# Conservation of momentum and conservation of energy details

• B
• annamal

#### annamal

If we have a ball with mass m dropped from a height h down to the ground, how come we can't set the conservation of energy equation just as the velocity of the ball turns 0.
mgh = 0
If instead the ball were moving with an initial velocity v, would the equation be
##mgh + \frac{1}{2}mv^2 = 0##?
If there were an external force that caused the ball to drop with initial velocity v, would conservation of energy still be applicable since there was an external force?

How come with conservation of momentum laws, you can't have one mass m colliding with the second mass ##m_2## and then both masses stopping together after the collision as in:
##mv_1 = 0##. How come momentum is not conserved here?

It is certainly possible for two 1Kg masses traveling towards each other at the same speed to collide with each other with no rebound. Since momentum is a vector, the original momentum is zero - with the momentum of each mass cancelling the momentum of the other out. And, of course, the final momentum would also be zero.

As far as energy goes, this would be an "inelastic collision", with the energy going to heat up the resulting 2Kg mass.

• Delta2, PeroK and Lnewqban
It is certainly possible for two 1Kg masses traveling towards each other at the same speed to collide with each other with no rebound. Since momentum is a vector, the original momentum is zero - with the momentum of each mass cancelling the momentum of the other out. And, of course, the final momentum would also be zero.

As far as energy goes, this would be an "inelastic collision", with the energy going to heat up the resulting 2Kg mass.
For my momentum example, I am talking about one mass colliding with another mass at rest and both having 0 velocities after the collision

For my momentum example, I am talking about one mass colliding with another mass at rest and both having 0 velocities after the collision
That is not possible because it violates conservation of momentum. If you think you’ve found such a situation, you’ve misanalyzed it somehow.

• malawi_glenn and Delta2
how come we can't set the conservation of energy equation just as the velocity of the ball turns 0.
mgh = 0
We can. Because ##m## and ##g## are both non-zero and constant, ##mgh=0## means that ##h=0## - we’re just choosing to label the initial height of the ball as zero and the height of the ground as some negative number instead of calling the ground height zero and saying that the initial height is a positive number.

Calculate it either way (to make it easy, stick with case where the ball starts with velocity zero) and you’ll see that the total energy (sum of kinetic and potential) is conserved as the ball falls.

• Lnewqban
That is not possible because it violates conservation of momentum. If you think you’ve found such a situation, you’ve misanalyzed it somehow.
Ok, but I would like to understand why it violates conservation of momentum. Is there a net external force? What is the external force? Is it the frictional force? But the frictional force between the 2 masses exists in inelastic collisions as well (which does have momentum conserved)

We can. Because ##m## and ##g## are both non-zero and constant, ##mgh=0## means that ##h=0## - we’re just choosing to label the initial height of the ball as zero and the height of the ground as some negative number instead of calling the ground height zero and saying that the initial height is a positive number.

Calculate it either way (to make it easy, stick with case where the ball starts with velocity zero) and you’ll see that the total energy (sum of kinetic and potential) is conserved as the ball falls.
No, that is not what I mean. In my conservation of energy problem, I am setting the ground with height h = 0; If we have a ball dropped with height h > 0, the initial energy state is mgh. I want the final energy state to be when the ball hits the floor at 0 velocity, so the final energy state is 0, but if I set those equal mgh = 0...but I wanted h > 0. I don't understand why that equation cannot work.

I don't understand why that equation cannot work.
Because mechanical energy (KE + PE) is not conserved in an inelastic collision. The kinetic energy of the ball becomes internal energy of the ball-floor system when it comes to rest.

Let me know if I'm understanding you properly.

• Lnewqban
Because mechanical energy (KE + PE) is not conserved in an inelastic collision. The kinetic energy of the ball becomes internal energy of the ball-floor system when it comes to rest.

Let me know if I'm understanding you properly.
Yes you are understanding me properly. Conservation of energy doesn't work which makes sense, but conservation of momentum does not work either in a case where the ball is dropped with a certain velocity and collides with the floor with 0 velocity, mv = 0... how come?

Something large and quite massive is moving differently after the collision...momentum is conserved...so what is now moving?

Something large and quite massive is moving differently after the collision...momentum is conserved...so what is now moving?
So if there is nothing moving after the collision, momentum is not conserved? Is there an external force (since that is when momentum is not conserved)?

Yes you are understanding me properly. Conservation of energy doesn't work which makes sense, but conservation of momentum does not work either in a case where the ball is dropped with a certain velocity and collides with the floor with 0 velocity, mv = 0... how come?
That is the misanalysis that I thought you might be making.
There are two masses here: the mass of the ball and the mass of the earth. Just before the collision the ball is moving towards the Earth at speed ##v## and the Earth is not moving: the total momentum is ##0\times m_E+mv## where ##m## is the mass of the ball, ##v## is the velocity of the ball immediately before the collision, and ##m_E## is the mass of the earth. When the ball hits the Earth it exerts a force on the earth. This force accelerates the Earth by a small amount (small because the force is very small and ##m_E## is very large) so the speed of the Earth changes by a small amount that we'll call ##\Delta v_E##. Thus after the collision the Earth's momentum is not ##0\times m_E##, it is ##\Delta v_Em_E## and when you remember to include this term you will see that momentum is conserved.

