2 masses collide, one has spring. -Collision, Momentum, Spring energy

[azurken]

Homework Statement

Mass 1 at 10m/s collides into mass 2 at rest, which has a spring attached to it. The second mass has a spring at 200N/m and natural length L0 = 0.1m. At the instant they collide, the spring is compressed to its max amount and the masses move with the same speed V. Determine the delta X between the objects at this instant. Both momentum and mech energy are conserved.

M1=0.4kg
M2=0.8kg
V1=10 m/s
V2=0 (at rest)
K= 200N/m
L0=0.1m
dX=??

Homework Equations

Vf=(m1v1+m2v2)/(m1+m2)

SPE=0.5Kx^2

KE=0.5mV^2

dX=L-L0

The Attempt at a Solution

Vf=(.4)(10)/(1.2)=3.333

So there are three instances here of the collision, when the m1 goes to m2 is the first. Then M1 and M2 being the same object at that one instant. And after. I don't need to measure after so I can just do.

KE1+PE1=KE2+PE2

which gives me KE1=.5*m1*v1^2

No contact made at PE1 so there's none here.

KE2 = (.5)*(m1+m2)*(Vf)^2

Then PE2 and at this point there's contact and since I know it's compressing the Delta X should be negative (which is still strange) when it all works out.

PE2 = (.5)*(200N/m)*(dX)^2

All this will give me KE1=KE2+PE2 to

20m/s = [6.666(Kg)(m^2)/(s^2)] + (.5)*(200N/m)*dX^2

20m/s - [6.666(Kg)(m^2)/(s^2)] = (.5)*(200N/m)*dX^2

.2 m/s - 0.666 (kgm^2)/(s^2) = dX^2

Yes the units don't match up. I can't clear it but if I ignore all that, a -(.466) sqrooted is -.683 as the dX.

 

[azurken]

Alright so yeah. After doing a couple other trials, I'll basically get numbers larger than the L0 (natural length) for dX. Which can't be true because it can only compress so much. It cannot be more than 0.1m.

 

[voko]

The arrangement is not clear. Is the spring connected to anything else? What is the orientation of the spring with regard to the other mass? A picture would help immensely.

 

[azizlwl]

Homework Statement

Mass 1 at 10m/s collides into mass 2 at rest, which has a spring attached to it. The second mass has a spring at 200N/m and natural length L0 = 0.1m. At the instant they collide, the spring is compressed to its max amount and the masses move with the same speed V. Determine the delta X between the objects at this instant. Both momentum and mech energy are conserved.

M1=0.4kg
M2=0.8kg
V1=10 m/s
V2=0 (at rest)
K= 200N/m
L0=0.1m
dX=??

Homework Equations

Vf=(m1v1+m2v2)/(m1+m2)

SPE=0.5Kx^2

KE=0.5mV^2

dX=L-L0

The Attempt at a Solution

Vf=(.4)(10)/(1.2)=3.333

So there are three instances here of the collision, when the m1 goes to m2 is the first. Then M1 and M2 being the same object at that one instant. And after. I don't need to measure after so I can just do.

KE1+PE1=KE2+PE2

which gives me KE1=.5*m1*v1^2

No contact made at PE1 so there's none here.

KE2 = (.5)*(m1+m2)*(Vf)^2

Then PE2 and at this point there's contact and since I know it's compressing the Delta X should be negative (which is still strange) when it all works out.

PE2 = (.5)*(200N/m)*(dX)^2

All this will give me KE1=KE2+PE2 to

20m/s = [6.666(Kg)(m^2)/(s^2)] + (.5)*(200N/m)*dX^2

20m/s - [6.666(Kg)(m^2)/(s^2)] = (.5)*(200N/m)*dX^2

.2 m/s - 0.666 (kgm^2)/(s^2) = dX^2

Yes the units don't match up. I can't clear it but if I ignore all that, a -(.466) sqrooted is -.683 as the dX.

You have calculate KE1 in bold.
Then why change the unit?