Pre-U Pre-U 9795/2 2014 June — Question 9 11 marks

Exam BoardPre-U
ModulePre-U 9795/2 (Pre-U Further Mathematics Paper 2)
Year2014
SessionJune
Marks11
TopicPower and driving force
TypeWork-energy method on incline
DifficultyChallenging +1.2 Part (i) is a straightforward constant acceleration kinematics problem (2 marks). Part (ii) requires setting up and solving a differential equation from P=Fv with variable acceleration, involving separation of variables and integration—more demanding than typical A-level mechanics but follows a standard Pre-U/FM pattern without requiring novel insight.
Spec3.02d Constant acceleration: SUVAT formulae3.02f Non-uniform acceleration: using differentiation and integration6.02k Power: rate of doing work6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product
An engine is travelling along a straight horizontal track against negligible resistances. In travelling a distance of 750 m its speed increases from 5 m s\(^{-1}\) to 15 m s\(^{-1}\). Find the time taken if the engine was
  1. exerting a constant tractive force, [2]
  2. working at constant power. [9]
Question 9:
(ii) ---
9 (i)
AnswerMarks
(ii)1
750 = (5 + 15)t ⇒ t = 75
2
Let P denote the constant power of the engine.
P dv P ∫t =∫15
Newton II: =m ⇒ dt vdv
v dt m 0 5
15
Pt v2
( )
⇒ =  =100
m 2
 
5
P dv P ∫750 dx=∫15
Also =mv ⇒ v2dv
v dx m 0 5
750P v3 15  3250
⇒ =   = 
m  3   3 
5
P 13 900
⇒ = t and t = or 69.2 seconds
AnswerMarks
m 9 132
If a [= ] found, need second
15
equation for M1
Or:
1
2 2
Energy: 750F = m(15 – 5 )
2
F 10m
15=5+ t ⇒ t = =75
m F
P
For F = seen
v
If ∆KE used, must equate to
AnswerMarks
PtM1
A1
(B1)
(B1) [2]
B1
M1
A1
A1
A1
M1
A1
A1
A1 [9]
AnswerMarks Guidance
Page 6Mark Scheme Syllabus
Pre-U – May/June 20149795 02
10 (i)
(ii)
AnswerMarks
(iii)0.05λ
Hooke’s Law: = 2.5g ⇒ λ = 250 N [AG]
0.5
Let e m be the extension in the new equilibrium position.
250e
4g = ⇒ e = 0.08 m
0.5
250(0.08+x )
Newton II: 4& x &=4g− ⇒ &x&=−125x
0.5
(a) Amplitude of new motion is 0.08 – 0.05 = 0.03 m
(b) Maximum speed = 125 × 0.03 = 0.335 ms –1 .
2π 2π
(c) T = = = 0.562 seconds.
ω 125
1 π 2
(d)  +sin−1  = 0.206 seconds
AnswerMarks
125 2 3Find new e
Needs 0.08√
&x&=10−125x
SR: and then
x = y – 0.08 : B2
cwo
√ on a or ω
√ on ω
2 2
sin –1 (“ ”) /ω or cos –1 (“ ”)/ω
3 3
AnswerMarks
Deal with quadrantsM1A1 [2]
M1
A1
M1
A1 [4]
B1 [1]
B1√ [1]
B1√ [1]
M1
M1
A1 [3]
11 (i)
(ii) (a)
AnswerMarks
(b)gx 2 tan 2α – 2V 2 x tan α + (gx 2 + 2V 2 y) = 0 or equiv
2
For boundary of accessible points B – 4AC = 0
⇒ 4V 4 x 2 – 4gx 2 (gx 2 + 2V 2 y) = 0 (*)
1 ( )
⇒ 2gV 2 y = V 4 – g 2 x 2 ⇒ y= V 4−g 2 x 2 [AG]
2gV 2
dy gx 2
Or: =xsec2α − tanαsec2α
dα 2V 2
V 2 1 ( )
= 0 ⇒ x= ⇒ y= V 4−g 2 x 2
gtanα 2gV 2
1 ( )
Putting y = – h and V 2 = gh ⇒ −h= g 2 h 2−g 2 x 2
2g 2 h
⇒ x = 3h
Substituting for x in (*) gives
3gh 2 tan 2α – 2 3gh 2 tan α + 3gh 2 – 2gh 2 = 0
⇒ 3 tan 2α – 2 3 tan α + 1 = 0
1 π
⇒ tan α = ⇒ α = 30° or
AnswerMarks
3 6y must be part of quadratic
Allow ∆ ≥ 0
“Quadratic equation” required
in question, so B0
y = h: M1A0
−B
Or quote tan α =
AnswerMarks
2AB1
M1
A1
A1 [4]
B0M1
A1
A1
[max 3]
M1
A1
A1 [3]
M1
A1
A1 [3]
AnswerMarks Guidance
Page 7Mark Scheme Syllabus
Pre-U – May/June 20149795 02
Notation: Speed of wind relative to: ground, w; cyclist at 15ms –1 , u; cyclist at 10ms –1 , v
Angle of wind velocity with due west, θ (see diagram in method α)
Question 9:
--- 9 (i)
(ii) ---
9 (i)
(ii) | 1
750 = (5 + 15)t ⇒ t = 75
2
Let P denote the constant power of the engine.
P dv P ∫t =∫15
Newton II: =m ⇒ dt vdv
v dt m 0 5
15
Pt v2
( )
⇒ =  =100
m 2
 
