Finding final speed with two unknowns (Vf1 & Vf2)

In summary, the two balls, with masses of 0.540 kg and 0.440 kg, collide head-on with the 0.540 kg ball initially moving east at a speed of 3.70 m/s and the 0.440 kg ball at rest. By using the equations for kinetic energy and momentum, and factoring out the masses, it is possible to find the final speeds and directions of both balls after the perfectly elastic collision. After simplifying and solving, it is determined that the final speed of the 0.540 kg ball will be 3.7 m/s east, while the final speed of the 0.440 kg ball will be 7.4 m/s east.
  • #1
tjohn101
93
0

Homework Statement


A ball of mass 0.540 kg moving east (+x direction) with a speed of 3.70 m/s collides head-on with a 0.440 kg ball at rest. If the collision is perfectly elastic, what will be the speed and direction of each ball after the collision?

ball originally at rest m/s east
ball originally moving east m/s east

Known/Unknowns:
m1= .540 kg
Vi1= 3.70 m/s
Vf1= ?
m2= .440 kg
Vi2= 0 m/s
Vf2= ?

Homework Equations


[tex]
\frac{1}{2}m_1vi_1^2 + \frac{1}{2}m_2vi_2^2 = \frac{1}{2}m_1vf_1^2 + \frac{1}{2}m_2vf_2^2
[/tex]

[tex]
m_1\vec{vi}_1 + m_2\vec{vi}_2 = m_1\vec{vf}_1 + m_2\vec{vf}_2
[/tex]

Did I miss any?

The Attempt at a Solution


I've tried plugging everything into the above equations, but the two unknowns have me stuck. I know I'm supposed to use basic Algebra to give each unknown it's own equation and solve from there, but I have honestly forgotten how. Any help on how to solve for one of the unknowns is greatly appreciated. Thank you!
 
Last edited:
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  • #2
Rewrite the relevant equations by using v1i, v1f, v2i and v2f.
Collect the terms congaing m1 in one side and m2 on the other side in both the equations.
Take the ratio of left hand side and right hand side and equate.
Now see whether you can proceed further.
 
  • #3
No. I tried to create separate equations and attempted to solve for Vf1 by substituting Vf2 with an equation for Vf2 (sorry if it doesn't make sense) but I only came up with zero...
 
  • #4
It is always better to show your calculations, so that we can check where you went wrong. After rearranging the two equations , you get
v1i^2 -v1f^2 = v2f^2 - v2i^2...(1)
v1i -v1f = v2f - v2i...(2)
Factorize both the sides in (1) and take the ratios of left hand side and right hand side of the equations.
 
  • #5
rl.bhat said:
It is always better to show your calculations, so that we can check where you went wrong. After rearranging the two equations , you get
v1i^2 -v1f^2 = v2f^2 - v2i^2...(1)
v1i -v1f = v2f - v2i...(2)
Factorize both the sides in (1) and take the ratios of left hand side and right hand side of the equations.

Oh okay I will. I just thought it would've been a little extreme as they took up the whole page in my binder. 8l

Anyways, I constantly end up getting this and it doesn't work:

(1/2m1vi1 - 1/2m1vf1)/(1/2m2)=Vf2

I have a feeling it's completely wrong

EDIT: Turns out I WAS completely wrong. I solved the problem finally. :D

I did it by doing exactly what rl.bhat gave me and then solved for both v2f and v1f by getting each one alone.
I eventually got v1f=v2f + 3.7 and v2f=3.7 - v1f.
Then I plugged v1f into the original equation (second one listed under my "relevant equations" and solved for V1f. Then, I plugged that answer into the V2f=3.7 - v1f and found the answer.
 
Last edited:
  • #6
tjohn101 said:
I did it by doing exactly what rl.bhat gave me and then solved for both v2f and v1f by getting each one alone.
I eventually got v1f=v2f + 3.7 and v2f=3.7 - v1f.
Then I plugged v1f into the original equation (second one listed under my "relevant equations" and solved for V1f. Then, I plugged that answer into the V2f=3.7 - v1f and found the answer.


I must be missing something. Shouldn't you take into account the masses into that equation? I think that it would be correct if m1=m2, but they are not.
 
  • #7
Vector_Joe said:
I must be missing something. Shouldn't you take into account the masses into that equation? I think that it would be correct if m1=m2, but they are not.

Don't feel bad, I thought the same thing. Here's why you wouldn't:

You start out with the two equations I have listed as relevant equations (first is kinetic energy, second is momentum). Both of them will be used.

Next, you rewrite the momentum equation by doing m1Vi1 - m1Vf1 = m2Vf2 - m2Vi2 and factoring out the mass. This leaves you with m1(Vi1 - Vf1) = m2(Vi2 - Vf2)

Then you do the same thing for the kinetic energy equation and get m1(Vi1^2 - Vf1^2) = m2(Vi2^2 - Vf2^2)
Which becomes m1(Vi1 - Vf1)(Vi1 + Vf1) = m2(Vi2 - Vf2)(Vi2 + Vf2)

Now that you have the two equations, you divide the momentum equation by the kinetic equation.

(m1(Vi1 - Vf1) = m2(Vi2 - Vf2)) / (m1(Vi1 - Vf1)(Vi1 + Vf1) = m2(Vi2 - Vf2)(Vi2 + Vf2))

Both masses cancel out and so do the similar equations. This leaves you with Vi1 + Vf1 = Vi2 + Vf2

Then just solve from there.

It looks complicated but it really isn't that bad. : )
 

1. What is the equation for finding final speed with two unknowns?

The equation for finding final speed with two unknowns (Vf1 & Vf2) is Vf1 + Vf2 = Vf.

2. How do you solve for Vf if both Vf1 and Vf2 are unknown?

In order to solve for Vf, you will need to have at least one other known variable, such as the initial speed (Vi) or the acceleration (a). You can then use the equation Vf = Vi + at to solve for Vf.

3. Can you use this equation if only one of the final speeds is unknown?

Yes, you can still use the equation Vf1 + Vf2 = Vf if only one of the final speeds is unknown. You will just need to rearrange the equation to solve for the unknown final speed.

4. Is there a specific unit for final speed in this equation?

The unit for final speed in this equation will depend on the units used for the initial speed (Vi) and the acceleration (a). It is important to make sure that all units are consistent when using this equation.

5. Can this equation be used for any type of motion?

Yes, this equation can be used for any type of motion as long as the initial speed (Vi) and the acceleration (a) are known or can be calculated. It is a general equation that can be applied to various scenarios.

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