I think I understand this problem, but I'd really appreciate someone confirming for me...

A sliding block of 40kg moves at a speed of 10m/s across a frictionless surface. If 10kg block is placed on top of the original block, what will it's new speed be?

So, I found the original KE of the 40kg block using KE=1/2mv^2 and got 2,000J. I then used 2000J=1/2(40kg + 10kg)v^2 to find v=8.9 m/s. This seems appropriate to me, however my book has only listed 8m/s and 10m/s as possible answer choices. So either I'm mistaken or my book has a typo.

Any suggestions?

Thank you!

Momentum conservation:

No external forces :

mVi = MVf

Try another conservation, namely, conservation of momentum.
Then, remember carefully what the condition is where kinetic enegy consevation is correct.

Aww, yes. It's an inelastic collision. KE isn't conserved. But momentum is: m1v1 m2v2=(m1+m2)vf

Thank you, I guess I just needed someone to remind me not to be silly :P

Hi there,
Since there is no friction between the block and the floor it is sliding along, it doesn't matter how much downward force you apply upon the original box as it will not slow the box down; and by downward force I mean simply applying more mass on top of the original block of mass. In real life this increased weight would slow the box even more since the force from friction that would slow it down is the product of the weight and the coefficient of friction for the surface it is sliding on.

Again though, since the block is sliding on a frictionless surface the only thing that would slow it down would be any horizontal component of force that acts to speed it up or slow it down, but any vertical components of force have no effect on the speed of all the mass because there is nothing that slows it down; adding more mass just makes it heavier.

Simply adding more mass onto something that can move at a constant velocity with no forces acting upon it does nothing to slow down or speed up the moving mass. In other words by making the box heavier, with no friction there is no increase in the counterforce from friction to slow it down by any amount.

Does adding this extra mass give the whole system additional kinetic energy; yes. In real life the answer would probably be no, because adding more mass would increase the force from friction causing it to slow down at a greater rate than with less mass. That is what's neat about working with ideal things like this. By simply adding more and more mass on top of a block that is sliding along a floor with no counter force from friction, then one can keep increasing the kinetic energy of the block by adding all this extra mass, since it in no way slows the block down.

I guess the closest example I can think of is a block of matter sliding along a sheet of ice. As it comes near you jump on the block for a ride. Now let’s say you weigh 3 times as much as the block itself the KE before you jumped on was 0.5mV^2, now with you on it the mass is now (mass of block + mass of you) and the mass of you = 3*mass of block, so the new mass is (mass of block + 3*mass of block) = (m + 3m) = 4m. For the kinetic energy to remain the same it would have to be = 0.5*4m*(V/2)^2, or in other words you may expect the velocity to drop by half when you jump on, but as I'm sure you can reason, if this block is sliding along on an ice sheet and you jump on, there will be only a small change in the velocity due to such a small amount of friction between the block and the ice sheet.

Long story short, if a block of matter is sliding along a frictionless surface then no amount of downward force will slow the block as by simply adding more weight has no effect on the speed of the block since there is no counter force from friction there to slow it down.

Since the block was initially moving at 10 m/s it will keep moving at 10 m/s. So there's no typo in your book, it's kind of a "trick" question. I noticed you used the conservation of momentum to get 8 m/s; that is a trick don't fall for it. Momentum is conserved in an inelastic collision; but the collision in this case is from the mass being dropped onto the sliding block, so m1v1 = m2v2 where v1 is the downward velocity of the block, and there is no downward velocity. Furthermore there is no collision that the block makes with anything while it is sliding, so the conservation of momentum is not the proper concept to use for this problem as the only collision is that of the 10kg mass being dropped onto the 40kg mass, and indeed momentum is conserved since 40kg*0 = 50kg*0, becasue there is no change in downward velocity from the collision.

Hope that helps.

Many Smiles,
Craig

Hello again,
I kind of rushed through why the conservation of momentum shouldn't be used in this problem, I will now explain in more detail why not.

Let Vx = horizontal velocity and Vy = vertical velocity, or
V = Vxi + Vyj where V is a velocity vector broken into its horizontal and vertical components and i and j are the unit vectors pointing in the appropriate directions.

