Challenge Questions

I plan to reply in earnest later when not being paid to be productive. Potential and Kinetic Energy equations would be a hard sell to my boss. But until then perhaps the poster would define what kind of vector the velocity of 5 m/s is? Directly Up the plane, or down the plane? Or some angle inbetween? Also i am assuming the speed of 12 m/s is directly down the plane as it is referred to as a speed. Are we neglecting friction? and rotational inertia?

At the end of the plane its speed is 12 m/s.

Maybe i'm missing something, but isn't that the same as the speed at the end of the ramp...

i'll go out on a limb here, but i believe the question answeres itself, which leads me to think i have missed something.

-12m/s

oops, i did misundersand the question... I should read more carefully next time.

P.S. In view of previous contributions: in saying that frictional losses don't change things, I have been assuming that the 5 m/s rock is already rolling and follows the same path. In the absence of friction, the path doesn't matter so long as the final height is the same and the other conditions I mentioned still apply.
If the 5 m/s rock is not already rolling there would be additional frictional losses; the initial kinetic energy would also not include any rotational kinetic energy - both effects would normally reduce the final velocity (again between 12 and 13 m/s).

Curiously, if the coefficient of sliding friction between the rock and the plane exactly matches the frictional value needed to keep it rolling (starting from rest), that too could result in a final velocity of 13 m/s.

You get a GA from me Fyz.

I tackled it much more elaborately but got the same result:-

Assuming a rolling sphere ( Rotational Inertia=2/5mr² ) starting from 0 velocity

Potential Energy = Kinetic translational + kinetic Rotational

mgh = ½mv² + ½Iω²
mgh = ½mv² + ½(2/5mr2)v²/r² = 7/10m12²

Now if initial velocity is 5m/s, we have

Same Potential Energy + Initial kinetic(T+R) = Final Kinetic(T+R)

7/10m12² + 7/10m5² = 7/10mv²

Therfore 12² + 5² = v²

V ⁼ 13m/s

The speed gained due to gravity would remain the same. 12 m/s plus the initial speed 5 m/s = 17 m/s.

Mr. Gee

That was my initial thought as well, but the math does not bear it out. Two different methods have each arrived a the same answer of 13 m/s.

For bonus points, can anyone explain why the above solution is incorrect?

Because Energy is proportional to the square of the velocity, You cannot simply add the individual velocities 12m/s to 5m/s to get an answer of 17m/s.

You must add 12 squared to 5 squared and then get the square root of the result.

12^{2 +} 5² = 144 + 25 = 169.

The square root of 169 is 13.
Thus, when the initial velocity is 5m/s, the final velocity now is 13m/s.

In general, if the initial velocity was u m/s, then the final velocity would be:-

v = √(u² + 12²)

N.B.This simplified math does rely on the fact that both translational kinetic energy and rotational kinetic energy are proportional to the translational velocity squared.

Sorry Folks, this is my post. I logged in on different computer.

I think Fyz already explained why 17m/s would not be correct, by showing that 13m/s would be correct.

Another way of looking at it: Suppose the ramp slope is such that the acceleration is 6m/s². Time on the ramp (time to accelerate to 12 m/s) would be 2 seconds. Then the ramp would need to be 12 meters long, (2 seconds at an average speed of 6m/s).

For a starting speed of 5m/s: V² = V₀² + 2AX. So V² = 25 + 2x6x12 = 169. Therefore, V = 13. So, at an average speed of 9 m/s the 12 foot trip takes 1 1/3 seconds. To check: the initial speed (5 m/s) plus the increment in speed (1 1/3 x 6m/s = 8 m/s) equals the final speed, 13 m/s.

In the second case, the stone is on the ramp for too short a period to accelerate to 17 m/s.

But the real explanation is Fyz's , I think. We know the change in kinetic energy resulting from the ride down the ramp must be the same in both cases. So in one case, the kinetic energy goes from 0 to 144 (times a number) and in the other case it goes from 25 to 169 (times a number).

That's probably to muddy to qualify for a bonus point.

It's interesting how we all look at this in different ways. I did this (imagining an extra length of starting slope) ;

v = √ (2gh) to get h= 7.339 for 12 m/s

h= 1.274 for 5 m/s

Total h = 8.613 → → v= 13 m/s

Not as nice as Fyz' equation juggle , but it leaves me slightly puzzled. 38% of velocity (5m/s) is reached after 15 % of the drop. I seriously need to go back to 101 !

