# Homework Help: Relative velocity of blocks

1. Apr 29, 2013

### Saitama

1. The problem statement, all variables and given/known data
(see attachment)

2. Relevant equations

3. The attempt at a solution
Consider the system at time t (see attachment, I am representing the blocks as point masses)
The component of acceleration along the string is $a_0\cos \theta$. This is the acceleration of the other two blocks at the sides. The component of acceleration along the horizontal direction is $a_0 \sin \theta \cos \theta$.
$$v\frac{dv}{dx}=a_0 \sin \theta \cos \theta$$
As $x=l\sin \theta$, $dx=l\cos \theta d\theta$
$$vdv=a_0 \sin \theta \cos^2 \theta d\theta$$
The problem is, what should be the limits for $\theta$? And is my working even correct?

Any help is appreciated. Thanks!

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• ###### blocks3.png
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2. Apr 29, 2013

### Staff: Mentor

I would consider this in the frame of block B. You get a very common physical setup, which can be solved without differential equations.

The limits for θ are given by the initial layout and the final one (collision).

3. Apr 29, 2013

### Saitama

So there will be pseudo force acting on the other two blocks? But then how would I calculate the acceleration in the horizontal direction?

Are the limits 0 to pi/2?

4. Apr 29, 2013

### Staff: Mentor

Once you see the analogy, it is easy ;). You can even treat it as real force...

Sure.

5. Apr 30, 2013

### Saitama

Sorry, I am still lost on this one.

But that doesn't give me the right answer. I had the following expression:
$$\int_{0}^v vdv=\int_{0}^{\pi/2} a_0 \ell \sin \theta \cos^2 \theta d\theta$$
$$\frac{v^2}{2}=\frac{a_0 \ell}{3}$$
This is incorrect as per the answer key.

6. Apr 30, 2013

### Staff: Mentor

Both blocks will feel a fictious force of m*a0 "downwards" acting on them.
Just consider that as real force, and you get a
pendulum in a gravitational field.

7. Apr 30, 2013

### Saitama

How is that a pendulum in a "gravitational field" when the motion takes horizontally?

And why does my DE gives the wrong answer? :(

8. Apr 30, 2013

### Staff: Mentor

The motion is the same as if the setup would be in a gravitational field (with gravity acting "sidewards"). It is not in a relevant real gravitational field, of course.

I don't know. What is v (in which frame)? Why did you introduce x?

The question asks to find the relative velocity of A and C, while you just calculated the velocity of A (or C). That factor of 3 should not appear, however.

9. Apr 30, 2013

### Saitama

Ah yes, I get your point. It turns out to be same as if the motion takes place in gravity.
$$ma_0\ell=\frac{1}{2}mv^2 \Rightarrow v=\sqrt{2a_0 \ell}$$
Hence the relative velocity before striking is $2\sqrt{2a_0 \ell}$

Thank you mfb!

v is in inertial frame.

Acceleration is vdv/dx so I had to introduce x and express it in terms of $\theta$.

10. May 1, 2013

### Staff: Mentor

That is complicated. In particular, you don't get the relative velocity with that, and "v" as integration limit does not make sense (it is a vector).

Anyway, treating it like a pendulum is way easier.