coconut64
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Size:  169.3 KB for 2b) the answer is 900n but I don't understand how it's worked out. I used the I=FT method. I worked out the impulse which is 2880 but in the markscheme impulse is simply 3000*2.4. I don't really get what happened here. Thanks
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coconut64
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Size:  312.8 KB here is my working. Thanks.
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atsruser
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(Original post by coconut64)
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I don't follow your calculation. The required impulse to bring the coupled units to a halt is

I = mv-mu = 3000 \times 2.4 - 3000 \times 0 = 3000 \times 2.4 \\ = 7200 \text{ N s} .
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Notnek
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(Original post by coconut64)
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You have worked out the impulse applied to A in the collision. But part (b) is about the force applied after the collision.

You should be working out the impulse applied to the combined mass that brings the trucks to rest.
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coconut64
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(Original post by atsruser)
I don't follow your calculation. The required impulse to bring the coupled units to a halt is

I = mv-mu = 3000 \times 2.4 - 3000 \times 0 = 3000 \times 2.4 \\ = 7200 \text{ N s} .
Is this for particle b or a ? Thanks
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Zacken
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(Original post by coconut64)
Is this for particle b or a ? Thanks
"coupled unit" means that B and A are now struck together and treated as one particle.
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coconut64
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(Original post by Zacken)
"coupled unit" means that B and A are now struck together and treated as one particle.
I understand this but where does 3000*0 come from?? Thanks
The particles are at rest after 8s. So would it be I=3000(0) - 3000(-2.4)?
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Zacken
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(Original post by coconut64)
I understand this but where does 3000*0 come from?? Thanks
Halt: 0 \, \text{ms}^{-1}. So m \times \text{velocity at halt} = 3000 \times 0 = 0.
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coconut64
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(Original post by Zacken)
Halt: 0 \, \text{ms}^{-1}. So m \times \text{velocity at halt} = 3000 \times 0 = 0.
The particles are at rest after 8s. So would it be I=3000(0) - 3000(-2.4)?
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Zacken
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(Original post by coconut64)
The particles are at rest after 8s. So would it be I=3000(0) - 3000(-2.4)?
Which is precisely the same thing as what atsuser wrote, just in a different format.
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coconut64
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(Original post by Zacken)
Which is precisely the same thing as what atsuser wrote, just in a different format.
So is that right?? Only that makes sense to me...
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Zacken
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(Original post by coconut64)
So is that right?? Only that makes sense to me...
Uh, yeah - seems fine to me.
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coconut64
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(Original post by Zacken)
Uh, yeah - seems fine to me.
Cheers.
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atsruser
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(Original post by coconut64)
The particles are at rest after 8s. So would it be I=3000(0) - 3000(-2.4)?
There's a sign error in my earlier post - I swapped v and u - but it doesn't make much difference to the principle, or to the result. You're still trying to destroy 7200 N s of momentum by applying a frictional force, regardless of which way the vector is pointing.
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atsruser
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(Original post by coconut64)
The particles are at rest after 8s. So would it be I=3000(0) - 3000(-2.4)?
I can't see why you've written -2.4 - you haven't explicitly shown a +ve direction in your diagram, and I would have thought that it would be more natural to take "right = +ve" here.

It's a good idea always to label the direction of +ve vectors clearly before you start work - it's easy to become confused otherwise.
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coconut64
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(Original post by atsruser)
I can't see why you've written -2.4 - you haven't explicitly shown a +ve direction in your diagram, and I would have thought that it would be more natural to take "right = +ve" here.

It's a good idea always to label the direction of +ve vectors clearly before you start work - it's easy to become confused otherwise.
The direction of impulse would be to the right sorry I can't really make it clear by typing it out.Thanks.
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