Inelastic collisions, springs and ramps

[AlkaPhys]

Homework Statement

A block of mass M is compressed a distance x against a spring which has a spring constant k.
When released, the block slides along a frictionless surface until it undergoes a completely inelastic collision with a block of mass 3M. The two blocks travel together some distance and then travel up a ramp where they eventually stop.

1. What is the speed of the two blocks after the collision?
2. How high will they travel up the ramp?
I should give my answer in terms of k,x,g,M

Homework Equations

(mgy), gsin(theta)? mgcos?

The Attempt at a Solution

First, when they collide, the speed of m slows down to 1/4 m when it collides in elastically with 3M. The PE of the spring is 1/2(kx^2). I think I would use mgcos(theta) to find the height. The problem with (mgy) is that, I'm not looking for the gravitational PE, I'm looking for how high it goes. My teacher says that mgcos(theta) is not the way, but I can't find a solution. Can anyone help?

 

[tiny-tim]

Hi AlkaPhys!

(have a theta: θ and try using the X² tag just above the Reply box )

… I should give my answer in terms of k,x,g,M
…
I think I would use mgcos(theta) to find the height. The problem with (mgy) is that, I'm not looking for the gravitational PE …

Yes, you are looking for the gravitational PE …

you need to use conservation of energy, starting with the energy of the two blocks immediately after the collision.

 

[AlkaPhys]

Thanks for replying!

The PE _grav would be 4(mgy) because of the combined weight, but this is the PE at the highest point. This is where it's 100% PE and 0 KE. Since KE is 1/2mv², the mass would be 4m, the velocity would be 1/4m?as it starts to go up the ramp, the KE falls, and the PE grows. how would I find how high the mass goes before the KE drops to 0?

 

[AlkaPhys]

Also, I found that in a situation like this, with inelastic collision, energy is not conserved, but momentum is.

 

[tiny-tim]

Yes, that's right …

momentum (and angular momentum) is always conserved in collisions, but energy never is unless the question says so.

Before the collision, energy is conserved, and after the collision, energy is conserved,

but during the collision, only momentum is conserved …

so you can't use the same total energy both before and after the collision, for after you need to start again, with the v you got from momentum.

(I don't really follow your previous question … just use KE + PE = constant)