The question is: What are the limits for theta and is the working correct?

[Saitama]

Homework Statement

(see attachment)

Homework Equations

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##.
[tex]v\frac{dv}{dx}=a_0 \sin \theta \cos \theta[/tex]
As ##x=l\sin \theta##, ##dx=l\cos \theta d\theta##
[tex]vdv=a_0 \sin \theta \cos^2 \theta d\theta[/tex]
The problem is, what should be the limits for ##\theta##? And is my working even correct?

Any help is appreciated. Thanks!

 

[mfb]

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).

 

[Saitama]

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

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

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

Are the limits 0 to pi/2?

 

[mfb]

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

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

Are the limits 0 to pi/2?

Sure.

 

[Saitama]

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

Sorry, I am still lost on this one.

Sure.

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

 

[mfb]

Sorry, I am still lost on this one.

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

Spoiler

pendulum in a gravitational field.

 

[Saitama]

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

Spoiler

pendulum in a gravitational field.

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

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

 

[mfb]

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.

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

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.

 

[Saitama]

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.

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

Thank you mfb!

I don't know. What is v (in which frame)?

v is in inertial frame.

Why did you introduce x?

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

 

[mfb]

v is in inertial frame.

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.