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# A level maths- mechanics M1 watch

1. Anyone here plzzz need help with mechnics....
2. ....Okay? What exactly do you need help with? You haven't provided us with a question.
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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???
5. HELOOOOOOOOOOOOOOOOOOOO any one???????????????????
6. I NEED HELP!!! plsssss someone help me with mechanics I am failing
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
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.
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.
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
11. Try this, I hopefully simplified enough. Ignore my terrible handwriting.
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
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
14. (Original post by TheTechnoGuy)
Beat me to it, aha
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. [6]

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
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.
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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