Solving for Velocity in a Momentum and Kinetic Energy Problem

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Homework Help Overview

The discussion revolves around a problem related to momentum and kinetic energy, specifically focusing on the relationships between the velocities of two colliding objects. The original poster presents equations derived from the problem and expresses confusion regarding the algebraic manipulations involved.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the algebraic rearrangements of equations and question the validity of certain substitutions. There is a focus on clarifying the correct use of variables and the implications of a potential typo in the equations presented.

Discussion Status

The conversation is ongoing, with some participants providing guidance on identifying errors in the equations. There is an acknowledgment of the need for clarity regarding the definitions of variables, particularly the ratio of masses, and how they relate to the velocities in the context of the problem.

Contextual Notes

Participants note that "k" represents the ratio of the masses of the two objects involved in the collision, which is central to understanding the problem. There is also mention of the assumption that one object is stationary and the other is moving, which may influence the approach to solving for the velocities.

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Breaking rules of Algebra?

Homework Statement


I am copying this problem from my textbook. It is about momentum and Kinetic energy

By rearranging the top equation, we have PART A
v₁'=v₁‐kv₂'

v₁'²+kv₂'²=v₁²

Now we can substitute the top equation into the bottom: PART B

(v₁‐kv₂')²+kv₂'²=v₁'²
v₁²‐2kv₁v₂'+k²v₂'²+kv₂'²=v₁'²

After the cancellations, we are left with PART C

‐2v₁+kv₂'+v₂'=0






The Attempt at a Solution



My problem lies in part B. According to Part A, v₁'=v₁‐kv₂'
But in part B, (v₁‐kv₂')²+kv₂'²=v₁'², see the problem? v₁‐kv₂' has been plugged into v₁'² and it reappeared again on the right hand side.

Right I need help. KEEP IN MIND THAT "K" IS THE RATIO OF M2/M1
ALSO PLEASE SHOW ME HOW TO GET TO PART C
 
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What do you mean it reappeared again on the right hand side? It looks fine. I'm not sure about the jump to part C though...
 


Pengwuino said:
What do you mean it reappeared again on the right hand side? It looks fine. I'm not sure about the jump to part C though...

Shouldn't it have been this?

(v₁‐kv₂')²+kv₂'²=v₁²
 


Yes, you're right, I need glasses. It has to be unprimed on the right side. Sounds like a typo (prime's are easy to mess up :smile:). The jump to part C makes sense now.
 


Pengwuino said:
Yes, you're right, I need glasses. It has to be unprimed on the right side. Sounds like a typo (prime's are easy to mess up :smile:). The jump to part C makes sense now.

What should I do? What the general solution?

I am suppose to get a result of this

v₁'=(m₁‐m₂)v₁/(m₁+m₂)
v₂'=2m₁v₁/(m₁+m₂)
 


Where are these masses coming from? You have part C, it should be immediate what v2' is and from the first equation, you can determine v1'.
 


Pengwuino said:
Where are these masses coming from? You have part C, it should be immediate what v2' is and from the first equation, you can determine v1'.

It's a momentum question. Basically it states an object with velocity collides a stationary object that is heavier than the object clashing in. The result is both objects repelling and heading the opposite direction of each other. I am suppose to find the velocity using the conversation of energy - KE.
 


I mean what is the equation for the masses? I assume it's k = something about masses.
 


Pengwuino said:
I mean what is the equation for the masses? I assume it's k = something about masses.

Sorry, I don't know what you are talking about? K (in post 1) is the ratio of M2/M1
 
  • #10


So k=m2/m1. Part C shows you exactly what v2' is so you should be able to solve for it immediately with no tricks or substitutions other then k=m2/m1. To solve for v1', use the very first equation you have after you solve v2'.
 
  • #11


Pengwuino said:
So k=m2/m1. Part C shows you exactly what v2' is so you should be able to solve for it immediately with no tricks or substitutions other then k=m2/m1. To solve for v1', use the very first equation you have after you solve v2'.

Can you show me the algebra...? Because I am still stuck on Part B.
 
  • #12


You were correct in noticing it was a typo in the third and subsequently 4th line. The [tex]v_1 '^{^2 }[/tex] should have been a [tex]v_1 ^2[/tex] on the right side. The line after it is also incorrect. You have the right idea to plug in v1' from the first equation. Now expand the left side and its a simple cancellation. Subtract [tex]v_1 ^2[/tex] from both sides and you get part C and you can solve for v2' immediately.
 

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