Classical mechanics: Momentum and velocity

[REVIANNA]

A small block of mass m slides on a long horizontal table when it encounters a wedge of mass 2m and height h

The wedge can also slide along the table. The mass slides up the wedge all the way to the top and then slides back down, never getting over the top. All surfaces are frictionless.

Consider three moments in time:

• t1 – when the block is on the table, before sliding up the wedge;
• t2 – when the block is at the top of the wedge;
• t3 – when the block is on the table, after sliding down the wedge.

The speed of the wedge is:

The greatest at t3 (CORRECT)
Zero at both t1 and t3
The same but not zero at both t1 and t3
Zero at t2

My answer to part 1 is justified because:

The momentum of the wedge-and-block system is conserved.

The Attempt at a Solution

first of all I didn't understand how the momentum is conserved? momentum is only conserved when no external forces act on the system( here: block and wedge).But gravity(ext) (mgsin(theta)) is definitely acting on the block.

secondly I understand why the speed will be greatest at t3 (because acceleration will be down the wedge and it will increase the velocity)
I don't understand why the other option "ZERO AT t2" is wrong.won't it momentarily stop at the top of the wedge.

thirdly,the hint to this problem said "We could write an expression for the total momentum of the system at t1, t2, and t3." How would we do that?

I really can't wrap my head around the fact that the wedge is moving too. I dealt with a similar problem (no wedge but a slab) which used the concept of centre of mass. Can we "conveniently" use it here.

How is the height of the wedge useful?

 

[haruspex]

Gravity acts on both vertically. It cannot affect horizontal momentum of the block+wedge system.
The question asks about the maximum speed of the wedge, not the block.
For the momentum expressions, you need to find some velocities. You have the information that the block just makes it to the top of the wedge.
At t2, what is the relationship between the velocity of the block relative to the ground and the velocity of the wedge?
What equation for velocity do you get from energy conservation?

 

[REVIANNA]

I understood that the momentum is conserved in the x direction. N is the force on the wedge by the block which is also the force on the block by the wedge.
https://drive.google.com/file/d/0B_5Dy6Va9eR8bWhYcVRJcHM1QkE/view?usp=sharing
https://drive.google.com/file/d/0B_5Dy6Va9eR8bjl5SHIwd0RaRmc/view?usp=sharing
now using the velocity at t2 and the net force(Nsin(theta)) I can find the final velocity at t3.
am I correct?

 

[PeroK]

I understood that the momentum is conserved in the x direction. N is the force on the wedge by the block which is also the force on the block by the wedge.
https://drive.google.com/file/d/0B_5Dy6Va9eR8bWhYcVRJcHM1QkE/view?usp=sharing
https://drive.google.com/file/d/0B_5Dy6Va9eR8bjl5SHIwd0RaRmc/view?usp=sharing
now using the velocity at t2 and the net force(Nsin(theta)) I can find the final velocity at t3.
am I correct?

You definitely don't need to look at the forces between block and the wedge for this problem. Because the surfaces in contact are both accelerating, neither represents an inertial reference frame in which you can apply $F = ma$

Instead, you should think about conservation of momentum in the x-direction and, if necessary, conservation of energy (kinetic and potential).

Note that when the block "stops" on the wedge, that means the wedge and the block are moving to the right at the same speed. To an inertial observer, when the block stops moving to the right, it will already be sliding back down the block (if, indeed, the block ever stops).

I think this problem is tricky to visualise, but it's worth trying to wrap your head round it!

 

[REVIANNA]

Instead, you should think about conservation of momentum in the x-direction and, if necessary, conservation of energy (kinetic and potential).

how should I think about momentum and energy conservation without velocity and therefore without force

To an inertial observer, when the block stops moving to the right, it will already be sliding back down the block

I did not understand this. does this mean that when the block stops (with respect to the wedge) it will actually be moving as viewed from the ground because the wedge will be moving .

and really why will the wedge move at all t3 has it not lost contact with the ground?

 

[PeroK]

how should I think about momentum and energy conservation without velocity and therefore without force

You don't need to analyse the contact forces in order to use conservation of momentum and of energy.

