Elementary physics: Which body will reach the wall first?

[wilvar]

The two bodies shown at the picture are moving along the surfaces marked as A and B with no friction and no rotation and with the same initial velocity. Which one will arrive first at the wall? We may assume that the second body never looses contact with surface B. Note that this question was posed to students with no knowledge of integrals or derivatives. Any well explained solution is most welcome.

 

[robb_]

Is this in a g field?

 

[berkeman]

Moved to homework forums. Welcome to the PF, wilvar, but please be careful not to post homework or coursework in the general forums.

What are your thoughts so far on this?

 

[berkeman]

BTW, an easier way to picture it is for the bottom path to be down at a 45 degree angle for the first half, and then back up at a 45 degree angle for the 2nd half. It makes visualizing the velocity components easier. You can also do the full kinematic equations easier if you want to check the intuitive answer that you get pretty quickly with this view.

 

[arunma]

Since the second object doesn't end up gaining any net kinetic energy through the trip, it seems that they'll reach the wall on the far end at the same time.

 

[Delzac]

Nope they will not reach the far end at same time.

Think Conservation of energy. Think about their Kinetic energy at different points in time.

 

[wilvar]

First of all this happens in a normal g-field. This is not homework or coursework. Its a multiple choise question from a national competition of Physics which deals with basic physics laws and i don't like the way it is written. Now the two bodies will not reach at the same time to the wall since the second one moves for a period of time with higher velocity than the first one. The idea of simulating the movement with inclined planes is of course already been tried out and leads no where. The real problem is to proove with elementary maths (no derivatives or integrals) that while the second body moves through the curved hole is gaining velocity at the horizontal axe of its movement, thus it is completing the journey in less time than the first one. The problem is quite complicated since one has to take into account that body B does not loose contact with the surface and doesn't bounce at the bottom of the hole. These pose restrictions for the allowed values of v0 and the radius of the hole which are not mentioned in the question, so how can we give only one solution? This is why I'm confused as far as the solution goes.
P.S. Even with differential calculus the question is hard to be answered. Τhe most simple explantion i can give goes like this: at the bottom of the hole from conservation of energy one depicts that the horizontal velocity of the second body will be equal to Sqr(u0^2 +2gh), where h is the depth of the hole, hence at the bottom of the hole ux > u0. Since velocity is a continuous function this means that while moving in the curve ux remains larger than u0, so the second body moves horizontally faster than the first one and reaches the wall first.What do you think about that argument?

 

[turdferguson]

This is similar to the argument that a boat traveling in still water will travel faster than a boat that travels the first half of the trip upstream and the second half downstream.

In both cases, let total distance equal 2D. Let v be the speed of the boat and c be the speed of the current. Boat 1 is traveling a distance of 2D at a velocity of v. Its time is then 2D/v. Boat 2 is traveling a distance D at velocity v+c and a second distance D at velocity v-c. Its total time is then (D/v+c) + (D/v-c). Finding a common denominator results in (Dv-Dc+Dv+Dc/v²-c²) or 2Dv/v²-c²

Now its clear that Boat 1 in still water's time of 2Dv/v² (2D/v) is less than Boat 2 in the current's time of 2Dv/v²-c². The denominator is smaller and the time is greater. On average, it is traveling slower

QED with no calc necessary

 

[HallsofIvy]

Nope they will not reach the far end at same time.

Think Conservation of energy. Think about their Kinetic energy at different points in time.

Think "velocity is a vector quantity". B's horizontal velocity is not changed by going down and then back up the depression. Since the time to reach the wall depends only on horizontal velocity both will reach the wall at the same time.

 

[wilvar]

You are wrong. The vx velocity while the body moves along the curve is larger than v0, (increasing to an amount while going down and decreasing to its original value v0 while going up). So the second body travels the horizontal distance of the gap faster than the first body.

 

[nrqed]

Think "velocity is a vector quantity". B's horizontal velocity is not changed by going down and then back up the depression. Since the time to reach the wall depends only on horizontal velocity both will reach the wall at the same time.

It's true that after having come out of the depression, the horizontal velocity will be the same as what it was before getting in the depression. But while it is in the depression, the horizontal velocity will obviously be larger than when it is on the flat section. Therefore, the ball going through the depression will reach the end first.

Patrick

 

[Delzac]

Anyway even if we know that horizontal velocity is larger in the depression, how can we be sure that the extra length(added by the depression) is not significant enough to cause B to reach the wall at the same time as A or even later?

 

[AlephZero]

Good question - the answer depends only on conservation of energy, Newtons 2nd law, and the fact the laws of Newtoninan mechanics are symmetrical when time is reversed.

