Question 3:
3 | First M1 for resolving vertically (up or down) for B, with correct no. of
terms.
First A1 for a correct equation.
Second M1 for resolving parallel to the plane (up or down) for A, with
correct no. of terms.
A2 for a correct equation (-1 each error)
OR: M2 A3 for the whole system equation - any method error loses all the
marks.
B1 for perpendicular resolution
Third M1 for sub for R in F = µR
Fourth DM1, dependent on first and second M marks, for eliminating a.
Fourth A1 for 8g/3, 26.1 or 26 (N). (392/15 oe is A0)
4.
(a)
(b) | 1
Use of s=ut+ at2
2
1
−2t+ gt2 (+ or – 50)
2
1
20t− gt2 (+ or – 50)
2
1 1
50 = −2T + gT2 +20T − gT2 =18T
2 2
50
T = =2.777....=2.8 or better
18
h=20×T −4.9×T2 =17.74....≈17.7 (18 to 2 s.f.)
(use of 2.8 gives 17.584) | M1
A1
A1
M1
A1
(5)
M1A1
(2)
[7]
Notes on Question 4
Q4(a) | First M1 for use of s = ut + 1/2at2 (or use of 2 suvat formulae AND
eliminating v, to give an equation in s and t). N.B. M0 if they use s = 50 or
u = 0 or v = 0)
First A1 with u = 2 and a = -g or -9.8 to obtain a distance, possibly with 50
added or subtracted. (2 and 4.9 must have opposite signs)
Second A1 with u = 20 and a = -g or -9.8 to obtain a distance, possibly
with 50 added or subtracted. (2 and 4.9 must have opposite signs)
Second M1 dependent on first M1 for a correct equation obtained correctly
in T only.
Third A1 for 25/9 oe, 2.8 or better
Q4(b) | First M1 for substituting their T value (allow –ve changed to +ve but A
mark is then unavailable) into an appropriate equation
First A1 for 17.7 or 18 (m). (A0 if they then add 50)
5.
(a)
(b)
(c) | u+v 2+v
s= t 10= ×3.5
2 2
20 26
v= −2= =3.71 (m s-1)
3.5 7
26
−2
a= v−u = 7 = 24 =0.490 (m s-2)
t 3.5 49
Normal reaction : R=0.6gcos25°
Resolve parallel to the slope : 0.6gsin25°−µ×R=0.6×a
µ=0.41 or 0.411 | M1A1
A1
(3)
M1A1
(2)
B1
M1A2
A1
(5)
[10]
Notes for Question 5
Q5(a) | First M1 for producing an equation in v only.
First A1 for a correct equation
Second A1 for 26/7 oe, 3.7 or better (ms-1)
Q5(b) | M1 for producing an equation in a only.
A1 for 24/49, 0.49 or better (ms-2)
Q5(c) | B1 for R = 0.6gcos25o
M1 for resolving along the plane, correct no. of terms etc.
A2 (-1 each error) R and a do not need to be substituted
Third A1 for 0.41 or 0.411
6.
(a)
(b)
(c) | Use of r =r +vt
0
(−4i+2j)+(3i+3j)t =(−4+3t)i+(2+3t)j
(6i+ j)+(−2i+nj)t =(6−2t)i+(1+nt)j
Position vectors identical ⇒−4+3t =6−2t AND 5t =10,
Either equation
2+3×2=1+2n,
n=3.5
Position vector of P is (−4+6)i+(2+6)j=2i+8j
Distance OP = 22 +82 = 68 =8.25(km) | M1
A1
(2)
B1
M1
A1
DM1
A1
(5)
M1A1
M1A1
(4)
[11]
Notes for Question 6
Q6(a) | M1 for clear attempt to use r + tv (M0 if r and v reversed)
0 0
A1 for answer in any form.
Q6(b) | B1 for (6i + j) + (-2i + nj)t seen or implied
First M1 for equating their i- cpts and their j- cpts. (must have both
equations in terms of same t)
First A1 for a correct equation (either)
Second M1 dependent on first M1 for producing an equation in n only.
