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J10
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#21
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#21
what did everyone get for V and alpha?
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J10
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#22
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#22
what did you state for the force acting on the truck?
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J10
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#23
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#23
predictions for grade boundaries?
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ollyhal
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#24
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#24
(Original post by J10)
what did everyone get for V and alpha?
uh I think I got 20.6 and 39.1
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J10
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#25
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#25
anybody get 44.something for 7c?
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ollyhal
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#26
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#26
(Original post by J10)
anybody get 44.something for 7c?
yeah buddy! I think...
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J10
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#27
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#27
(Original post by ollyhal)
yeah buddy! I think...
coooool.
What do you reckon the grade boundaries will be like?
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ollyhal
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#28
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#28
Hopefully like 50 for an A we can dream!
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stuart_aitken
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#29
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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 = 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) = 8.54ms
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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ollyhal
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#30
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#30
(Original post by 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) = 8.54ms
And just to be sure, I also wrote "relative to the water, the boat will be going at 10ms"

Opinions on that?
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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J10
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#31
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#31
(Original post by 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) = 8.54ms
And just to be sure, I also wrote "relative to the water, the boat will be going at 10ms"

Opinions on that?
I got both of those for colliding question.

How many marks was it worth?
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bugsuper
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#32
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#32
(Original post by 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) = 8.54ms
And just to be sure, I also wrote "relative to the water, the boat will be going at 10ms"

Opinions on that?
The question specified that the velocity of the boat relative to the water was still 4ms and that the current remained constant throughout the movement. I think all it could to was to change the angle of travel so that the resultant velocity was across the path AB
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ollyhal
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#33
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#33
(Original post by J10)
I got both of those for colliding question.

How many marks was it worth?
6 marks
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Gary
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#34
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#34
What did people get for the time for 7c? where it said it was travelling north west?
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stuart_aitken
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#35
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#35
And for the one with the golf ball, I got:

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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J10
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#36
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#36
(Original post by Gary)
What did people get for the time for 7c? where it said it was travelling north west?
44.something?
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STraynor
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#37
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#37
(Original post by Gary)
What did people get for the time for 7c? where it said it was travelling north west?
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?
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stuart_aitken
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#38
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#38
(Original post by 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.
Yep that's obvious, now that I think about it. Of course it's gonna appear slower if it's going against the water. DOH!
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Gary
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#39
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#39
(Original post by J10)
44.something?

(Original post by 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?
mhmm i think im the only one got t=24....

quite alot got 12
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ollyhal
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#40
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#40
(Original post by Gary)
mhmm i think im the only one got t=24....

quite alot got 12
definitely got 12s for t!
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