So if there is nothing moving after the collision, momentum is not conserved?
After the collision something is always moving - there is no way that ##\Delta v_E## can be zero. However In practice it is so small - don't take my word for this, calculate it for yourself! Assume a one kilogram ball moving at ten meters per second and Google will tell you ##m_E## - that we don't worry about it.

• Lnewqban and hutchphd
Conservation of energy doesn't work which makes sense,
Conservation of mechanical energy doesn't work. (But if you include all forms of energy, it does.)
but conservation of momentum does not work either
As explained by @Nugatory and others: You have two bodies colliding -- the ball and the earth. Compare the total momentum of that system before and after the collision and you'll find that momentum is conserved just fine. (Of course, the Earth is quite a bit more massive than the ball!)

• Lnewqban
How come with conservation of momentum laws, you can't have one mass m colliding with the second mass ##m_2## and then both masses stopping together after the collision as in:
##mv_1 = 0##. How come momentum is not conserved here?

Conservation of mechanical energy doesn't work. (But if you include all forms of energy, it does.)

As explained by @Nugatory and others: You have two bodies colliding -- the ball and the earth. Compare the total momentum of that system before and after the collision and you'll find that momentum is conserved just fine. (Of course, the Earth is quite a bit more massive than the ball!)
What if 2 balls collide each with mass m, one with initial velocity and one at rest, and both end up with 0 velocity.

The conservation of momentum doesn't make sense like this: mv = 0. How come?

• PeroK
What if 2 balls collide each with mass m, one with initial velocity and one at rest, and both end up with 0 velocity.

The conservation of momentum doesn't make sense like this: mv = 0. How come?
Unless outside forces are involved, it's not going to happen. Assuming a perfectly inelastic collision, where the balls stick together, there will be a non-zero velocity after the collision.

• Lnewqban
What if 2 balls collide each with mass m, one with initial velocity and one at rest, and both end up with 0 velocity.

The conservation of momentum doesn't make sense like this: mv = 0. How come?
It seems as if you are not reading the replies you have already received - exactly this question was answered in several psts above.

Unless outside forces are involved, it's not going to happen. Assuming a perfectly inelastic collision, where the balls stick together, there will be a non-zero velocity after the collision.
Ok, got it.
My last question was if there were an external force that caused the ball to drop with initial velocity v, would conservation of energy still be applicable since there was an external force? Or would the conservation of energy be applicable only after the external force stopped causing the force?

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Or would the conservation of energy be applicable only after the external force stopped causing or force?
Exactly. Mechanical energy will be conserved once that external force is removed. (Note that in this example we are treating gravity as an internal force within the ball-earth system.)

Conservation of mechanical energy doesn't work. (But if you include all forms of energy, it does.)
For conservation of mechanical energy for dropping a ball on the ground, I have 3 scenarios for conservation of energy:
1) For the scenario right before the ball hits the ground ##mgh = 0.5mv^2##
2) For the scenario while the ball hits the ground using the potential spring force of the ball and ground ##mgh = 0.5kx^2 + 0.5k_2{x_2}^2##
3) For the scenario right after the ball hits the ground ##mgh = 0.5mv^2##

All 3 are correct, right, especially scenario 2? At all instances of time, mechanical energy is conserved without a nonconservative force?

For conservation of mechanical energy for dropping a ball on the ground, I have 3 scenarios for conservation of energy:
1) For the scenario right before the ball hits the ground ##mgh = 0.5mv^2##
2) For the scenario while the ball hits the ground using the potential spring force of the ball and ground ##mgh = 0.5kx^2 + 0.5k_2{x_2}^2##
3) For the scenario right after the ball hits the ground ##mgh = 0.5mv^2##

All 3 are correct, right, especially scenario 2? At all instances of time, mechanical energy is conserved without a nonconservative force?
Looks like in scenario 2 you are imagining the collision of ball and ground to be a perfectly elastic collision, so that mechanical energy is conserved. And you are further modeling the spring PE of ball and floor. With the assumption that the collision is elastic, your thinking is OK.

You can write the conservation equation like this: GPE + KE + SPE1 + SPE2 = constant
GPE = gravitational PE; SPE = spring/elastic PE for the ball and the floor.

Before it hits the floor and after it rebounds, SPE1,2 = 0, and you'll get your scenarios 1 and 3. When the ball is at its lowest point, and ball and floor are maximally compressed, the KE is zero and you'll get your scenario 2.

Looks like in scenario 2 you are imagining the collision of ball and ground to be a perfectly elastic collision, so that mechanical energy is conserved. And you are further modeling the spring PE of ball and floor. With the assumption that the collision is elastic, your thinking is OK.