5
P dv P ∫750 dx=∫15
Also =mv ⇒ v2dv
v dx m 0 5
750P v3 15  3250
⇒ =   = 
m  3   3 
5
P 13 900
⇒ = t and t = or 69.2 seconds
m 9 13 | 2
If a [= ] found, need second
15
equation for M1
Or:
1
2 2
Energy: 750F = m(15 – 5 )
2
F 10m
15=5+ t ⇒ t = =75
m F
P
For F = seen
v
If ∆KE used, must equate to
Pt | M1
A1
(B1)
(B1) [2]
B1
M1
A1
A1
A1
M1
A1
A1
A1 [9]
Page 6 | Mark Scheme | Syllabus | Paper
Pre-U – May/June 2014 | 9795 | 02
10 (i)
(ii)
(iii) | 0.05λ
Hooke’s Law: = 2.5g ⇒ λ = 250 N [AG]
0.5
Let e m be the extension in the new equilibrium position.
250e
4g = ⇒ e = 0.08 m
0.5
250(0.08+x )
Newton II: 4& x &=4g− ⇒ &x&=−125x
0.5
(a) Amplitude of new motion is 0.08 – 0.05 = 0.03 m
(b) Maximum speed = 125 × 0.03 = 0.335 ms –1 .
2π 2π
(c) T = = = 0.562 seconds.
ω 125
1 π 2
(d)  +sin−1  = 0.206 seconds
125 2 3 | Find new e
Needs 0.08√
&x&=10−125x
SR: and then
x = y – 0.08 : B2
cwo
√ on a or ω
√ on ω
2 2
sin –1 (“ ”) /ω or cos –1 (“ ”)/ω
3 3
Deal with quadrants | M1A1 [2]
M1
A1
M1
A1 [4]
B1 [1]
B1√ [1]
B1√ [1]
M1
M1
A1 [3]
11 (i)
(ii) (a)
(b) | gx 2 tan 2α – 2V 2 x tan α + (gx 2 + 2V 2 y) = 0 or equiv
2
For boundary of accessible points B – 4AC = 0
⇒ 4V 4 x 2 – 4gx 2 (gx 2 + 2V 2 y) = 0 (*)
1 ( )
⇒ 2gV 2 y = V 4 – g 2 x 2 ⇒ y= V 4−g 2 x 2 [AG]
2gV 2
dy gx 2
Or: =xsec2α − tanαsec2α
dα 2V 2
V 2 1 ( )
= 0 ⇒ x= ⇒ y= V 4−g 2 x 2
gtanα 2gV 2
1 ( )
Putting y = – h and V 2 = gh ⇒ −h= g 2 h 2−g 2 x 2
2g 2 h
⇒ x = 3h
Substituting for x in (*) gives
3gh 2 tan 2α – 2 3gh 2 tan α + 3gh 2 – 2gh 2 = 0
⇒ 3 tan 2α – 2 3 tan α + 1 = 0
1 π
⇒ tan α = ⇒ α = 30° or
3 6 | y must be part of quadratic
Allow ∆ ≥ 0
“Quadratic equation” required
in question, so B0
y = h: M1A0
−B
Or quote tan α =
2A | B1
M1
A1
A1 [4]
B0M1
A1
A1
[max 3]
M1
A1
A1 [3]
M1
A1
A1 [3]
Page 7 | Mark Scheme | Syllabus | Paper
Pre-U – May/June 2014 | 9795 | 02
Notation: Speed of wind relative to: ground, w; cyclist at 15ms –1 , u; cyclist at 10ms –1 , v
Angle of wind velocity with due west, θ (see diagram in method α)
An engine is travelling along a straight horizontal track against negligible resistances. In travelling a distance of 750 m its speed increases from 5 m s$^{-1}$ to 15 m s$^{-1}$. Find the time taken if the engine was

\begin{enumerate}[label=(\roman*)]
\item exerting a constant tractive force, [2]

\item working at constant power. [9]
\end{enumerate}

\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2014 Q9 [11]}}
This paper (12 questions)
View full paper
Q1 8 Q2 8 Q3 8 Q4 10 Q5 13 Q6 13 Q7 8 Q8 9 Q9 11 Q10 12 Q11 10 Q12 10