First consider the motion from the start to the bottom of the depression. The horizontal component of force on the body is always >= 0 (measuring positive from left to right). Therefore the horizontal speed is >= the speed on the straight track.
Therefore the particle will reach the bottom of the depression sooner.

You know the particle does not leave the track, therefore there is no energy loss by impact when it "bounced". By conservation of energy, the particle will have the same energy before and after the dip, and therefore the same velocity.

By symmetry, consider the reversed motion of the particle. By the same argument as before, the time from the centre of the dip to the wall is less than for the straight track.

NB this argument uses the symmetry of Newtonian mechanics with time-reversal - it doesn't depend on the shape of the track being symmetrical about the lowest point.

 

[Doc Al]

Usually this problem is posed with the depressed path having a long, flat stretch at the bottom of the depression. In that case it's easy to answer, since the particle taking that path has a higher speed for a longer period of time.

But this case requires more thought (unless I'm missing something obvious), since the horizontal portion of the depression is negligible.

But while it is in the depression, the horizontal velocity will obviously be larger than when it is on the flat section. Therefore, the ball going through the depression will reach the end first.

That's not obvious to me. At any point in the depression, the speed of the particle is:
$$V = \sqrt{V_0^2 + 2gh}$$

But the horizontal component is only:
$$V_h = V \cos\theta$$

Depending upon the angle and depth, V_h may well be < V_0.

The problem is more interesting than it looks. (Of course, I haven't had my coffee yet this morning.)

 

[AlephZero]

At any point in the depression, the speed of the particle is:
$$V = \sqrt{V_0^2 + 2gh}$$

But the horizontal component is only:
$$V_h = V \cos\theta$$

Depending upon the angle and depth, V_h may well be < V_0.

The body is moving along a surface and does not lose contact with it. That means V_h cannot be less than V_0

If it were sliding along a wire, the situation is different and you are right, the time of travel could be slower for a deep depression. See also my previous post.

 

[Doc Al]

The body is moving along a surface and does not lose contact with it. That means V_h cannot be less than V_0

Why is that? Do you disagree with the formula I provided for velocity (which is just energy conservation).

If it were sliding along a wire, the situation is different and you are right, the time of travel could be slower for a deep depression.

I'd think it might be slower for a shallow depression

 

[AlephZero]

Of course your velocity formula is right.

Because the object is sliding on a surface, the contact force must be normal to the surface and can only act outwards from the surface to the object. While the object is traveling down the slope there is no way there can be horizontal component of force that can reduce the horizontal component of velocity. It can only stay the same or increase.

To deal with the motion up again, consider the situation with time reversed.

If the object were sliding along a wire, the situation is different because there can be a horizontal component of force in either direction. For example if the wire went vertically downwards at some point, clearly the horizontal velocity would be zero at that point.

If the shape of the surface falls away faster than a parabola (whose shape depends on the initial speed) then the object will lose contact with it.

 

[Doc Al]

Of course your velocity formula is right.

Because the object is sliding on a surface, the contact force must be normal to the surface and can only act outwards from the surface to the object. While the object is traveling down the slope there is no way there can be horizontal component of force that can reduce the horizontal component of velocity. It can only stay the same or increase.

The tricky part is the transition from the horizontal to the depression.

Consider a particle with initial speed 10 m/s and a depression with slope of 45 degrees. When the particle drops 1 m below its original height, what is its velocity? And what is the horizontal component of that velocity?

 

[nrqed]

Usually this problem is posed with the depressed path having a long, flat stretch at the bottom of the depression. In that case it's easy to answer, since the particle taking that path has a higher speed for a longer period of time.

But this case requires more thought (unless I'm missing something obvious), since the horizontal portion of the depression is negligible.

That's not obvious to me. At any point in the depression, the speed of the particle is:
$$V = \sqrt{V_0^2 + 2gh}$$

But the horizontal component is only:
$$V_h = V \cos\theta$$

Depending upon the angle and depth, V_h may well be < V_0.

The problem is more interesting than it looks. (Of course, I haven't had my coffee yet this morning.)

It's obvious that v_h will increase if you look at the forces. The normal force cancels the component of the gravitational force that is perpendicular to the surface. That leaves the component of gravity parallel to the surface. This will necessarily increase both v_x and v_y.

The issue of increased distance is more subtle and I admit that I have to think about that a bit more.

Regards

Patrick

 

[AlephZero]

The tricky part is the transition from the horizontal to the depression.

Consider a particle with initial speed 10 m/s and a depression with slope of 45 degrees. When the particle drops 1 m below its original height, what is its velocity? And what is the horizontal component of that velocity?

The question says "We may assume that the second body never looses contact with surface B" so that example isn't covered by the question.

If the initial velocity was 10m/s then the initial depression can't be "steeper" than the curve the the particle would make in free flight: x = 10t, y = -4.9t^2. Otherwise it would lose contact.