Once a rolling rock comes down from Point A to Point B only due to gravity, the net kinetic energy gained would be the same (between A and B) irrespective of whether it starts from rest or an initial velocity.Let us first calculate the Kinetic energy of the mass of 10-kg travelling with a velocity of 12m/s. To this we add the initial Kinetic energy (due to velocity of 5m/s.) Then we get the new Kinetic energy (1/2*m*V*V). Find the new velocity by simple mathematics -which will be 13-m/s

It does not mention the steepness of the plane. So I imagine the acceleration a = 10

Let Ep= potential energy, a the acceleration, h the height)

Then Ep = m*a*h = 10*10*h = 100*h

Let Ek be the kinetic energy, v speed

But Ek = .5*10*v*v = 50*12*12 = 720

So 720 = 100*h

So h=7.2

Let u be the begin speed, v be the end speed

With an initial velocity of 5 we have v*v = u*u +2*a*h

So v*v = 5*5 + 2*10*7.2 = 25 + 144 = 169

So v = 13

The Answer is 13m/s and here is a verification of it:

From the first State Vi=0m/s and Vfinal=12m/s we can compute, using the conservation of energy theorem, the height where the rock was released from and it was found to be equal to 7.34m.

1/2mV^2=mgh => h=(1/2V^2)/g

Going to the second state where Vi=5m/s. The Sum of Kinetic energy and Potential energy at the top of the ramp is balanced with the sum of Kinetic energy and Potential energy at the bottom of the ramp(where potential energy is assumed to be = 0)

1/2m(Vi^2)+mgh= 1/2m(Vfinal^2)

=> Vfinal = 13m/s

Interesting but I suspect this is a trick equation. As there are no specifics, I too have to assume there is no friction and that the acceleration from zero to 12 is a result of the acceleration of gravity. Then it should be safe to assume that the rock will increase in speed by 12 as it travels that same length irreguardless of the initial speed. I proport that the answer is simply 5 + 12 = 17 m/s.

Change in speed = acceleration x time. Starting at 5 m/s means it takes less time to go down the ramp, so the speed increase should be less than 12 m/s. (And, as noted in ##8, 11, 14, this solution would also give a final change in the kinetic energy that exceeds the change in potential energy)

Ok, I failed to consider the time element, but the acceleration due to gravity is still constant. So yes the speed will be below 17m/s.

I do not understand why somebody up above was trying to hit in the Pythagorean theorem.

What that was intended to say is that the numbers 5, 12 were apparently chosen deliberately on the basis that sqrt(5²+12²) is a whole number - suggesting that the originator of the challenge thought about this, and intended the question to be interpreted as being about energy conservation.
Because such numbers allow simple construction of right-angles, and this was their first known specific application, geometers sometimes called them "Pythagorean numbers". However, I admit it was a slip-up on my part to use that term without qualification, because number theorists use this term with a different meaning - i.e. numbers that are equal to the sums of their factors (and this is now the most common meaning for the term).

....numbers that are equal to the sums of their factors....

6, 28, 496, 8128 etc are perfect, but where did they come into all this ????

An accident of linguistics - I wrote "Pythagorean numbers"; unfortunately, since I first learned about such things, this has become ambiguous, and the current term that uniquely defines what I meant is "Pythagorean triples". (But Wikipedia and MathResource still accept the original name with its original meaning)

The problem is that Pythagoras and his followers had some mythical views about number, and it seems that any one of these quasi-religious associations is currently co-opted into the category of Pythagorean Numbers. Once one of these became semi-accepted, the situation has grown like Terpsichore, and virtually any number or group of numbers with properties relevant to the Pythagoreans has laid claim to being "Pythagorean". For example:
All perfect squares
Perfect numbers
Amicable number pairs
The Tetraktys
Pythagorean quadruples (e.g. 1,2,2,3)
. (In spite of the absence of evidence that Pythagoras or his followers had direct interest - and, curiously, I don't believe that the slightly less uninteresting "Cubic Quartets" have been so classified)

I didn't think Pythagoras was directly associated with "perfect numbers", though it's unlikely he even discovered Pythagorean Triples (Maybe the Berlin Papyrus, about 2000 BC). I must away to my Tarantella class ............

The (3, 4, 5) triple was of course well known and in use for right angles well before the time of Pythagoras. Whether he himself had anything to do with adding additional triples or with the generalisation of the sums of squares (or any other interest in Pythagorean triples) is at best dubious, as is any personal involvement with any other of the geometric or numerical discoveries that were attributed to him. In reality, anything called Pythagorean can at most be traced to those who followed his number-mystical philosophy - and even then some of it is likely to have been co-opted from other sources. In my personal view (and it is only an opinion), the aspect of his numerical reputation where he is most likely to have had involvement is the harmonic ratios of musical intervals (in modern (=medieval) terminology: octave = 2:1, fifth = 3:2, fourth = 4:3*). The reason I give credence to this is that it (and the related Tetraktys) feature so prominently in the mystique. The ideas of perfect numbers have long been ascribed to the Pythagoreans, but (SFIK) it is quite possible they were later bolts-on ascribed to them by the neo-Pythagoreans. Either way, the term "Pythagorean number" has over the last few years become hugely ambiguous, and I should have used "Pythagorean triple" had not old habits intervened.

N.B. The most appropriate (albeit partial) web-summary of information on the Pythagoreans I could find due to Drakos.

