# Edexcel M4 Exam Discussion Thread [June 5th] Watch

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#81

(Original post by

For question 4 on M4 June 2011 why are the vector triangles of the mark scheme not head to tail? The 'head' of the true velocity of the Wind meets the 'head' of the hiker.

Also the mark scheme doesn't show the vector approach I used but I assume in any relative motion question that either a vector or geometrical approach is permitted?

**Gawain**)For question 4 on M4 June 2011 why are the vector triangles of the mark scheme not head to tail? The 'head' of the true velocity of the Wind meets the 'head' of the hiker.

Also the mark scheme doesn't show the vector approach I used but I assume in any relative motion question that either a vector or geometrical approach is permitted?

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#82

**Gawain**)

For question 4 on M4 June 2011 why are the vector triangles of the mark scheme not head to tail? The 'head' of the true velocity of the Wind meets the 'head' of the hiker.

Also the mark scheme doesn't show the vector approach I used but I assume in any relative motion question that either a vector or geometrical approach is permitted?

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#83

**Gawain**)

For question 4 on M4 June 2011 why are the vector triangles of the mark scheme not head to tail? The 'head' of the true velocity of the Wind meets the 'head' of the hiker.

Also the mark scheme doesn't show the vector approach I used but I assume in any relative motion question that either a vector or geometrical approach is permitted?

**both methods are allowed.**

Vector component methods are conceptually easier but if the angles are ugly they are a nightmare.

(I remember well after my degree I still could not understand geometric methods. But one summer a year 12 indicated to me that he was planning to take a mechanics module which involved this stuff.

Necessity is the mother of invention. It took about a week of frustration and I eventually understood what is going on.

I can assure you geometric methods are far superior for practically every problem and it is a real shame you are so close to your exam.)

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#84

(Original post by

Do you mind outlining you other approach to these questions? I've tried doing some of these questions purely by vectors, but the resulting quartic equation I get is practically unsolvable in exam conditions.

**ThatPerson**)Do you mind outlining you other approach to these questions? I've tried doing some of these questions purely by vectors, but the resulting quartic equation I get is practically unsolvable in exam conditions.

So

Also

Meaning:

Therefore:

Now sub back in and use arctan to get a bearing of 25 and Pythagoras to get |v|=7.6.

June 2012 Q4

Therefore

Subbing back into the i components gives

Which gives theta=55.2, this equates to a bearing of 34.8. Also subbing theta back into our time expression gives t=0.646 which is 39 minutes.

For the first question I found the vector method to be quicker but for the second I did it geometrically first.

Does anyone have any tips for making sure your vector arrows are on the correct sides? There seem to be various different permutations especially if one of the speeds is initially unknown. Is it to do with the relative starting positions?

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#85

(Original post by

**Gawain**)
Spoiler:

Show

June 2011 Q4)

So

Also

Meaning:

Therefore:

Now sub back in and use arctan to get a bearing of 25 and Pythagoras to get |v|=7.6.

June 2012 Q4

Therefore

Subbing back into the i components gives

Which gives theta=55.2, this equates to a bearing of 34.8. Also subbing theta back into our time expression gives t=0.646 which is 39 minutes.

For the first question I found the vector method to be quicker but for the second I did it geometrically first.

Does anyone have any tips for making sure your vector arrows are on the correct sides? There seem to be various different permutations especially if one of the speeds is initially unknown. Is it to do with the relative starting positions?

So

Also

Meaning:

Therefore:

Now sub back in and use arctan to get a bearing of 25 and Pythagoras to get |v|=7.6.

June 2012 Q4

Therefore

Subbing back into the i components gives

Which gives theta=55.2, this equates to a bearing of 34.8. Also subbing theta back into our time expression gives t=0.646 which is 39 minutes.

For the first question I found the vector method to be quicker but for the second I did it geometrically first.

Does anyone have any tips for making sure your vector arrows are on the correct sides? There seem to be various different permutations especially if one of the speeds is initially unknown. Is it to do with the relative starting positions?

Think I might use this instead of geometry in the exam. I get quite confused when drawing the diagram for more complicated questions.

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#86

What kind of difficulty do you expect for that M4? Just did 2013 one and lost ~10marks

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#87

For me M4 is far harder than M5, M4 is easy physics but intense maths computation and it's so easy to have mistake during the workout. M5 is just hard physics, simple maths

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#88

Help! spent 30min trying to understand 2013 Q7b, how does the scalar product on MS work?

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#89

(Original post by

Help! spent 30min trying to understand 2013 Q7b, how does the scalar product on MS work?

**cogito.**)Help! spent 30min trying to understand 2013 Q7b, how does the scalar product on MS work?

We can use fact that the components of velocity parallel to the wall don't change to get b in terms of a.

x.y gives us the component of x that acts in the direction y.

To get the component of the initial velocity (b,0) parallel to (2,1) we do (b,0).(2,1)=2b. Similarly for the component of the final velocity we do (a,a).(2,1)=3a.

Therefore 2b=3a.

Now e= component of final velocity perpendicular to the wall / component of initial velocity perpendicular to the wall.

We do the same thing as above to get the components , (b,0).(-1,2)=-b and (a,a).(-1,2)=a

So e=a/b=2/3.

