Mechanics qs

http://filestore.aqa.org.uk/sample-papers-and-mark-schemes/2017/june/AQA-MM1B-QP-JUN17.PDF


can someone help me with qs 2b

http://filestore.aqa.org.uk/sample-papers-and-mark-schemes/2017/june/AQA-MM1B-QP-JUN17.PDF


can someone help me with qs 2b

See https://www.thestudentroom.co.uk/showthread.php?t=5440602

hi thanks for replying

i still dont understand how to do it

Do you not understand conservation of momentum? If you don't, see page 12 of http://www.mrbartonmaths.com/resources/a%2520level%2520revision/M1_Not_Formula_Book.pdf. Note that here we are not using momentum to solve a collisions problem, but just in a more general context.

thanks for replying
yes i do understand it but i just dont know how to answer this specific qs and i dont really understand the markscheme

Well, write down the masses and velocities...

i dont see what im doing wrong
momentum before = momentum after
(-0.4)(100)+(2)(20)= (80)(v)+(20)(2)+(20)(2)

v= -1

can you tell me what im dong wrong

a) 0.4ms^-1 - I hope you got this

So we know that the trolley and the man and the sandbag are moving with (100kg) are moving with 0.4ms^-1
If the sandbag travels 2ms^-1 relative to the trolley/man which travels opposite to the bag, then what will the speed of the sandbag be?
So work out the momentum before *
Then work out the momentum after**
And equate them.

a) 0.4ms^-1 - I hope you got this

So we know that the trolley and the man and the sandbag are moving with (100kg) are moving with 0.4ms^-1
If the sandbag travels 2ms^-1 relative to the trolley/man which travels opposite to the bag, then what will the speed of the sandbag be?
So work out the momentum before *
Then work out the momentum after**
And equate them.
Reply with your answers and I can try to guide you with your solution

You're not using the relative velocity correctly - it's 2 ms^-1 relative to the trolley, so go and look up what relative velocity means if you've forgotten.

Bear in mind that while the speed for part (a) is $0.4 \ \text{ms}^{-1}$, the actual velocity is $-0.4 \ \text{ms}^{-1}$, so it's necessary to be careful with signs.

my teacher has never mentioned this term to us during the year

the relative velocity is confusing me

Then your teacher is incompetent, as they missed out a key part of your syllabus, and so you should complain to your headmaster/headmistress. If you have a textbook, read the section within it on relative velocity, or if you don't have a textbook, read page 9 of this PDF: https://www.mrbartonmaths.com/resources/a%20level%20revision/M1.pdf

ok i think i understand what relative velocity
so for this qs the person is moving -0.4 ms-1 but will see the bag move at at a different speed (is this relative velocity)

a) 0.4ms^-1 - I hope you got this

So we know that the trolley and the man and the sandbag are moving with (100kg) are moving with 0.4ms^-1
If the sandbag travels 2ms^-1 relative to the trolley/man which travels opposite to the bag, then what will the speed of the sandbag be?
So work out the momentum before *
Then work out the momentum after**
And equate them.
Reply with your answers and I can try to guide you with your solution

so if the person is moving -0.4ms-1 this person will see the object moving at 2ms-1
but it is actually moving at -0.4 +2 =1.6