# A level maths- mechanics M1

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#2

....Okay? What exactly do you need help with? You haven't provided us with a question.

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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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(Original post by

....Okay? What exactly do you need help with? You haven't provided us with a question.

**eden3**)....Okay? What exactly do you need help with? You haven't provided us with a question.

plssssssss

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#8

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#9

(Original post by

I NEED HELP!!! plsssss someone help me with mechanics I am failing

**Hard work**)I NEED HELP!!! plsssss someone help me with mechanics I am failing

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#10

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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#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

Final Answer: 3200sinacosa/g

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

Final Answer: 3200sinacosa/g

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#13

(Original post by

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

**HΕseki**)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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#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

Add the masses: 1500+900=2400kg

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

the coupling, stating whether this is a tension or a thrust. [6]

Newton's second Law F=ma

Add the masses: 1500+900=2400kg

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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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 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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#17

(Original post by

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.

**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.

4p + 12q = 68

3p + 5q = 33

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

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