What Conservation Laws Apply in a 1-D Collision Involving a Spring?

[beatem]

Homework Statement

Everything is in symbols.

So a block of mass M1 moves on a frictionless surface with velocity v0. There is a mass M2 at rest sitting there with a spring and platform (each of negligible mass) attached to it. The spring constant is k.

What quantities are conserved throughout the collision?

Write equations that express the conservation laws during the collision when the spring is compressed to a distance X.

Homework Equations

momentum (m1v1 + m2v2...)
energy
spring equation E = 1/2kx^2

The Attempt at a Solution

I'm mostly just stuck on how to approach this because of the spring. Obviously momentum is conserved regardless of the type of collision, but I'm just not entirely sure which way to go.

I'm assuming that when the mass M1 hits the spring and it compresses to some maximum length, the velocity of the first mass becomes zero and will turn around when the spring decompresses.

I think both momentum and kinetic energy are conserved throughout the whole collision, but that during the collision itself, it is inelastic, so energy is not conserved. The KE of mass 1 turns into PE in the spring with 1/2kx^2. But then what happens? Does it divide in half and send equal kinetic energy to both m1 and m2?

 

[Delphi51]

Welcome to PF, beatem!
M1 might not stop; I think you should keep a v1 for its velocity.
Looks to me like energy is conserved during the collision - no conversion to heat (if that were not the case, some information about the loss would have to be given). How the energy is divided among the ke of M1, the ke of M2 and the spring depends on the relative masses and the k, so you'll have all these quantities in your answer. You are just asked to write equations, not solve them for anything.

 

[venkatg]

Let M1,M2 be masses and
V1 = initial velocity of mass M1
V2 = final velocity of mass M1
U2 = final velocity of mass M2
K = spring stifness
X = deformation in spring after collision that is given

Conservation of momentum:

Total Initial momentum = Total final momentum
M1 * V1 = M1 * V2 + M2 * U2 ----------------------------- (1)

Conservation of Energy:
Total K.E before collision = Total K.E after collision

(1/2) * M1 *V1^2 = (1/2) * M1 * V2^2 + (1/2) * M2 * U2^2 + other losses ------(2)

But (1/2) * M1 * V2^2 = (1/2)* K*(X^2) ---------- (3) as the spring after the collision has given its energy to the Mass M1

From (3) we can find V2. Put that into (1) and get U2.

Put V2,U2 into (2). And find the "other losses".

(We can find the coefficient of restitution using this)

 

[beatem]

Thanks so much for your help guys! This is making so much more sense - I didn't even really know where to start, but this works well.

My professor told me to have masses equal to each other, from this point, so:

I got V2 = x*sqrt(k/m) and U2 = V0 - x*sqrt(k/m).

I just have one last question:

"Solve the equations in part 2 for the speeds of the blocks v1 and v2 during the collision when the spring is compressed to a distance X (the masses assumed equal now)."

Are these just V2 and U2 described above?

And would this involve another momentum equation later, in which the final v1 and v2 (called V2 and U2 here) from before the collision become the initial v1 and v2 for another momentum equation that takes place after the collision?:

m1V2 + m2U2 (momentum of after the collision) = m1V2' + m2U2' ? (another phase?)

I think energy would still be conserved too, right? (so this would be another equation to help figure out V2' and U2' if this is what I should be doing?)