# A few conceptual momentum questions

SkyrimKhajiit
Hello everyone,

I've been going through my school's freshman physics course pretty smoothly, but momentum seems to be a difficult concept for me to master. I understand all the equations and how the variables relate to one another, but we've gotten to the law of conservation of momentum and a lot of extremely basic questions (that refer back to Newton's Third Law) are making me question my understanding.

I'm not really understanding the conservation of momentum regarding the fact that "before" momentum is the same as "after" momentum. Is it a millisecond before the event? 2 minutes before event? Here's an example of a problem included in the lesson on momentum conservation (answer is shown on the website, but I'm just looking for further in-depth explanation):

http://www.physicsclassroom.com/class/momentum/Lesson-2/Momentum-Conservation-Principle - (you can see all the question/answer sets if you scroll to the bottom :) )

#4: If a ball is projected upward from the ground with ten units of momentum, what is the momentum of recoil of the Earth? Do we feel this? Explain.

• This problem basically sums up what I was saying. How do I know if the ball was at rest and then thrown up (delta p would of course be 10 in this case) vs observing the ball right before and after the event?
• I remember doing a gun recoil example in class - my teacher said that when the bullet is fired, while the bullet does have a high velocity and therefore high momentum, the gun has a total momentum of 0 ? (is gun momentum+bullet momentum not >1?).
• So yes, the answer on the website says 10 units downward to match the upward momentum - but why? How would you express that in an equation/formula?

Here's another one:

#1: When fighting fires, a firefighter must use great caution to hold a hose that emits large amounts of water at high speeds. Why would such a task be difficult?

• I understand that by the third law, if the hose is pushing the water forward, the water must be pushing the hose back. But how does this relate to momentum conservation? How would you express the situation with variables like p, m, v, etc. (or would you not?)

I have many more questions, but I don't want to spam, so this will be my last one:

A clown is on the ice rink with a large medicine ball. If the clown throws the ball forward, then he is set into backwards motion with the same momentum as the ball's forward momentum. What would happen to the clown if he goes through the motion of throwing the ball without actually letting go of it? Explain.

http://www.physicsclassroom.com/class/momentum/Lesson-2/The-Law-of-Action-Reaction-(Revisited)

• Okay, so momentum before the event is 0, correct?
• I also know that because he is going into backward motion with the same momentum as the ball's forward momentum (let b=ball and c=clown): pc=pb. But where do I go from there? I know they have the same change in momentum, but how would that explain what would happen if the clown "goes through the motion of throwing the ball without actually letting go"?

Homework Helper
Gold Member
"before" momentum is the same as "after" momentum. Is it a millisecond before the event? 2 minutes before event?
as long as there are no external forces acting on the system under consideration, momentum is constant.
How do I know if the ball was at rest and then thrown up (delta p would of course be 10 in this case) vs observing the ball right before and after the event?
I don't understand what distinction you are making. The change in momentum of the ball (whether it was at rest or not) is equal and opposite to the change in momentum of the rest of the system (earth+thrower).
my teacher said that when the bullet is fired, while the bullet does have a high velocity and therefore high momentum, the gun has a total momentum of 0 ?
Maybe some misunderstanding here. For simplicity, suppose the gunner is standing on ice. If gun and bullet are at rest before pulling the trigger, the gun+gunner's momentum as the bullet leaves the muzzle is equal and opposite to the bullet's momentum.
• the answer on the website says 10 units downward to match the upward momentum - but why? How would you express that in an equation/formula?
How would you write a conservation of momentum equation? You've surely seen one.
When fighting fires, a firefighter must use great caution to hold a hose that emits large amounts of water at high speeds. Why would such a task be difficult?
This one is not straightforward. If the hose is held straight, it would seem there's no difficulty. The water is not undergoing a change of momentum whilst in the hose. The problems start when the fireman attempts to direct the jet to different targets, putting a bend in the hose. Momentum is a vector, so negotiating that bend involves a change of momentum, and the fireman may experience a considerable sideways force. With that in mind, even when trying to hold the hose straight it may be hard to control since small movements may become magnified.
What would happen to the clown if he goes through the motion of throwing the ball without actually letting go of it?
Think through how this would feel. You get the ball up to some speed. As you are about to release it, you're not pushing any more, so there's not much force between hand and ball. Then you suddenly decide to hang on to the ball. What will you feel?