I did not understand this. does this mean that when the block stops (with respect to the wedge) it will actually be moving as viewed from the ground because the wedge will be moving .

Yes. So, you do understand it!

and really why will the wedge move at all t3 has it not lost contact with the ground?

If you imagine the wedge has a large mass, then the block will slide up, back down and effectively reverse its direction. At t3 it will be moving backwards. Using conservation of momentum will tell you what happens when the wedge is not so massive and is significantly accelerated by the block. In this case, the wedge has a mass of only $2m$, so you need to do some calculations using conservation of momentum.

Hint: this problem is not that different from a collision. In terms of the end result.

 

[REVIANNA]

At t3 it will be moving backwards.

you mean back to the block ? what will cause anything to go against gravity.
I will attempt this question again after leaning thoroughly about conservation of energy and momentum.
till then Thank you .

 

[REVIANNA]

Note that when the block "stops" on the wedge, that means the wedge and the block are moving to the right at the same speed.

I understood that at $t_2$ the velocity of the wedge wrt the ground is equal to that of the block wrt the ground.
so I wrote these: at time $t_2$ and $t_3$
conservation of momentum :

$(m*v_2)-(2*m*v_2)=(m*v_b)-(2*m*v_w)$

and conservation of energy :

$(0.5*m*v^2)+(0.5*2*m*v^2)+(m*g*h)=(0.5*m*v_b^2)+(0.5*2*m*v_w^2)$

am I correct?

Well this still doesn't prove that the velocity of the wedge is maximum at $t_3$.

If you imagine the wedge has a large mass, then the block will slide up, back down and effectively reverse its direction. At t3 it will be moving backwards.

you mean block moves back to the wedge ? what will cause anything to go against gravity.

 

[haruspex]

$(m*v_2)-(2*m*v_2)=(m*v_b)-(2*m*v_w)$

I assume the LHS is the total momentum at t₂, and the RHS the total at some later time (t₃). So why the minus signs?

Well this still doesn't prove that the velocity of the wedge is maximum at t₃.

Not immediately, no - you would have more work to do. E.g. you could consider the more general condition that it has only descended by height y from the top, solve for the wedge velocity, and find what y <=h maximises it. Or you could avoid the algebra and construct a line of reasoning.

you mean block moves back to the wedge ?

No, back from the wedge. When it reaches ground level again, it might be moving to the left or to the right, depending on relative masses. But it will be moving less to the right than the wedge is.
PeroK meant that the sliding back down could be a reversal of the block's original direction.

 

[REVIANNA]

@haruspex
I had a look at the details of the answer and I understood

1. the energy is conserved and the momentum is conserved in the x direction
2. the initial momentum of the wedge is zero and the the block's positive.(at t_1)
3. the moment the block is at the top of the wedge it stops wrt the ground and shares the same horizontal velocity as that of the wedge.
4. than the block slides down and its velocity is lower than what it began with because it got the wedge moving and the wedge's velocity is higher than at t_2

I have calculated the velocities in terms of the block's initial velocity. And it is higher than at t_2 .
force is not considered while solving so I don't understand this-
"regardless of the direction of the block's motion the normal force always increases the momentum of the wedge and decreases that of the block".

which normal force?
because there is a normal force on the wedge due to the block and one on the block due to wedge (that's the reason why the momentum is conserved in the first place as they are internal to the system)

Is the increase(for wedge) and decrease(for block) just because the N is to the right(for wedge) and to the left for block.

 

[haruspex]

3. the moment the block is at the top of the wedge it stops wrt the ground and shares the same horizontal velocity as that of the wedge.

Stops wrt the wedge. Note that it also has no vertical velocity, so has the same total velocity as the wedge.

Is the increase(for wedge) and decrease(for block) just because the N is to the right(for wedge) and to the left for block.

Yes.

 

[REVIANNA]

Stops wrt the wedge. Note that it also has no vertical velocity, so has the same total velocity as the wedge

sorry I meant wrt to the wedge it stops but as seen from the table it is moving with the wedge's velocity.

and thanks!