Spring Compression: 2.5kg Block + 4.5kg Block

[Oomair]

Homework Statement

A 2.5 kg block slides along a frictionless tabletop at 6.0 m/s toward a second block (at rest) of mass 4.5 kg. A coil spring, which obeys Hooke's law and has spring constant k = 860 N/m, is attached to the second block in such a way that it will be compressed when struck by the moving block

Homework Equations

The Attempt at a Solution

i first used M1V1 = M2V2, M2 is the block at rest, i calculated the speed of block at rest and i got 3.33 m/s, and then i use the energy equations to find the spring compression

.5 m1(6)^2 = .5k(x^2) + .5m2(3.33)^2, i solve for x, but it comes out wrong, anything I am missing here?

 

[EngageEngage]

You forgoet to mention what the question was asking, but...You should be able to forget the first blcok after the collision. I would just use momentum conservation to find the velocity of the second block after the collision. Then the kinteic energy associated with this velocity will be converted completely to the potential energy to be stored in the spring once its compressed.

 

[EngageEngage]

in other words, 1/2mv^2 = 1/2kx^2 (all for the second block), where v is the velocity of the second block after the collision, or the start point of the new system.

Edit: you might also want to see if the collision is completely elastic or not, because this would change your answer

 

[Doc Al]

i first used M1V1 = M2V2, M2 is the block at rest, i calculated the speed of block at rest and i got 3.33 m/s,

This assumes that all the momentum of the first block is transferred to the second. That's not true.

I assume you are trying to find the maximum spring compression. Hint: How will the velocities of the two blocks be related when the spring is at maximum compression?

(But you are correct to be using conservation of momentum and energy. So you are close.)

 

[Oomair]

The book is saying that the collision is perfectly elastic, but I am confused on how i would relate the speeds in terms of energy

 

[Doc Al]

Yes, the collision is perfectly elastic. You'll need that fact later.

When the spring is at maximum compression, what must be the relative speed of the two blocks?