will not allow me to properly take VF out of the brackets and root. I either end up with a negative under the root, or end up cancelling VF or end up with 0/M
 

Attachments

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  • #7
$$ \frac{1}{2}mv_o^2 = \frac{1}{2}mv^2+ \frac{1}{2}Mu^2 $$

$$mv_o = mv + Mu$$

Two equations, two unknows ##v,u##. Whatever velocity you are trying to solve for eliminate it from the other equation by substitution.

For instance use momentum to put ##v## in terms of ##u##. Then sub into energy, expand and solve the resulting quadratic in ##u##.

EDIT: I agree with @kuruman your initial height is not correct for a PE = 0 Datum you have selected.
 
Last edited:
  • #8
You say that ##v_0=\sqrt{2gh_i}## and I agree.
Then you say ##h_i=0.5\cos10^{\text{o}}.## I disagree. If that is correct, then if the angle is ##90^{\text{o}}##, i.e. the pendulum is released with the string in the horizontal position, ##h_i=0.## Is that reasonable?
pineapplebanana said:
I dont understand how F=ma is going to work in this case as its for force? and not energy?
I don't either. This is not a F=ma problem. The force that one ball exerts on the other lasts for a very short time and is not constant. You need to write two equations, one for momentum conservation and one for energy conservation before and after the collision.

I will get you started. From momentum conservation
##mv_0=mv_f+MV_f\implies v_f=v_0-\dfrac{M}{m}V_f.##
Put that in the energy conservation equation and solve for ##V_f##. Note that ##\frac{1}{2}mv_0^2=mgh_i##, so be sure you have the correct expression for ##h_i.##
 
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  • #9
kuruman said:
You say that ##v_0=\sqrt{2gh_i}## and I agree.
Then you say ##h_i=0.5\cos10^{\text{o}}.## I disagree. If that is correct, then if the angle is ##90^{\text{o}}##, i.e. the pendulum is released with the string in the horizontal position, ##h_i=0.## Is that reasonable?

I don't either. This is not a F=ma problem. The force that one ball exerts on the other lasts for a very short time and is not constant. You need to write two equations, one for momentum conservation and one for energy conservation before and after the collision.

I will get you started. From momentum conservation
##mv_0=mv_f+MV_f\implies v_f=v_0-\dfrac{M}{m}V_f.##
Put that in the energy conservation equation and solve for ##V_f##. Note that ##\frac{1}{2}mv_0^2=mgh_i##, so be sure you have the correct expression for ##h_i.##
hi is where the pendulum is released from before the collision, so why wouldn't it be 0.5cos10? since it is up slightly from the ground? i guess hi could maybe be 0.5 - 0.5cos10?
 
  • #10
pineapplebanana said:
hi could maybe be 0.5 - 0.5cos10?
Quite
 
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  • #11
pineapplebanana said:
hi is where the pendulum is released from before the collision, so why wouldn't it be 0.5cos10? since it is up slightly from the ground? i guess hi could maybe be 0.5 - 0.5cos10?
by conservation of energy equation do you mean KEi + PEi = KEf +PEf, or do you mean just KEi = KEf. because i've been trying the second one as you can see in the most recent screen shot i have posted, but I just for whatever reason cannot wrap my head around getting the VF by itself
 
  • #12
pineapplebanana said:
the pendulum is released where it makes an angle of 10 degrees to the vertical like in my drawing. not 80, not 90
Anyhow, like has been said... the initial height is wrong w.r.t. the datum. Correct that, then work on solving the system of equations.
 
  • #13
erobz said:
Anyhow, like has been said... the initial height is wrong w.r.t. the datum. Correct that, then work on solving the system of equations.
i dont even care about that part anymore. I just want to know how to actually finish the problem. its not due for homework or anything, I am just studying for an exam, but this problem is sending me for a trip. Having the right or wrong number doesn't change the algebra.
 
  • #14
Then you have to solve the system of equations in #7.
 
  • #15
I realize that. and I get stuck here:
which is why i am asking for help
1699586524757.png
 
  • #16
pineapplebanana said:
I realize that. and I get stuck here:
which is why i am asking for help
View attachment 335107
##v_f## is wrong on the right side. You have a velocity equal to a momentum... and can you use ##u,v##? ##v_f## and ##V_f## in your chicken scratch is not helping. We use LaTeX Guide to express our mathematics on the site. Give it a try...if anything it might distract you long enough to reset your vision on this problem.
 
  • #17
With the substitution I suggested in post #8 the energy conservation equation becomes
$$\frac{1}{2}mv_0^2=\frac{1}{2}m\left(v_0-\frac{M}{m}V_f\right)^2+\frac{1}{2}MV_f^2$$ Expand the square and gather terms. What do you get?
 
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  • #18
pineapplebanana said:
I realize that. and I get stuck here:
which is why i am asking for help
View attachment 335107
Your final equation there (bottom right) is dimensionally inconsistent.
Go back to the preceding line, substitute for ##v_f## in there and gather the ##V_F## terms together.

For future reference, there is a shortcut when momentum and KE are both conserved in a 1D collision. This can be obtained by combining those two conservation equations: ##v_{1f}-v_{2f}=v_{2i}-v{1i}##. It avoids the quadratics. It is a special case of the more general "Newton's Experimental Law" https://en.wikipedia.org/wiki/Coefficient_of_restitution.
 
  • #19
pineapplebanana said:
i dont even care about that part anymore. I just want to know how to actually finish the problem. its not due for homework or anything, I am just studying for an exam, but this problem is sending me for a trip. Having the right or wrong number doesn't change the algebra.
It doesn't male much sense to attempt a problem like this without first studying a simple elastic collision on its own, without the complication of a pendulum. Or, in fact, studying an object swinging on a pendulum. Given that you got that wrong by using the cosine directly.

There are many sources online where you can find an analysis of a totally elastic collision in one dimension. I suggest you find one of those and "wrap your head round the algebra" involved.
 
  • #20
Thread closed for Moderation...
 
  • #21
pineapplebanana said:
i dont even care about that part anymore. I just want to know how to actually finish the problem. its not due for homework or anything, I am just studying for an exam, but this problem is sending me for a trip. Having the right or wrong number doesn't change the algebra.
Thread will remain closed. OP requested that their account be deleted and this thread be taken down, so apparently it wasn't just "studying" for that exam. The thread will remain visible.

Thanks all who tried to help this OP.
 

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