education.nsw.gov.au · web viewx = d dx 1 2 v 2 = d dx 9 2 -3 x 2 + x 4 2 =-6x+2 x 3 =2 x 3 -6x...
TRANSCRIPT
MEX-M1 Sample questions
M1.1 Simple Harmonic Motion
Introducing simple harmonic motion1. 2014 Mathematics Extension 1 HSC, Question 7
A particle is moving in simple harmonic motion with period 6 and amplitude 5. Which is a possible expression for the velocity, v, of the particle?
a. v=5 π3cos ( π3 t)
b. v=5cos( π3 t)c. v=5 π
6cos ( π6 t)
d. v=5cos( π6 t)Solution: Standard format of SHM (no phase shift)
x=a sin (nt )+ca=5
period=2 πn
6=2 πn
n=π3
x=5sin( π3 t)+cv= x=π
3×5×cos ( π3 t)
¿ 5π3cos( π3 t)∴ a is the correct answer.
© NSW Department of Education, 2019 1
2. 2014 Mathematics Extension 1 HSC, Question 12 part a.A particle is moving in simple harmonic motion about the origin, with displacement x metres. The displacement given by x=2sin 3 t, where t is time in seconds. The motion starts when t=0.a. What is the total distance travelled by the particle when it first returns to the origin?b. What is the acceleration of the particle when it is first at rest?
Solution:
a. 2+2=4m, given the amplitude is 2m.Explanation:x=02sin 3t=0
sin 3 t=03 t=0 , π ,2π…
t=0 , π3, 2π3…
The particle first returns to the original after π3 seconds.
Total displacement is2+2=4m
b. When at rest, x=0x=6cos3 t6cos3 t=0cos3 t=0
3 t=π2 (first solution)
t=π6
The particle is first at rest when t=π6 seconds
x=−18sin 3 t
whent=π6
x=−18sin(3× π6 )
¿−18sin( π2 )¿−18
2 MEX-M1 Sample questions
∴The acceleration of the particle when first at rest is−18ms−2
© NSW Department of Education, 2019 3
3. 2013 Mathematics Extension 1 HSC, Question 12 part e.A particle moves along a straight line. The displacement of the particle from the origin is x, and its velocity is v. The particle is moving so that v2+9x2=k, where k is a constant. Show
that the particle moves in simple harmonic motion with period 2π3 .
Solution:
v2+9 x2=k
v2=k−9 x2
12v2=1
2k−92x2
x= ddx ( 12 v2)= d
dx ( 12 k−92 x2)=−9 x
For SHM, x=−n2(x−b)
∴ x=−9 x moves in SHM with n=3andb=0
Period=2πn
=2 π3 as required.
Determining the speed during SHM
4. A particle is moving along the x-axis in simple harmonic motion. The displacement of the particle is x metres. The particle is at rest when x=−2 and when x=8. It takes 6 seconds to travel from x=−2 to x=8. What is the maximum speed of the particle?
a.56π ms-1
b.43π ms-1
c.53π ms-1
d.103π ms-1
Solution: Draw a diagram to represent the situation.
4 MEX-M1 Sample questions
∴T ( period)=12 s and n=2 πT
=π6
Centre of motion is hallway between −2m and 8m, i.e. at 3m
x=−n2 ( x−3 )
ddx ( 12 v2)=−n2 (x−3 )
12v2=−n2∫ (x−3 ) . dx
12v2=−n2( x22 −3 x)+C
v2=−n2 (x2−6 x )+Cwhen x=8, v=00=−n2 (82−6×8 )+C
C=16n2∴ v2=n2 (16+6 x−x2 ) s=√v2=√n2 (16+6 x−x2 )The maximum speed is at the centre of the motion, when x=3∴ s=√( π6 )
2
(16+6×3−32 )¿ 56 π m s−1∴a is the correct answer.
Solving Problems with SHM5. 2016 Mathematics Extension 1 HSC, Question 13 part a.
