# Block on movable wedge

1. Mar 5, 2005

### davidY

simple question: a movable block on a movable wedge on some frictionless ground. we know all masses, angles, and there is no friction. the block starts on the wedge at some height h, and slides down the wedge. the block moves forwards, and the wedge moves backwards. what is the velocity of the wedge, the moment before the block hits the ground - i.e. block still on wedge?

i tried this question. however, i could only get the velocity of the wedge, after the block is totally off the wedge by conservation of energy and momentum.

2. Mar 5, 2005

### Gamma

I would use conservation laws between the initial point and the point where the block has travelled a vertical distance of h. You need the velocity of the wedge just before the block leaves the wedge. That means it is still on the wedge. I think you are worried about the small difference in h caused by the width of the wedge. I don't think that will make a huge difference.

3. Mar 5, 2005

### maverick280857

Well momentum is conserved in the horizontal direction if you take the wedge and block as the system under consideration and so you can find find a general expression relating displacements independent of velocity.

Finding the velocity at any height h of the block on the wedge isn't as easy as finding it when it is at the bottom of the wedge. However, coupled with momentum conservation in the horizontal direction that shouldn't be a problem either.

By the way if you have to find displacements, remember that if momentum is conserved in a particular direction then the coordinate of the mass center of the system referred to that direction remains fixed relative to an inertial observer. This is just a restatement of momentum conservation, but by chosing coordinates wisely, this constraint on the system is faster and easier than direct momentum conservation.

Cheers
vivek

4. Mar 5, 2005

### davidY

yeah, i want the total velocity of the little block. however here are the conservation equations:

Energy Initial = mgh
Energy Final = 1/2 M(velocity of wedge)^2 + 1/2m(velocity in horizontal direction of little block)^2 + 1/2 m(velocity in vertical direction of little block)^2

P Inital = 0
P final = M(velocity of wedge) + m(velocity in horizontal direction of little block)

however, it is not true that tan(angle of wedge)=(velocity in vertical direction)/(velocity in horizontal direction).

as this assumption is only true when the wedge is unmoving, however since this is a free wedge, this is not true in the centre of mass reference frame.

File size:
1.8 KB
Views:
179
5. Mar 7, 2005

### Staff: Mentor

But with respect to the wedge the velocity of the block is parallel to the wedge. Try expressing your equations in terms of the speed of the block with respect to the wedge. And note that the wedge only moves horizontally.

6. Jul 16, 2011

I've also come across this problem in the past few days and it has me puzzled.

For a wedge angle 'alpha', how is the final velocity of wedge and block calculated ?

Last edited: Jul 16, 2011
7. Jul 16, 2011

### BruceW

A professor gave this question when I first started my undergraduate physics degree. I remember it took me ages to work it out.

davidY - your equation for the energy is incorrect. This is because there is a force from the ground upwards on the wedge. This force is transmitted to the small block (which has a component of motion opposite to this force), therefore negative work is being done on the system. (I.e. in this system, energy is not conserved).
In this problem, you don't need to use energy conservation.

I just calculated the problem (hopefully correctly) by using the coordinate system where the objects are initially at rest (the most simple way I can think of).
To give you a hint, the way I did it, I used 4 equations:
2 were from resolving the forces on the small block (which become 1 when you combine them to get rid of the normal force). Another equation was conservation of horizontal momentum. The last equation was the condition that the small block remained touching the wedge.

And from these, I did a lot of rearranging to get the acceleration of the block in terms of the acceleration due to gravity (g), the ratio of the two masses, and the angle of the wedge.
And from there you would just need to use initial conditions to get the speed of the small block when it reached the bottom of the wedge.
One more hint: to check your answer, see what your equation approximates to when the mass of the wedge is much greater than the mass of the block. If you've done it right, it will approximate the case of a wedge which is fixed to the floor.

8. Jul 17, 2011

### IDOGAWACONAN

I come up with a kind of method... I don't know if it work

First your equations are wrong ,the right one should be:

Energy Final = 1/2 M(velocity of wedge)^2 + 1/2m(velocity in horizontal direction of little block)^2 + 1/2 m(velocity in vertical direction of little block)^2+mg* h...the h is not the initial one but the h 'now'.

And tan(angle of wedge)=(velocity in vertical direction)/(velocity in horizontal direction-velocity of wedge).

We can express all these velocity with the velocity in vertical direction of little block.And then replace it with dh/dt),then you get a differential equation.Solve it and you'll get all.

9. Jul 17, 2011

### IDOGAWACONAN

And of course if you just want to know the relationship between velocity with h but not with t,you don't have to replace it with dh/dt to make a differential equation,it's easier to solve.

10. Jul 17, 2011

Thank you for your thoughts however,

Bruce,
It is a closed system. Energy must be conserved.

IDOGAWACONAN,
2 equations, 3 unknown velocities ... not enough for a solution.
What is the basis for the "tan(angle of wedge) ..." equation ?
Perhaps conservation of momentum as mentioned by Bruce, is a 3rd equation ?