There are two possible cases for introducing the 10kg mass onto the sliding 40kg mass. The first is that the 10kg mass is simply placed on top of the 40kg mass while it is sliding and therefore the initial state of "things" is :

m1Vy1 + m1Vx1 + m2Vy2 + m2Vx2
where m1 is the 10kg mass and m2 is the 40kg mass, if the mass m1 is simply placed on top of mass m2 then it has no initial vertical velocity component at all and the sliding mass has no vertical velocity component, so the momentum before placing the 10kg mass on the 40kg mass must equal the momentum of the "system" after placing the mass m1 on m2, written as:

m1Vx1 + m2Vx2 = (m1 + m2)*Vx

because the 10kg mass must have the same x component of velocity in order to be placed on top of mass m2, if it didn't then mass m2 would pass right by not giving anyone the chance to place it on top of the mass m2, so mass m1 must be "brought" up to the same speed in order to be nicely placed on top of mass m1, therefore Vx1 = Vx2 = Vx.

So simply placing the mass on top of the sliding mass results in both masses moving together as one with the same speed as they had initially in the x direction.

The second case is if the mass m1 is dropped onto mass m2 the initial state is:

m1Vy1 + m2Vx2

Now mass m1 is not moving at the same speed as mass m2 but rather is simply dropped onto it from some height above it; giving it a vertical velocity component only. This initial state must equal the final state of the system:

m1Vy1 + m2Vx2 = (m1 + m2)Vx where Vx = Vx2, this shows that both masses are travelling at the same speed after one is dropped onto the other; a downward travelling mass m1 that strikes a horizontally moving mass m2 in no way changes the horizontal speed as the impact came vertically; so now both masses are moving with a velocity that is the same as the mass m2 had before being struck on its top by the falling mass m1 (remember there is no friction to slow things down). So what happened to the vertical velocity component of the mass m1? It delivered its energy into mass m2 causing an impulse upon mass m2, look at it like this, when the mass m1 hits the mass m2 it comes to a stop and imparts a downward force upon the mass m2, so the above equation can be written as:

m1Vy1 + m2Vx2 = Vxm1 + Vxm2 + 0*m1 +0*m2 or
m1(Vy1 - 0) + m2Vx2 = (m1 +m2)Vx

This last equation states that the change in momentum of the 10kg mass upon impacting the 40kg mass has no effect on the horizontal velocity component of both masses when in contact and moving with each other.

Now a change in momentum yields an impulse on the 40kg mass, that is mass*delta(V) = Force*delta(t)

That is what happened to the 10kg mass; it underwent a change in momentum which yielded an impulse upon the 40kg mass that acted downwards. So we can replace the change in momentum of mass m1 with the impulse function it applies to the mass m2, re-writing yields:

F(t2 - t1) + m2Vx = (m1 + m2)Vx

That force vector acts only to push the 40 kg mass into the floor it is sliding on and it exists only for a short period of time, namely the time it takes to go from some non zero valued vertical velocity to a stop; which will be pretty quick.

So by dropping the 10kg mass onto the 40kg mass all that happens is a force above and beyond the weight of the 10kg mass was applied downwards onto the 40kg mass for a brief moment in time and then it stops and moves right along with the 40kg mass whose velocity never changed as no horizontal force components were ever applied to the 40kg mass. The energy the 10kg mass had just before it struck the 40kg mass went into the 40kg mass and is absorbed and converted into thermal energy. It heats up the sliding block by a tiny amount, but it's energy has no effect on the speed the two masses will travel at in the horizontal direction since its force was applied entirely in the vertical direction.

Again in real life with friction, the two masses would slow down even more than the 40kg mass would alone because the additional mass would create a larger counter force from friction slowing the two bodies down more than one alone would slow down due to friction, but with no friction and no energy or force applied to the 40kg mass in the horizontal direction, which is the direction the 40kg mass is moving in, then there can be no change in the velocity of the two masses in the horizontal direction; as adding an extra 10kg to a moving 40kg mass on a friction free surface simply causes it to weigh more, but has no effect on the horizontal component of the velocity of this heavier piece of matter.

