Collision between two masses with spring

[colmes]

Homework Statement

Mass 1 weighs 10kg and has a velocity of 5m/s [E], Mass 2 weighs 5 kg and has a velocity of 8m/s[W]. Additionally, mass 2 has a spring attached to it with k = 2000N/m. The two masses collide head-on in a perfectly elastic collision. What is the compression of the spring and what is the velocity of the two masses after the collision?

Homework Equations

m1v1 + m2v2 = (m1+m2)v'
Energy of Spring = 0.5k(dx)²
0.5mv'² + 0.5k(dx)² = 0.5m1(v1)² + 0.5m2(v2)²

The Attempt at a Solution

I was able to calculate the compression of the spring by using conservation of momentum and energy but I am not sure how I should find the velocities of the two masses after the collision. Should I use these equations below?

 

[ehild]

It is an elastic collision at the end, is it not? Energy is conserved, but that of the spring is zero when the masses are separated. So you can use the equations for the velocities, valid for elastic collision. But the last sentence uses singular "what is the velocity of the two masses after the collision?", that might be the common velocity v' at the maximum compression of the spring.

ehild

 

[haruspex]

Not sure I understand the set-up. Is the spring in between the masses? On the other side of mass 1, attached to something else?
What does this mean:

What is the compression of the spring ... after the collision?

If the spring is involved in the collision, when does the collision end?

 

[ehild]

The spring is attached to one of the masses, and to nothing else.

Yes, the word "collision" has been used in different senses here in PF already. Collision is assumed instantaneous, is it not? But sometimes the interaction between the "colliding" objects take some time, and the collision begins when the objects start to interact (one object touches the spring on the other object) and ends when they do not interact any more The first object detaches from the spring).

ehild

 

[haruspex]

The spring is attached to one of the masses, and to nothing else.

So its other end is free? And it is not in the line between the masses? No mass is given for it, so I fail to see how it comes into the calculation. We might as well be told one mass is red.

 

[ehild]

Well, the text is not accurate, but there are quite a few such problems ... The spring must be in the line between the masses, which collide head on, how would it be compressed otherwise?

ehild

 

[haruspex]

Well, the text is not accurate, but there are quite a few such problems ... The spring must be in the line between the masses, which collide head on, how would it be compressed otherwise?

That's reasonable, but that led to my other question in post #3: when is "after the collision"? Wouldn't that be when the spring is no longer compressed at all? Maybe it means at closest point of the two masses, i.e. maximum compression. But then, is it the velocity of the masses at that time which is also wanted?

 

[ehild]

I think, the last (rather sloppy) sentence of the problem was meant as "What is the maximum compression of the spring during the collision and what are the velocities after the collision". The velocities are not constant during the time the masses interact with the spring.

ehild

 

[haruspex]

I think, the last (rather sloppy) sentence of the problem was meant as "What is the maximum compression of the spring during the collision and what are the velocities after the collision". The velocities are not constant during the time the masses interact with the spring.

ehild

Yes, those would be the natural questions to ask.

 

[ehild]

Yes, but the last question of the original text is in singular .


ehild

 

[SilverSharpie]

Not liking the wording on this questions at all but here goes...
Going with conservation of energy, at the point where the two objects are just about to collide the KE there should equal the PE when the spring is fully compressed. (no non-conservative forces, so no loss of energy, and when the spring is fully compressed there would be no KE)

KE_i = 1/2 m₁v_1,i² + 1/2 m₂v_2,i²
KE_i = 1/2 (10kg)(5m/s)² + 1/2 (5kg)(8ms)²
KE_i = 285 J

So if KE intial is 285 J, then PE when spring is fully compressed = 285 J.

No PE due to gravity or other sources. So all PE in spring.

PE = 1/2 k x² = 285 J → x = √[(2 * PE)/k]
∴ x = 0.53 m

or so i think...

 

[nasu]

If you consider the energy in lab reference frame, not all the KE is converted into PE.
The center of mass of the system is moving even when the spring is at maximum compression.
The velocity of the CM (nonzero in this problem) is not changed by internal forces.

 

[SilverSharpie]

In a perfectly elastic collision as the problem seems to be referring to all KE would be transferred to PE. (There would also be no sound or friction, among several other conditions that I know not enough about to discuss)

But if this problem is for a intro level physics course I'm pretty sure it's safe to use the law of conservation of energy and use 'ideal' conditions.

 

[nasu]

So you think that the momentum is not conserved in this collision?
What is the total momentum before the collision? What will be if the two balls have zero velocity (their KE is zero)?

Your statement applies only if the total momentum is zero. Condition always true in the center of mass frame but not always in the lab frame.

 

[ehild]

Not liking the wording on this questions at all but here goes...
Going with conservation of energy, at the point where the two objects are just about to collide the KE there should equal the PE when the spring is fully compressed.

You know it wrong. Conservation of momentum applies. The spring is compressed by the two objects approaching. When they move with the same velocity (the velocity of the CM) the spring is not compressed further.. The initial KE of the masses is converted to the KE of the whole system + PE of the spring.

ehild