Perfectly Elastic Collision Problem

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Homework Help Overview

The problem involves a perfectly elastic collision between two balls of different masses and initial velocities. The original poster seeks to determine the final speed of the lighter ball after the collision, using principles of conservation of momentum and kinetic energy.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need for simultaneous equations due to having two unknowns. They explore the conservation of momentum and kinetic energy as necessary equations for solving the problem.

Discussion Status

Some participants have provided guidance on applying conservation principles, while others express confusion about the equations and their application. The original poster has attempted to manipulate the equations but has encountered difficulties, leading to further questions about the correctness of their steps.

Contextual Notes

The original poster mentions a derived equation for perfectly elastic collisions that applies when one mass is at rest, indicating a potential constraint in their understanding of the problem setup.

jonnejon
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Homework Statement


A 110 g ball moving to the right at 4.0 m/s catches up and collides with a 400 g ball that is moving to the right at 1.2 m/s.
If the collision is perfectly elastic, what is the speed of the 110 g ball after the collision?

Homework Equations


KEi=KEf
1/2m1v1i^2 + 1/2m2v2i = 1/2m1v1f^2 + 1/2m2v2f^2


The Attempt at a Solution


Tried plugging it in but I don't really understand the concept so I am let with 2 final velocity. Please help.

110g < 400g so 110g should move left and 400g should move right, so does that make v2f=0?
 
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You have two unknowns and therefore need two simultaneous equations. What else is conserved besides energy?
 
Momentum and kinetic energy?
 
jonnejon said:
Momentum and kinetic energy?
Correct! So you already have one equation (conservation of energy), can you now write a second equation using conservation of momentum? You should then have a system of two simultaneous equations.
 
Nope, I don't understand how. There is an equation my textbook derived for perfectly elastic collisions but it is only if the second mass is at rest.
 
jonnejon said:
Nope, I don't understand how. There is an equation my textbook derived for perfectly elastic collisions but it is only if the second mass is at rest.
What does the principle of conservation of momentum state?
 
Pi=Pf mivi=mfvf
 
jonnejon said:
Pi=Pf mivi=mfvf
Correct, so can you now apply conservation of momentum to your problem?
 
So, I solved for m2vf2 in the momentum equation and plug that into the energy equation and solve for vf1 but I got it wrong.
 
  • #10
jonnejon said:
So, I solved for m2vf2 in the momentum equation and plug that into the energy equation and solve for vf1 but I got it wrong.
Could you detail your steps?
 
  • #11
solve for v2f in momentum equation:
v2f = (m1v1i + m2v2i - m1v1f) / m2
v2f = (v1i/m2 + v2i - v1f/m2)
plug it in the energy equation:
1/2m1v1i^2 + 1/2m2v2i^2 = 1/2m1v1f^2 + 1/2m2(v1i/m2 + v2i - v1f/m2)^2
solve for v1f?
 
  • #12
jonnejon said:
solve for v2f in momentum equation:
v2f = (m1v1i + m2v2i - m1v1f) / m2
v2f = (v1i/m2 + v2i - v1f/m2)
Your step from the first line to the second is incorrect. Where did the m1 go?
 
  • #13
Opps, I thought it cancels out because of the subtraction sign.

1/2m1v1i^2 + 1/2m2v2i^2 = 1/2m1v1f^2 + 1/2m2 (m1v1i/m2 + v2i - m1v1f/m2)^2
1/2m1v1i^2 + 1/2m2v2i^2 = 1/2m1v1f^2 + 1/2m2 (m1^2v1i^2/m2^2 + v2i^2 - m1^2v1f^2/m2^2)
1/2m1v1i^2 + 1/2m2v2i^2 = 1/2m1v1f^2 + (m1^2v1i^2 - m1^2v1f^2)/2m2 - m2v2i^2/2
1/2m1v1i^2 + 1/2m2v2i^2 = 1/2m1v1f^2 + (m1^2 (v1i^2 - v1f^2)/2m2) - m2v2i^2/2
(1/2m1v1i^2 + 1/2m2v2i^2)(2m2)/(m1^2) + m2v2i^2/2 = 1/2m1v1f^2 + (v1i^2 - v1f^2)

Man, this equation just getting to complicated. Am I even doing it right?
 
  • #14
Yes it is getting rather complicated. I really can't decipher what you've written, you could use latex to make things easier to read,

Conservation of Energy
[tex]m_1v_{1i}^2+m_2v_{2i}^2 = m_1v_{1f}^2+m_2v_{2f}^2[/tex]

Conservation of Momentum
[tex]m_1v_{1i}+}m_2v_{2i} = m_1v_{1f}+m_2v_{2f}[/tex]
 
  • #15
I tried doing that but it was just too messy for me. I actually found out how to do it. There are perfectly elastic collision equations in my textbook for velocities at rest. But they said you can use the Galilean transformation of velocities to make the 2nd object's velocity into zero and then find the final velocity.

Used:
v' = v - V

Thanks for your help though.
 
  • #16
in elastic collision momentum and energy is conserved
 

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