The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

OCR (non mei) M1 Friday 17th June 2016

Scroll to see replies

Original post by physicskid123
look it's cos a = (4^2 + 4^2 - 6^2) / 2*4*4 so cos a = -.125 and so a is 97.18


Resultant of the 2 forces was 6N..

Original post by Jackandrew1
Just a quick question for question 4 part iii. In the unofficial mark scheme it mentioned that it had already been verified. However in the question it said that after B collides with CD that there are no more collisions after this point. This means that surely B can not have a velocity of zero as A would then collide with B. Therefore shouldn't the velocity of B be at least greater than 4ms^-1 (which was the previous speed of A).


Where's the unofficial markscheme??
Original post by swagmister
Where's the unofficial markscheme??


Theres a link for it on the first page

https://docs.google.com/document/d/12BU7INUV7qu4dzLXYXeycROONeWs2txffn--DGS9mIo/edit#
Hey, I don't think it matter but I think questions 2 and 3 on the mark scheme were the other way round, in case there is any referencing confusion :smile: correct me if wrong
Reply 44
Avatar for ajamesg
ajamesg
1
Original post by Parallex
Resultant of the 2 forces was 6N..


How many marks was this whole question worth?
Original post by Parallex
Resultant of the 2 forces was 6N..

1fde3bbfc36505960e4f2926775101d5.png here u go my friend.
Reply 46
Avatar for a_obrien
a_obrien
Original post by Jackandrew1
Just a quick question for question 4 part iii. In the unofficial mark scheme it mentioned that it had already been verified. However in the question it said that after B collides with CD that there are no more collisions after this point. This means that surely B can not have a velocity of zero as A would then collide with B. Therefore shouldn't the velocity of B be at least greater than 4ms^-1 (which was the previous speed of A).


Yes I agree. I can't remember my answer but I know that I had momentum before = 0.2 x 4 + 0.4v because the minimum possible velocity of B after the collision would have to be 4ms-1 to avoid further collisions.
(edited 7 years ago)
where is the unofficial mark scheme???????????
Original post by physicskid123
1fde3bbfc36505960e4f2926775101d5.png here u go my friend.


What's that trying to show? If you're going to say I'm wrong at least post your working that leads you to a resultant force of 6N, instead of putting numbers in to an online calculator without even knowing what they're showing.
Original post by Parallex
What's that trying to show? If you're going to say I'm wrong at least post your working that leads you to a resultant force of 6N, instead of putting numbers in to an online calculator without even knowing what they're showing.


it's to show the resultant aka if you put in 2 sides of 4cm and 4cm (which are our 2 forces of 4N) and you put in the resultant force of 6N you get the angles shown

I'm done here mate bye seeu soon xxx
Original post by physicskid123
it's to show the resultant aka if you put in 2 sides of 4cm and 4cm (which are our 2 forces of 4N) and you put in the resultant force of 6N you get the angles shown

I'm done here mate bye seeu soon xxx


Like I said, draw me a triangle with the angle between the 2 4N forces being 97.18 degrees and show me a resultant of 6N.
Reply 51
Avatar for ajamesg
ajamesg
1
Original post by Parallex
Like I said, draw me a triangle with the angle between the 2 4N forces being 97.18 degrees and show me a resultant of 6N.


I'm sorry to say that he is right:

if you use the sine rule, you get the same answer (97.181 degrees)

as you get x = 4sin(97.181)/sin(41.41) and that gives you 6
Original post by Parallex
Like I said, draw me a triangle with the angle between the 2 4N forces being 97.18 degrees and show me a resultant of 6N.

QQ
grade boundary predicitions????
IMG_7485.jpg here is my solution
(edited 7 years ago)
Reply 55
Avatar for ajamesg
ajamesg
1
IMG_0005.jpgthis is the proof that it is 97.18 degrees
Original post by daniellon
For question 5 I said the direction is to the right. Can I get away with it


Yes almost certainly
was P 12 on q5?
Original post by ajamesg
IMG_0005.jpgthis is the proof that it is 97.18 degrees


You don't get the resultant by drawing a line between the 2, use the parallelogram rule for adding vectors.
Reply 59
Avatar for ajamesg
ajamesg
1
Original post by Parallex
You don't get the resultant by drawing a line between the 2, use the parallelogram rule for adding vectors.


yes you are right. how many marks was the whole question worth?

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you