# Perfectly Elastic Collision Problem

jonnejon

## 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?

Staff Emeritus
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You have two unknowns and therefore need two simultaneous equations. What else is conserved besides energy?

jonnejon
Momentum and kinetic energy?

Staff Emeritus
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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.

jonnejon
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.

Staff Emeritus
Gold Member
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?

jonnejon
Pi=Pf mivi=mfvf

Staff Emeritus
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Pi=Pf mivi=mfvf
Correct, so can you now apply conservation of momentum to your problem?

jonnejon
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.

Staff Emeritus
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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.

jonnejon
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?

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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?

jonnejon
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?

Staff Emeritus
Gold Member
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
$$m_1v_{1i}^2+m_2v_{2i}^2 = m_1v_{1f}^2+m_2v_{2f}^2$$

Conservation of Momentum
$$m_1v_{1i}+}m_2v_{2i} = m_1v_{1f}+m_2v_{2f}$$

jonnejon
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