Question about momentum and kinetic energy.

[Pie1345]

Homework Statement

Two blocks with masses 1 kg and 4 kg respectively are moving on a horizontal frictionless surface. The 1 kg block has a velocity of 12 m/s and the 4kg block is ahead of it, moving at 4 m/s. The 4 kg block has a massless spring attached to the end facing the 1 kg block. The spring has a force constant k equal to 1,000 M/m. a) what is the max compression of the spring after the collision b) what are the final velocities of the blocks after the collison has taken place?

Homework Equations

Conservation of momentum
Conservation of kinetic energy (plus energy stored in spring)[/B]

The Attempt at a Solution

I first assumed the collision would be inelastic and I found the velocities of both blocks to be 5.6 m/s. Then I set kinetic energy before collision equal to kinetic energy after collision PLUS the energy stored in the spring.

My answer for the first part was 23 cm (0.23m) and my answer for the second part was 0.8m/s and 7.2 m/s. [/B]

 

[gneill]

Hi Pie1345,

Welcome to Physics Forums!

In your attempt you should explain your reasoning as to why you first assumed the collision to be inelastic. There's a particular reason why this approach will work for determining the spring constant. Can you make a suitable argument?
(By the way, your result of ~23 cm looks reasonable to me).

For the final velocities, take a look to make sure that the signs of the values are correct. Will both blocks still be moving in the same direction? Again, the magnitudes of your values look okay to me.

 

[rude man]

Absent a picture i don't see how a "collision" can take place. The spring will be compressed to some maximum; is that the definition of the moment of "collision"?

 

[jbriggs444]

Absent a picture i don't see how a "collision" can take place. The spring will be compressed to some maximum; is that the definition of the moment of "collision"?

As I read the question, the "collision" takes place during the interval when the two blocks are close enough so that the spring is compressed.

a) asks for the maximum spring compression during this interval. The term "after" perhaps meaning that this is "after" the collision begins.
b) asks for the final velocities after this interval. This time "after" meaning after the collision ends, the blocks having rebounded due to the spring.

 

[rude man]

As I read the question, the "collision" takes place during the interval when the two blocks are close enough so that the spring is compressed.

a) asks for the maximum spring compression during this interval. The term "after" perhaps meaning that this is "after" the collision begins.
b) asks for the final velocities after this interval. This time "after" meaning after the collision ends, the blocks having rebounded due to the spring.

OK, let's go with that. Stay tuned!

 

[rude man]

Well, I came up with a solution in terms of v1, v2, m1, m2, k, etc. but I did not crunch the actual numbers.

1. What can you say about total momentum during the entire process?
2. What can you say about the total energy, same process?
3. What can you say about the relative velocities of the two masses when the spring is fully compressed?
4. Can you use total energy and total momentum after spring is again relaxed to compute final velocities?

This problem is challenging IMO.