# Elastic collision problem

#### Jatsix30

1. Homework Statement

A 4 kg block moving to the right at 6 m/s collides elastically with a 2 kg block moving to the right at 3 m/s. Find the final velocities and the initial and final energies. Show all your work. [ignore external forces]

2. Homework Equations

In an elastic collision momentum inital = momentum final and KEi = KEf
momentum = mv
KE= .5mv^2

3. The Attempt at a Solution

I used these equations and did some algebra but ended up getting too many variables. Am i missing some info here?

Thanks.

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I used these equations and did some algebra but ended up getting too many variables. Am i missing some info here?
You should end up with a system of two equations with two unknowns, i.e. the final velocities.

#### cristo

Staff Emeritus
I used these equations and did some algebra but ended up getting too many variables. Am i missing some info here?
Show the working that you have done. I assure you, there is enough information given in the question to solve the problem.

#### billiards

You need to know what the energy is at the start for each particle add them together and then equal it to the total energy after the collision.

#### Jatsix30

algebra

ok, well knowing that p initial = p final, i knew that

m1v1i + m2v2i = m1v1f + m2v2f

Moving the variables around so that i would have the masses as common variables, i got

m1 (v1i - v1f) = m2 (v2f - v2i) -- equation 1

at this point i worked with the kinetic energy equations, knowing that KE initial = KE final.

.5m1(v1i)^2 + .5m2(v2i)^2 = .5 m1(v1f)^2 + .5 m2(v2f)^2

after factoring out .5 and moving the variables so that i had the masses as common variables, i got

m1(v1i^2 - v1f^2) = m2 (v2f^2 - v2i^2) -- equation 2

at this point i divided this equation (equation 2) by the equation i got after working (equation 1) with the momentum equations to eliminate the masses. (this is something my teacher said to do, though i'm not really sure how it is helpful.)

After eliminationg the masses on each side, i got

(v1i^2 - v1f^2) / (v1i - v1f) = (v2f^2 - v2i^2) / (v2f - v2i)

knowing that the numerators in each case were differences of two squares, i factored that and canceled out one factor on each side with the denominator, leaving

v1i + v1f = v2f + v2i

and this is where i got stuck.

#### cristo

Staff Emeritus
ok, well knowing that p initial = p final, i knew that

m1v1i + m2v2i = m1v1f + m2v2f

Moving the variables around so that i would have the masses as common variables, i got

m1 (v1i - v1f) = m2 (v2f - v2i) -- equation 1

at this point i worked with the kinetic energy equations, knowing that KE initial = KE final.

.5m1(v1i)^2 + .5m2(v2i)^2 = .5 m1(v1f)^2 + .5 m2(v2f)^2

after factoring out .5 and moving the variables so that i had the masses as common variables, i got

m1(v1i^2 - v1f^2) = m2 (v2f^2 - v2i^2) -- equation 2

at this point i divided this equation (equation 2) by the equation i got after working (equation 1) with the momentum equations to eliminate the masses. (this is something my teacher said to do, though i'm not really sure how it is helpful.)
You are using the correct method upto here, however I would not divide one equation into the other. (I can see this is just confusing you, and besides you have the two equations which will give the solution).

You are given the values of the masses, and the initial velocties, so put them into eqn 1. This will then be an expression in v2f and v1f. Do a similar thing for the energy equation, yielding a second expression in v1f and v2f.

These simultaneous equations can then be solved for the unknown variables. (i.e. use eqn 1 to substitute for v2f in eqn 2, enabling you to calculate v1f).

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