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A level Maths Mechanics help

Please can someone help!!!!

2696ED9E-2DAC-4889-87C9-87D1E78961B9.jpg.jpeg

I’ve tried really hard to do Question 3 by myself but I can’t work out how to do it. Any help would be so appreciated xxx
(edited 6 years ago)
Reply 1
Original post by littlebitthick
Please can someone help!!!!

2696ED9E-2DAC-4889-87C9-87D1E78961B9.jpg.jpeg

I’ve tried really hard to do Question 3 by myself but I can’t work out how to do it. Any help would be so appreciated xxx

First you need to find the resultant force on the particle. Any ideas?
Original post by Notnek
First you need to find the resultant force on the particle. Any ideas?


So add the vectors together to get 6i -j

Then using Pythagoras get the square root on 37
Reply 3
Original post by littlebitthick
So add the vectors together to get 6i -j

Then using Pythagoras get the square root on 37

Well you could leave it as a vector and then use F = ma to find the acceleration vector. The question doesn't ask for the magnitude of the acceleration but it isn't completely clear.

Once you have the acceleration as a vector you can use SUVAT with vectors to find the initial velocity vector u.

Try this and post your working if you get stuck.
(edited 6 years ago)
Original post by Notnek
Well you could leave it as a vector and then use F = ma to find the acceleration vector. The question doesn't ask for the magnitude of the acceleration but it isn't completely clear.

Once you have the acceleration as a vector you can use SUVAT with vectors to find the initial velcocity vector u.

Try this and post your working if you get stuck.


My problem is that if I use a= square root of 37
Then my initial velocity becomes negative and that confuses me.
Does that just mean that the initial velocity of the particle is the opposite direction to that of the acceleration??
Reply 5
Original post by littlebitthick
My problem is that if I use a= square root of 37
Then my initial velocity becomes negative and that confuses me.
Does that just mean that the initial velocity of the particle is the opposite direction to that of the acceleration??

Yes it would mean that. But you shouldn't be finding magnitudes - leave everything as vectors and use v = u + at where v is the vector 0i + 0j.
Oh okay. I’ve never done it like that before. So by Putting the vector 6i-j into F=ma when the particle is 2kg, you would just end up with 3i -1/2 j as ur acceleration vector??
Reply 7
Original post by littlebitthick
Oh okay. I’ve never done it like that before. So by Putting the vector 6i-j into F=ma when the particle is 2kg, you would just end up with 3i -1/2 j as ur acceleration vector??

Yes that's right. F=ma and SUVAT work for vectors so you should stick with vectors when doing these kinds of questions.

A question may ask for e.g. the magnitude of the velocity (i.e. the speed) but if not just leave answers as vectors.
Thank you so much for all your help!!!! xxx

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