What if the directions of the initial and final velocity are different in the conservation of mechanical energy equation?

Question

I saw a problem, and saw that it could be easily solved using conservation of mechanical energy. So I wrote my equation:

$$mgh_1 + \frac12mu^2 = mgh_2 + \frac12mv^2$$ Where $u$ is initial velocity, $v$ is final velocity, $h_1$ is the initial height ,$h_2$ is the final height and $m$ is the mass.

But what do I do if the mass follows a parabolic path? Let’s say the initial velocity is entirely horizontal and the final velocity is points along the tangent of my initial and final points. If I solve for the final velocity, will I get the component of the final velocity in the $y$ direction? Should I input the the $x$ component of the initial velocity or its $y$ component ( which is $0$)?

If I plug in the actual resultant initial velocity(which in this case is entirely horizontal) will I get the resultant final velocity? If not, why? The term $\frac12)mv^2$ is kinetic energy. So if I input my actual resultant initial velocity and solve for the final velocity by the M.E. Conservation equation , I should get the resultant final velocity, right? Since kinetic energy cannot have ‘components’ Isn’t kinetic energy $\frac12mv^2$(resultant).

It doesn’t make sense if I say that that a body has, for eg. $70\,\rm{J}$ of Kinetic energy in the $x$-direction and $25\,\rm{J}$ in the $y$-direction? Let’s say I use the conservation of M.E. equation to solve for the final velocity in vertical circular motion. Initial point: $P_1$, final point: $P_2$. If I input my initial velocity as some value ( pointing along the tangent at $P_1$) and solve for the final velocity, assuming I know the other variables, will I get the resultant final velocity pointing along the tangent at point $P_2$ or will I get some component of the final velocity pointing in the direction of the initial velocity at $P_1$???

On YouTube, people were were solving with this equation and getting resultant final velocities for complex paths also. ( like half a loop, many half and quarter loops,etc.etc.) However, in projectile motion, they always input initial velocity in the y direction and solve for final velocity in the y direction. Since the velocity in the $x$ direction remains constant, they find the resultant final velocity by Pythagorean theorem and find the angle theta with the vertical... I don’t understand?

Answer 1

You have a lot of text here that I will not parse through, but I get your confusion I believe.

In the definition of kinetic energy $K=\frac12mv^2$, $v$ is the speed of the object. i.e. $v=\sqrt{\mathbf v\cdot\mathbf v}=\sqrt{v_x^2+v_y^2}$ in two dimensions. So another way to write out the kinetic energy is $K=\frac12m(v_x^2+v_y^2)$. This is not breaking up the energy into components like you would with a vector, since kinetic energy is a scalar. However, if you want to mathematically break up the energy into "energy due to horizontal motion" and "energy due to vertical motion" you could do that I suppose: $K=K_x+K_y=\frac12mv_x^2+\frac12mv_y^2$. However, this is not a usual treatment of kinetic energy, and it might give the false impression that kinetic energy has components.

So, for parabolic motion, your energy conservation equation still holds. $u$ is the initial speed of the object, $v$ is the final speed (based on what you are considering to be "initial" and "final"). $u$ and $v$ are not velocity vectors, just their magnitudes.

Answer 2

Just use Pythagoras to write squared vector magnitudes in terms of components $$u^2 = |u|^2 = u_x^2 + u_y^2$$ $$v^2 = |v|^2 = v_x^2 + v_y^2$$ and note that there is no change in the horizontal component of velocity.