# Impulse question (mechanics M1) Watch

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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

(Original post by

Hi, can someone explain why the Impulse of 14 N s turns to negative and what are the lines either side of it please?

**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?

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#3

**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?

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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#4

(Original post by

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

**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

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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#5

(Original post by

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?

**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?

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.

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#6

(Original post by

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.

**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.Thanks for clearing that up .

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#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.

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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#8

(Original post by

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.

**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.

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.

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#9

(Original post by

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:

**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:

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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#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!

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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#11

(Original post by

I also don't like the solution given for this question - it's just going to confuse students who read it.

**notnek**)I also don't like the solution given for this question - it's just going to confuse students who read it.

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