Motion Problem with two moving masses

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  • #1
Gank
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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

  • #2
ehild
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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
 
  • #3
Gank
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Is it momentum? If so then mVi = mVf + Mw
 
  • #4
ehild
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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
Gank
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Do I do this simultaneously? By substituting one of the equations into the other?
 
  • #6
ehild
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Yes, do it.

ehild
 
  • #7
Gank
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What variable should I be eliminating?
 
  • #8
ehild
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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
 
  • #9
Gank
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So from that I get:

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

Where do I progress from here?
 
  • #10
ehild
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Factorize Vi^2-vf^2 and simplify.
 
  • #11
Gank
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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?
 
  • #12
ehild
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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
 
  • #13
Gank
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So if I rearrange that equation in terms of M I find that M=m(Vi - Vf)/(Vi + Vf).
 
  • #14
Gank
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Ah I see that m is Larger than M
 
  • #15
Gank
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So I find that w/Vf= 2m/(m-M) is that correct?
 
  • #16
ehild
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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
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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
Gank
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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
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The magnitude of Vf is greater than W. Yes, 1≥2m/(M-m). What does it mean for m?


ehild
 
  • #20
Gank
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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
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Yes, but you need to state the upper limit of m. Reformulate your statement with m as the subject .

ehild
 
  • #22
Gank
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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
 
  • #23
ehild
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That is correct. So the upper limit for m is M/3.

ehild
 
  • #24
Gank
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What are your ways of solving these sort of problems?
 
  • #25
ehild
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You have seen my ways:smile:.
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
 
  • #26
Gank
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Why is it that the co-efficient of restitution is given by

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

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