# I am confused in work-energy theorem, Why the ball doesn't go to the same height from which it fall down? (sorry if my grammer isn't correct) [closed]

Here's the figure, ball fall from height of 2 m (at point A) But reaches the height of 1m at point P

• 1. Please do not just put a question into the title, but also write an actual question in the body of the question. 2. Please add enough context to your question so that answerers can understand what the problem is: Where is that picture from? What is it supposed to depict? Is there friction involved? What are the points $A$ and $C$? What is this supposed to have to do with the work-energy theorem? Commented Sep 29, 2023 at 23:11

I'm not sure what your question has to do with the work energy theorem, but the ball doesn’t reach the same height as its original height because it is still in motion (has kinetic energy) at the 1 m height.

For conservation of mechanical energy its kinetic energy plus gravitational potential energy at the 1 m height equals its gravitational potential energy at the initial 2 m height. Just before impacting the ground, when its height is 0 m, its kinetic energy will equal its initial gravitational potential energy.

Hope this helps.

The key is to note that point C is a ramp, not a hill. The trajectory of the ball as it leaves the ramp is dependent very much on the angle of the ramp - a shallow ramp will result in a low, long trajectory, while a steep ramp will result in a high, short trajectory. It should be no surprise that the height of the ball depends on the angle you launch it.

The ball has sufficient energy to roll down a hill and then back up a hill of the same height, stopping at the top with zero kinetic energy. But that's not what happens here, the ball never stops. Since the ball always has some kinetic energy, it can't attain the same height, as that would require an increase in the total energy.

• Why does it not go as a projectile till kinetic energy becomes zero and fall freely later on ? How is it constrained to follow that fixed path without energy loss. Commented Sep 29, 2023 at 19:39
• @Aurelius The ball has horizontal velocity and no way to get rid of it when in freefall. When the ball is on the track, it can slow down horizontally as well as vertically, since the normal force from the track has a horizontal component. Without the track, only gravity acts on the ball, which only has a vertical component. Had the ramp launched the ball straight up, it would return to height A since none of the kinetic energy would be "used up" in the horizontal component of the ball's motion. Commented Oct 2, 2023 at 11:35

Assuming no dissipation (friction, air drag), energy theorem tells you that the total mechanical energy is constant.

At the initial condition, the mechanical energy is completely gravitational potential energy $$$$E_1 = m g h_1$$$$ on the upper point of the trajectory there is the non-zero (if the trajectory is not vertical) contribution of kinetic energy as well $$$$E_2 = m g h_2 + \dfrac{1}{2} m v^2$$$$ so that $$h_2 = h_1 - \dfrac{v^2}{2 g}$$.

As an example, if the section OC was horizontal and the ball kept going on horizontal surface, it would have zero height w.r.t. that plane keeping constant horizontal velocity $$v = \sqrt{2gh}$$.

Your ball would accelerate on its way down and have a velocity: $$v1=v0+gt$$ (that is g right? If no air drag is assumed, it should be. $$v0,t$$ respectively being the initial velocity and the time it takes to reach the bottom) at the bottom. And then change direction due to a collision with that surface and keep moving up till its velocity vector changes direction again due to the deceleration it experiences as a result of gravity. But the velocity it has at that peak would still be greater than $$v0$$ and so (as previous posters pointed out) it cannot have as high a potential energy as it initially did since its kinetic energy is higher than it initially was. If no friction existed, $$mgh+0.5m*v*v = constant$$

Hey here in this question considering no dissipative forces present(friction, buoyancy,air resistance etc...) On applying conservation of mechanical energy...

We get, mgh=mgh'+1/2mv²

See here I have added 1/2mv² as well, as the particle is in projectile motion it is having a velocity parallel to ground which moves it forward so a kinetic energy is present even at its top-most point.

The particle is having a projectile motion( on considering a simple case on Earth of a ball having a velocity and leaving a ramp)...Since we get an extra term of 1/2mv² the final height h' is lesser than the initial height h.

Hope this helps you...