The tide can be modelled using simple harmonic motion. At a particular location, the high tide is 9 metres and the low tide is 1 metre. At this location the tide completes full periods every 25 hours. Let t be the time in hours after the first high tide today.
a. Explain why the tide can be modelled by the function x=5+4cos ( 4 πt25 )Solution:From reference sheet: x=acos (nt+α )+cHigh tide = maximum = 9Low tide = minimum = 1
∴a=9−12
=4
Period, T=252; (two tides per day)
252
=2 πn
∴n=4 π25
t=0 at high tide so α=0
© NSW Department of Education, 2019 5
∴9=4cos( 4 π25 ×0)+CC=5
∴ x=5+4cos ( 4 πt25 )
b. The first high tide tomorrow is at 2 am. What is the earliest time tomorrow at which the tide is increasing at the fastest rate?
Solution:
x=5+4cos ( 4 πt25 )dxdt
=−4 ∙ 4 π25sin ( 4 πt25 )
d2xd t 2
=−16π25
∙ 4 π25cos( 4πt25 )
Increasing fastest when d2 xd t 2
=0
cos ( 4 πt25 )=0∴ 4 πt25
=π2, 3π2,⋯
t=258, 758,⋯
Tide is decreasing (going out) when t=258 as
dxdt
=−16 π25
sin ( 4 π25 × 258 )=¿−16 π25
¿
∴ Tide is increasing (coming in) when t=758
=9 38 as
dxdt
=−16 π25
sin ( 4 π25 × 758 )=¿ 16 π25
¿
So 938 hours after 2 am makes the time 11:22
12 am (11:22 :30 am)
(or 1138 hours after midnight).
6 MEX-M1 Sample questions
6. The deck of a ship was 2.4m below the level of a wharf at low tide and 0.6m above wharf level at high tide. Low tide was at 8.30 am and high tide at 2.35 pm. Find when the deck was level with the wharf, if the motion of the tide was simple harmonic.
Solution:
Amplitude: a=2.4+0.62
=1.5m
Let 8.30am be t=0, the low tide
The wharf is 0.6m below the high tide, height=1.5−0.6=0.9
Let x=−acos (nt )
Period is twice the time from 8:30am to 2:35pm,
6hours∧5minutes=21900 seconds
T=2×21900=43800
n=2 πT
= 2π43800
x=−1.5cos ( 2π43800
t )Solve x=0.9
−1.5cos( 2π43800
t)=0.9cos ( 2π
43800t )=−0.9
1.5
2π43800
t=cos−1(−0.91.5 )t=43800
2πcos−1(−0.91.5 )
¿ 438002π
×2.21…
¿15435.8375…seconds
¿4 hours17minutes(correct¿the nearestminute)
Which refers to 12 :47 pm that day.
© NSW Department of Education, 2019 7
8 MEX-M1 Sample questions
M1.2: Modelling motion without resistance
Introduction to mechanics7. 2018 VCE Specialist Mathematics examination 2, Section A Question 16
The diagram below shows a mass being acted on by a number of forces whose magnitudes are labelled. All forces are measured in newtons and the system is in equilibrium.
The value of F2 is
a. √22
(8+3√3 )
b. 11√22
c. 3√22
d. 7.78e. 7.0
Solution: To be in equilibrium vertical and horizontal forces must have a net of zero.
Find the vertical component of the force of 3 newtons.
sin 30= w3
w=3×sin 30
¿1.5newtons
Vertical forces:
© NSW Department of Education, 2019 9
F2V=1.5+4
¿5.5newtons
sin 45=5.5F2
F2=5.5sin 45
¿ 5.51√2
¿5.5√2
¿ 11√22
∴b is the correct answer.
8. 2018 VCE Specialist Mathematics examination 1, Question 1Two objects of masses 5 kg and 8 kg are attached by a light inextensible string that passes over a smooth pulley. The 8 kg mass is on a smooth plane inclined at 30 ° to the horizontal. The 5 kg mass is hanging vertically, as shown in the diagram below.
a. On the diagram above, show all forces acting on both masses.
b. Find the magnitude, in ms−2, and state the direction of the acceleration of the 8 kg mass.
