Man jumping onto a see-saw to launch bricks vertically

[lioric]

Thank you all very much
I believe I have learned a great deal from this.
This question is something I m making for my students to work on.
May I finish the question properly and post it for you guys to try it out before I give it to my students?

 

[jbriggs444]

May I finish the question properly and post it for you guys to try it out before I give it to my students?

Certainly.

 

[jbriggs444]

Brick from height of h= 10m
PE = mgh = √(KE /0.5 x m)= v of brick

This is extremely sloppy. As written, the above equation states that a velocity is equal to an energy.

What I expect that you mean to have written would equate the final PE of the brick with the initial KE of the brick. e.g. $mgh=\frac{1}{2}mv^2$. Solving for the initial velocity of the brick, one would then get $v=\sqrt{2gh}$.

I also have a pet peeve about writing down variable names without having defined them first. But I'll forgive that in the name of brevity.

Man with unknown v
Vbrick x distance x mass of brick / Distance x mass of man = v of man

Here you seem to be equating the angular momentum of the brick with the angular momentum of the man. That is exactly what I had warned you NOT to do.

The physical meaning of such an equation would be that the interaction between man, brick and seesaw is such that the man stops in place immediately after impact [and before hitting the ground!] and that his prior angular momentum is transferred entirely to the brick.

Such an interaction results in a gain of kinetic energy as a result of the collision. The only way this works is if there is an explosive device under the man's shoes or if he pushes off very strongly in order to bring himself to an immediate stop.

Since you had suggested that mechanical energy is conserved, what I had in mind was equating the initial kinetic energy of the man with the final kinetic energy of man and brick combined.

Conservation of mechanical energy:
$$\frac{1}{2}MV_i^2=\frac{1}{2}MV_f^2 + \frac{1}{2}mv_f^2$$

If you want to go with an inelastic collision then you can apply a simpler equation:

Lever law:
$$v_f=V_f\frac{d_{brick}}{d_{man}}$$
Where V_f is the velocity of the man post-impact.

Either way you can then factor in conservation of angular momentum:
$$mv_fd_{brick} + MV_fd_{man} = MV_id_{man}$$

Where $v_f$ and $V_f$ are the velocities of brick and man post collision and $V_i$ is the velocity of the man post-fall but pre-collision.

We can use
v²=u² + 2as to find the height of the man jumping off

Yes, indeed. Though you might want to re-think the use of the same variable ("v") for both brick and man.

 

[lioric]

Certainly.

Thank you very much
It was truly a delight to work with such brilliant minds

 

[PeroK]

Can I ask why angular momentum is conserved? The rod is made of rigid material which I interpret to mean that if the man end is moving down with instantaneous speed $v$, the brick end is moving up with the same speed. The magnitude of the linear acceleration is the same at both ends for the same reason. Now when the man end hits the ground and stops, the upward speed at the brick end drops to zero very very rapidly meaning an acceleration pointing down starting from an initial value of zero. At the point where the magnitude of the brick end's increasing downward acceleration becomes greater than g, the bricks lose contact and are launched in the air. It is, therefore, safe to assume that the launching speed of the bricks is the same as the speed of the man end just before it hits the ground. In the preceding time interval, after the man lands on the end of the seesaw and before that end hits the ground, angular momentum is not conserved because of the external torque of gravity about the fulcrum.

 

[kuruman]

This is what I wrote at the beginning of this post
Now even I'm confused

Sorry if contributed to your confusion; I had a different model in mind.

 

[lioric]

Sorry if contributed to your confusion; I had a different model in mind.

It's ok. Because of your question I now know when the angular momentum is conserved and when it is not.
I love learning stuff here.
See you guys soon with the final draft of this question