Problem regarding centre of mass and linear momentum.

[sankalpmittal]

Homework Statement

A block of mass "m" is placed on a triangular block of mass "M" , which in turn is placed on a horizontal surface as shown in figure. Assuming frictionless surfaces , find the velocity of the triangular block when the smaller block reaches the bottom end.

Figure : http://postimage.org/image/cv0qo69kd/

Homework Equations

If no external net forces act on a system :
Conservation of linear momentum : P_f=P_i
Also if net external forces are zero , and initially system is at rest then , by concept of centre of mass ,
m₁x₁=m₂x₂
m₁Δx₁=m₂Δx₂

The Attempt at a Solution

Now Here is what I tried so far :

Let's take two blocks as a system , no net external force acts horizontally , centre of mass will not change its position in that direction. But how to apply conservation of momentum here ? I am totally confused. This is question too different from others I tried.

I did 58 questions of centre of mass , and this 59th one has bashed my brain hard.

Please help !

Thanks in advance...

 

[tms]

If no external net forces act on a system :

What about gravity?

 

[sankalpmittal]

What about gravity?

Re-read. I said no net external force acts on the system in horizontal direction below "The attempt at a solution".

 

[Leong]

square root of [2mgh/(M+m)]?

 

[fgb]

square root of [2mgh/(M+m)]?

I have not solved the problem, but shouldn't the answer be a function of θ? For θ=90°, for instance, we know the answer should be zero, since m would be in freefall and would not "push" the triangular block.

 

[fgb]

$v = \sqrt{\dfrac{2mgh}{M+\frac{M^2}{m\cos^2{\theta}}}}$, maybe?

Not sure my answer is correct, but here are a few hints: Use conservation of linear momentum (horizontal) to relate V (velocity of big block) to the HORIZONTAL component of v (velocity of small block).

Use also conservation of energy. Notice that in the end you have KE for both blocks, and KE depends on the magnitude of velocity (not only the horizontal component. How can you relate v to v_horizontal?)

UPDATE: Notice that $v \rightarrow 0$ as $\theta \rightarrow 90°$ :)

 

[Leong]

I have not solved the problem, but shouldn't the answer be a function of θ? For θ=90°, for instance, we know the answer should be zero, since m would be in freefall and would not "push" the triangular block.

What you said is very true.

 

[sankalpmittal]

square root of [2mgh/(M+m)]?

Not even near.

I have not solved the problem, but shouldn't the answer be a function of θ? For θ=90°, for instance, we know the answer should be zero, since m would be in freefall and would not "push" the triangular block.

You're correct.

$v = \sqrt{\dfrac{2mgh}{M+\frac{M^2}{m\cos^2{\theta}}}}$, maybe?

Not sure my answer is correct, but here are a few hints: Use conservation of linear momentum (horizontal) to relate V (velocity of big block) to the HORIZONTAL component of v (velocity of small block).

Use also conservation of energy. Notice that in the end you have KE for both blocks, and KE depends on the magnitude of velocity (not only the horizontal component. How can you relate v to v_horizontal?)

UPDATE: Notice that $v \rightarrow 0$ as $\theta \rightarrow 90°$ :)

Almost near to correct answer put thinking straightforward your answer is also incorrect. Numerator part of your answer completely matches with the correct answer , however denominator part does not. Thanks for replies anyway.

Ok , perhaps you might have done some careless mistakes as your logic's fine. I will use your logic to see if I get the correct answer. Will get back on ya.

Edit : In image , angle is alpha , and its intact.

 

[sankalpmittal]

Hii fgb ,

I tried all those hints , and get the same answer as yours , exactly same..

My first attempt was as follows :

Let the velocity of small block be v₁ at bottom and the triangular block be traveling due left by v₂ ,

Applying conservation of linear momentum at bottom ,

mv₁cos(α) = Mv₂
v₁=Mv₂/m

Considering conservation of mechanical energy for the system :

mv₁²/2 + Mv₂²/2 = mgh

Now putting v₁=Mv₂/m in above equation , I got the answer same as yours yet not the correct answer.

