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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?

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2 Answers 2

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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.

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  • $\begingroup$ So if I input the initial velocity in the x direction in the conservation of M.E. Equation, will I get the final velocity in the x direction when I solve for it? Or the resultant velocity? What about using conservation of M.E. In circular motion. If I input the initial velocity pointing along the tangent at a point 1, will I get the velocity pointing along the tangent at another point 2? $\endgroup$
    – Curious
    May 15, 2020 at 15:03
  • $\begingroup$ @Curious I am afraid I don't understand what you are asking me. In your energy conservation equation $u$ and $v$ are speeds. Not velocity components. $\endgroup$ May 15, 2020 at 15:04
  • $\begingroup$ True. Here is the problem I was solving : will you pls take a look? m.youtube.com/watch?v=tnp7fEdlndk if I solve using the conservation of M.E. , I then use the initial speed in (1/2)mu^2 which is 4.46 m/s to find v. But I got a different answer $\endgroup$
    – Curious
    May 15, 2020 at 15:10
  • $\begingroup$ I have to use 4.4 m/s as ‘u’ , right? Since speed does not care about direction. Then I should have got v as the final speed( which is always ‘resultant’ since it has no components.) but I didn’t get the same answer as given at the end of the problem. If you do not wish to click on the link: just type “ (Part 2 of 2) An introductory projectile motion problem with an initial horizontal velocity.” on YouTube. Please help... $\endgroup$
    – Curious
    May 15, 2020 at 15:22
  • $\begingroup$ @Curious PSE is not intended to be an "exercise help site". My answer here as addressed your conceptual question of how speed relates to the scalar quantity of kinetic energy. If you are having trouble with a specific calculation there are other sites to go to. $\endgroup$ May 15, 2020 at 15:26
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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.

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