11. Jul 17, 2011

### Quinzio

Let's call $a$ the horizontal acceleration of the wedge.
The tilted surface will accelerate *perpendicular to itself* by $a\ sin \alpha$.
So the block will experience an acceleration perpendicular to the surface $g\ cos\ \alpha -a\ sin \alpha$.
Thus the force perpendicular has to be $m(g\ cos\ \alpha -a\ sin \alpha)$.
This is the force $F$ that will push the wedge. Horizontally acceleration will be only: $F\ sen\ \alpha / M$.
This will again give an expression for $a$.
So we equate $a = {m \over M} sen\ \alpha(g\ cos\ \alpha -a\ sin \alpha)$.

Ok, so with this last equation, we have the acceleration of the wedge.
Then we can find again the acceleration of the wedge surface *perpendicular to itself*.
This will be as well the acceleration of the block perpendicular to the surface.
Now we know all the accelerations on the block. We can find how long does it take to hit the ground

12. Jul 17, 2011

### IDOGAWACONAN

There aren't only 2 equations:
1.Energy Final = 1/2 M(velocity of wedge)^2 + 1/2m(velocity in horizontal direction of little block)^2 + 1/2 m(velocity in vertical direction of little block)^2+mg* h
2.velocity in horizontal direction of little block *m=-velocity of wedge*M
3.tan(angle of wedge)=(velocity in vertical direction)/(velocity in horizontal direction-velocity of wedge).
(velocity in horizontal direction-velocity of wedge) is the relative velocity,similiar to (velocity in horizontal direction) when the wedge is still.

13. Jul 17, 2011

It sounds very reasonable, although you have "a" on both sides. Taking a across gives quite a complex expression.
I assume the acceleration on the block down the slope is
g sin $\alpha$ - a cos $\alpha$
... however in reality, the direction in which the block moves is NOT at the angle of the slope but rather a greater angle. It would only move at the angle of the slope if the wedge didn't move. It would appear that this would effect your calculation ?

I suppose the test is whether the sum of the two kinetic energies is equal to the initial potential energy.

14. Jul 17, 2011

### BruceW

It can be easily shown that energy is not conserved in this system. I will assume that energy is conserved, and then show that the resulting equation must be incorrect. First, the equation for energy:
$$Energy = \frac{1}{2}m_w{\dot{x}_w}^2 + \frac{1}{2}m_b{\dot{x}_b}^2 +\frac{1}{2}m_b{\dot{y}_b}^2 + m_bgy_b$$
and the equation for conservation of horizontal momentum:
$$m_w \dot{x}_w + m_b \dot{x}_b = 0$$
Now, differentiate the energy equation with respect to time, gives:
$$0 = m_w \ddot{x}_w + m_b \ddot{x}_b + m_b \ddot{y}_b + m_b g \dot{y}_b$$
and differentiate the conservation of horizontal momentum equation with respect to time gives:
$$0 = m_w \ddot{x}_w + m_b \ddot{x}_b$$
Combining these two equations gives:
$$m_b \ddot{y}_b + m_b g \dot{y}_b = 0$$
and the initial velocity is zero, so the initial acceleration is zero, so neither of the objects would move!
This proves that using conservation of energy results in an unrealistic answer, which is because energy is not conserved, it is not a closed system!

Edit: the w stands for wedge and the b stands for block in the equations.

15. Jul 17, 2011

There is obviously an error in your equation above. If the masses are equal, it implies the acceleration of the block is equal and opposition to the acceleration of the wedge, regardless of the angle of the wedge.

If conservation of kinetic energy did not hold, heat would be produced ... but the system is frictionless and cannot produce such heating. It is a closed system and the 1st Law of Thermo must hold.

16. Jul 18, 2011

### rl.bhat

Consider the motion of thee block on the movable wedge. Consider the co-ordinate axis such that x-axis is parallel to the inclined plane of the wedge.
When the wedge moves through a distance X horizontally, block moves through a distance y perpendicular to the wedge surface such that y = X*sinθ.
Hence ay = A*sinθ .........(1), where A is the acceleration of the wedge.
Now consider the force equations.
For wedge: N*sinθ = M*A......(2)
For block: mg*sinθ = m*ax.........(3)
mg*cosθ - N = m*ay.....(4)
substitute the value of ay from eq.(1) in eq.(4) and find N.
Substitute this value in eq.(2) to find the acceleration A of the wedge.

17. Jul 23, 2011

### BruceW

Woops, I now realise that energy is conserved in this system. I got this wrong:
Which is wrong because actually, differentiating the equation for energy gives:
$$0 = m_w \dot{x}_w \ddot{x}_w + m_b \dot{x}_b \ddot{x}_b + m_b \dot{y}_b \ddot{y}_b + m_b g \dot{y}_b$$

I've found this problem can be done in two possible ways, (which both get the same answer).
First way) use conservation of energy and horizontal momentum along with the fact that the block stays on the surface of the wedge.
Second way) simply resolve the forces on the two objects, and use the fact that the block stays on the wedge.
For either method, you need to do a bit of differentiating and rearranging to get the answer. (Which is a slightly complicated expression).
Should I write it down, to help anyone who stumbles on this thread in the future, or is that considered bad etiquette?