*but apparently having no interest in including major and minor thirds 5:4 and 6:5 respectively

P.S. I didn't know you were a fan of sting

Oh no.....it's back to vegetables. The whole bean story is worth a thread (or maybe even a string) in it's own right.

Don't tempt me into A MidSumner Nights Dream ! The Rye humour might be too much.

You are forgetting that the higher initial velocity gets you to the end of the ramp much quicker, so there is less time for acceleration to increase the velocity.

distance = time x average velocity

so s = t*(12/2) = t*((5+Vf)/2)

which gives Vf = 7

Final velocity is 7 m/s.

Now you're talking - start off quicker to end up slower!

I see your point, but I doubt your logic - will it really be slower than a zero velocity start? It isn't intuitively obvious how this could be...

I think I see where all the '13' people are coming from.

12M/s is the end speed resulting from the drop in height, Not the acceleration.

If you think of it as a dropping ball. the accelleration is constant, the distance the ball travels each second increases therefore the time it takes to go a measured distance decreases as the time increments.

By starting at 5 m/s you move down the drop path, giving less time to go a measured distance. the distance is set by the length (height) of the ramp. As the ramp height (in time) shortens (because your going faster) the acceleration period decreases, giving less acceleration.

Thus the increase in speed for a given distance is less the faster your initial velocity.

Ther rock isn't slowing down, nor is the acceleration decreasing. The acceleration period decreases (BUT WILL ALWAYS BE POSITIVE)!

As an example consider a more extreme starting position of 100m/s velocity. The piddling amount of time available (acceleration period) will add stuff all to the speed even though the acceleration rate is unchanged.

I hope this helps clear things up

You've noted that the times will be different, but, then given them both the same name: "t"; and, then cancelled them to give your result.

thanks, you are correct; i should have used a t1 & t2 and introduced another relationship to account for their inequality.

so now, s = T1*(12/2) = T2*((Vf+5)/2)

but, since t = v/a in general, T1 = 12/a & T2 = (Vf-5)/a

so (12/a)*(12/2) = ((Vf-5)/a)*((Vf+5)/2)

and since acceleration really is constant, yields...

Vf = 13

which matches the energy-based equalities result

I suppose that one could argue that if ever there was a demonstration that there is a need for the anonymous post option, this post might be it. Anonymity allows guests to go where others fear to tread, promoting thinking well "outside the box".

(Dear guest: please understand that I, and most others here, would not poke fun at your answer, tempting though it may be, if you had a name. As a guest, you have no reputation to harm. I'm sure you can see why your post could cause some chuckles: you are no doubt aware of the basic math rule that you should check your answer for plausibility. You are probably also aware that entire NASA missions have gone haywire as a result of things as simple and obvious as swapping metric units for ... hmmm, let's just call them "Imperial" units so we can blame it on the Brits. Everyone makes silly mistakes -- I could write a book about my own.)

"let's just call them "Imperial" units so we can blame it on the Brits"

hmmm: I think we Brits were forced to abandon imperialism some time ago - both as regards empire (though some still harbour delusions) and as regards engineering units. For example, we even quote the pitch of IC package pins in metric units (2.54-mm, 1.016-mm, etc)

Could the reason that the USA persists be a desire that one will perpetuate the other?

"we even quote the pitch of IC package pins in metric units (2.54-mm, 1.016-mm, etc)"

Metric units? Is that an improved version of the Wentworth sizes on older English vehicles?

Whitworth!

Worth-what?

I am going to stick my neck out a little here and agree with you. Because I have decided that if the rock was pushed fast enough down the ramp, it would stop. Is that right?

Probably, because if it were pushed down the ramp at speed = c, its mass would increase to = ∞ at which point the ramp would collapse, and movement on it would cease. Might be some movement beneath it, however...

A different meaning for "at the end of the ramp"?

On the other hand: taking the most naive pedantic view, and assuming a gradient of more than 1/1000, the downward curvature of the motion would be very small*, so the ramp should not collapse. It would be attracted towards the rock - but only after the rock had passed it, so that would not affect the rock.
On the other hand, we are told the rock is rolling, so it would disintegrate at much lower speeds than light; also, we are not told there is no atmosphere, so it might well burn out; but as for stopping before the end of the ramp, I think not.

*Operating over much greater distances, the vastly larger mass of the sun bent starlight by only 1.7 arc-seconds in Eddington's experiment

Some words of warning

This is a fairly simple challenge on the face of it, and I am quite sure the intended answer is 13m/s as many have obtained by various methods.

All "correct" answers so far are incomplete however. They all end up by concluding V=√169 which of course has two roots; +13 (the "right" root) and -13 (the "wrong" root). To be thorough, one must explain away the negative root (which implies the boulder is rolling up the ramp). Remember Velocity V has direction; Energy doesn't

This may seem to be somewhat pedantic for this question, but more than one time I have been guilty of choosing the "wrong" root and got in trouble.

True, but the question also asks for speed (scalar) not velocity (vector) so I think we're pretty safe.