It may be a little "warmer" after having energy dumped into it from the 10kg mass (if dropped onto the 40kg mass), but all the energy of the 10kg mass went into heating the 40kg mass up just a little and had no effect on its speed. It will simply ride on top of the 40kg mass, causing it to weigh more, but with no friction, it does not change the speed of the two masses together.

I hope that is clearer. I've used this same problem on a number of exams I have given out to my students for just this reason. I want to know that they understand concepts like force, velocity, mass, momentum, impulse, inertia, speed, vectors and their components, acceleration, friction and all the other things that can be thought about from just this one problem; just so you know the answer is 10 m/s not 8 m/s which is what you'll get if you assume that by placing extra weight onto something that is moving with no frictional forces acting upon (or any forces for that matter) it that this extra weight can slow the two masses down; it doesn't, it just makes the whole thing heavier. Which again in real life would slow it down some, but in an ideal world of no friction to counter the motion of the 40kg mass then it doesn't matter how much extra mass is placed on top of it as this simply increases its weight, but has no effect on its horizontal velocity.

Does that make sense? I hope so, it is an important concept to understand.

Many Smiles,
Craig

"because the 10kg mass must have the same x component of velocity in order to be placed on top of mass m2, "

Well, I think this is the point where my opinion differs from yours.
In my opinion, in problems like this, we should consider the mass to be placed without any speed. We can, of course, put (for example) a bottle on a moving car without giving the bottle any speed, can't we?
(If the weight has its initial speed equal to the base speed, the final speed will be certainly 10m/s. I agree with you in this point.)

Now, to avoid confusion, I'll call the 40kg mass 'base' and the 10kg one 'weight'.
In this problem, the weight is put on the base without speed (v=0).
First, the weight will slide on the base, but as soon as the weight touches the base, frictional force will act between them. It decreases the speed of the base, and increses the speed of the weight. Because frictional force is not a consevative force, the objects lose some of their kinetic enegy (and they become a little warmer).
The statement of the problem says nothing about the friction between the weight and the base, but it seems natural to assume there is friction.(If there isn't, then there is no force acting between them--it's all the same before and after the weight is put!)

Hi there,
You say, "In my opinion, in problems like this, we should consider the mass to be placed without any speed. We can, of course, put (for example) a bottle on a moving car without giving the bottle any speed, can't we?"

Nothing was stated about the weight sliding once being set upon the base. Would it slide in real life, it may or it may not, it depends on how much friction exists between the base and the weight and how fast the base is moving at. Since these are unknown values within the problem it is more proper to take what you do know; that a weight is placed upon the base. I wouldn't just assume that the weight would slide upon the base providing a counter force that would slow the base down by some small amount, as you don't know if the weight will slide upon being set on the base or not.

The quote above is your opinion and if simply placed on top of the moving base with no velocity depending on how much friction exists between the weight and the base will decide if it slides or not. If it does slide then this would produce an additional counter force on the base slowing it by a small amount and I was clear in stating that as long as no horizontal components of force are applied to the base then its velocity would remain unchanged since if it slides it acts as a counter force that would slow it down.

This base would have to be moving slow enough for you to place the weight upon the base for the weight to have no initial velocity; your example of placing a bottle upon a moving car would be the same. The car would have to be moving pretty slowly for you take and simply place the bottle on top of the car, if it were moving at 60 m.p.h good luck trying to set the bottle on top of the car at that speed.

Since we don't know the speed of the base, the static or kinetic coefficient of friction between the base and the weight, if the weight is placed upon the base or is dropped on it or if the weight had the same velocity as the base and then was placed upon it; it is a far safer assumption to make that the base mass of 40kg is somehow simply increased to 50 kg; and then ask the question by what amount does the velocity change when a mass is increased from 40kg to 50 kg while sliding upon a frictionless surface.

Since an increase in mass would only slow the base down if there is friction between the base and the floor, and there is not, the base velocity must remain unchanged.

