# A level maths- mechanics M1

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#1
Anyone here plzzz need help with mechnics....
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4 years ago
#2
....Okay? What exactly do you need help with? You haven't provided us with a question.
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#3
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#4
so the bit I am confused about is why did they times it by 2? and also ehy h=0 and they divided by the acceleration why???
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#5
HELOOOOOOOOOOOOOOOOOOOO any one???????????????????
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#6
I NEED HELP!!! plsssss someone help me with mechanics I am failing
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#7
(Original post by eden3)
....Okay? What exactly do you need help with? You haven't provided us with a question.
I posted the question so can u help now?
plssssssss
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4 years ago
#8
(Original post by Hard work)
I posted the question so can u help now?
plssssssss
Calm down.. I haven't done A-levels since last June so it's taking me a while to figure out.
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4 years ago
#9
(Original post by Hard work)
I NEED HELP!!! plsssss someone help me with mechanics I am failing
I'm sorry I only understand the alternative answer.
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4 years ago
#10
(Original post by Hard work)

Vertical Motion (initial speed) = 40sina

Recall the equation s=ut+1/2at^2 but as it is decelerating it becomes s=ut-1/2at^2.

Note that in this case a is interchangable with g as it is acceleration due to free fall.

So next you rearrange for t:

s=0 as there is no displacement so you can rule that out.

Add 1/2gt^2 on both sides leaves 1/2gt^2=ut
The t's on both sides cancel out leaving 1/2gt=u
Times both sides by 2 to give you gt=2u
Then divide both sides by g so t=2u/g
Then just substitute T= 2x40xsina/g

T=80sina/g
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4 years ago
#11
Try this, I hopefully simplified enough. Ignore my terrible handwriting.
2
4 years ago
#12
As for the second half of the question it is just Speed=Distance/Time
So we rearrange for D giving D=ST where D is Rm.

Use cos to calculate the horizontal motion (initial speed) which is 40cosa.

R=ST so you now have R=40cosaxT
You know that T=80sina/g from the first part of the question so you just substitute giving...
40xcosax80xsina/g
40x80=3200
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4 years ago
#13
(Original post by Hōseki)
Vertical Motion (initial speed) = 40sina

Recall the equation s=ut+1/2at^2 but as it is decelerating it becomes s=ut-1/2at^2.

Note that in this case a is interchangable with g as it is acceleration due to free fall.

So next you rearrange for t:

s=0 as there is no displacement so you can rule that out.

Add 1/2gt^2 on both sides leaves 1/2gt^2=ut
The t's on both sides cancel out leaving 1/2gt=u
Times both sides by 2 to give you gt=2u
Then divide both sides by g so t=2u/g
Then just substitute T= 2x40xsina/g

T=80sina/g
Beat me to it, aha
0
4 years ago
#14
(Original post by TheTechnoGuy)
Beat me to it, aha  1
4 years ago
#15
(ii) Calculate the acceleration of the car and trailer. Calculate also the force in
the coupling, stating whether this is a tension or a thrust. 

Newton's second Law F=ma

Driving force = 600N
F=ma
600N=2400kgxa
A=0.25m/s^2

For calculating the Tension you also use F=ma
Where F is the tension.
T=900kgx0.25m/s^2
T=225N
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#16
Thanks so much guys really appreciate it.
I have another question I am struggling with,
1. Ann and Beryl are both pushing a piano. Ann pushes in the direction 4i + 3j andBeryl in the direction 12i + 5j, where i and j are the standard unit vectors.Together they produce a force of (68i + 33j) N so that p(4i + 3j) + q(12i + 5j) = 68i + 33j, where p and q are constants.Calculate the force with which Beryl pushes.
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4 years ago
#17
(Original post by Hard work)
Thanks so much guys really appreciate it.
I have another question I am struggling with,
1. Ann and Beryl are both pushing a piano. Ann pushes in the direction 4i + 3j andBeryl in the direction 12i + 5j, where i and j are the standard unit vectors.Together they produce a force of (68i + 33j) N so that p(4i + 3j) + q(12i + 5j) = 68i + 33j, where p and q are constants.Calculate the force with which Beryl pushes.
I believe it's a simultaneous -
4p + 12q = 68
3p + 5q = 33

Take it from here and find the values for p and q.
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