# Motion Problem with two moving masses

Gank
A block of mass m slides down a ramp. At the bottom, it strikes a block of mass M, which is at rest on a horizontal surface. If the collision is elastic and friction can be ignored, determine the upper limit on mass m if it is to rebound from M, slide up the ramp, stop, slide back down the ramp and collide with M again.

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

## Answers and Replies

Homework Helper
A block of mass m slides down a ramp. At the bottom, it strikes a block of mass M, which is at rest on a horizontal surface. If the collision is elastic and friction can be ignored, determine the upper limit on mass m if it is to rebound from M, slide up the ramp, stop, slide back down the ramp and collide with M again.

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

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

Homework Helper
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

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

Homework Helper
Yes, do it.

ehild

Gank
What variable should I be eliminating?

Homework Helper
You have the equations
m(Vi2-Vf2)=MW2
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

Gank
So from that I get:

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

Where do I progress from here?

Homework Helper
Factorize Vi^2-vf^2 and simplify.

Gank
So now I get that m=M(Vi + Vf)/(Vi - Vf). So this means that m is smaller than M by a factor of (Vi + Vf)/(Vi - Vf).

Is this the limit?

Homework Helper
No, you need the limit in terms of M.

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

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

Gank
Ah I see that m is Larger than M

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

Homework Helper
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

Homework Helper
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

Gank
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).

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

ehild

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

Homework Helper
Yes, but you need to state the upper limit of m. Reformulate your statement with m as the subject .

ehild

Gank
So m must be less than or equal to one third of M if it is to catch up with M and collide a second time

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

ehild

• 1 person
Gank
What are your ways of solving these sort of problems?

Homework Helper
You have seen my ways .
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

Gank
Why is it that the co-efficient of restitution is given by

(Va - Vb)/(Vb - Va) and not (Va - Vb)/(Va - Vb)?