Second A1 for n = 3.5 oe
Q6(c) | First M1 for clear attempt to find pv of P, using their t and/or n value(s)
First A1 for 2i + 8j
Second M1 for attempt to find magnitude of their p
Second A1 for √68, 2√17, 8.2 or better (km)
7
(a)
(b)
(c) | Use of v2 =u2 +2as
142 =202 −2a×100
Deceleration is 1.02(m s-2)
Horizontal forces on the car: ±Tcosθ−300=750×−1.02=−765
T = - 1550/3
The force in the tow-bar is 1550/3, 520 (N) or better (allow –ve answer)
Horizontal forces on the truck: ±Tcosθ−500− R=1750×−1.02
Braking force R = 1750 (N)
ALT: Whole system: 800+R=2500×1.02
R = 1750 | M1
A1
A1
(3)
M1A2 f.t.
A1
(4)
M1A2 f.t.
A1
(4)
[11]
M1A2 f.t.
A1
Notes for Question 7
Q7(a) | M1 for a complete method to produce an equation in a only.
First A1 for a correct equation.
Second A1 for 1.02 (ms-2) oe. must be POSITIVE.
Q7(b) | M1 for considering the car ONLY horizontally to produce an equation in T
only, with usual rules. i.e. correct no. of terms AND T resolved:
±T cosθ−300= 750 x -1.02
A2 ft on their a for a correct equation (300 and a must have same sign); -1
each error (treat cos 0.9 as an A error)
A1 for 1550/3 oe, 520 or better (N) N.B. Allow a negative answer.
Q7(c) | M1 for considering the truck ONLY horizontally to produce an equation,
with usual rules. i.e. correct no. of terms AND T resolved:
±T cosθ−500−R=1750 x -1.02
A2 ft on their T and a for a correct equation (500, a and R must have same
sign); -1 each error (treat cos 0.9 as an A error)
A1 for 1750 (N).
OR
M1 for considering the whole system to produce an equation in R only,
with usual rules. i.e. correct no. of terms.
A2 ft on their a for a correct equation (a and R must have same sign) -1
each error
A1 for 1750 (N).
N.B. If 300 and 500 are given separately, penalise any sign errors only
ONCE.
8.
(a)
(b) | R
2R
0.2 m 0.8 1-x x m
A B
C D
50g
Vertical equilibrium:R+2R=50g,
Moments about C: 50g×0.8=(1.8−x)×2×R
3×0.8=3.6−2x, x=0.6
S
4S
0.2 m 0.8 0.6 0.4 m
A B
C E
mg
50g
S, 4S
Vertical equilibrium: S+4S =(50+m)g =5S
Moments about B: 50g×1=4S×0.4+S×1.8=3.4S
5
50× =(50+m)
3.4
m = 400/17, 24, 23.5 or better | M1A1
M1A1
DM1A1
(6)
B1
M1A1
M1A1
DM1
A1
(7)
[13]
Notes for Question 8
Q8(a) | In both parts consistent omission of g’s can score all the marks.
First M1 for vertical resolution or a moments equation, with usual rules.
(allow R and N at this stage)
First A1 for a correct equation (with N = 2R substituted)
Second M1 for a moments equation in R and one unknown length with
usual rules.
Second A1 for a correct equation.
Third M1, dependent on first and second M marks, for solving for x
Third A1 for x = 0.6.
S.C. Moments about centre of rod: R x 0.8 = 2R(1 – x) M2 A2
Q8(b) | B1 for S and 4S placed correctly.
First M1 for vertical resolution or a moments equation, with usual rules.
(allow S and 4S reversed)
First A1 for a correct equation.
Second M1 for a moments equation in S (and m) with usual rules.
Second A1 for a correct equation.
Third M1, dependent on first and second M marks, for eliminating S to
give an equation in m only.
Third A1 for m = 400/17 oe or 24 or better.
N.B. SC If they use the reaction(s) found in part (a) in their equations, can
score max B1M1A0M1A0DM0A0.