# AQA Maths Mechanics 1B Jan 23rd 2013 Watch

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#24

(Original post by

what did everyone get for V and alpha?

**J10**)what did everyone get for V and alpha?

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#27

(Original post by

yeah buddy! I think...

**ollyhal**)yeah buddy! I think...

What do you reckon the grade boundaries will be like?

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#29

Right, that was the hardest Mechanics exam I've done. The only two that I'm really unsure of are:

For the colliding particles:

Momentum A = 5*4 = 20

Momentum B = 4*-3 = -12

It was given that the final velocity of A was 0.6ms, however we weren't given direction.

So I figured that particle A, of mass 5kg, therefore had a final momentum of 5*(+/-0.6) = 3 or -3.

With momentum at -3, it would mean particle A had seen a change in momentum of 20-3 = 17

Therefore particle B would have a new momentum of -12+17 = 5

Giving it a final velocity of 5/4 =

OR, particle B would have a new momentum of -12+23 =11

Giving it a final velocity of 11/4 =

For the boat on the river... I first considered that it needs to double it's velocity against the water, to compensate for the 3ms acting against it. So velocity against the water needs to be 6.

Then I considered that a velocity of 6ms combined with a velocity of 4ms across the river will result in the wrong angle of approach. This means that if the boat travels at 6m against the flow, it will need have a different velocity perpendicular to it, to maintain the 53 degree angle.

So... the velocity

So, to maintain the exact same angle of approach, and have the same velocity, resulting in exactly the same return path, I figured that it would need to be 6ms against the river, resulting in an observed velocity of 3ms, combined with the required 8ms at that angle. So, its observed velocity is sqrt(3^2 + 8^2) =

And just to be sure, I also wrote "relative to the water, the boat will be going at 10ms"

Opinions on that?

For the colliding particles:

Momentum A = 5*4 = 20

Momentum B = 4*-3 = -12

It was given that the final velocity of A was 0.6ms, however we weren't given direction.

So I figured that particle A, of mass 5kg, therefore had a final momentum of 5*(+/-0.6) = 3 or -3.

With momentum at -3, it would mean particle A had seen a change in momentum of 20-3 = 17

Therefore particle B would have a new momentum of -12+17 = 5

Giving it a final velocity of 5/4 =

**1.25ms**OR, particle B would have a new momentum of -12+23 =11

Giving it a final velocity of 11/4 =

**2.75ms**For the boat on the river... I first considered that it needs to double it's velocity against the water, to compensate for the 3ms acting against it. So velocity against the water needs to be 6.

Then I considered that a velocity of 6ms combined with a velocity of 4ms across the river will result in the wrong angle of approach. This means that if the boat travels at 6m against the flow, it will need have a different velocity perpendicular to it, to maintain the 53 degree angle.

So... the velocity

*across*the river is found in 6tan53.1 = 8msSo, to maintain the exact same angle of approach, and have the same velocity, resulting in exactly the same return path, I figured that it would need to be 6ms against the river, resulting in an observed velocity of 3ms, combined with the required 8ms at that angle. So, its observed velocity is sqrt(3^2 + 8^2) =

**8.54**msAnd just to be sure, I also wrote "relative to the water, the boat will be going at 10ms"

Opinions on that?

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#30

(Original post by

Right, that was the hardest Mechanics exam I've done. The only two that I'm really unsure of are:

For the colliding particles:

Momentum A = 5*4 = 20

Momentum B = 4*-3 = -12

It was given that the final velocity of A was 0.6ms, however we weren't given direction.

So I figured that particle A, of mass 5kg, therefore had a final momentum of 5*(+/-0.6) = 3 or -3.

With momentum at -3, it would mean particle A had seen a change in momentum of 20-3 = 17

Therefore particle B would have a new momentum of -12+17 = 5

Giving it a final velocity of 5/4 =

OR, particle B would have a new momentum of -12+23 =11

Giving it a final velocity of 11/4 =

For the boat on the river... I first considered that it needs to double it's velocity against the water, to compensate for the 3ms acting against it. So velocity against the water needs to be 6.

