lioric said:
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.