You can write the conservation equation like this: GPE + KE + SPE1 + SPE2 = constant
GPE = gravitational PE; SPE = spring/elastic PE for the ball and the floor.

Before it hits the floor and after it rebounds, SPE1,2 = 0, and you'll get your scenarios 1 and 3. When the ball is at its lowest point, and ball and floor are maximally compressed, the KE is zero and you'll get your scenario 2.
Yes. I am talking about an elastic scenario.

In scenario 2, would the impulse = ##\int_{t1}^{t2}(0.5k{x(t)}^2 + 0.5k_2{x_2(t)}^2)dt## or just J = ##\int_{t1}^{t2}(0.5k{x(t)}^2)dt## where k is the spring constant of the ball, and x(t) is the ball's compression, or should it be ##k_2## the spring constant of the wall and ##x_2(t)## the ground's compression? J = -mv -mv = -2mv

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Yes. I am talking about an elastic scenario.

In scenario 2, would the impulse = ##\int_{t1}^{t2}(0.5k{x(t)}^2 + 0.5k_2{x_2(t)}^2)dt## or just J = ##\int_{t1}^{t2}(0.5k{x(t)}^2)dt## where k is the spring constant of the ball, and x(t) is the ball's compression, or should it be ##k_2## the spring constant of the wall and ##x_2(t)## the ground's compression?
Realize that the ground exerts a force on the ball while they are in contact. You'd need to integrate that force over time to get the impulse it provides. More complicated than you'd like, I suspect. But you know what the answer will be:

J = -mv -mv = -2mv
That's all you really need. The impulse equals the change in momentum.

Realize that the ground exerts a force on the ball while they are in contact. You'd need to integrate that force over time to get the impulse it provides.
According to my class notes, the impulse (change in momentum) for the ball, is just the force the ball exerts on the ground integrated for a certain time duration.

According to my class notes, the impulse (change in momentum) for the ball, is just the force the ball exerts on the ground integrated for a certain time duration.
The impulse imparted to the ball is the force exerted on the ball integrated over time. Of course, per Newton's 3rd law, the ball and ground exert equal and opposite forces on each other.

• Lnewqban
Yes. I am talking about an elastic scenario.
If you include an analysis of the collision itself, then you have a transformation from gravitational potential energy, to kinetic energy, to elastic potential energy, back to kinetic energy and back to gravitational potential energy.

• Lnewqban
The impulse imparted to the ball is the force exerted on the ball integrated over time. Of course, per Newton's 3rd law, the ball and ground exert equal and opposite forces on each other.
So the force exerted on the ball by the ground equals the force exerted by the ball on the ground, so we were just talking about a sign difference?

So then if the impulse is the force exerted on the ball by the ground, J = ##\int_{t1}^{t2}(k_2{x_2(t)})dt##, where ##k_2## is the spring constant of the ground and ##x_2(t)## is the ground's compression

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So the force exerted on the ball by the ground equals the force exerted by the ball on the ground, so we were just talking about a sign difference?
A sign difference and which object is receiving the force.

So then if the impulse is the force exerted on the ball by the ground, J = ∫t1t2(k2x2(t))dt, where k2 is the spring constant of the ground and x2(t) is the ground's compression
But this is difficult to calculate because the displacement is very very small and the spring constant is huge. But, happily, we know Newton's Law #3

.

A sign difference and which object is receiving the force.

But this is difficult to calculate because the displacement is very very small and the spring constant is huge. But, happily, we know Newton's Law #3

.
So you're saying the equation should beJ = ## \int_{t1}^{t2}(-k{x(t)})dt## where k is the spring constant of the ball, and x(t) the ball's compression?

• hutchphd
So the force exerted on the ball by the ground equals the force exerted by the ball on the ground, so we were just talking about a sign difference?
Yes. And signs matter. Recall that momentum is a vector. If we take up as positive, the upward force of the ground exerts a positive impulse on the ball. Thus the impulse, which equals the change in momentum, is (+mv) - (-mv) = +2mv.

• hutchphd
Rather than focus on "forces", it might be better to focus on the
"transfer of momentum [from one object to another]"...
one object gaining momentum ##\Delta \vec p_1\stackrel{Newton2nd}{=} \Delta \vec p_{1\mbox{ from 2}}##
because the other object transferred momentum to it: ##\Delta \vec p_{2\mbox{ from 1}} \stackrel{Newton3rd}{=} - \Delta \vec p_{1\mbox{ from 2}}## .

Yes, it deals with the impulse and the time-integral of the force...
but, as others have tried to emphasize,
it's more than a formula, a "change", or a sign.
It's about what object is applying the force (the source of the force.. source of the momentum-transfer)
and what object is receiving that force (the target of the force.. target of the momentum-transfer).
Think of it as an accounting problem.

• jbriggs444