# Elastic collision with conservation of energy and momentum?

• zell_D
In summary, the problem involves two trains, one moving and one stationary, colliding with each other and using their springy bumpers to bounce off without damage. The problem asks to identify the isolated system and solve for the final velocity of each train in terms of the initial velocity of the moving train and the masses of the two trains. Using the equations for conservation of momentum and conservation of energy, the final velocities can be expressed in terms of the initial velocity and the masses of the trains. The equations can then be solved simultaneously using algebraic methods such as FOIL to find the final velocities of the two trains.
zell_D

## Homework Statement

a moving train collides with a train that is not moving, and the trains use their springy bumpers to bounce off of each other without damage. Assume diff masses for train A and B. Identify your isolated system, solve for the final velocity of EACH train in terms of INITIAL VELOCITY of the initially moving train and the 2 masses

## Homework Equations

P=mv
Conservation of momentum: Pfinal=Pinitial
Assumption of no external forces, spring as part of the system. k not addressed

## The Attempt at a Solution

pi=[mA](viA) (since B is at rest)
pf=[mA](vfA)+[mB](vfB)

pi=pf
[mA](viA)=[mA](vfA)+[mB](vfB)
since A's final velocity defined as in the negative direction in my system
[mA](viA)=-[mA](vfA)+[mB]()

vfA = [[mB]vfB-[mA](viA)]/mA
vfB = [[mA]vfA+[mA](viA)]/mB

my question this is the furthest i got to. I do not know if i am right or wrong. But it seems like that's not the furtherst i can solve this problem. since the final velocity variable still are present in both

zell_D said:

## The Attempt at a Solution

pi=[mA](viA) (since B is at rest)
pf=[mA](vfA)+[mB](vfB)

pi=pf
[mA](viA)=[mA](vfA)+[mB](vfB)
Good.
since A's final velocity defined as in the negative direction in my system
[mA](viA)=-[mA](vfA)+[mB]()
OK, but no need for this step. It's OK to have negative values for V.

vfA = [[mB]vfB-[mA](viA)]/mA
vfB = [[mA]vfA+[mA](viA)]/mB
This won't help. What you need is a another independent equation to use in addition to conservation of momentum. Hint: What else is conserved? (See the title of this thread!)

energy is conserved...

so KEiA+KEiB=KEfA+KEfB?
KE initial for B = 0 so
KEiA=KEfA+KEfB

(1/2)mA[vi^2]=(1/2)mA[vfA^2]+(1/2)mB[vfB^2]

but wouldn't i end up with both final velocities again?

zell_D said:
energy is conserved...

so KEiA+KEiB=KEfA+KEfB?
KE initial for B = 0 so
KEiA=KEfA+KEfB

(1/2)mA[vi^2]=(1/2)mA[vfA^2]+(1/2)mB[vfB^2]
Good.

but wouldn't i end up with both final velocities again?
You must combine both equations--solve them simultaneously. (This one and the one you had before from momentum conservation.)

Just take one of the final equations from your last post and substitute into this one. Then you can solve for both unknowns.

are the conservation of momentum equations right? since i made the velocity negative and all

oh and do i use FOIL since there are going to be squares lol haven't done algebra in like forever -_-

zell_D said:
are the conservation of momentum equations right? since i made the velocity negative and all
As long as you remember how you defined the sign.

oh and do i use FOIL since there are going to be squares lol haven't done algebra in like forever -_-
Sounds like a plan to me.

## 1. What is an elastic collision?

An elastic collision is a type of collision between two objects where both energy and momentum are conserved. This means that the total kinetic energy and momentum of the system before and after the collision remain the same.

## 2. How does conservation of energy apply to elastic collisions?

In an elastic collision, the total kinetic energy of the system (sum of the kinetic energy of both objects) remains constant. This means that the kinetic energy of the objects before the collision is equal to the kinetic energy after the collision.

## 3. What is the role of momentum in an elastic collision?

Momentum is also conserved in an elastic collision. This means that the total momentum of the system (sum of the momentum of both objects) remains the same before and after the collision. This is represented by the equation: m1v1 + m2v2 = m1v1' + m2v2', where m and v represent mass and velocity, respectively.

## 4. How is an elastic collision different from an inelastic collision?

In an inelastic collision, some of the kinetic energy is lost and converted into other forms of energy, such as heat or sound. However, in an elastic collision, all of the kinetic energy remains in the form of kinetic energy after the collision.

## 5. What are some real-life examples of elastic collisions?

A common example of an elastic collision is billiard balls colliding on a pool table. Other examples include bouncing basketballs, collisions between atoms in a gas, and collisions between particles in particle accelerators.

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