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Mechanics M1: Post your doubt here

Silent Hunter

Silent Hunter

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sagar65265

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Silent Hunter said:
thanks but can you go through its MS and see..... they have sone somthings like sin 1 and sin 5 etc to get new resistive work done? whys that done?

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_ms_41.pdf
Click to expand...

The work done by the resistive force is the magnitude of the resistive force multiplied by the distance the force moves through; for the frictional/resistive force, this distance is the hypotenuse of each triangle.

The hypotenuse of the left - hand side triangle is given by:

sin(5) = 45/hyp
hyp = 45/sin(5)

So the work done by the resistive force going up the first slope is the magnitude of the force multiplied by the length of the hypotenuse = 45/sin(5).

But question 2 deals with the second triangle, so the length of the right - hand side triangles hypotenuse is given by:

sin(1) = 45/hyp
hyp = 45/sin(1)

So there are three forces acting on the car, the driving force, the frictional force and the gravitational force. Each one of these forces transfers some energy;

i) The frictional force transfers energy from the car to the surroundings, e.g. in the form of thermal energy, where the air particles around the car heat up, etc so the work done by the frictional force is negative work.
ii) The driving force transfers chemical energy in the fuel to kinetic energy of the car so the work it does is a positive value - 1660 kJ as given in the question.
iii) The gravitational force this time has a component down the slope and thus in the direction of motion, so it transfers energy from the potential energy of the earth-car system to the kinetic energy of the car, so it does positive work.

So the sin(1) and the sin(5) values only appear to find out how much work the resistive forces have done, so that the other values required can be calculated.

Hope this helped!
Good Luck for all your exams!
 
Silent Hunter

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sagar65265 said:
The work done by the resistive force is the magnitude of the resistive force multiplied by the distance the force moves through; for the frictional/resistive force, this distance is the hypotenuse of each triangle.

The hypotenuse of the left - hand side triangle is given by:

sin(5) = 45/hyp
hyp = 45/sin(5)

So the work done by the resistive force going up the first slope is the magnitude of the force multiplied by the length of the hypotenuse = 45/sin(5).

But question 2 deals with the second triangle, so the length of the right - hand side triangles hypotenuse is given by:

sin(1) = 45/hyp
hyp = 45/sin(1)

So there are three forces acting on the car, the driving force, the frictional force and the gravitational force. Each one of these forces transfers some energy;

i) The frictional force transfers energy from the car to the surroundings, e.g. in the form of thermal energy, where the air particles around the car heat up, etc so the work done by the frictional force is negative work.
ii) The driving force transfers chemical energy in the fuel to kinetic energy of the car so the work it does is a positive value - 1660 kJ as given in the question.
iii) The gravitational force this time has a component down the slope and thus in the direction of motion, so it transfers energy from the potential energy of the earth-car system to the kinetic energy of the car, so it does positive work.

So the sin(1) and the sin(5) values only appear to find out how much work the resistive forces have done, so that the other values required can be calculated.

Hope this helped!
Good Luck for all your exams!
Click to expand...

thanks but why cant we use simply 360 kj ? i mean they said that its constant throughout ?
 
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sagar65265

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Silent Hunter said:
thanks but why cant we use simply 360 kj ? i mean they said that its constant throughout ?
Click to expand...

Sorry I didn't answer your initial question, I got ahead of myself:(

I'm pretty sure they mean that the work done by the force resisting motion on the journey from A to B was 360 kJ; the work done by the force resisting motion on the journey from B to C should be an entirely different value, because the distance AB is different from the distance BC, so it's only the magnitude of the force that resists motion which is constant throughout, e.g. equal to 500N (or something else, 500N is not the correct value, it's just a random one!) throughout the entire journey from A to C, etc.

In other words, 360 kJ is the work done by the resistive force only from A to B, whereas this question concerns the journey from B to C, during which the resistive force would have done a different amount of work.
The paper saying "The resistance to motion is constant throughout the whole journey." means that the resistive force is constant; the work that this force is done is different from the force itself, so it's not the work done that's constant, it's the force itself that is constant.

Hope this helped!
Good Luck for all your exams!
 
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Jiyad Ahsan said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_41.pdf
Q4(ii), it says in the ms that direction changes at 12 secs.. how do you calc that? i cant seem to get it
Click to expand...


When the direction changes, the value of the velocity is negative; you then need to get the value of t when the velocity slows to zero and changes direction. You can get this time by setting v=0 in the given equation to get,


0.75t^2 − 0.0625t^3 = 0
t^2(0.75 - 0.0625t) = 0

Dividing both sides by t^2,

0.75 - 0.0625t = 0
0.75 = 0.0625t
12 = t

So when t is greater than 12, i.e. after 12 seconds, the velocity will be less than zero, i.e. negative. When t = 12 seconds, the velocity will become zero; after this, it will have a "negative" velocity.

Hope this helped!
Good Luck for all your exams!
 
Jiyad Ahsan

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sagar65265 said:
When the direction changes, the value of the velocity is negative; you then need to get the value of t when the velocity slows to zero and changes direction. You can get this time by setting v=0 in the given equation to get,


0.75t^2 − 0.0625t^3 = 0
t^2(0.75 - 0.0625t) = 0

Dividing both sides by t^2,

0.75 - 0.0625t = 0
0.75 = 0.0625t
12 = t

So when t is greater than 12, i.e. after 12 seconds, the velocity will be less than zero, i.e. negative. When t = 12 seconds, the velocity will become zero; after this, it will have a "negative" velocity.

Hope this helped!
Good Luck for all your exams!
Click to expand...
i swear i did the same thing but somehow the answer i got was 30 :/ i dnt know what i did..
thanx anyway :)
 
qffdhruba

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minie23 said:
Thank you soooo much (Y)
Click to expand...

Hey this is in reply to this:

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s12_qp_43.pdf

Please explain me no. 7(iii) (iv)

first off....

after B hits the ground....there is no tension.....so A decelerates due to gravity as it moves upward....
to find the time it takes to reach max height we do:

2.6/10 = 0.26s (2.6 because tht was the speed of the particles when B hit the ground)

to find the extra distance travelled upwards after B hits the ground:

s = 2.6(0.26) + 0.5(-10)(0.26)^2 = 0.338m

after reaching the max height, it starts to fall again...due to gravity...and returns to a height it was when B hit the ground.....

u can find the time taken to return to this height by solving for t in:

0.338 = 0.5(10)t^2 t = 0.26

so the total time it was in motion:

0.26 + 0.26 + 0.5 = 1.02s
(0.5 while B was moving downwards to the ground, 0.26s while A was moving up after B hit the ground, 0.26s after A has reached its maximum height and returns to a height it was when B hit the ground)

The distance it travels:
0.338*2 + 0.65 = 1.326m

(0.65 befr B hit the ground, 0.338 after B hit the ground, 0.338 again to return to a height it was when B hit the ground)

Graph uploaded.....

Hope i helped....:)
 

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