Loop problem

1. Jul 25, 2009

lone21

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

A 2.00 kg block is at rest at the top of a frictionless inclined plane. The block
is released and collides with another 1.00 kg block at the bottom of the ramp. The two
blocks experience a perfectly elastic collision and the 1.00 kg block goes frictionless loop
with a radius of 5.00 m. After exiting the loop, the blocks slide along a surface where the
1. If the 1.00 kg block just makes it around the loop, what was the speed of the block?
2. What was the speed of the 2.00 kg block that was sliding down the ramp just before
3.What was the initial height of the 2.00 kg block before it started to move?
4. How far will the two blocks slide before coming to rest after exiting the loop?

2. Relevant equations
K=1/2mVi^2
Velocity around the loop = sqrt(rg)
1/2mVi^2 + mgh = 1/2mVf^2 + mgh
Elastic collision
m1V1i+m2V2i= m1V1f+m2V2f

3. The attempt at a solution
I solved the first part of the problem by taking the velocity around the loop
which is 7m/s
then using the other equation solved for the velocity when the block is not in the loop
so it becomes 1/2 (1) (7m/s)^2 + (1)(9.8)(10) = 1/2 (1)Vf^2 + (1)(9,8)(0)
see the initial velocity will be 7m/s at the height of the loop and i want to figure out the final velocity when the height is at zero.
I solve for vf and get 15.65m/s

After this i get stuck, I know that the veloctiy at zero before and after the loop are the same for the 1kg block, but I don't know how to solve for the Vinitial or Vfinal in the elastic collision equation.
2kgV1i=1kg(15.65m/s) - 2kgV2f
also theres the conservation equation
1/2m1(V1i)^2=1/2m1(V1f)^2+1/2m2(v2f)^2

I'm not sure where I should go from here, I've tried substituting for variables but I still get stuck with bad answers, please help

2. Jul 26, 2009

Redbelly98

Staff Emeritus
Welcome to Physics Forums.

Looks like a simple typo here, or a mistake. For the final momentum, you add, not subtract, the momenta of each mass.

Substituting is definitely the way to go here, since we have 2 equations in the 2 unknowns v1i and v2f. Can you show how you worked it out, then maybe we can spot what is probably an arithmetic error?

3. Jul 27, 2009

lone21

Thats not a typo,
the reason for it being negative is because that block will be going in the opposite direction and therefore it would be subtracted.
I'm not sure if that's correct ,but in the examples given to me by my professor thats what seemed correct.
I know substitution will work, but the equations are whats getting me, because in the book it says to add ,but in my examples it subtracts.

4. Jul 28, 2009

Redbelly98

Staff Emeritus
You can subtract the final momentum of the 2kg if you wish, and define "positive" as going in the opposite direction for that particular velocity. I just find it easier, when I don't know the final answer, to say that one direction is positive for all velocities involved.

But, going ahead with your definition ...

I'm looking at these two equations more carefully:

• In the momentum equation, I think you mean to have v1f at the end, since you wrote v1i for the 2kg mass on the left side.
• In the energy equation, you have not used any of the known quantities: the two masses, and the known v2f for the m2=1kg mass.
After taking care of those issues, you could get an expression for v1i in the momentum equation, and then substitute that into the energy equation.

Last edited: Jul 28, 2009