I do agree with you completely that even if simply placed upon the base the weight has an inertia that has to be overcome to get it moving along with the base and this would require taking away some of the kinetic energy from the base imparting it into the weight to overcome the inertia of the weight, but having given this problem myself a number of times on exams to my students for well over a decade now, what I am concerned with, as is the originator of this theoretical problem, is whether or not the student knows that a downward force alone acting upon some moving mass on a frictionless surface, has no effect on the speed of that mass.

The question was, "A sliding block of 40kg moves at a speed of 10m/s across a frictionless surface. If 10kg block is placed on top of the original block, what will it's new speed be?"

Without knowing the friction between the weight and the base we can't be sure if it will slide or simply stay put if placed upon the base; and it is an important thing to know because if the friction is small enough then the weight will slide some after being placed upon the base before stopping on top of the base. If it slides it will "suck" more energy out of the base than if it simply stayed in place when put there because the energy lost to the work done by friction (F"dot"d) upon the weight plus that needed to get the weight to overcome its inertia and moving along with the base is greater than that needed to simply overcome the inertia of the weight.

Since we don't know this and it will change the velocity of the base by differing amounts depending on whether the weight slides once placed upon the base or simply stays there on top of the base, then this must not be the point of the question. The point is whether or not the moving block slows if its mass is increased on a frictionless surface, and the answer to that question is no.

Since the question is looking for a definite answer (an actual value), assuming the weight will affect the velocity of the base from friction between the base and the weight, regardless of whether the weight slides or not once placed upon the base, requires both the static and kinetic coefficients of friction between the base and the weight to solve for the answer; neither of which are given. Since they are not given, what the question really wants to know is that if a 40kg block is moving at 10 m/s and its mass is increased to 50kg by what amount does the velocity of the base change by; it doesn’t change its speed at all.

Like I said I've been giving this problem to college students on exams for over a decade and by not providing enough information to solve for a change in velocity due to the friction between the base and the weight, then one should assume that what is meant is if the base mass is simply increased by 10kg how much does the velocity change by. Again since it is moving upon a frictionless surface any additional downward force does not translate into a counter force in any way upon the masses in any direction they are allowed to move along. So no change in velocity occurs.

I want to be clear, even moving on a frictionless surface if you place a 10kg weight onto this 40kg base the base is "forced" to "grab" onto the weight and bring it up to speed with it slowing it down in the process, but not knowing either the static or kinetic coefficients of friction one cannot answer if the weight first slides upon being placed onto the base or if the static coefficient is great enough, that the weight will sit there not moving with respect to the base.

In either case the base still must "grab" onto this additional weight which would slow it down, but if it doesn't slide it takes less energy away from the base to get it moving along with the base. If it does slide then additional energy is lost in the form of thermal energy into the weight and base due to it sliding, plus it still needs the same amount of energy the non-sliding weight took to get up to speed with the base. If the weight doesn't slide all the energy taken from the base is used in bringing the weight up to some velocity, if the weight slides, some of the bases energy is lost to heat (thermal energy), which slows the base down just a little plus energy is taken by the weight to bring it up to some speed with the base that will be less than if the block didn't slide once set upon the base; becasue no sliding means base energy lost goes into getting weight up to a speed that is close to what the base had before the two met. If it slides some base energy is simply lost to heat and therefore togther the two masses would move even slower than if the block didn't slide upon coming into contact with the base.

We do not know if the weight will slide or not and even if we did without the coefficients of friction to let us know how much energy is given away to bring the weight up to speed and or given up to thermal energy due to it possibly sliding, then one must assume the point of the question is not whether the friction between the weight and the base slows the two down, but rather by simply increasing the mass of some moving object on a frictionless surface does it's velocity change? The answer to that question is no.


Is that better? If so, I think now you'll see why I worded my previous replies the way I did. Again, don't get me wrong, I know that the addition of this weight to the base will cause it to slow, but given that the needed information is missing/unknown, then one should assume the point of the question is not whether the velocity of the weight and base slows due to the combining of the two, but rather does the base slow by increasing its mass with the weight - real life - yes; frictionless surface - no.

