Finding speed of objects after elastic collison

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Homework Help Overview

The discussion revolves around determining the final speeds of two objects after an elastic collision, utilizing principles of conservation of momentum and kinetic energy. The participants are exploring the relationships between initial and final velocities of the objects involved in the collision.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using conservation equations to express final velocities in terms of each other. There is a focus on how to isolate variables in the equations provided. Some participants express confusion about the direction of velocities and the implications of the collision being perfectly elastic.

Discussion Status

Several participants have provided equations and attempted to manipulate them to solve for the final velocities. There is an ongoing exploration of the relationships between the variables, with some guidance offered on how to express one variable in terms of another. However, there is no explicit consensus on the final values or methods being used.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can share or the methods they can use. There are indications of potential misunderstandings regarding the setup of the problem and the assumptions about the initial conditions of the objects involved.

joe426
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Homework Statement



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Homework Equations



1/2m1v12 + 1/2m2v22 = 1/2m1v1f + 1/2m2v2f
m1v1 + m2v2 = m1v1f + m2v2f
v1 - v2 = -(v1f - v2f)

The Attempt at a Solution


So I solved the momentum of conservation for the final velocity of object 1. I then plug that equation into the third equation listed above, as v1f. But the thing is the equation i get for v1f has v2f in it and that is what I am solving for and this is where I am stuck.
 
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I don't think any of the velocity equals 0 at any point. Initially they are moving in the same direction. Then since it says its perfectly elastic, one object will bounce back and have a velocity in the opposite direction. I think the 60g tennis ball will bounce back in the negative direction but I still can't figure out how to find the velocities after the collision :(
 
Show your whole work please.

ehild
 
First use conservation of momentum
( m1v12 + m2v22 - m2v2f2 ) / m1 = v1f2

Then plug this v1f into this helper equation, the third equation listed in first post.
v2f = v1 - v2 + [( m1v12 + m2v22 - m2v2f2 ) / m1 ]1/2


And I'm stuck because I can't get both v2f to one side so i can solve for it. then once i get v2f i plug it back into the first equation, [( m1v12 + m2v22 - m2v2f2 ) / m1 ], and solve for v1f
 
Use the last two equations. Plug in the numbers first. Express v1f from the third equation.

ehild
 
joe426 said:
1/2m1v12 + 1/2m2v22 = 1/2m1v1f + 1/2m2v2f
m1v1 + m2v2 = m1v1f + m2v2f
v1 - v2 = -(v1f - v2f)

Use the last two equations. Plug in the data first, express vif form the last equation, and substitute into the second one.

ehild
 
v1f = -v1 + v2 - v2f

Plugged that into,
(m1v1 + m2v2 - m1v1f) / m2 = v2f

v2f = -.368m/s
 
joe426 said:
v1f = -v1 + v2 - v2f

Check it.
 
ehild said:
Check it.

v1f = -v1 + v2 - v2f is really v1f = -v1 + v2 + v2f. ..

(m1v1 + m2v2 - m1v1f) / m2 = v2f turns into (m1v1 + m2v2 + m1v1f) / m2 = v2f

so v2f = .307m/s
 
  • #10
joe426 said:
v1f = -v1 + v2 - v2f is really v1f = -v1 + v2 + v2f. ..

(m1v1 + m2v2 - m1v1f) / m2 = v2f turns into (m1v1 + m2v2 + m1v1f) / m2 = v2f

Check.

ehild
 
  • #11
Plug in the numbers for v1 and v2: v1f = v2f - 1.35.
Plugging in the data into the equation m1v1+m2v2=m1v1f+m2v2f and also substituting v2f - 1.35 for v1f: 0.2535 = 0.06(v2f - 1.35)+0.09v2f.
Expand and collect the terms with v2f, solve.

ehild
 
  • #12
Its 3.75 or 1.35?
 
Last edited:
  • #13
Munzi5 said:
Its 3.75 or 1.35?

What is 3.75 or 1.35?

ehild
 

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