(I'm not being very precise here - the real constraint is the curvature of the depression at any point as a function of the velocity of the particle at that point, but the exact details are irrelevant to the argument).

Finding the speed by conservation of energy is fine, but it doesn't tell the whole story. Going back to your $v_h \cos \theta$ argument, the condition that the particle stays on the surface will give a constraint on the possible values of theta (it might be an interesting calculus of variations problem to work out the constraint explicitly) and the consequence will be that$v_h \cos \theta > v_0$.

If the horizontal velocity changes, there must be a horizontal force to change it. As the previous post said, if you draw a free body diagram it's clear that on the way down the depression, you can't have a force that reduces the horizontal velocity.

 

[AlephZero]

The issue of increased distance is more subtle and I admit that I have to think about that a bit more.

Hopefully, more thought will show the increased distance is irrelevant. The time to get to the other side depends only on the horizontal velocity and the horizontal distance (which does not vary).

 

[BTM1]

HallsofIvy's answer is the correct one. Whatever horizontal acceleration is picked up going down the hill is canceled by the deceleration going back up. The shape of the depression is irrelevant as long as the elevation is the same before and after the depression. There is simply no net horizontal force.

 

[AlephZero]

Once consequence of all this which I do find counter-intuitive is: suppose the depression is starts as a parabola, so the particle just skims down it without losing contact. Then there is a small radius curve at the bottom and a parabola going back up.

In that case the time taken would be almost identical, because the only part where the horizontal velocity was faster would be the small curve at the bottom of the depression. But, assuming the diagram in the question means the curve is sufficiently smooth, there must always be a part at the bottom where the horizontal velocity increases, because at the lowest point, the depression IS horizontal, and Doc Al's formula correctly says the horiz. velocity is greater at that point.

 

[Doc Al]

My bad

The question says "We may assume that the second body never looses contact with surface B" so that example isn't covered by the question.

You're right. My example is bogus and is not physically realizable without the particle losing contact with the surface.

Whatever horizontal acceleration is picked up going down the hill is canceled by the deceleration going back up. The shape of the depression is irrelevant as long as the elevation is the same before and after the depression. There is simply no net horizontal force.

It's true that once the particle leaves the depression its acceleration is again zero, but that doesn't change the fact that its horizontal speed is faster while in the depression.

 

[wilvar]

when the body enters the curve we can assume that: mgcos(θ) - N = mu^2/R, where R is the radious of its curved trajectory. Since it doesn't loose contact must be: u0^2 < gRcos(θ). From conservation of energy: u^2 = = u0^2 + 2gR(1 - cos(θ)) and ux = ucos(θ) for the horizontal velocity. Now its easy to proove that while in the curve: ux > u0. I'm looking for a simpler solution than this guys.

 

[dextercioby]

There's no net force acting on the moving body in the horizontal direction for any of the 2 cases presented. Therefore, $v_{0}$ is conserved and the time of arrival in the second case is equal to the one in the first.

 

[AlephZero]

There's no net force acting on the moving body in the horizontal direction for any of the 2 cases presented.

True, in the sense that $$\int f_h\,dx = 0$$

Therefore, $v_{0}$ is conserved and the time of arrival in the second case is equal to the one in the first.

That doesn't follow. Suppose f_h = 1 for time 0 to 1, f_h = -1 for time 1 to 2.

Then $$\int f_h\,dx = 0$$ and $$\int f_h\,dt = 0$$ but the mean horizontal velocity in the dip is $$v_h + \frac{1}{2m}$$ where m is the mass of the particle.

 

[dextercioby]

Due to the symmetry of the problem, the average of the vertical component is 0. Therefore the average of the velocity vector in the curved region is $v_{0} \vec{i}$.

So i don't understand how the "mean velocity" could be the one you wrote...

 

[Doc Al]

There's no net force acting on the moving body in the horizontal direction for any of the 2 cases presented.

What do you mean? While the particle slides along the horizontal section, there is no net force on the particle. But there's certainly a net force on it as it slides down (and up) the depression--it accelerates.

 

[AlephZero]

Due to the symmetry of the problem, the average of the vertical component is 0. Therefore the average of the velocity vector in the curved region is $v_{0} \vec{i}$.

So i don't understand how the "mean velocity" could be the one you wrote...

Let the time to go through the dip be from t=0 to t=T and the particle have unit mass.

Agreed $$\int_0^T f\,dt = 0$$

But $$v(t) = v(0) + \int_0^t f\,dt$$

And the mean velocity is $$1/T\int_0^T v(t)\,dt$$

The condition that $$\int_0^T f\,dt = 0$$ doesn't imply the mean velocity is v(0). It only implies the final velocity v(T) = v(0).