Now here comes the second attempt :

I realized that as small block comes down it will have two velocity vectors. One v₁cos(α) and the other v₂ , it will also be traveling backwards at peak point and thus , I obtained :

mv₁cos(α) = Mv₂ + mv₂
v₁=v₂(M+m)/mcos(α)

Then I again used conservation of mechanical energy equation. Still I got the numerator part correct but denominator part did not match.

May be others can reply. I am clueless.
Some hints will do...

 

[ehild]

$v = \sqrt{\dfrac{2mgh}{M+\frac{M^2}{m\cos^2{\theta}}}}$, maybe?

$v = \sqrt{\dfrac{2mgh}{M+\frac{M^2}{m\cos^2{\alpha}}}}= \sqrt{\dfrac{2mgh}{M+\frac{M^2}{m}(1+\tan^2{\alpha})}}$

ehild

 

[sankalpmittal]

$v = \sqrt{\dfrac{2mgh}{M+\frac{M^2}{m\cos^2{\alpha}}}}= \sqrt{\dfrac{2mgh}{M+\frac{M^2}{m}(1+\tan^2{\alpha})}}$

ehild

Hii ehild !

I can not understand. The answer posted by fgb is wrong and does not match with by textbook's answer , no matter how it is changed by applying trigonometry.

Edit : Please see my working in post #9. Where did I do wrong ?

 

[fgb]

sankalpmittal, could you post the correct answer? I know it is kind of cheating, but it might be useful to find out where we are getting it wrong :P

 

[sankalpmittal]

sankalpmittal, could you post the correct answer? I know it is kind of cheating, but it might be useful to find out where we are getting it wrong :P

The answer given is ,

[{2m²ghcos²α}/{(M+m)(M+msin²α)}]^1/2

 

[ehild]

The velocity of the block with respect to the slope is parallel to the slope, the velocity in the rest frame of reference is not.

Let be u the speed of the block with respect to the slope, traveling with velocity v₂ then the velocity in the rest frame of reference is v_x=v₂+ucosα, the y component is v_y=-usinα. You have to calculate the KE of the block from these v_x and v_y.

ehild

 

[sankalpmittal]

The velocity of the block with respect to the slope is parallel to the slope, the velocity in the rest frame of reference is not.

Let be u the speed of the block with respect to the slope, traveling with velocity v₂ then the velocity in the rest frame of reference is v_x=v₂+ucosα, the y component is v_y=-usinα. You have to calculate the KE of the block from these v_x and v_y.

ehild

I followed your method ,

m(v_x²+v_y²)/2 + Mv₂²/2 = mgh

And

mv_x = Mv₂

Putting ,v_x=v₂+ucosα
And v_y=-usinα in above two equations , I got the answer as :

[{2m²ghcos²α}/{M²+m²sin²α+Mm-3Mmsin²α}]^1/2

This denominator part does not match the correct answer.

 

[ehild]

You denoted the speed of the slope by v2. In that case vx=-v2+ucos(alpha).

ehild

 

[sankalpmittal]

You denoted the speed of the slope by v2. In that case vx=-v2+ucos(alpha).

ehild

Phew ! Now I got the correct answer ! Thanks a lot for help , ehild !

Uhh , I can see that I lost my entire day doing my question...

Also , seeing the back of my textbook , its given a hint for this question. It asks me to find the acceleration of the slope and the relative acceleration of the small block. You gave me entirely different method , thanks once again.

 

[ehild]

These "moving constraints" problems are really tricky. Using conservation of energy and conservation of momentum looked the easiest method to apply, and it was your method:)

ehild

 

[Leong]

These "moving constraints" problems are really tricky. Using conservation of energy and conservation of momentum looked the easiest method to apply, and it was your method:)

ehild

You're good. You can see the different reference frames and introduce relative velocity.