10 MEX-M1 Sample questions
cos60= b8g
8 g×cos60=bb=4 gNet Force=5 g−4 g=g N up the planeF=mam=5+8=13g=13a
a= g13
∴ The 8kg mass moves up the plane with an acceleration of g13
9. 2017 VCE Specialist Mathematics examination 2, Section A Question 17Forces of 10 N and 8 N act on a body as shown below.
The resultant force acting on the body will, correct to one decimal place, have
a. magnitude 15.6 N and act at 26.3° to the 10 N force.b. magnitude 9.2 N and act at 49.1° to the 10 N force.c. magnitude 15.6 N and act at 33.7° to the 10 N force.d. magnitude 9.2 N and act at 70.9° to the 10 N force.e. magnitude 15.6 N and act at 49.1° to the 10 N force.
© NSW Department of Education, 2019 11
Solution: Add the vectors tip to tail:
Use the cosine rule:R2=102+82−2×10×8× cos120¿100+64−160×−0.5¿244R=√244¿15.6N (1dp)Use the sine rule:sin θ8
= sin120√244
θ=sin−1( 8sin 120√244 )≈26.3 °∴a is the correct answer.
10. 2016 VCE Specialist Mathematics examination 2, Section A Question 14Two light strings of length 4 m and 3 m connect a mass to a horizontal bar, as shown below. The strings are attached to the horizontal bar 5 m apart.
Given the tension in the longer string is T 1 and the tension in the shorter string is T 2, the ratio
of the tensions T1T2
is
a.35
b.34
12 MEX-M1 Sample questions
c.45
d.54
e.43
Diagram with additional information: The triangle is right angles, T 1 makes an angle of
tan1( 34 ) with the horizontal bar.
Draw a force diagram:
The angle between mg and T 1 is 90−tan−1 34 (complementary angles)
The angle betweenmg and T 2 is tan−1 34 (angle sum of a triangle)
tan( tan−1 34 )=T 1
T 234=T 1T 2∴ b is the correct answer.
© NSW Department of Education, 2019 13
Determining and applying equations of motion (constant acceleration)
11. 2018 VCE Specialist Mathematics examination 2, Section A Question 15A constant force of magnitude P newtons accelerates a particle of mass 8 kg in a straight line from a speed of 4ms–1 to a speed of 20ms–1 over a distance of 15m.
The magnitude of P is
a. 9.8b. 12.5c. 12.8d. 100e. 102.4
Solution:
v2=u2+2ax
202=42+2×a×15400=16+30a
384=30a
a=38430
F=ma
¿8× 38430
¿102.4
∴ e is the correct answer.
12. 2017 VCE Specialist Mathematics examination 1, Question 9
A particle of mass 2 kg with initial velocity 3 i+2 jm s−1
experiences a constant force for 10
seconds. The particle’s velocity at the end of the 10-second period is 4 3i−18 j ms−1
.
a. Find the magnitude of the constant force in newtons.Solution: Acceleration is the change in velocity over the change in time.
x=(43 i−18 j)−(3i+2 j )
10=40 i−20 j
10=4 i−2 j
F=ma=2× (4 i−2 j )=8i−4 j|F|=√82+(−4 )2
¿√80¿4 √5
b. Find the displacement of the particle from its initial position after 10 seconds.
14 MEX-M1 Sample questions
x=4 i−2 j
x=∫ 4 i−2 j dt¿4 t i−2 t j+c
x=3 i+2 j when t=0
3 i+2 j=c
x=4 t i−2 t j+3i+2 j= (4 t+3 ) i−(2 t−2) j
x=∫0
10
((4 t+3 )i−(2 t−2) j )dt¿ [ (2 t2+3 t )i−(t2−2t ) j ]0
10
¿ ((2×102+3×10 )i−(102−2×10 ) j)−0¿230i−80 j
Determining and applying equations of motion (non-constant acceleration)
13. 2016 VCE Specialist Mathematics examination 2, Section A Question 15A variable force of F newtons acts on a 3 kg mass so that it moves in a straight line. At time t seconds, t ≥0, its velocity v metres per second and position x metres from the origin are given by v=3 – x2.
