Poll of the Day > Physics question regarding momentum and impulse

(+) Yeah! (+)

My answer key differs from what I'm getting for this question. I'd like to hear what some of you think.



I'm getting 442 by the way.
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i don't have a fucking clue, bro.
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dyodhw
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What's the answer then?
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Blighboy posted...

dyodhw

I'm the teacher.

shadowsword87 posted...

What's the answer then?

The answer key says 431 which I'm very confident is wrong.

Impulse is a change in momentum, or mass times the change in velocity. Cart 1 will have its velocity change more towards the left. Therefore the impulse should be to the left.

As for the magnitude of the impulse, impulse can also be found by taking the force and multiplying it by the time. The force on each cart should be the same (Newton's third law) and the time of contact will also be the same, therefore the magnitude of the impulse should be the same.

The only reason I'm questioning it is because it's a resource provided by the government and it is extremely rare for them to be wrong in their educational resources.
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I'm not 100% sure I know when the time is which is where the confusion is happening.

Breaking down the formula, impulse = (delta)F * (delta)t.
There isn't an initial force going to the right, during the collision there's a force going to the left, and then no force. So it goes, 0 -> (whatever value) left -> 0
If you choose to read the time right after the collision, you are reading a left force going to 0, which is an impulse to the right.
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I always thought you were a social studies teacher. Never knew you were doing science like me

I see what you're saying, but the question does say "when the two carts are in contact".
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green dragon posted...

I always thought you were a social studies teacher. Never knew you were doing science like me

Mostly math but I've also got physics. I've taught some social before.
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how can you say that anything goes laterally when they're symmetrical and moving down a straight line? how can you say that anything goes to one side but not the other?

nevermind, I see, left and right are forward and backward, not turning... does that mean the diagonals are up and down I guess?
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Sahuagin posted...

does that mean the diagonals are up and down I guess?

Nah, those are the "x"s and dots.
X is in, and a dot is out.
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yeah from the little I've read (I don't even have grade 10 physics) I don't see how that could be correct. impulse is supposed to be a measurement of how much the force modified the momentum. 3 seems to mean that the momentum of cart 1 increased to the right? that can't be the case. the momentum of cart 1 is moved to the left, not the right.
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Sahuagin posted...

yeah from the little I've read (I don't even have grade 10 physics) I don't see how that could be correct. impulse is supposed to be a measurement of how much the force modified the momentum. 3 seems to mean that the momentum of cart 1 increased to the right? that can't be the case. the momentum of cart 1 is moved to the left, not the right.

Agreed. A left impulse will cause the velocity of Cart 1 change to the left. There is no way that the velocity of Cart 1 will change to the right.

As for the magnitude of the impulse, I am still confident that they will be the same for both carts.
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The force acting upon cart 1 at the point of contact is the momentum of cart 2. Ergo, the direction of the impulse on (not of) cart 1 is the direction of the momentum of cart 2.

The relative size of the magnitude between the impulse of (not on) cart 1 (i.e., the force it applies to cart 2) is twice as large as the impulse of (not on) cart 2 (i.e. the force it applies to cart 1).
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dainkinkaide posted...

The force acting upon cart 1 at the point of contact is the momentum of cart 2. Ergo, the direction of the impulse on (not of) cart 1 is the direction of the momentum of cart 2.

The momentum of Cart 2 is to the left, so the impulse on Cart 1 should also be to the left. That's what you're saying, correct?

dainkinkaide posted...

The relative size of the magnitude between the impulse of (not on) cart 1 (i.e., the force it applies to cart 2) is twice as large as the impulse of (not on) cart 2 (i.e. the force it applies to cart 1).

I see where you're coming from. I'll look into it again.
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one thing I don't like about the question is the phrasing "For the time interval where the two carts are in contact". an instant in time is not really an interval; delta-t would be zero. I guess the interval must be from before there was collision to immediately after the collision?

the more I think of it, the more it makes sense that whatever the first two numbers are, they must be the same. impulse is like accumulating force. it's going to be a vector in the same direction as the force, just a question of magnitude. (think of a space vehicle in orbit. apply thrust, that is your force. the impulse is the accumulation of that force in the direction(s) that it is applied.)

for the magnitude of the impulse... it seems like it might be double or half. I'm a bit confused now by the phrasing "impulse of" VS "impulse on".