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#90

(Original post by

The impulse acts in the direction perpendicular to the wall which is (-1,2).

We can use fact that the components of velocity parallel to the wall don't change to get b in terms of a.

x.y gives us the component of x that acts in the direction y.

To get the component of the initial velocity (b,0) parallel to (2,1) we do (b,0).(2,1)=2b. Similarly for the component of the final velocity we do (a,a).(2,1)=3a.

Therefore 2b=3a.

Now e= component of final velocity perpendicular to the wall / component of initial velocity perpendicular to the wall.

We do the same thing as above to get the components , (b,0).(-1,2)=-b and (a,a).(-1,2)=a

So e=a/b=2/3.

**Gawain**)The impulse acts in the direction perpendicular to the wall which is (-1,2).

We can use fact that the components of velocity parallel to the wall don't change to get b in terms of a.

x.y gives us the component of x that acts in the direction y.

To get the component of the initial velocity (b,0) parallel to (2,1) we do (b,0).(2,1)=2b. Similarly for the component of the final velocity we do (a,a).(2,1)=3a.

Therefore 2b=3a.

Now e= component of final velocity perpendicular to the wall / component of initial velocity perpendicular to the wall.

We do the same thing as above to get the components , (b,0).(-1,2)=-b and (a,a).(-1,2)=a

So e=a/b=2/3.

Posted from TSR Mobile

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#91

Another quick question, how do you know which direction two balls will rebound? Or it does not matter when calculating the final velocity components?

Firstly i thought the sign positive/negative could determine the direction at the end but i found the Newton's experimental law requires the separation velocity and it's hard to get it around.

(Original post by

The impulse acts in the direction perpendicular to the wall which is (-1,2).

We can use fact that the components of velocity parallel to the wall don't change to get b in terms of a.

x.y gives us the component of x that acts in the direction y.

To get the component of the initial velocity (b,0) parallel to (2,1) we do (b,0).(2,1)=2b. Similarly for the component of the final velocity we do (a,a).(2,1)=3a.

Therefore 2b=3a.

Now e= component of final velocity perpendicular to the wall / component of initial velocity perpendicular to the wall.

We do the same thing as above to get the components , (b,0).(-1,2)=-b and (a,a).(-1,2)=a

So e=a/b=2/3.

Firstly i thought the sign positive/negative could determine the direction at the end but i found the Newton's experimental law requires the separation velocity and it's hard to get it around.

**Gawain**)

The impulse acts in the direction perpendicular to the wall which is (-1,2).

We can use fact that the components of velocity parallel to the wall don't change to get b in terms of a.

x.y gives us the component of x that acts in the direction y.

To get the component of the initial velocity (b,0) parallel to (2,1) we do (b,0).(2,1)=2b. Similarly for the component of the final velocity we do (a,a).(2,1)=3a.

Therefore 2b=3a.

Now e= component of final velocity perpendicular to the wall / component of initial velocity perpendicular to the wall.

We do the same thing as above to get the components , (b,0).(-1,2)=-b and (a,a).(-1,2)=a

So e=a/b=2/3.

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#92

(Original post by

What kind of difficulty do you expect for that M4? Just did 2013 one and lost ~10marks

**cogito.**)What kind of difficulty do you expect for that M4? Just did 2013 one and lost ~10marks

I just updated my site so they should be corrected( I hope ...)

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#93

**useful material for the upcoming exams**http://www.thestudentroom.co.uk/show....php?t=3353445

http://www.thestudentroom.co.uk/show....php?t=3361867

http://www.thestudentroom.co.uk/show....php?t=3361905

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#94

Yes I'm currently doing Q15 on the oblique impact pdf and i couldn't understand the concept of NEL.

Is the after impact velocity have to be certain direction when drawing the diagram? And the "separation velocity" means the larger one minus the smaller one but in which direction since we don't know them at the beginning.

Is the after impact velocity have to be certain direction when drawing the diagram? And the "separation velocity" means the larger one minus the smaller one but in which direction since we don't know them at the beginning.

(Original post by

thanks you for pointing out the typos

I just updated my site so they should be corrected( I hope ...)

**TeeEm**)thanks you for pointing out the typos

I just updated my site so they should be corrected( I hope ...)

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#95

(Original post by

Yes I'm currently doing Q15 on the oblique impact pdf and i couldn't understand the concept of NEL.

Is the after impact velocity have to be certain direction when drawing the diagram? And the "separation velocity" means the larger one minus the smaller one but in which direction since we don't know them at the beginning.

**cogito.**)Yes I'm currently doing Q15 on the oblique impact pdf and i couldn't understand the concept of NEL.

Is the after impact velocity have to be certain direction when drawing the diagram? And the "separation velocity" means the larger one minus the smaller one but in which direction since we don't know them at the beginning.

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#98

(Original post by

is this restitution?

**TeeEm**)is this restitution?

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#99

(Original post by

yes the e=separation/approaching nearly killed me...

**cogito.**)yes the e=separation/approaching nearly killed me...

Why don't you speak to you teacher?

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#100

(Original post by

this is introduced in M2 and is extended in M4.

Why don't you speak to you teacher?

**TeeEm**)this is introduced in M2 and is extended in M4.

Why don't you speak to you teacher?

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