Many Smiles,
Craig

There seems to be a problem with this

That is what's neat about working with ideal things like this. By simply adding more and more mass on top of a block that is sliding along a floor with no counter force from friction, then one can keep increasing the kinetic energy of the block by adding all this extra mass, since it in no way slows the block down.

We would be violating the principle of conservation of energy in this case. Even ideally we cant keep increasing the mass and not have the velocity affected as the energy would tend to infinity.

Suppose we gently place the block on the lower one. The block has zero vel in the horz direction initially and hence zero K.E. When it moves along with the lower one, its vel increases to that of the system, and thus its K.E. is now non zero. Where does this extra energy come from?

physicsquest, you already know what i will say, still........
- friction force that acts between blocks is working here to check relative velocities between blocks and ultimately making them move with a common velocity. the kinetic energy of upper block is coming due to work done by friction on it in the forward direction.(the same is acting back for lower block.)


had it been a frictionless interface(BETWEEN THE TWO BLOCKS) then they will never attain common velocity and in that case lower block will continue to move at 10 m/s .
however , in real life there is friction, and this friction makes the bodies move together finally with a common velocity, . That velocity will be 8m/s .and this decrease of kinetic energy is because some part is expended as elastic energy of deformation at the surface of contact . Often these deformation ressults in heat generation-as bonds are broken and new bonds formed at the junction.

The energy that the weight takes on comes from the base (and I will address the comment you bolded in your reply shortly).

Not knowing what will occur when the weight is placed onto the base, will it slide and take base energy and convert it into heat (wasted energy) or will the weight stay in the place where it was placed on the base. An important thing to know if an actual value is desired and it is. The answer to the question is either 8 m/s or 10 m/s according to the OP's MCAT study book and according to many teacher edition books that I've seen the problem in. I don't think that all of my books and the OP's book all have typos as the answer.

If you use kinetic energy to find the velocity change you get about 8.94 m/s (KE = 0.5*40*100 = 0.5*50*V^2, solve for V), that's neither 8 nor 10 m/s. So there must be a different interpretation for the problem. That is my point. Without the information needed to know if energy is lost to heat, which means we need to know the weights static and kinetic coefficients of friction to first find out if the weight when placed on the base moves and if it does how far so we can calculate the work done by the force of friction on the weight and base; which is equal to the amount of energy that was converted into heat over the time period it took to move the weight over the top of the base.

Since we don't know if the added weight will slide and dissipate energy in the form of heat plus take the energy it needs from the base so it will move along with the base when it stops sliding, or if it will simply take the energy it needs to move along with the base without sliding, there are a number of different possible velocities the two masses may end up at.

Since this information is purposely left out, it's proper to interpret the question as a mass of 40 kg moving along a frictionless surface with a velocity of 10 m/s, what happens if that mass is increased to 50 kg?

So my interpretation of the problem is like this: the base is sliding along on an ice sheet and you skate up near it and place 10kg of mass on top of this moving base. Since all you did was increase the weight by adding this mass, it doesn't slow down the base mass on a frictionless surface.

Of course in real life there will always be some fiction between a moving base with a mass of 40kg and if you go over and add 10 kg on top of the 40kg mass its weight increases, which increases the frictional counter force to the motion of now both masses and they both slow, but we are talking about an ideal surface with no friction. If you skate up near the moving base and gently place a 10kg mass on top of the 40 kg mass you don't change the overall velocity of the two masses compared to the initial velocity of the base mass.

Energy is still conserved because you are delivering on top of the moving 40kg base a 10kg weight that also has a velocity of 10 m/s, how else can you gently place the weight on top of the base unless you're standing right over it able to gently place the weight onto the base.

So I didn't mean to imply that simply dropping non moving masses onto a moving mass will keep increasing the energy of the combined mass system to the point of heading towards infinite energy; but rather the more mass you bring up to speed with the base and place it gently onto the base will keep increasing the energy of the combined mass system. That much is true.

