- #1

- 31

- 0

So I have found that:

m(Vi)^2 = m(Vf)^2 + M(W)^2

And that m can only catch M if the velocity Vf is greater than W so the limit becomes Vf=W.

Thanks

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Gank
- Start date

- #1

- 31

- 0

So I have found that:

m(Vi)^2 = m(Vf)^2 + M(W)^2

And that m can only catch M if the velocity Vf is greater than W so the limit becomes Vf=W.

Thanks

- #2

ehild

Homework Helper

- 15,543

- 1,912

So I have found that:

m(Vi)^2 = m(Vf)^2 + M(W)^2

And that m can only catch M if the velocity Vf is greater than W so the limit becomes Vf=W.

Thanks

The block M is in rest initially, so the other block certainly catches it.

You need to give the limit for m compared to M, so the block of mass m travels backwards after collision. There is one more conservation law you need to take into account.

ehild

- #3

- 31

- 0

Is it momentum? If so then mVi = mVf + Mw

- #4

ehild

Homework Helper

- 15,543

- 1,912

Is it momentum? If so then mVi = mVf + Mw

Yes. Now find the relation between Vi and Vf, using both conservation laws.What is the condition of m so that it can catch M when slides down the ramp second time, that is |Vf|≥W?

ehild

Last edited:

- #5

- 31

- 0

Do I do this simultaneously? By substituting one of the equations into the other?

- #6

ehild

Homework Helper

- 15,543

- 1,912

Yes, do it.

ehild

ehild

- #7

- 31

- 0

What variable should I be eliminating?

- #8

ehild

Homework Helper

- 15,543

- 1,912

m(Vi

and

m(Vi-Vf)=MW

Does not it look easier to eliminate W ? Square the second equation and divide the first one with it.

ehild

- #9

- 31

- 0

So from that I get:

m=M(Vi^2 - Vf^2)/(Vi - Vf)^2

Where do I progress from here?

m=M(Vi^2 - Vf^2)/(Vi - Vf)^2

Where do I progress from here?

- #10

ehild

Homework Helper

- 15,543

- 1,912

Factorize Vi^2-vf^2 and simplify.

- #11

- 31

- 0

Is this the limit?

- #12

ehild

Homework Helper

- 15,543

- 1,912

Isolate Vf. It must be negative in order that m return back to the slope. What does it mean for m compared to M?

m goes up on the slope and then returns back. Without friction, it reaches the bottom with the same speed. It can catch M if that speed |Vf| exceeds the speed of M .

So you have to find bot Vf and W in terms of m/M and Vi, and find m/M from the condition |Vf|≥W.

ehild

- #13

- 31

- 0

So if I rearrange that equation in terms of M I find that M=m(Vi - Vf)/(Vi + Vf).

- #14

- 31

- 0

Ah I see that m is Larger than M

- #15

- 31

- 0

So I find that w/Vf= 2m/(m-M) is that correct?

- #16

ehild

Homework Helper

- 15,543

- 1,912

Ah I see that m is Larger than M

No, just the opposite. m must go up to the slope so it has to move backwards after the collision, That means negative Vf. Vf<0.

ehild

- #17

ehild

Homework Helper

- 15,543

- 1,912

So I find that w/Vf= 2m/(m-M) is that correct?

That is al right. Now you have to ensure that the block of mass m catches the larger block.

ehild

- #18

- 31

- 0

Is this where Vf≥w? as vf≥w, then w/vf must be less than or equal to one. So then 1≥2m/(M-m).

- #19

ehild

Homework Helper

- 15,543

- 1,912

The magnitude of Vf is greater than W. Yes, 1≥2m/(M-m). What does it mean for m?

ehild

ehild

- #20

- 31

- 0

That M must be greater than or equal to 3m if m is to catch up with M and collide a second time?

- #21

ehild

Homework Helper

- 15,543

- 1,912

ehild

- #22

- 31

- 0

- #23

ehild

Homework Helper

- 15,543

- 1,912

That is correct. So the upper limit for m is M/3.

ehild

ehild

- #24

- 31

- 0

What are your ways of solving these sort of problems?

- #25

ehild

Homework Helper

- 15,543

- 1,912

Read the problem carefully. What is given, and what is the question.

Identify the problem. It was elastic collision. I collected the relevant equations.

I usually make a figure. It helps to imagine what happens. Here, there was a mass uphill, and the other on the ground. The mass on the hill went down, gained some velocity Vi and collided with the other one, and it got a negative velocity and went back on the hill.

I derived the velocities (Vf and W) after the collision. In order that m returned to the slope, its velocity must have been negative after the collision. To get negative velocity, m had to be smaller than M.

There was no friction so the kinetic energy was the same when the mass m returned as it was when it started to rise. Now it is two objects, m catching M. It is only possible when its speed is greater than that of M.

The most important thing is to imagine what happens.

ehild

Share:

- Replies
- 3

- Views
- 2K