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Linear Motion under a
Variable Force
Mark Scheme 3
Level
International A Level
Subject
Maths
Exam Board
CIE
Topic
Linear Motion under a Variable Force
Sub Topic
Booklet
Mark Scheme 3
Time Allowed:
62 minutes
Score:
/51
Percentage:
/100
Grade Boundaries:
A*
>85%
A
777.5%
B
C
D
E
U
70%
62.5%
57.5%
45%
<45%
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1
2
3
0.6vdv/dx = 0.3x
M1
Newton’s Second Law with a = vdv/dx
0.6v2/2 = 0.3x2/2(+ c)
A1
From ∫0.6vdv = ∫0.3xdx
[x2/2] 80 = [v2] v2
M1
Uses limits of finds constant
v = 6 ms–1
A1
(i) 0.25dv/dt = –3t
[4]
[4]
M1
Newton’s Second Law, – sign essential
v = –12t2/2 (+ c)
A1
Accept uncancelled form
0 = 12 × 32/2 + c
M1
Appropriate use of v = 0, t = 3
Initial speed = 54 ms–1
A1
(ii) ∫dx = ∫(54 – 6t2)dt
M1
x = [54t – 6t3/3] 30
A1
x = 108 m
A
[4]
Goes beyond c = 54
Separates variables, integrates v
candidates value [v in (i)]
(i) 0.2vdv/dx = –0.4x
M1
Newton’s Second Law, – sign essential
v2/2 = –2x2/2 (+ c)
A1
Accept uncancelled form
0 = –2 × 2.52/2 + c → c = 6.25
M1
KE = 0.2 × 6.25 = 1.25 J
A1
(ii) 22/2 = –2x2/2 + 6.25
x = 2.06
[4]
v = 3.54 ms–1
v = 2 in accurate integral attempt at
limits or finding arbitrary constant e.g.
in (i)
M1
A1
[7]
[3]
[2]
[6]
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4
(i
0.05dv/dt = 0.05g – 0.01v
M1
dv/dt = 10 – 0.2v
A1
∫
dv/(10 – 0.2v) =
∫
dt
M1
–ln(10 – 0.2v)/0.2 = t (+ c
A1
t = 0, v = 0, hence c = –5ln10
M1
ln(10 – 0.2v)/10 = 0.2t, 1 – 0.02v = e–0.2t
v = 50 – 50e–0.2t
A1
Uses Newton’s Second Law
–4.60517
[6]
(ii) dx/dt = 50 – 50e–0.2t
x=
∫
(50 – 50e–0.2t)dt
M1
x = 50t + 50e–0.2t/0.2 (+c)
A1
h = [50t + 50e–0.2t/0.2] 02
M1
h = 17.6
A1
Or uses h = 0, t = 0 to evaluate
c = (–250) and then finds h(2)
[4]
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5
(i
0.4δv/δt = 0.2v2
M1
∫v
A1
−2
∫
δv = −0.5 δt
Newton’s Second Law with a = δv/δt
–v–1 = –0.5t (+ c)
t = 0, v = 8, hence c = –0.125
M1
v = 1/(0.125 + 0.5t) = 8/(1 + 4t) AG
A1
(ii) δx/δt = 8/(1 + 4t)
[4]
M1*
∫
x = 8 δt / (1 + 4t)
x=
8
4
ln(1 + 4t) (+ c)
t = 1.5, x =
8
4
*ln(1 + 4 × 1.5
OP = 3.89 m
6
(i
∫
2
[
Accept c = 0 assumed
D*
M1
Or limits used
A1
a = 0 when x = 2.5
vdv/dx = 15 – 6x
vdv = (15 – 6x)dx
∫
A1
]
v /2 = 15x − 3x 2 (+ c)
v = 6.12
8
4
[ln(1 + 4t] 1.5
0
4
B1
M1
A1
M1
For use of limits 0 and 2.5 or evaluating
c(=0)
A1
[5]
(ii) Solves 15x – 3x 2 = 0
a (= 15 – 6 × 5) = –15ms −2
x = 5. Accept assumption c = 0.
M1
A1
[2]
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7
(i
dv/dt = –2.5k v
∫
v
v
0.5
−0.5
dv = –2.5k
B1
∫
dt
M1
/0.5 = –2.5kt (+ c)
A1
t = 0, v = 9 hence c = 6 and
t = 2, v = 4 hence k = 0.4
v = (6 – t) 2 /4 = (t – 6) 2 /4
4
M1
∫
(ii) x =
(t – 6) 2 /4dt
3
x = (t – 6) /(3 × 4) (+ c)
t = 0, x = 0 hence c = 18
x(3) = 18 – (3 – 6) 3 /12
x(3) = 15.75
0.4dv/dt = –k v
LHS = 0.8 v
v = (6 – t)/2
Uses correct limits
A
A1
[5]
M1
∫ (6 – t)
2
/4dt
A1
M1
–(6 – t) 3 /(3 × 4) (+ c)
Or uses limits 0, 3
A1
Accept 15.7 or 15.8
–dx
M1
From mvdv/dx = –k v
v 2 = – x ( + c)
A1
OR
1
∫ v 2 dv =
2
3
3
∫
x = 18 – 23 v
x = 15.75
3
2
M1
Using v = 9, x = 0 so c = 18
A1
Put t = 3 to find v = 2.25
[4]