Lastly, again, since we do not know how the weight is placed onto the base, whether or not it slides and dissipates energy from the base in the form of heat, neither coefficients of friction which could tell us if the weight stays in place or slides (assuming it is simply placed on top of the base with an initial velocity of 0, something else not given); and if we knew that the weight did slide we'd have to know by what distance to find the energy lost to heat, then we could find out how much is left within the base to get the two masses moving at some slower velocity.

Since weight is a vector quantity that in this case points downwards, and in our case is an ideal surface with no friction so an increase in downward force has no accelerating effect of the base mass; other than to increase its weight which is countered by the floor so it doesn't fall through the floor nor jump up above it.

So my point is, given the nature of the question, which lacks crucial information if ones wants to know the energy loses to heat if the weight slides when first placed on it, how far it slides which would tell us how much energy is lost to heat, or if it even slides at all; I believe the intention of the question is more to the effect that if you have a mass moving on a frictionless surface and you add some more mass to it (by moving right up side by side with the base mass to place additional mass upon it) what is the velocity of this new mass system.

To be honest, it's kind of a trick question, because we know that if a stationary mass is placed onto a mass that is moving, this additional mass, in order to move with the moving mass, has to have its inertia overcome so it will "drain" the base mass of some of its energy to come up to a mutual speed for the two masses that is less than the speed of the initially moving mass by itself.

Given how the question is worded, we don't know how the weight is placed onto the base mass, we don't know if the weight will slide dissipating energy in the form of heat therefore taking energy away from the whole system and giving it away to keep the surrounding air a little warmer, or if it even slides at all; if we knew the coefficients of friction then we could tell whether or not the weight if simply placed onto the base mass slides or not, hence wasting additional energy or not.

Since we are not given the information needed to find out how much energy may be lost to friction if the weight slides when placed onto the base mass (which is also something we don't know if it occurs or not, that is does it slide or does it stay where placed) and what is required from the question is an actual value for an answer, not simply does it slow down or not, then the only way I see to interpret the question is to ask myself, if this mass is moving and I come up to this moving mass and place some more mass on top of it what is its new velocity? It's still 10 m/s.

I'll agree it's a badly worded question, but it is purposely worded badly as the idea is for someone to look at the question and ask themselves, "does this 10kg mass when placed on top of the moving 40 kg mass stay where I placed it or does it slide to some extent wasting some of the 40kg masses energy as heat? Since I don't have enough information to answer this question knowing that adding this additional mass will slow it down by some amount depending on whether or not it slides when first placed onto the base mass then I should assume that the question really is, by applying an additional downward force onto the 40kg mass, what is its new velocity".

Since there's no friction any increased downward force has no effect on the speed of the masses; but given the replies I've seen to my explanations of this problem I think I will reword the problem for future use on exams I give to people; as I hoped including this question on an exam would cause people to ask themselves what I just did in the previous paragraph. After 10 years of using this question on exams, I can count on both hands the number of people who have given me the answer of 10 m/s. That alone should be a sign that the question is confusing and doesn’t really make people ask themselves those questions I spoke of but rather seems to only confuse them.

I have so many classes I teach, I often overlook patterns like this. So having said that, to all of you that have responded with further questions as to how this can be correct, I thank you, as I now understand where the confusion comes from and I now believe it may be too great of a leap to draw the same conclusion about the question that I know it as; knowing that, it just seems natural to me to say to myself what they mean is if this 40kg mass is moving at 10 m/s and you come up alongside this moving mass and add another 10kg of mass on top of it what is the new velocity of the two masses; but after a decade of being a college professor I should know by now that what seems to follow logically to me, is by no means what everyone else feels logically follows.

As long as you understand what I was trying to get at, which is the same thing the question is trying to get at (granted I now think it does a poor job in its implication), then all’s well that ends well; and I will reword the question for future exams.

Many Smiles,
Craig

The energy that the weight takes on comes from the base .

This should mean that the base loses energy which means it slows down.
We cannot end up with more energy than we started out with.