Then I considered that a velocity of 6ms combined with a velocity of 4ms across the river will result in the wrong angle of approach. This means that if the boat travels at 6m against the flow, it will need have a different velocity perpendicular to it, to maintain the 53 degree angle.

So... the velocity

So, to maintain the exact same angle of approach, and have the same velocity, resulting in exactly the same return path, I figured that it would need to be 6ms against the river, resulting in an observed velocity of 3ms, combined with the required 8ms at that angle. So, its observed velocity is sqrt(3^2 + 8^2) =

And just to be sure, I also wrote "relative to the water, the boat will be going at 10ms"

Opinions on that?

**stuart_aitken**)Right, that was the hardest Mechanics exam I've done. The only two that I'm really unsure of are:

For the colliding particles:

Momentum A = 5*4 = 20

Momentum B = 4*-3 = -12

It was given that the final velocity of A was 0.6ms, however we weren't given direction.

So I figured that particle A, of mass 5kg, therefore had a final momentum of 5*(+/-0.6) = 3 or -3.

With momentum at -3, it would mean particle A had seen a change in momentum of 20-3 = 17

Therefore particle B would have a new momentum of -12+17 = 5

Giving it a final velocity of 5/4 =

**1.25ms**OR, particle B would have a new momentum of -12+23 =11

Giving it a final velocity of 11/4 =

**2.75ms**For the boat on the river... I first considered that it needs to double it's velocity against the water, to compensate for the 3ms acting against it. So velocity against the water needs to be 6.

Then I considered that a velocity of 6ms combined with a velocity of 4ms across the river will result in the wrong angle of approach. This means that if the boat travels at 6m against the flow, it will need have a different velocity perpendicular to it, to maintain the 53 degree angle.

So... the velocity

*across*the river is found in 6tan53.1 = 8msSo, to maintain the exact same angle of approach, and have the same velocity, resulting in exactly the same return path, I figured that it would need to be 6ms against the river, resulting in an observed velocity of 3ms, combined with the required 8ms at that angle. So, its observed velocity is sqrt(3^2 + 8^2) =

**8.54**msAnd just to be sure, I also wrote "relative to the water, the boat will be going at 10ms"

Opinions on that?

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#31

**stuart_aitken**)

Right, that was the hardest Mechanics exam I've done. The only two that I'm really unsure of are:

For the colliding particles:

Momentum A = 5*4 = 20

Momentum B = 4*-3 = -12

It was given that the final velocity of A was 0.6ms, however we weren't given direction.

So I figured that particle A, of mass 5kg, therefore had a final momentum of 5*(+/-0.6) = 3 or -3.

With momentum at -3, it would mean particle A had seen a change in momentum of 20-3 = 17

Therefore particle B would have a new momentum of -12+17 = 5

Giving it a final velocity of 5/4 =

**1.25ms**

OR, particle B would have a new momentum of -12+23 =11

Giving it a final velocity of 11/4 =

**2.75ms**

For the boat on the river... I first considered that it needs to double it's velocity against the water, to compensate for the 3ms acting against it. So velocity against the water needs to be 6.

Then I considered that a velocity of 6ms combined with a velocity of 4ms across the river will result in the wrong angle of approach. This means that if the boat travels at 6m against the flow, it will need have a different velocity perpendicular to it, to maintain the 53 degree angle.

So... the velocity

*across*the river is found in 6tan53.1 = 8ms

So, to maintain the exact same angle of approach, and have the same velocity, resulting in exactly the same return path, I figured that it would need to be 6ms against the river, resulting in an observed velocity of 3ms, combined with the required 8ms at that angle. So, its observed velocity is sqrt(3^2 + 8^2) =

**8.54**ms

And just to be sure, I also wrote "relative to the water, the boat will be going at 10ms"

Opinions on that?

