M1 impulse question

OP

here is my working. Thanks.

Reply 2

atsruser

11

Original post by coconut64

here is my working. Thanks.

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}

.

Reply 3

Notnek

21

Original post by coconut64

here is my working. Thanks.

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.

Reply 4

OP

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

Reply 5

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.

Reply 6

OP

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)?

(edited 8 years ago)

Reply 7

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

.

Reply 8

OP

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)?

Reply 9

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.

Reply 10

OP

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...

Reply 11

Original post by coconut64

So is that right?? Only that makes sense to me...

Uh, yeah - seems fine to me.

Reply 12

OP

Original post by Zacken

Uh, yeah - seems fine to me.

Cheers.

Reply 13

atsruser

11

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.

Reply 14

atsruser

11

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.

Reply 15

OP