Like r.samanta , i too would go for the answer 8 ,as in this case we only assume that momentum is conserved (which is always true) and that K.E. is not ( which is possible when masses stick together) . We do not end up with a total energy greater than what we started out with which i think is totally unacceptable ( as there is no mechanism or source which jacks it up)

I never said there is some way for energy to "magically" keep increasing or that we can end up with more energy than when we started out, in fact directly from my last reply I wrote:

"So I didn't mean to imply that simply dropping non moving masses onto a moving mass will keep increasing the energy of the combined mass system to the point of heading towards infinite energy; but rather the more mass you bring up to speed with the base and place it gently onto the base will keep increasing the energy of the combined mass system. That much is true."

Also the conservation of momentum comes directly from energy being conserved. Energy is ALWAYS conserved; KE in inelastic collisions is not conserved, but the total energy, in all its forms, of the system is conserved. All of its loses may not always be accounted for by some people but it is always conserved. I can very quickly show the conservation of momentum that you are using is derived from an assumed conserved KE and PE; and yet you know that KE is not conserved in an inelastic collision as you said so yourself. In fact momentum is the rate of change in energy for some "thing" with respect to a infinitesimally small change in velocity.

If KE and PE alone are conserved (and they are not) assuming they are and deriving the momentum equation from this assumption yields:
0.5*40*V1^2 + 50gh = 0.5*50*V2^2 + 50gh where the left side is the KE and PE of the "system" before the 10kg mass is placed upon the 40kg base and the right side represents the KE and PE of the system after the meeting of the two masses.

Now take the derivative of each side with respect to a small change in velocity:

m1*V1 = m2*V2

Now what you want to do is say that m1 = 40kg, V1 = 10 m/s, and m2 = 50kg and when you solve for V2 you get 8 m/s, but this "conservation of momentum" equation that you are using comes directly from an assumed conserved KE and PE alone.

Let me pose it differently. Energy in all forms is conserved regardless of the type of collision. So let's account for all of it and then derive the conservation of momentum equation from that:
0.5m1V1^2 + m1gh + 0.5m2V2^2 + m2gh1 = 0.5(m1 + m2)V^2 + m1gh + m2gh2 + Kfm2X
where: h is the initial and final height of the 40kg mass (let's call it the base), h1 is the initial height the 10kg mass (let's call this the weight), h2 is the final height of the weight with respect to the floor (just the height of the base), Kf is the coefficient of kinetic friction, X is the distance the weight "may" slide when placed upon the moving mass, and V is the final velocity of the two masses together. Rearranging this equation so that the left side shows only the kinetic energy from the base yields:

0.5*m1V1^2 = 0.5*(m1 + m2)V^2 + m2g(h2 - h1) + Kfm2X

This equation will hold true and in fact the additional 400 Joules of energy, assuming they do move with a final velocity of 8 m/s is either accounted for by the addition of potential energy into the weight by being placed upon the base, and/or is dissipated by the weight sliding when first placed upon the base.

Now let’s derive the momentum equation accounting for all energy transfers and possible losses from the above equation. Taking the derivative again of each side:

m1V1 = (m1 + m2)V + Kfm2g(dX/dV) and solve for the final velocity:
V = [m1V1 - Kfm2g(dX/dV)]/(m1 + m2)

Now looking at this final equation for conserved momentum, again the only way the answer can be 8 m/s is if the weight in no way slides once placed upon the base, but in fact, in order for the final answer to be 8 m/s the weight either has to slide some distance, or the weights PE must be increased by placing it upon the base; you don't know which or if it is even a combination of the two, so why assume it is all going into the weight by raising its PE? The KE of the base initially was 2000J, if after the two masses are moving at 8 m/s then the total KE of the two masses is 1600J. So 400J of energy must have gone somewhere, or isn't accounted for, since energy doesn't simply disappear.

Simply knowing that the final kinetic energy of the two using your 8 m/s value is less than the initial KE energy of the base means that that we must not be accounting for all the energy losses (or gains) after the combining of the two masses. You can one, assume that the weight when placed upon the base does not move and slows the two masses to 8 m/s; if you do then you must also assume that the additional 400J of energy needed to keep the total energy of the system conserved was given to the system by increasing the PE of the weight by raising it and placing it upon the base (nothing of which was ever mentioned in the questions description). Or secondly, assume that the "missing energy" was completely dissipated by the weight when it was placed upon the base by sliding some distance transforming some of the KE of the base into thermal energy (wasted heat) for the base and the weight. Or lastly, that it is some combination of the two. The weight took on some amount of potential energy and also slid once placed upon the base and that accounts for the missing 400J of energy.

