# Impulse question (mechanics M1)Watch

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Thread starter 2 years ago
#1

Hi, can someone explain why the Impulse of 14 N s turns to negative and what are the lines either side of it please?
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2 years ago
#2
(Original post by jojo55)

Hi, can someone explain why the Impulse of 14 N s turns to negative and what are the lines either side of it please?
The lines mean modulus or magnitude. Since impulse is a vector, it has a direction (for M1 in with motion in straight lines the direction is just whether the value is positive or negative) and magnitude. When working with the collision in terms of A, the impulse it receives is in the direction that B is moving in, which is to the left or along the negative x axis so the value is negative.
1
2 years ago
#3
(Original post by jojo55)
Hi, can someone explain why the Impulse of 14 N s turns to negative and what are the lines either side of it please?
Where has this solution come from? It smacks of someone not understanding the question.

In answer to your question, the vertical lines represent the modulus function

|x| = x for x>= 0 and
|x| = -x for x < 0.

The graph looks like a "v" with the point at the origin.

If you have |y| = 10 say, then either y=10, or y=-10
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2 years ago
#4
(Original post by ghostwalker)
Where has this solution come from? It smacks of someone not understanding the question.

In answer to your question, the vertical lines represent the modulus function

|x| = x for x>= 0 and
|x| = -x for x < 0.

The graph looks like a "v" with the point at the origin.

If you have |y| = 10 say, then either y=10, or y=-10
It is from the Maths Genie website.

If they've used modulus, so it can be either -14 or 14? Why have they chose -14 as I don't get that part?
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2 years ago
#5
(Original post by makin)
It is from the Maths Genie website.

If they've used modulus, so it can be either -14 or 14? Why have they chose -14 as I don't get that part?
Impulse is a vector quantity.

A exerts an impulse on B of 14Ns

So, by Newton's laws, action and reaction....

B exerts an impulse on A of -14Ns.

Considering A. Momentum after = momentum before + impulse, so:

2v = 2x5 - 14

Hence v=-2.

Edit: There is no need to be using the modulus function here.
1
2 years ago
#6
(Original post by ghostwalker)
Impulse is a vector quantity.

A exerts an impulse on B of 14Ns

So, by Newton's laws, action and reaction....

B exerts an impulse on A of -14Ns.

Considering A. Momentum after = momentum before + impulse, so:

2v = 2x5 - 14

Hence v=-2.

Edit: There is no need to be using the modulus function here.
Ah I see it now. I did not know that B exerts an impulse on A of -14Ns.

Thanks for clearing that up .
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2 years ago
#7
He put two sets of working,
One for -14 and one for +14.

Using +14 for impulse, the velocity of A goes to twelve an increase from 2, as it just collided this doenst makes sense, it shouldve slowed down. (Imagine a car crashing and then speeding up as a result of the crash) So the Impulse has a value of -14.
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2 years ago
#8
(Original post by ghostwalker)
Impulse is a vector quantity.

A exerts an impulse on B of 14Ns

So, by Newton's laws, action and reaction....

B exerts an impulse on A of -14Ns.
This is right, based on the diagram and assuming +ve to the right, but I'm not quite sure of your line of reasoning. I would:

a) draw a big +ve arrow pointing rightwards
b) point out that since then by the direction of travel, the exerted by B on A is directed -vely, and so then is the impulse, so I=-14 N s.

However, based on the wording of the question, we could also draw a mirror image diagram, and the signs would change.

Edit: There is no need to be using the modulus function here.
I agree. Whichever way round we draw the diagram, we can figure out the direction of impulse and do a single calculation.
1
2 years ago
#9
(Original post by atsruser)
This is right, based on the diagram and assuming +ve to the right, but I'm not quite sure of your line of reasoning. I would:
Yes, I am assuming the direction of motion of the initial mass (A) is the positive direction, though I didn't state it.

Although the question only says the magnitude of the impulse of A on B is 14 Ns, I think it's clear, with the above, that as a vector it is +14 Ns.

And the rest follows.
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2 years ago
#10
I also don't like the solution given for this question - it's just going to confuse students who read it.

The impulse exerted on B by A has magnitude 14 so it must be +14 pointing to the right so you just work out the change in momentum for particle B taking right as positive and set it equal to +14. Or you do the reverse for the impulse on B by A. Simple!
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2 years ago
#11
(Original post by notnek)
I also don't like the solution given for this question - it's just going to confuse students who read it.
I await more threads arising due to the Maths Genie website.
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