SkyrimKhajiit

I don't understand what distinction you are making. The change in momentum of the ball (whether it was at rest or not) is equal and opposite to the change in momentum of the rest of the system (earth+thrower).

How would you write a conservation of momentum equation? You've surely seen one.

I didn't imagine a thrower since the ball is coming out of the ground and there is recoil on Earth. I just don't understand what law says that there has to be an equal and opposite recoil (didn't say force --> third law). Would it have to do with impulse (F*t)?

And the only equations we've gotten specifically for conservation of momentum are that delta p1=-delta p2. Of course we still have our original momentum and impulse equations, but all we're told about conservation of momentum is the law itself (total momentum is conserved --> equal)

Think through how this would feel. You get the ball up to some speed. As you are about to release it, you're not pushing any more, so there's not much force between hand and ball. Then you suddenly decide to hang on to the ball. What will you feel?

Well I thought there would still be some motion since you're hanging there with the ball in front of you - my teacher says that since you didn't throw it, there is no forward momentum, and therefore no backward momentum..?

Homework Helper
Gold Member
the ball is coming out of the ground and there is recoil on Earth
It was a long link, and you did not specify which part you were referring to. Makes no difference whether it's a human thrower or a mortar or whatever.
specifically for conservation of momentum are that delta p1=-delta p2
OK, and you can write those ##\Delta p##s in terms of masses and and initial and final velocities?
I thought there would still be some motion since you're hanging there with the ball in front of you
Again, suppose you're standing on ice. As you push the ball forwards, your feet will slide backwards. When about to release the ball, the ball has forward momentum and you have equal and opposite momentum. But then you hold on to the ball and bring it to rest. In so doing, the forward pull of the ball stops you sliding back. You and the ball are now at rest, with the ball in your outstretched arms. Your feet are a bit further back than originally, but no longer moving.
Finally, you bring the ball back to your chest. Your feet slide forward again. Not only has the total momentum been zero throughout, the combined mass centre of you and ball has not moved throughout.

SkyrimKhajiit
Sorry, I guess we had a bit of miscommunication...

Again, suppose you're standing on ice. As you push the ball forwards, your feet will slide backwards. When about to release the ball, the ball has forward momentum and you have equal and opposite momentum. But then you hold on to the ball and bring it to rest. In so doing, the forward pull of the ball stops you sliding back. You and the ball are now at rest, with the ball in your outstretched arms. Your feet are a bit further back than originally, but no longer moving.
Finally, you bring the ball back to your chest. Your feet slide forward again. Not only has the total momentum been zero throughout, the combined mass centre of you and ball has not moved throughout.

But wow, this explains it perfectly, thank you.

And regarding the ball/Earth situation, just to make sure, momentum before the event=0, correct? Therefore, 0=p1+p2? (Let u=unknown units) So 0=10u+p2-->-10u=p2. Is that how to somewhat logically prove it? I'm not worried about what threw it, but rather why it is that momentums should be equal and opposite (which law).

Thanks.

Homework Helper
Gold Member
And regarding the ball/Earth situation, just to make sure, momentum before the event=0, correct? Therefore, 0=p1+p2? (Let u=unknown units) So 0=10u+p2-->-10u=p2. Is that how to somewhat logically prove it? I'm not worried about what threw it, but rather why it is that momentums should be equal and opposite (which law).

Thanks.
Yes.

e^(i Pi)+1=0
And regarding the ball/Earth situation, just to make sure, momentum before the event=0, correct? Therefore, 0=p1+p2? (Let u=unknown units) So 0=10u+p2-->-10u=p2. Is that how to somewhat logically prove it? I'm not worried about what threw it, but rather why it is that momentums should be equal and opposite (which law).

Thanks.