It follows that
a. F=−2 xb. F=−6 xc. F=2 x3−6 xd. F=6 x3−18 xe. F=9 x−3 x2
Solution
v=3−x2
v2=9−6x2+x4
12v2=9
2−3x2+ x
4
2
x= ddx ( 12 v2)= d
dx ( 92−3 x2+ x4
2 )=−6 x+2x3=2x3−6 x
F=ma=3 (2 x3−6 x )=6 x3−18 x∴ d is the correct answer.
© NSW Department of Education, 2019 15
M1.3: Resisted motion
Analysing motion through a resisting medium14. An object is moving on a smooth horizontal plane at 20ms−1in a resisted medium. The
resistance force acting on the particle is proportional to the square of the speed and is the only force acting on the particle throughout the duration of its motion. After 10 metres, the particle moves at 4ms−1.
a. Show that the model for velocity in terms of displacement is equal to v=20e−x ln 510
Solution:F∝v2
F=nv2
F=man v2=manmv2=a
r v2=a ,wherer=nm
a=v dvdx
r v2=v dvdx
rv=dv /dx1rv
=dxdv
1vdv=r dx
∫ 1v dv=∫rdx
ln v=rx+cx=0 , v=20ln 20=0 r+cc=ln 20x=10 , v=4ln 4=10r+cSubstitute c= ln 20ln 4=10r+ ln 20ln 4−ln 20=10 r
ln 420
=10 r
110ln 15=r
16 MEX-M1 Sample questions
r= ln5−1
10=−ln5
10
ln v=−ln 510
x+ln 20
ln v=−x ln 510
+ln 20
v=e−x ln 510 + ln 20
v=e ln 20e− x ln510
v=20e−x ln 510
b. Calculate the velocity of the object after 12 metres.
v=20e−12 ln510
¿2.8991…ms−1
c. Develop a model for displacement as a function of time, describing the motion of the particle.
v=20e−x ln 510
dxdt
=20e−x ln 510
dtdx
= 120
ex ln 510
∫1dt= 120∫e
x ln 510 dx
t= 12 ln5
ex ln 510 +c
x=0 , t=0
0= 12 ln 5
+c
c= −12 ln 5
t= 12 ln5
ex ln 510 − 1
2 ln5
t= 12 ln5
(ex ln 510 −1)
2 t ln 5=ex ln 510 −1
ex ln 510 =2 t ln5+1
x ln 510
=ln (2 t ln5+1 )
© NSW Department of Education, 2019 17
x=10 ln (2t ln 5+1 )ln5
18 MEX-M1 Sample questions
d. Develop a model for velocity as a function of time.
x=10 ln (2t ln 5+1 )ln5
= 10ln5
ln (2 t ln 5+1 )
x= 10ln 5
× 2 ln 52 t ln5+1
¿ 202t ln 5+1
e. State the limitations of the model
The exponential function is always greater than zero, i.e. e− x>0. It approaches the horizontal asymptote y=0 which is when x=0, but never reaches it. Given that the model for velocity
v=20e−12 ln510 comprises of a function of this type, this model will never reach v=0. Therefore
this model indicates that the object will never become stationary or stop, whereas in reality it will. This model is appropriate to describe the motion of the object until it becomes close to stopping.
© NSW Department of Education, 2019 19
Solving problems involving motion upwards and downwards in a resisting medium
15. 2018 Mathematics Extension 2 HSC, Question 14 part b.A falling particle experiences forces due to gravity and air resistance. The acceleration of the particle is g –k v2, where g and k are positive constants and v is the speed of the particle. (Do NOT prove this.)
