Notes

Q1 a) Velocity = rate of change of displacement (with respect to time ) (1)
b) i) 70 km / h = 70 x 1000/3600 m/s = 19.44 m/s
Ek = 1/ mv^2 = ½ x 130 x 19.44^2
= 2.46E4 J (3)
ii) Volume is 8 x smaller (2 x 2 x 2)
Density is same
So mass is 8 x smaller = 130/8 = (16.25 kg) (2)
TOTAL: 6

Easy question to kick off. Unit conversion Ok. Last part may confuse.

2 Moments with a force at an angle. Expecting a lot of blanks here.
a) Weight x d = T cos 40 x 0.75
So d = 5.1 cos40 x 0.75 / (1.2 x 9,81) = 0.249m (3)
b) Resultant force horizontally must be zero. Tension has a component to
Left so force at support must have a component to right and cant be vertical. (1)
TOTAL: 4


3 a) i) brittle / elastic / obeys Hooke’s law (2)
ii) steeper line that flattens out. (2)
b) i) YM = stress / strain
so strain = stress / YM
= 1.8E7 / 2.0E11 = 9.0E-5 (2)
ii) stress = F / A
so F = stress x A = 1.8E7 x π x (2.6 x 10-2) ^2 = 3.82E4 N (2)
iii) Weight = 2Tsin12 (resolve forces vertically) = 1.59E4N (3)
TOTAL 11
Last bit caused the usual problems
Some messed up the area calc

4 a) i) No horizontal forces act so no horiz acc. (1)
ii) Line that starts horiz and reaches ground nearer to wall.
Takes same time to fall
- because vertical acc is same and vert distance is same (3)
b) Drop ball through a height of one metre and measure time. Repeat and average.
s = ut + 1/2 gt^2 u = 0 so g = 2h/t^2 (3)
c) i) Constant acc down slope (negative)
Slows down (at steady rate)
Reaches max height v=0 at t=1.5s
Rolls back at increasing speed to bottom (at constant acc) (3)
ii) Max height = area under graph (or use s= 1/2 (u +v) t )
= 1/2 x 1.5 x 4.0 = 3.0m (3m will lose a sig fig mark) (2)
TOTAL 12
a) is the tricky bit and many will get aii wrong

5 a) The switch from table tennis ball to tennis ball threw a lot! Poor proof reading.
i) Acc = acc of free fall or 9.81
Only force acting is weight / drag = 0 when v=0 (2)
ii) AT max velocity drag = weight (1)
iii) Gold ball has higher terminal vel.
because mass bigger, wieght is bigger so needs more drag so higher v (3)
b) i) T 25m/s drag = 2000N (from graph - did you spot kN on axis?)
so resultant F = 3200 - 2000 = 1200N
so a = F/m = 1200/8000 = 0.15 ms-2 (3)
ii) Either look up max speed when drag = 3200N = 33 ms-1
or look drag at 40ms-1 drag = 5000N which is bigger than driving force
so would slow down (1)
c) Seat belt allows driver to slow down over a longer time
so acc = dvdt is less
so Force = ma is less.
Using a wide seat belt gives lower pressure (=F/A) on ribs so less damage (3)
TOTAL 13
I think the drag graph will throw many. Tricky.

6a) WD = force x distance moved in direction of force. (1)
b) Some debate about what "what happens to the WD " means.
I just wrote "it becomes heat" due to friction
Some wrote WD increases at a uniform rate - which may be OK (1)
c) 1W = 1J/1s (1)
d) Rate of doing work = gain in gravpe per unit time = mgh/t
h = 60m (pythagoras)
so rate = 5200 x 9.81 x 60 / (1.5 x 60) = 3.4E4 Js-2 (3)
Efficiciency = Pout/ Pin x 100% = 3.4E4/170E3 x100 = 20% (1)
TOTAL7
b is confusing and d is tricky.

7 If you confused length and extension you got wiped out here.
a) k = F/x = 3.0 / (8.0 - 2.0)E-2 = 50 Nm-1 (1)
b) New extension = 10cm
Energy = 1/2 kx^2
so change in energy = 1/2 x 50 x 10E-2 ^2 - 1/2 x 50 x 6.0E-2 ^2 = 0.16J (3)

Object has a force of 5.0N up and 3.0N down so resultant F = 2.0N up
mass = 3.0/9.81 = 0.306kg
so acc = F/m = 2.0/0.306 = 6.54 ms-2
Most got this wrong. SHould have drawn the forces acting
TOTAL 7

Tricky paper.

My predicted boundaries

100% 55
A 45
B 40
C 35
D 30
E 25

Still got all to play for with the EWP paper. Get revising!
Col