# Kinematics contradicting conservation of energy?

I was having some conceptual difficulty reconciling my intuitive understanding of kinematics with conservation of energy, so I made up a short problem that tested my intuitions:

Suppose I define an initial point to be 20m above the ground. An object leaves this initial point along Path A, which is straight down. Another object leaves the initial point along Path B, which is parallel to the ground. Both objects have the same mass, and the same initial speed of $3 \frac{m}{s}$. What is the speed of each object as it hits the ground?

My calculations for Path A:

$$y_{f} = y_{i} + v_{i}t + \frac{1}{2}a_{y}t^{2}$$ $$0 = 20m - 3\frac{m}{s}t - \frac{1}{2}(9.8\frac{m}{s^{2}})t^{2}$$ $$t = \frac{3 + \sqrt{9 - 4(20)(\frac{1}{2})(-9.8)}}{40} = 0.576s$$ $$v_{f}=v_{i}+at$$ $$v_{f} = -3\frac{m}{s} - (9.8\frac{m}{s^{2}})(0.576s) = -8.64\frac{m}{s}$$ $$s = 8.65 \frac{m}{s}$$

My calculations for Path B: $$y_{f}=y_{i}+v_{i}t+\frac{1}{2}a_{y}t^{2}$$ $$0 = 20m+0t+\frac{1}{2}(-9.8\frac{m}{s^{2}})t^{2}$$ $$t = \sqrt{\frac{20m}{\frac{1}{2}(9.8\frac{m}{s^{2}})}} = 2.02s$$ $$v_{f} = v_{i}+at$$ $$v_{fy} = 0-9.8\frac{m}{s^{2}}(2.02s) = -19.799\frac{m}{s}$$ $$v_{fx} = 3\frac{m}{s}$$ $$s = |\vec{v}| = \sqrt{(19.799\frac{m}{s})^{2} + (3\frac{m}{s})^{2}} = 20.02\frac{m}{s}$$

This result clearly defies conservation of energy. Both objects start at the same initial height, and so have the same potential energy. They both end their paths at the same height, and so end with the same potential energy. But they both have different velocities, and so different kinetic energies.

The thing is, this is exactly what my intuitions would predict. How much an object accelerates in total is just a function of the time it spends accelerating. The object traveling along Path A has a fast downwards velocity, so it only has a short time to accelerate. Its final speed is just the sum of its initial speed and however much it accelerates in that time.

The object traveling along Path B has no initial downwards velocity, and so has plenty of time to accelerate. The total final speed of this object is the vector sum of its initial velocity and however much it gained while accelerating downwards, which is a much larger number than the total acceleration of the first object.

So what's going on? The calculations break conservation of energy, so I must be doing something wrong. And even if there's a flaw in my calculations and the numbers actually do come out the same, my intuition still says otherwise.

• Where does the division by $40$ come from in your first calculation? (When you compute a root of the quadratic equation.) Feb 19 '16 at 0:32
• It seems like the fact that it wasn't a quadratic equation in standard form, with the squared term first, made me think that the bottom should be 2(20). I recomputed it and came up with 2.35s, which is longer than the time for Path B, which doesn't seem right. And doing the next calculation gives me a speed of 26m/s, which is still different. Feb 19 '16 at 0:40

The two particles have the same energy. It's easier to use the equation $$v^2 = u^2 + 2a\Delta y$$

For particle A, all motion is in the y-direction, so $$v^2 = v_y^2 = 3^2 + 2(9.8)(20)$$ $$v = 20.02 ms^{-1}$$

You went wrong in your application of the quadratic formula. $$t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ $$a = -0.5\cdot 9.8 = -4.9$$ $$b = -3$$ $$c = 20$$

Taking the negative sqrt solution and correcting $2a$, $t=1.74 s$

Plugging this into the $v = u + at$ formula yields $20.05 ms^{-1}$ for me, and the difference is just rounding error.

For particle B, your calculation was correct.

So now we've clarified that conservation of energy holds, but now to answer why.

The thing is, this is exactly what my intuitions would predict. How much an object accelerates in total is just a function of the time it spends accelerating.

Correct, but the kinetic energy of a particle is not dependent on how much it has accelerated. It only depends on the instantaneous velocity of the particle.

Here the final velocity in the y-direction for particle A will be higher than that of particle B. The final velocity in the x-direction for particle B will be higher than that of particle A (equally so, in fact). Calculating the final speed of both would yield the same answer, and so the kinetic energy will remain equal.