Prove that, after falling from rest through a distance, h, the speed of the particle is
√ gk (1−e−2kh ).
Solution:
x=g – k v2using x=v dvdx
v dvdx
=g– k v2
dvdx
=g – k v2
vdxdv
= vg – k v2
∫1dx=−12k∫
−2kvg– k v2
x=−12kln (g – k v2 )+c
x=0 , v=0
0=−12kln (g –k ×02 )+c
c= 12kln g
x=−12kln (g – k v2 )+ 1
2kln g
x=−12kln( g – k v2g )
Let x=h and v=s
h=−12kln( g –k s2g )
−2kh=ln ( g –k s2g )e−2kh=g – k s2
g
20 MEX-M1 Sample questions
ge−2kh=g – k s2
k s2=g−ge−2kh
k s2=g (1−e−2 kh )
s2= gk
(1−e−2kh )
s=±√ gk (1−e−2kh ) but s is acting in a positive direction (the direction of motion)
∴ s=±√ gk (1−e−2kh )as required.
16. 2013 Mathematics Extension 2 HSC, Question 15 part d.A ball of mass m is projected vertically into the air from the ground with initial velocity u. After reaching the maximum height H it falls back to the ground. While in the air, the ball experiences a resistive forcek v2, where v is the velocity of the ball andk is a constant. The equation of motion when the ball falls can be written as
m v=mg−k v2(Do NOT prove this.)
a. Show that the terminal velocity vT of the ball when it falls is √ mgkSolution:Terminal velocity occurs when acceleration is zero, i.e. ˙a=v=0mg−k vT
2=0
k vT2=mg
vT2=mg
k
vT=±√mgkvT=√ mgkb. Show that when the ball goes up, the maximum height H is
H=vT2
2gln (1+ u
2
vT2 )
Solution: Produce equations of motion for the upwards trajectory:
F=ma∨m v
© NSW Department of Education, 2019 21
m v=−mg−k v2
m v=−mg(1+ k v2mg )m v=−mg(1+ v
2
vT2 )
v=−g(1+ v2
vT2 )
v=−gvT2 (vT2+v2)
using v=vdvdx
v dvdx
=−gvT2 (vT2+v2)
dvdx
=−gvT2 ( vT2 +v2v )
dxdv
=−vT
2
g ( vvT2+v2 )
∫1dx=−vT2
g ∫ vvT2+v2
dv
x=−vT
2
2gln (vT2+v2)+c
x=0 , v=u
0=−vT
2
2gln (vT2 +u2 )+c
c=vT2
2gln (vT2+u2)
x=−vT
2
2gln (vT2+v2)+
vT2
2 gln (vT2+u2 )
x=vT2
2g (ln (vT2+u2 )−ln (vT2 +v2 ))
¿vT2
2g ln( vT2 +u2
vT2+v2 )
x=H whenv=0
H=vT2
2g ln( vT2+u2
vT2+0 )
¿vT2
2g ln( vT2
vT2 +
u2
vT2 )
22 MEX-M1 Sample questions
H=vT2
2gln(1+ u
2
vT2 ) as required.