I'm not sure of the formulas and wording, but I think that cart 1 has (or applies) twice the force of cart 2. cart 1 receives half the force that cart 2 receives. (call velocity 1 unit, then force is 500 and 250 or something like that.) they are affected for the same duration, so cart 2 will have its momentum modified by twice what cart 1's momentum is modified by. my intuition is that cart 1's velocity would be halved (-250 units), whereas cart 2's velocity would be reversed (-500 units).

am now reading about Newton's third law... seems to say that really they both get the sum of the forces or something, so in that case, it would be 750 units each? or whatever the value is, they'd have the same force applied, their own reversed at them, plus the other's. "impulse of" VS "impulse on" wouldn't matter, they both had the same force applied, and their momentums would have undergone similar modifications (just opposite directions). that feels counter-intuitive though. a semi-truck hitting a motor-cycle doesn't have its own force reflected back at it...
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dainkinkaide posted...

The relative size of the magnitude between the impulse of (not on) cart 1 (i.e., the force it applies to cart 2) is twice as large as the impulse of (not on) cart 2 (i.e. the force it applies to cart 1).

Try this application:

https://goo.gl/ieyY95

Regardless of an elastic or an inelastic collision, the impulse of each cart will be the same. Newton's third law should illustrate this. I'm not convinced that your perspective on this is an accurate interpretation of the question.

Either way, the government answer is wrong, because there is no way the answer for the second blank is 3.
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acceleration takes into account mass, so the smaller cart is accelerated more than the larger cart. but momentum takes mass into account too... so I guess it cancels out. the smaller cart is more "affected", but the momentum change is the same. the smaller cart is moving faster, but has less mass.
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faramir77 posted...

dainkinkaide posted...

The force acting upon cart 1 at the point of contact is the momentum of cart 2. Ergo, the direction of the impulse on (not of) cart 1 is the direction of the momentum of cart 2.

The momentum of Cart 2 is to the left, so the impulse on Cart 1 should also be to the left. That's what you're saying, correct?

Yes.

To put it in a less simplistic way: because the mass is constant, the change in momentum can be simplified to (mass * velocity after impact) - (mass * velocity before impact). Since the magnitude of velocity will obviously be lower after impact than before impact, the difference will be negative, which mirrors the direction of the resultant force. This means the direction of the impulse will be to the left.
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btw, what is this? is this high-school physics, or university physics?
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my answer is 7.
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dainkinkaide posted...

faramir77 posted...

dainkinkaide posted...

The force acting upon cart 1 at the point of contact is the momentum of cart 2. Ergo, the direction of the impulse on (not of) cart 1 is the direction of the momentum of cart 2.

The momentum of Cart 2 is to the left, so the impulse on Cart 1 should also be to the left. That's what you're saying, correct?

Yes.

To put it in a less simplistic way: because the mass is constant, the change in momentum can be simplified to (mass * velocity after impact) - (mass * velocity before impact). Since the magnitude of velocity will obviously be lower after impact than before impact, the difference will be negative, which mirrors the direction of the resultant force. This means the direction of the impulse will be to the left.

Alright, we agree on that. I still don't agree with your response for blank 3.
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Sahuagin posted...

btw, what is this? is this high-school physics, or university physics?

Physics 30 (12th grade) in the Alberta curriculum.
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faramir77 posted...

Physics 30 (12th grade) in the Alberta curriculum.

cool. I took (and barely passed) Chem 30 exactly 20 years ago, but I've never taken Physics anything ever, so it's this big unexplored scientific realm for me. always wanted to try taking Physics 201 at UofC, but never managed to find the right slot for it (and probably wouldn't have gone all that well anyway considering having no previous knowledge of it).
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faramir77 posted...

Alright, we agree on that. I still don't agree with your response for blank 3.

No, shit, you're right. I was forgetting to take the mass of the two carts into account and only thinking about velocity.
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dainkinkaide posted...

faramir77 posted...

Alright, we agree on that. I still don't agree with your response for blank 3.

No, shit, you're right. I was forgetting to take the mass of the two carts into account and only thinking about velocity.

Lol, it's all good.

Thanks for the input everyone. I'll use this question as a means of generating discussion amongst my students rather than as a form of assessment. The digital application I posted a few posts ago will be a good way of visually demonstrating it.
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both of the carts would fall off of the table! look how skinny it is.