4
$\begingroup$

Let us consider two hills of the same height, with one drastically steeper than the other. Ignoring friction and air resistance, is it true that if we release a ball at the top of the hills, their speed will be the same at the bottom?

I've considered equating the loss of the Gravitational Potential Energy with the gain in Kinetic Energy of the balls (of same mass). However, I feel as though there is something I'm missing there.

The discomfort in my thinking is this. In the steeper hill, though the component of Gravity acting in the direction of motion of the ball is substantially larger than in the case of the less steep hill, would that not also mean that the ball would travel for a much shorter period of time (in turn, giving it less time to accelerate due to gravity)?

I can also be completely wrong and I acknowledge that but any clarifications (with the supporting math behind it :) ) would be greatly appreciated!

$\endgroup$
2
  • 1
    $\begingroup$ Qualitative comment on the next to the last paragraph: the larger vertical component works for less time on the steeper hill. Quantitatively these tendencies compensate to give the same terminal velocity. $\endgroup$ Sep 9, 2022 at 16:02
  • $\begingroup$ Jesus, guys! Every single comment and answer (but not the OP's question) has used the word velocity. Didn't you all learn in kindergarten that what we are talking about here is speed? Downvotes for all of you! $\endgroup$
    – TonyK
    Sep 10, 2022 at 0:43

3 Answers 3

6
$\begingroup$

Since there's no friction and no drag, total energy is conserved. Its value is: $$E=\frac{1}{2}\,mv^2+mgz$$ with $z$ being the altitude. Its conservation between two instants is: $$\frac{1}{2}\,mv_1^2+mgz_1=\frac{1}{2}\,mv_2^2+mgz_2$$ As you can see, m disappears from this computation, and the difference in velocity only depends on the altitude difference. Starting with zero velocity: $$v_2=\sqrt{2g(z_1-z_2)}$$ which is independent of the slope.

$\endgroup$
0
2
$\begingroup$

With no friction and air resistance the ball will slide (not roll) down the incline. The steeper the incline the greater the acceleration due to gravity so the ball will reach the bottom sooner. But as has already been pointed out in the other answer, it's velocity and kinetic energy has to be the same due to conservation of mechanical energy ($\frac{1}{2}mv^{2}+mgh=constant$).

We can also show that the final velocity (and thus kinetic energy) will be the same at the bottom of the incline by kinematics regardless of the angle of incline. First the distance $s$ down the incline as a function of height is

$$s=\frac{h}{\sin\theta}\tag{1}$$

The distance $s$ traveled as a function of time $t$ for constant acceleration given an initial velocity of zero is

$$s=\frac{1}{2}at^2\tag{2}$$

equating (1) and (2)

$$\frac{h}{\sin\theta}=\frac{1}{2}at^2\tag{3}$$

$$t=\sqrt\frac{2h}{a\sin\theta}\tag{4}$$

the acceleration $a$ down the slope is

$$a=g\sin\theta\tag{5}$$

The velocity at the bottom of the slope is

$$v=at\tag{6}$$

Substituting (4) and (5) into (6)

$$v=g\sin\theta \sqrt\frac{2h}{g\sin^{2}\theta}=\sqrt\frac{g^{2}\sin^{2}\theta)(2h)}{g\sin^{2}\theta}=\sqrt {2gh}\tag{6}$$

Showing that the final velocity (and kinetic energy) at the bottom of the slope depends only on the height of the incline.

Hope this helps.

$\endgroup$
1
$\begingroup$

You may also want to calculate this more explicitely. Consider the equation of motion

$$s(t) \enspace = \enspace s_0 + v_0 \cdot t + a_0 \cdot \frac{t^2}{2} \quad ,$$

where $v_0$ is the initial velocity and $a_0$ is the initial acceleration (which will be constant along his path). If the hill is of height $h$, then the distance the ball will have to roll before reaching ground level is

$$s_0 = \frac{h}{\cos \alpha}$$

where $\alpha$ is the steepness of the hill (see Figure). Therefore, in this setting the ground level corresponds to $s = 0$.

enter image description here

The acceleration is dependend on the steepness as well and is of magnitude $g \cdot \cos \alpha$. We assume $v_0 = 0$. Inserting this into the equation of motion, one finds

$$s(t) \enspace = \enspace \frac{h}{\cos \alpha} - g \cdot \cos \alpha \cdot \frac{t^2}{2} \quad .$$

Note the negative sign in the second term. It is because we "start" at distance $s_0$ and want to go down to $s = 0$, so the acceleration is negatively directed. The time $t_1$ needed to reach ground level, i.e. at $s = 0$, is

$$s(t_1) = 0 \quad \Longrightarrow \quad t_1 = \sqrt{\frac{2h}{g \cdot \cos^2 \alpha}} \quad .$$

The velocity after said time is

$$s'(t_1) = g \cdot \cos \alpha \cdot t_1 = g \cdot \cos \alpha \cdot \sqrt{\frac{2h}{g \cdot \cos^2 \alpha}} = \sqrt{2hg}$$

which coincides exactly with the result "Miyase" obtained. It is independent of the steepness of the hill (represented by angle $\alpha$).

In other words, the acceleration on a steeper hill is larger than on a less steeper hill, but it acts on the body not as long as it does on the less steeper hill and so both balls end up with the same speed.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.