Why does the incline angle not affect how high a launched object will slide up a frictionless ramp?

Question

I am seeing a problem with the solution given in this book. How did the height of the box have nothing to do with the incline of the ramp? Intuitively it would seem the higher the incline the higher the box would go. Especially since there is no friction to be accounted for. I thought about it further and realized that if I were to consider solely the vertical component of the initial velocity (i.e.$10\sin (37^\circ)$) then the kinematic equations would give us $$h = \frac{(10\sin(37^\circ))^2}{2(9.81)} \approx 1.85.$$ Which is quite different than what the book gives. So what is bridging the gap? I in fact agree with their application of the conservation of mechanical energy but I am not intuitively convinced.

Any clarification would be appreciated.

EDIT: For anyone interested... this is how I personally convinced myself that the angle $\theta$ did not matter. First I notice that the height of the box can be exprerssed as $$h=|d|\sin \theta$$ where $d$ is the distance traveled by the box along the ramps incline. But by newtons second law one can see that the boxs' acceleration along the ramp is $$a_x = g\sin \theta.$$ But by the kinematic equations we see that $$v_f^2=v_0^2+2a_xd \Longrightarrow |d|=\frac{v_0^2}{2a_x}=\frac{v_0^2}{2(g\sin \theta)}.$$ Plugging this into our equation for $h$... $$h=|d|\sin \theta = \left(\frac{v_0^2}{2(g\sin \theta)} \right)\sin \theta =\frac{v_0^2}{2g}$$ hence rendering the incline angle superfluous.

I greatly appreciate everyones help.

Answer 1

The fact that the incline is frictionless allows us to use the conservation of energy approach, which is telling us that 'based on this amount of initial kinetic energy, the box will go this high'.

Take the instance where the 'incline' is $90^\circ$. The box is going straight upwards with initial velocity 10m/s. How high does it go? We can use kinematic equation $v_f^2-v_i^2 = 2ad$ where $v_f=0$ to find that $d=\frac{v_i^2}{2g}=5m$ just like it shows in the first part of the solution.

So what changes with the angle? The distance travelled along the incline depends on the angle. So again with the purely vertical approach, we find the height is 5m, and the distance traveled along the incline is the same 5m. But then take the incline at $10^\circ$. The box reaches 5m of height still, but has traveled along the incline a whopping 29m!

Answer 2

The reason that your application of the kinematic equations didn't work to find the height is because gravity is not the only force acting on the box in the vertical direction. The normal force $N$ from the ramp also has a vertical component, and so the vertical acceleration is not solely due to gravity.

To find what the vertical acceleration actually is, we note that we must have $a_y = a_x \tan \theta$ (with $\theta = 37^\circ$); this comes from the condition that the acceleration vector is tangent to the slope. Moreover, from the horizontal and vertical components of Newton's Second Law we have $$ -N \sin \theta = m a_x \qquad N \cos \theta - mg = m a_y $$ (with $x$ and $y$ increasing to the right and to the top of the page, respectively.) This is a set of three equations in three unknowns $N$, $a_y$, and $a_x$, which has as its solution $$ a_y = -g \sin^2 \theta, \qquad a_x = -g \sin \theta \cos \theta, \qquad N = m g \cos \theta. $$ Of note, the acceleration in the $y$-direction also gets a factor of $\sin^2 \theta$, which cancels out the factor of $\sin^2 \theta$ obtained from squaring $v_y$ and yielding $y_f = y_0 + v^2/(2g)$, just as we expected from energy conservation.

We can also see that the effects of the normal force are larger when the angle $\theta$ is small. For very gradual slopes, the normal force nearly cancels out the vertical acceleration, meaning that $a_y$ is small; but the initial velocity in the $y$-direction is also small. For very steep slopes, the normal force is much smaller, meaning that $a_y$ is closer to $g$; but the initial velocity in the $y$-direction is also bigger. The two factors (bigger/smaller initial vertical velocity and bigger/smaller vertical acceleration) cancel each other out, with the result that the final $y$-coordinate is independent of the angle.

Of course, this result is much easier to see from energy conservation! In fact, many (all?) problems like this can be solved, at least in principle, by the use of Newton's Laws. Energy conservation (and momentum conservation, for that matter) can be thought of as a "cheat code" for Newton's Laws that allows us to avoid hard work like I've done above.

Answer 3

The box is slowed down by the component of the weight parallel to the plane. If the plane is steeper (angle is larger) the acceleration is larger and the box stops after a short time. If the angle is small the acceleration is small and the box travels longer before it stops. But, on the other hand, you have to travel farther on a low angle plane to get to the same height. In the end, the math shows that these two effects compensate and you get to the same height no matter what the angle is. Of course, this is true if there is no friction. With friction, the height depends on the angle.

Answer 4

Your question is mostly rooted in what your intuition is telling you, so this answer tries to respond with an intuitive explanation - not a formulaically precise one.

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On a frictionless incline, the only braking force in play is gravity, which points straight down. This leads us to only need to consider the vertical component of an object in freefall (when launched at a given inclination) when answering a question about the altitude it will achieve.

This is presumably why you're assuming that the horizontal component can simply be ignored, but you've glossed over the fact that there is a (frictionless) slope here that is changing the behavior.

Because there is a slope involved, the entire velocity (horizontal and vertical) needs to be overcome before the box will drop back down. Therefore, this can be considered to be a battle between gravity and the full velocity of the box.

Intuitively it would seem the higher the incline the higher the box would go.

Your intuition is correct for an object in freefall (which was launched at an inclination), as it can happily maintain its horizontal component while at the same time its vertical component is being altered by gravity. The softer the inclination, the smaller the vertical component is (assuming the same total velocity regardless of inclination). The smaller the vertical component is, the quicker gravity overcomes it, leading to a lowered altitude compared to when it was launched at a higher inclination.

In the case of a launched object (no slope), the horizontal component is completely ignored as it remains unrelated to the vertical component. In this scenario, you do indeed only consider the vertical component.

Totally informal, but hopefully helpful way to think about it: even if gravity wins the battle and counteracts the vertical component, but the box is still moving sideways; the slope will turn that sideways movement into upwards movement, which then starts battling gravity again.
This is why the horizontal component does contribute towards the vertical component, when there is a slope that converts one into the other.

Because the slope is frictionless, that conversion from horizontal into vertical happens at 100% efficiency (no losses whatsoever), which is why we can happily assume that all of the horizontal velocity will be converted into the exact same amount of vertical velocity. Therefore, the box only comes to a halt when both vertical and horizontal velocity reach 0.

In essence, the slope ties the horizontal and vertical components together in a way that they live and die together. If you see this as a battle of forces it is a battle of gravity in one corner, and the combined horizontal and vertical forces (i.e. the total velocity of the box) in the other corner.

The only reason you'd need to consider the inclination is if you wanted to separate the horizontal and vertical components; which you simply don't need to do here.

Answer 5

The relation that you have written $$h = (v)^2/2g$$ is itself derived from conservation of energy but in that it is assumed that horizontal component of velocity to be zero. You can try the derivation yourself.

$$V = \sqrt{(u^2 + v^2)}$$ where u and v are horizontal and vertical components respectively and so V is total resultant velocity. Now applying energy conservation i.e. KE while throwing equals potential energy at highest point. $$ \frac{mV^2}{2} = mgh$$ Substitute V in terms of u and v and take u = 0, you will end up with the same equation as above. But in the case of incline, you have a significant u as well and hence the equation above for h can't be applied. Angle of incline simply doesn't contribute to height of ascend because of absence of friction.