Assuming your frame of reference is centered on the Earth so that it is considered stationary, then yes. It might be a lightbulb moment for you to realize that momentum is merely "stored force." It is the measure of force applied to an object over a time, as opposed to energy which is the force applied to an object over a distance.

"For every action there is an equal and opposite reaction." So if one object imparts a force on another object, the first object also feels that force.

F1 = F2
So, m1a1 = m2a2
Multiply both sides time to yeild m1v1 = m2v2 and we have our law for conservation of momentum.

SkyrimKhajiit
Assuming your frame of reference is centered on the Earth so that it is considered stationary, then yes. It might be a lightbulb moment for you to realize that momentum is merely "stored force." It is the measure of force applied to an object over a time, as opposed to energy which is the force applied to an object over a distance.

"For every action there is an equal and opposite reaction." So if one object imparts a force on another object, the first object also feels that force.

F1 = F2
So, m1a1 = m2a2
Multiply both sides time to yeild m1v1 = m2v2 and we have our law for conservation of momentum.
Thank you!

Now I've seen many people write v, not delta v. Is the v simply short for delta v, or is it actually velocity?

e^(i Pi)+1=0
v is some velocity and Δv is a change in velocity, but they tend to be used interchangeably. If you define your initial velocity to equal zero, then of course Δv really will equal v, multiply both sides by mass to say impulse = momentum. Or to be more precise, the sum of the impulses equals final momentum, meaning the sum of the forces over some time = momentum. Which is just a restatement of my earlier claim: momentum is the measure of net force applied to an object over a time.

SkyrimKhajiit
v is some velocity and Δv is a change in velocity, but they tend to be used interchangeably. If you define your initial velocity to equal zero, then of course Δv really will equal v, multiply both sides by mass to say impulse = momentum. Or to be more precise, the sum of the impulses equals final momentum, meaning the sum of the forces over some time = momentum. Which is just a restatement of my earlier claim: momentum is the measure of net force applied to an object over a time.

Thanks again.

So let's say v1 was not 0. Let's say it was 2m/s and v2 was 6m/s. Would that mean that it would have to be mΔv=mΔv (with 1 and 2 subscripts - sorry I just looked up how to use latex lol :D)

Last edited:
e^(i Pi)+1=0
What?

SkyrimKhajiit
What?

Homework Helper
Gold Member
Thanks again.

So let's say v1 was not 0. Let's say it was 2m/s and v2 was 6m/s. Would that mean that it would have to be mΔv=mΔv (with 1 and 2 subscripts - sorry I just looked up how to use latex lol :D)
Let me make a small correction to e^(i Pi)+1=0's posts. Using consistent signs for forces etc., if body 1 exerts force F2 on body 2 then body 2 exerts force F1 = -F2 on body 1. So if body 1, mass m1, undergoes a momentum change ##m_1\Delta v_1## as a result, body 2 undergoes a momentum change ##m_2\Delta v_2 = -m_1\Delta v_1##. Thus ##\Sigma m_i \Delta v_i = 0##.

e^(i Pi)+1=0
Thanks again.

So let's say v1 was not 0. Let's say it was 2m/s and v2 was 6m/s. Would that mean that it would have to be mΔv=mΔv (with 1 and 2 subscripts - sorry I just looked up how to use latex lol :D)

If they were acting on one another then yes. Initial total momentum of the system would sum to some non-zero number, x, and the final momentum would also sum to x.

SkyrimKhajiit
Let me make a small correction to e^(i Pi)+1=0's posts. Using consistent signs for forces etc., if body 1 exerts force F2 on body 2 then body 2 exerts force F1 = -F2 on body 1. So if body 1, mass m1, undergoes a momentum change ##m_1\Delta v_1## as a result, body 2 undergoes a momentum change ##m_2\Delta v_2 = -m_1\Delta v_1##. Thus ##\Sigma m_i \Delta v_i = 0##.

So I'm trying to learn the equations even though technically we haven't learned them. Would your first equation (m2Δv2=−m1Δv1) regard the change in momenta or the total momenta?