Question

A 1.1 kg block is placed at rest on a ramp at some height above the bottom of the track. It collides elastically with a 0.5 kg block at rest at the bottom of the track.(Figure 1) After colliding elastically both blocks make it safely around a loop-the-loop which has a radius 0.7 m. The track is frictionless.

Answer 1

I forgot the question. What is the minimum height the 1.1 kg block must be released above the bottom of the track so that both blocks safely traverse the loop-the-loop?

Answer 2

Use the conservation of mechanical energy to solve that, to get to 0.7m of altitute both must have a certain energy, and this energy is the minimum required for them to do the loop, and is also the same energy they must have at the beggining

Answer 3

Do you mind helping me start the problem?

Answer 4

Calculate the energy needed to reach the heigh of the loop, since you want the minimum energy, it should be only potential energy.

Answer 5

got it! thanks

Answer 6

@ivanmlerner not really correct. Both blocks have to have a minimum velocity at the top of the loop to negotiate it.

Answer 7

What do you mean by negociate it?

Answer 8

complete the loop

Answer 9

PEi = 1.1gh > 1.1g(.7) + .5g(.7) + 1/2 (1.1)(gr) +1/2(.5)(gr)

Answer 10

r=.7 I left it as r in the two right terms so you can see where it comes from...

Answer 11

I first found the velocity by using Vf^2=5gr. Then I used that to find the initial velocity. then i finally plugged it into the mgh=1/2M(Vi)^2 equation to get 12.5m for the height. Any mistakes?

Answer 12

Yes, you are correct for a circular loop, what I was wandering is, how much bigger does it have to be. In you formula you forgot that the maximum height is 2r, not r, and I didn't understood your expression for the kinectic energy.

Answer 13

yep I left out a 2:
PEi = 1.1gh > 2(1.1g(.7) + .5g(.7)) + 1/2 (1.1)(gr) +1/2(.5)(gr)

Answer 14

not really sure what you're doing there @physics_sucks
Vf is the velocity of...?
and why 5gr?

Answer 15

How did you get to that formula for the kinectic energy, I'm confused. Also, do you know anyway to find how much the speed of the block must be at the maximum height?

Answer 16

mV^2/r >mg

Answer 17

Sorry, still dont get it, where did this formula came from?

Answer 18

http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/cirvert.html#c1

Answer 19

Oh, I see, you used the centripetal force, nice.