# Calculating maximum heright of Mass 1 after elastic collision?

Calculating maximum heright of Mass 1 after elastic collision???

## Homework Statement

Here is the question:

Two blocks are free to slide along the ramp. The bloack of m1 = 5.00kg is released from the position at the top of ramp at height h = 5.00m above the flat part of the track. Protruding from its fron end is the north pole of an identical magnet embedded in the back end of the block mass m2 = 10.kp, initially at rest. The two blocks never touch.

Calculate the maximum height to which m1 rises after the elastic collision. ???

Here is a picture if it helps ya... ## The Attempt at a Solution

I am really not sure. I am not given an speeds.

I started out by trying to figure out vel or m1 as it slides down track by going:
mg/l = (5kg)(9.8)/5m = 9.8 m/s - But im betting thats wrong.

Im really stuck here. Can anyone help me with the first thing I should be trying to figuring out when looking at this problem specically ?

gneill
Mentor

I am really not sure. I am not given an speeds.
Because you're expected to find them from the given information...
I started out by trying to figure out vel or m1 as it slides down track by going:
mg/l = (5kg)(9.8)/5m = 9.8 m/s - But im betting thats wrong.
You'd win that bet The units in your equation do not yield m/s, so it cannot be correct.

The problem is an exercise in the application of the two fundamental conservation laws for motion; energy and momentum.

Use conservation of energy to determine the speed of the first block when it reaches the flat stretch of the ramp. Then you've got a classic elastic collision scenario to work through (both conservation laws involved). You'll want to find the kinetic energy of the first block after that collision, and use conservation of energy to find its return height.

As you stated here: "Use conservation of energy to determine the speed of the first block when it reaches the flat stretch of the ramp"

I am not sure how to do this?

gneill
Mentor

As you stated here: "Use conservation of energy to determine the speed of the first block when it reaches the flat stretch of the ramp"

I am not sure how to do this?

Check your text or class notes for sections on conservation of energy, especially as it pertains to gravitational potential energy and kinetic energy. Can you at least find the mathematical formulas for both forms of energy?

Is this correct?

Potential (U) at the top is = to Kinetic (K) at bottom.

So potential = mgh = (5kg)(9.8)(5m) = 245.0 J

Ke = 1/2mv^2 and solving for v^2

v^2 = 2Ke/m

v = squareroot(2Ke/m)

v = Sqrt((2)(245.0) / (5) = 9.9 m/s

Is that correct?

gneill
Mentor

Yes, that's correct. You might want to keep an extra decimal place or two for all intermediate calculations. This prevents rounding error from creeping into your results.

Yes, that's correct. You might want to keep an extra decimal place or two for all intermediate calculations. This prevents rounding error from creeping into your results.

Ahh, ok see what ur saying. :)

So the next step?

Since the smaller box bounces off (well, sorta since its magnetic) this would be an Elastic collision?

So, the formula I would use is: V1i - V2i = -(V1f - V2f)

EDIT: Wait no, that formula cant be right...Does the m2 mass ever move? - hmmm)

???

gneill
Mentor

Ahh, ok see what ur saying. :)

So the next step?

Since the smaller box bounces off (well, sorta since its magnetic) this would be an Elastic collision?

So, the formula I would use is: V1i - V2i = -(V1f - V2f)

???

You need to analyze the elastic collision to find the final velocity of the first block. To do that, you need two equations (since there are two unknowns: the final velocity of each block). You've given one relevant equation. The other (and arguably more important!) would be the equation expressing conservation of momentum for the collision.

I am a little lost here.

My next step would be what? To find the velocity of mass 1 right after the collision?

Im not sure which formula to use, since im trying to find final velocities of both masses ?

Is it:

V1f = (m1 - m2)/m1 + m2)V1u ? Would I use that one to find the Final velocity of the first mass?

gneill
Mentor

Yes, you want to find the velocity of the first block after the collision. That's really the only result from the collision that you need in order to proceed with the problem.

Rather than look for canned results (and having to memorize them for an exam), start with the general expression for conservation of momentum for the collision. You need one more equation, which you've already given, which is the expression relating the relative velocities of the colliding objects before and after collision. That's fine. You could also have used conservation of kinetic energy ((1/2)mv2 and all that), but that would be a bit more work.

Yes, this is what I am trying to do, rather than just memorize all the formulas.

This is the correct formula I am sure here: (I hope V1f = (m1 - m2)/m1 + m2)V1u + (2m2 / m1+ m2) V2i

Correct?

Now, you are saying insted of simply memorizing a formula like this, I should start with the general expression... Which is??? m1v1i + m2V2i = m1v1f + m2V2f ?

But, how would I do that? Like starting with the general expression, and then just come up with V1f = (m1 - m2)/m1 + m2)V1u + (2m2 / m1+ m2) V2i as I wrote out above?

EDIT: I got -3.29 m/s for mass 1 and 6.53 m/s for mass 2

Last edited:
gneill
Mentor

You start with the formulas that express the general conservation rules (you have two of them in this case), and solve them as simultaneous equations in two unknowns. That will result in the equations for the final velocities.

Your value for the final velocity of block one is okay. A small bit of rounding error, but close enough for engineering purposes Great! So I got both velocities correct...

Now, I really have no idea from the 2 velocities, how I find how far it will go up that ramp??

gneill
Mentor

Great! So I got both velocities correct...

Now, I really have no idea from the 2 velocities, how I find how far it will go up that ramp??

How did you find the initial velocity of block one when it fell from the top of the ramp? Do the same thing, only in reverse.

Umm :)

So, would I go

245.0J = (5)(-3.29)(x) solve for x?

gneill
Mentor

Umm :)

So, would I go

245.0J = (5)(-3.29)(x) solve for x?

Nope. 245.0J was the KE of the block when it first arrived at the bottom of the ramp. Throw that away, you're done with that part of the problem.

The CURRENT velocity of block one is (approximately) 3.3 m/s heading upslope. What kinetic energy does that represent?

What do you mean, What kinetic energy does that represent?

Would it not be Ke = 1/2MV^2

gneill
Mentor

What do you mean, What kinetic energy does that represent?

Would it not be Ke = 1/2MV^2

Yes, what's it's value?

If that KE is converted to PE for block one (M*g*h), how high will it reach?

Ahh I think I got it..

I got 27.06 for it...

So, 27.06 = (5)(9.8)(x) solve for x = .552

Is that correct???

gneill
Mentor

Ahh I think I got it..

I got 27.06 for it...

So, 27.06 = (5)(9.8)(x) solve for x = .552

Is that correct???

Yup. Close enough for government work Without rounding of intermediate results you would obtain x = 0.556m.