So if you are going to use the concept of conserved momentum, then you need to use the correct equation that accounts for the momentum of everything within the system. The equation you used only accounts for energy transfers, but no possible loses; and is derived from an assumed conserved amount of KE and PE, which you know is not conserved. This equation accounts for the momentum of the system before and after:
m1V1 = (m1 + m2)V + Kfm2g(dX/dV)

If the weight does not slide the frictional term can be dismissed and one can assume that the additional 400J of energy needed to account for the conservation of all the energy is accounted for by raising the weight imparting an additional 400J of PE into the weight. Then, yes, the final speed of the two masses would be 8 m/s.

If the weight slides at all, then some unknown amount of energy is lost to heat, and then we don't know if all of it was lost to heat or if simply some of it was and the remaining unaccounted for energy was put into the weight increasing its PE by some smaller amount such that the increased PE of the weight plus that dissipated by the weight sliding sums to the "missing" 400J of energy.

You have no idea if the weight will slide or not. Therefore one should not only look at the conservation of momentum, but also the conservation of all the energy within the system, since this is where the conservation of momentum is derived from and can be very useful in describing what may occur when two objects come together and KE is transferred, PE of either or both objects can be increased, and energy can be lost due to heat.

Looking at all the energy within the system, before the two masses meet, it is entirely within the base with 2000J of energy. After the two masses meet, if moving at 8 m/s like you say, their KE is 1600J, so for energy to be conserved we either had to dissipate 400J of energy from the 10kg weight sliding once it was placed upon the base mass (in which case the final velocity will be less than 8 m/s according to V = [m1V1 - Kfm2g(dX/dV)]/(m1 + m2)) or assume that the weight does NOT slide and therefore we imparted 400J of PE into the weight when we raised it up and placed it upon the base mass (again, this is a direct result of your assumption that no energy is lost to heat, but doesn't it seem strange to you that the only way for your answer to be 8 m/s is to assume that 400J of PE was introduced into the weight by placing it upon the base?), or lastly it is some combination of the two.

Again, why is it OK for you to assume that lifting the weight high enough such that it takes on 400J of energy to keep energy conserved is a better solution than assuming that the weight is brought up to speed with the base and is placed upon it? If we go with your final answer then we need to assume that we lifted the weight up from a low enough position to impart 400 J of energy into it to keep energy conserved, if we go with my assumption then we need to assume the weight is brought up to speed with the base mass and then placed upon it. Who's correct? We both are.

It's a matter of who assumes what when the two masses are brought together. And again, I've seen this problem so many times, that since different initial assumptions will always lead to different final velocities, the point of the question is far simpler. It is, what if a mass is sliding upon a frictionless surface and through some means a downward force acts on this mass and its weight increases, what is the final velocity? Simply increasing the downward force upon something that is moving on a friction free surface has no effect on that objects speed.

If it has to "grab" onto the mass, then yes it will slow the whole system down, so look at the total energy within the system before and after so you can see where possible energy gains or losses may come from/to, derive the conservation of momentum equations from the conserved energy equations, and then one can find the final velocity of the new two mass system. As I showed for this problem, it can only be 8 m/s if you assume that you had to raise the weight enough to impart 400J of PE into this weight in order for all the energy within the system to remain conserved (which is 4.2644 meters, or 13.9872 ft, above the floors surface, that makes it a pretty tall base mass, no?)

I guess I don't care if you believe the answer to be 8 or 10 m/s, but do you at least understand what the point I'm trying to make is?

Many Smiles,
Craig

There are two possible cases for introducing the 10kg mass onto the sliding 40kg mass. The first is that the 10kg mass is simply placed on top of the 40kg mass while it is sliding and therefore the initial state of "things" is.

__________________
Promotional Merchandise | promotional clothing | Promotional Products