Motion Problem with two moving masses

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In summary: It is a matter of convention. The kinetic energy is lost after the collision, so if Va > Vb, Vaf < Vbf. The minus sign comes from it. But it is a matter of convention, and it is just a convention. In summary, the conversation involved determining the upper limit on the mass m of a sliding block in order for it to rebound from a block of mass M, slide up and down a ramp, and collide with M again. Using the conservation of momentum and kinetic energy, we found that the limit for m is M/3 if the collision is elastic and friction can be ignored. We also discussed the importance of imagining the scenario and understanding conventions in equations.
  • #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
 
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  • #2
Gank said:
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
Is it momentum? If so then mVi = mVf + Mw
 
  • #4
Gank said:
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
Do I do this simultaneously? By substituting one of the equations into the other?
 
  • #6
Yes, do it.

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

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

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


ehild
 
  • #20
That M must be greater than or equal to 3m if m is to catch up with M and collide a second time?
 
  • #21
Yes, but you need to state the upper limit of m. Reformulate your statement with m as the subject .

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

ehild
 
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  • #24
What are your ways of solving these sort of problems?
 
  • #25
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
Why is it that the co-efficient of restitution is given by

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

What is a "Motion Problem with two moving masses"?

A "Motion Problem with two moving masses" refers to a physics problem that involves two objects of different masses moving simultaneously, where their positions, velocities, and accelerations are related to each other.

How do you solve a "Motion Problem with two moving masses"?

To solve a "Motion Problem with two moving masses", you need to use Newton's laws of motion and apply them to the two objects separately. You can also use equations such as the equation of motion, the equation for acceleration, and the equation for velocity to help you solve the problem.

What are the key variables involved in a "Motion Problem with two moving masses"?

The key variables involved in a "Motion Problem with two moving masses" are the masses of the two objects, their positions, velocities, and accelerations, as well as the forces acting on them. These variables are all interconnected and can be used to determine the motion of the objects.

Can a "Motion Problem with two moving masses" be solved using calculus?

Yes, a "Motion Problem with two moving masses" can be solved using calculus, specifically through the use of derivatives and integrals. These mathematical tools can help you find the instantaneous velocity and acceleration of the objects at a given point in time.

What are some real-life examples of "Motion Problems with two moving masses"?

Some real-life examples of "Motion Problems with two moving masses" include a car colliding with another car, a person pushing a shopping cart, or a rocket launching into space with multiple stages. These situations involve two objects of different masses moving in relation to each other, and their motion can be described using the principles of physics.

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