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boromir9111
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#41
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#41
(Original post by Clarity Incognito)
You're looking for P and Q. sin45 and cos45 have values that you can plug in. So do sin25 and cos25

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 \sin(45) = \cos(45) = \dfrac{1}{\sqrt{2}}
Ahh yes of course, so is it like this

 Tqsin25^{o} + \displaystyle\frac{1}{\sqrt{2}}T = 78.4 ???
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Clarity Incognito
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#42
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#42
(Original post by boromir9111)
Ahh yes of course, so is it like this

 Tqsin25^{o} + \displaystyle\frac{1}{\sqrt{2}}T = 78.4 ???
Yes, that could be a next step, substitute the value you get for the tension of P/Q into the other equation and then simplify to get the answer for either P and Q. Sub this value back into the other original to find the other force. I think you make life harder for yourself with all these extra letters, just call the forces P and Q, you don't need to call them Tq and Tp.
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boromir9111
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#43
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#43
(Original post by Clarity Incognito)
Yes, that could be a next step, substitute the value you get for the tension of P/Q into the other equation and then simplify to get the answer for either P and Q. Sub this value back into the other original to find the other force. I think you make life harder for yourself with all these extra letters, just call the forces P and Q, you don't need to call them Tq and Tp.
Yeah, it's confusing myself lol...... this value for Q doesn't work

 Qsin25 + \displaystyle\frac{1}{\sqrt{2}}Q = 78.4

 Q(sin25 + \displaystyle\frac{1}{\sqrt{2}}) = 78.4

 Q = 69.3N which is wrong, it should be 51.5N???
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Clarity Incognito
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#44
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#44
(Original post by boromir9111)
Yeah, it's confusing myself lol...... this value for Q doesn't work

 Qsin25 + \displaystyle\frac{1}{\sqrt{2}}Q = 78.4

 Q(sin25 + \displaystyle\frac{1}{\sqrt{2}}) = 78.4

 Q = 69.3N which is wrong, it should be 51.5N???
 Psin45 + Qsin25 = 8g

 Pcos45 = 10+Qcos25

 P = \dfrac{10+Qcos25}{cos45}

 \dfrac{sin45}{cos45} (10+Qcos25) + Qsin25 = 8g

 Q(cos25+ sin25) = 8g - 10

 Q = 51.47...
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boromir9111
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#45
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#45
(Original post by Clarity Incognito)
 Psin45 + Qsin25 = 8g

 Pcos45 = 10+Qcos25

 P = \dfrac{10+Qcos25}{cos45}

 \dfrac{sin45}{cos45} (10+Qcos25) + Qsin25 = 8g

 Q(cos25+ sin25) = 8g - 10

 Q = 51.47...
And problem solved lol.......thank you so much mate, much appreciated again :yep: ........ i shall +rep you when i can
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boromir9111
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#46
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#46
problem 10

question (iv) the horizontal component of the velocity is Vx = U.....so we need U to get V?? the mark scheme has the answer as 12.5/(0.565).....i got no idea how it gets that???

oh wait, x = ut.......x = 25(from the graph it touches at -5 and 20 so, a distance or range of 25 and time just do 2*0.565 and that's your answer......got it
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boromir9111
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#47
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Problem 11

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http://www.mei.org.uk/files/papers/m109jn_86dj.pdf how do i go about doing question 2 part (i) and (ii)????
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boromir9111
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#48
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#48
anyone?
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Ramjams
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#49
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#49
(Original post by boromir9111)
Problem 11

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http://www.mei.org.uk/files/papers/m109jn_86dj.pdf how do i go about doing question 2 part (i) and (ii)????
Think about it; the the question and graph provide the following information:

Initial Velocity: 6ms^-1
Final Velocity: ?
Acceleration: 4ms^-2 (for the first 2s)
Time: 2s

For (i) you should be able to use this information in an equation of motion to solve for final velocity.

Using a similar method to (i), for (ii)

Initial Velocity: Answer to (i)
Final velocity: -6
Acceleration: -8
Time: ?

For part (ii), use another equation of motion to solve for time, and add on 2 seconds to account for the first section of te graph.
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boromir9111
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#50
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#50
(Original post by Ramjams)
Think about it; the the question and graph provide the following information:

Initial Velocity: 6ms^-1
Final Velocity: ?
Acceleration: 4ms^-2 (for the first 2s)
Time: 2s

For (i) you should be able to use this information in an equation of motion to solve for final velocity.

Using a similar method to (i), for (ii)

Initial Velocity: Answer to (i)
Final velocity: -6
Acceleration: -8
Time: ?

For part (ii), use another equation of motion to solve for time, and add on 2 seconds to account for the first section of te graph.
Thanks
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hobnob
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#51
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Thread locked at the OP's request.
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