Inelastic Collision with Spring

[edud8]

Homework Statement

A .2kg mass traveling on a frictionless horizontal surface at a speed of 3 m/s. It hits a 1.3 kg mass at rest that is connected to a massless spring with a a spring constant of 100 Newtons per meter. The other end of the spring is fixed. Calculate the linear momentum and kinetic energy of the combined masses immediately after the impact.

Homework Equations

I can calculate the combined momentum of the masses but I don't know what to do with the spring.

The Attempt at a Solution

The inelastic equation formula is
(m1)(v1) + (m2)(v2) = (m1 + m2)(vf) where vf is the new velocity and then I would just plug in vf into:
KE = 1/2(m)(v)^2
as v and get the kinetic energy. My problem is I don't know what to do with the spring.

 

[rock.freak667]

So you know that (m1)(v1) + (m2)(v2) = (m1 + m2)(vf) from conservation of momentum.

You know m1,m2,v1 and v2 right? The question basically wants you to find the value of (m1+m2)(vf)

 

[edud8]

I get that but don't I have to do something with the spring?

 

[Hunterbender]

Not really. The spring simply transfer the energy. Think of it as a mediator.

KE of car 1 collide with car 2 ==> KE becomes PE in the spring ==> Spring Potential Energy pushes the on car 2 and becomes potential energy

(assuming all energy is conserve and there is no heat loss anywhere)

So spring is just there.Now, the fun question is: calculate the kinetic energy when the spring is being compress. Now then, you need to account for it, since the spring is potential rather than kinetic energy

 

[edud8]

So do i just set the kinetic energy of the masses equal to the potential energy of the spring like so:
(1/2)mv^2 = (1/2)(k)(x)^2 ==> .5(1.5)(.4)^2 = .5(100)(x)^2
==> .12J = 50x^2 so .0024 = x^2 ==> and Squareroot(.0024) = .0489m = x

Then I plugged it in F = kx and got 4.89N which is the force of the spring.
Now I'm stuck, a little help on how this helps me solve for the linear momentum or kinetic energy of the masses after impact.