How many marks was it worth?

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#32

**stuart_aitken**)

Right, that was the hardest Mechanics exam I've done. The only two that I'm really unsure of are:

For the colliding particles:

Momentum A = 5*4 = 20

Momentum B = 4*-3 = -12

It was given that the final velocity of A was 0.6ms, however we weren't given direction.

So I figured that particle A, of mass 5kg, therefore had a final momentum of 5*(+/-0.6) = 3 or -3.

With momentum at -3, it would mean particle A had seen a change in momentum of 20-3 = 17

Therefore particle B would have a new momentum of -12+17 = 5

Giving it a final velocity of 5/4 =

**1.25ms**

OR, particle B would have a new momentum of -12+23 =11

Giving it a final velocity of 11/4 =

**2.75ms**

For the boat on the river... I first considered that it needs to double it's velocity against the water, to compensate for the 3ms acting against it. So velocity against the water needs to be 6.

Then I considered that a velocity of 6ms combined with a velocity of 4ms across the river will result in the wrong angle of approach. This means that if the boat travels at 6m against the flow, it will need have a different velocity perpendicular to it, to maintain the 53 degree angle.

So... the velocity

*across*the river is found in 6tan53.1 = 8ms

So, to maintain the exact same angle of approach, and have the same velocity, resulting in exactly the same return path, I figured that it would need to be 6ms against the river, resulting in an observed velocity of 3ms, combined with the required 8ms at that angle. So, its observed velocity is sqrt(3^2 + 8^2) =

**8.54**ms

And just to be sure, I also wrote "relative to the water, the boat will be going at 10ms"

Opinions on that?

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#35

And for the one with the golf ball, I got:

Horizontal velocity = 38.4/2.4 =

Vertical velocity:

It's given that the ball stopped at 3m, so,

Using s=ut+(1/2)at^2

u=(s-(1/2)at^2)/t

u = (3-(1/2)-9.8*2.4^2)/2.4 = 13.01ms

Therefore V = sqrt(16^2 + 13.01^2) =

Angle = tan-1(13.01/16) =

For the question with the tractor and trailer, I can't remember my exact answers but it was something like:

Resistance ~580

Tension ~1000

Tension ~1000

Opinions?

Anybody remember the values of the question with the block on the slope?

Horizontal velocity = 38.4/2.4 =

**16ms**Vertical velocity:

It's given that the ball stopped at 3m, so,

Using s=ut+(1/2)at^2

u=(s-(1/2)at^2)/t

u = (3-(1/2)-9.8*2.4^2)/2.4 = 13.01ms

Therefore V = sqrt(16^2 + 13.01^2) =

**20.6ms**Angle = tan-1(13.01/16) =

**39.1 degrees**For the question with the tractor and trailer, I can't remember my exact answers but it was something like:

Resistance ~580

Tension ~1000

Tension ~1000

Opinions?

Anybody remember the values of the question with the block on the slope?

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#36

(Original post by

What did people get for the time for 7c? where it said it was travelling north west?

**Gary**)What did people get for the time for 7c? where it said it was travelling north west?

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#37

**Gary**)

What did people get for the time for 7c? where it said it was travelling north west?

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#38

(Original post by

For the momentum question I got those two answers however I am not one to say how to do the river crossing as I got it wrong but if you look back in the thread people have come up with 1.4ms.

**ollyhal**)For the momentum question I got those two answers however I am not one to say how to do the river crossing as I got it wrong but if you look back in the thread people have come up with 1.4ms.

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#39

(Original post by

44.something?

**J10**)44.something?

(Original post by

I got 12 seconds and my final answer was 59.2 I thought that the two particles were moving in the same direction so the total momentum was 32?

**STraynor**)I got 12 seconds and my final answer was 59.2 I thought that the two particles were moving in the same direction so the total momentum was 32?

quite alot got 12

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