AQA MECHANICS MM1B 20th JUNE 2017 UNOFFICIAL MARKSCHEME

These are my answers:

21) No idea about mew on the roof

For 21, Mu < 3/4.

As the roof was a slope, and the phone was sliding down, you can see that Fr < mgsin(a) (the component of the weight pulling the phone down the roof).

We know that R is the other component of weight, mgcos(a), so Fr = mu(mgcos(a).

This means mu(mgcos(a)) < mgsin(a)

mu < mgsin(a) / mgcos(a)
mu < sin(a)/cos(a)
mu < tan(a)

And as stated in the question, tan(a) = 3/4

mu < 3/4

It seemed like a lot of work for 2 marks though, but I'm pretty certain this is what they were looking for.

Different to a lot of people but on the last question, 7 marker I used this method:

S(x direction)= 8 metres
S(x direction)= ucosθt
ucosθ= 4cos(90-36.87) = 2.4ms-1

rearrange from 8=2.4(t) to make 't' the subject. t=(8/2.4)= 10/3. Time for phone to travel 8m in x-direction = 10/3seconds.

Working to find S(y direction)
S(y direction)= usinθt-1/2(g)(t^2)
t= 10/3 . usinθ= 4sin(90-36.87)= 3.2ms-1 . g=9.8ms-2

therefore S(y direction)= 3.2(10/3)-1/2(9.8)(10/3)^2= -43.8m. Therefore distance = 43.8m

Anyone get the same or where did I go wrong?

If this is the distance from the house, how the hell would a phone fall off a roof and land 43.8 metres away from the house.

Edit: I got 3.38m which sounds more realistic but id give you 7/7 for effort though, AQA probs won't tho :/

For the question about the genius on the trolley you said that the final velocity is 1.0. It's actually 0.9 because when he threw the second sandbag it was moving at 2 m/s RELATIVE TO HIM therefore it was moving at 2-0.4 RELATIVE to the ground which is 1.6 m/s. The total momentum of the sandbags is 20X2 +20X1.6= 72. The total mass of the man and trolley is 10 + 70 = 80. 72/80=0.9.
I didn't put negative values but I don't know if it should be negative or not.

I don't I don't know who is right or wrong I got -0.4 and -1 and this is correct according to my teacher but he does get wrong questions here and then ...oh well I guess you are correct ... there goes my 7 marks lol

I don't I don't know who is right or wrong I got -0.4 and -1 and this is correct according to my teacher but he does get wrong questions here and then ...oh well I guess you are correct ... there goes my 7 marks lol

10) Mew would be larger?

These are my answers:

No idea if they are wrong or right i literally did no revision for this cos i need a U in it to get an a* xD. So these could be wrong im not saying im right.

1) 46.2N
2) 23.1 N
3) 0.4m/s
4) 1.0m/s
5) 18 metres
6) 18.75 seconds
7) -2.5m/s^2
8) 19.6
9) 0.459
10) Mew would be larger?
11) 0.625
12) 330N
13) 1830N
14) 156N
15) 0.344
16) 0.0941
17) 4.57 m/s
18) this is a vector i think (5 3.2) m/s
19) 3.38m
20) 13.1 m/s
21) No idea about mew on the roof

Like i said these could be wrong so i'll edit if if they are, i'm not being like that sad kid who thinks they got it all right 😂

Lots of people it 1 for question 2 b, but I am pretty sure it is incorrect. They key is that the bag is travelling at 2 m s-1 relative to the velocity of the trolley, v - 2. This gives out equation:

80v + 20(v - 2) = 0.4 x 100

Edit:
It was also mew < 0.75 for the last question