# Those darn car on a slope problems

1. Jul 29, 2007

I need a bit of help finding a good source to use to help solve these problems. I am taking a test that is 20% physics (20% chem and the rest are life sciences) and I have not taken physics in over five years. I've been browsing through the internet for over two hours now and the best that I can find are lesson plans for teachers, but nothing that helps me get the basics of these types of problems. Tis very frustrating.

I would like some help either with what kind of formulas that I need to solve this type of problem or a place, either on this forum or out on the web, that can help me better understand what I am to do.

I'm not asking for anyone to solve homework for me, as I've already graduated from college, have no homework and am simply trying to understand a question in a certification exam. I may be making to much out of something really simple, but I'm getting very frustrated with this problem.

1. The problem statement, all variables and given/known data

First there is a picture of a ramp, 1.5m off the ground that ends in a loop that has a maximum height of 0.5m.

"A toy car is released at the top of the track shown above. The car goes down the slope and through the loop-de-loop. Negating friction, what is the speed of the car at the top of the loop?"

2. Relevant equations

I don't know! is this ultimately a vector problem or is it something far more complicated?

3. The attempt at a solution
(see explanation above)

2. Jul 29, 2007

### mgb_phys

The simplest approach to these problems is almost always conservation of energy.
What is the formula for potential energy and kinetic energy (hint wikipedia)
Remember energy is conserved - you must always have the same total of potential+kinetic at each position.

3. Jul 29, 2007

And yet, that seems to be my problem because for each formula for both kinetic and potential energy that I find I have to have the mass of the object, and yet the question mentions nowhere in it the mass of the toy car. This is what I keep bumping into while trying to find an answer to this problem.

4. Jul 29, 2007

### mgb_phys

If all the equations need the mass and one equation is equal to another, this gives you a bit of a clue about how to do it.

(potential + kinetic) at start = (potential + kinetic) at top of loop.

5. Jul 30, 2007

thus potential energy is found by m*g*h and kinetic is .5*m*v^2 (velocity squared). Correct?

Or do I use this other formula for Kinetic? E=p^2/2m ?

where p is the momentum?

I'm sorry for being dense I am trying to pull this all in.

bu tI ended up getting:

(x*9.8*1.5)+ (.5*y^2) = (x*9.8*0.5)+ (.5*y^2)

and that ends up zeroing out to 14.7x=4.9x and I know that that is not right!

6. Jul 30, 2007

*gasp* I got it! All the formulas having no mass means I can kick that part out.

The velocity at the start is zero (why didn't I think of that earlier?!)

Thus it becomes:

(9.8*1.5) + (.5 (0^2)) = (9.8*0.5) + (.5 (v^2))

14.7 = 4.9 + .5*v^2

9.8 = v^2 and thus v=4.4

*happy dance*

Thank you so very much for pointing me in the right direction! I'm in such a tizzy over taking this test that I'm over thinking things way too much!

Thank you again!

7. Jul 30, 2007

### mgb_phys

Since momentum p=mv you will see that the two equations for KE are the same, but stick with e = 0.5 mv^2 for now.

We have ( remember no ke at start )
0 + mg*1.5 = 0.5mv^2 + mg*0.5
mg*1= 0.5mv^2 the m cancels
2*g*1 = v^2 note the one metre
so v=sqrt(2*9.8)

Another tip for this kind of problem is check the units, here we end up with sqrt(accelaration*distance) which is m/s^2 * m = m^2/s^2 when you take the sqrt you end up with m/s which is what you want.
If this isn't clear remember that the 2*g is multiplied by 1metre.

Your not being dense, we just try not to give people the answer straight away - but sometimes it is difficult to give useful hints.