c. When the ball falls from height H it hits the ground with velocity w . Show that1w2
= 1u2
+ 1vT2
Solution: Produce equations of motion for the downwards trajectory:
F=ma∨m vm v=mg−k v2
m v=mg(1− k v2
mg )m v=mg(1− v2
vT2 )
v=g(1− v2
vT2 )
v= gvT2 (vT2−v2)
using v=vdvdx
v dvdx
= gvT2 (vT2−v2)
dvdx
= gvT2 ( vT2−v2v )
dxdv
=vT2
g ( vvT2−v2 )
∫1dx= vT2
g ∫v
vT2−v2
dv
x=−vT
2
2gln (vT2−v2 )+c
x=0 , v=0
0=−vT
2
2gln (vT2−02)+c
c=vT2
2gln (vT2 )
© NSW Department of Education, 2019 23
x=−vT
2
2gln (vT2−v2 )+
vT2
2gln (vT2 )
x=vT2
2g (ln (vT2 )−ln (vT2−v2) )
¿vT2
2g ln( vT2
vT2−v2 )
x=H when v=w
H=vT2
2g ln( vT2
vT2−w2 )
From part b, H=vT2
2gln (1+ u
2
vT2 )
vT2
2g ln(1+ u2
vT2 )= vT
2
2g ln( vT2
vT2−w2 )
1+ u2
vT2=
vT2
vT2−w2
vT2+u2
vT2 =
vT2
vT2−w2
(vT2+u2) (vT2−w2)=vT2 ⋅vT2
vT2 ⋅ vT2−vT2 ⋅w2+u2⋅ vT2−u2⋅w2=vT
2 ⋅vT2
−vT2 ⋅w2+u2 ⋅vT2−u2 ⋅w2=0
Divide through by vT2 ⋅w2 ⋅u2
−vT2 ⋅w2
vT2 ⋅w2⋅u2
+u2 ⋅vT2
vT2 ⋅w2⋅u2
− u2⋅w2
vT2 ⋅w2⋅u2
=0
−1u2
+ 1w2−
1vT2 =0
1w2
= 1u2
+ 1vT2 as required.
17. 2014 Mathematics Extension 2 HSC, Question 14 part c.A high speed train of mass m starts from rest and moves along a straight track. At time t hours, the distance travelled by the train from its starting point is x km, and its velocity isv km/h.
The train is driven by a constant forceF in the forward direction. The resistive force in the opposite direction is K v2, where K is a positive constant. The terminal velocity of the train is 300 km/h.
24 MEX-M1 Sample questions
a. Show that the equation of motion for the train is m x=F[1−( v300 )
2]m x=F−K v2
Terminal velocity occurs when acceleration is zero, i.e. x=0 when v=3000=F−K×3002
K ×3002=F
K= F3002
m x=F− F3002
v2
m x=F(1− v2
3002 )m x=F(1−( v
300 )2) as required.
© NSW Department of Education, 2019 25
b. Find, in terms of F and m, the time it takes the train to reach a velocity of 200 km/h.
m x=F(1−( v300 )
2)x= F
m (1−( v300 )
2)dvdt
=Fm (1− v
300 )(1+ v300 )
dtdv
=mF× 1
(1− v300 )(1+ v
300 )∫1dt=m
F∫1
(1− v300 )(1+ v
300 )dv
1
(1− v300 )(1+ v
300 )≡ A
1− v300
+ B
1+ v300
1≡ A(1+ v300 )+B(1− v
300)
when v=3001=2 A
A=12
when v=−3001=2 B
B=12
t= m2F∫
1
(1− v300 )
+¿ 1
(1+ v300 )
¿
t= 300m2F (− ln (1− v
300 )+ ln(1+ v300 ))+c
t= 150mF (− ln( 300−v
300 )+ln( 300+v300 ))+c
t=150mF
ln (300+v300300−v300
)+c¿ 150m
Fln( 300+v300−v )+c
26 MEX-M1 Sample questions
t=0 , v=0
0=150mF
ln( 300+0300−0 )+c0=150m
Fln1+c
c=0
t=150mF
ln( 300+v300−v )subsitute v=200
t=150mF
ln( 300+200300−200 )t=150m
Fln (5 ) hours
M1.4: Projectiles and resisted motion
Solving problems with projectiles in resisted motion18. An object of mass m is fired from ground level at um s−1 at an angle of 30° with air resistance
proportional to the square of the velocity.You may assume the following equations of motion:
Upwards trajectory: m⋅ yu=mg+k yu2
Downwards trajectory: m⋅ yd=mg−k yd2
Horizontal trajectory: m⋅ x=−k x2
uy=u sin30=u2ms−1
ux=ucos30=u√32
ms−1
Show that the maximum height reached is H= m2kln( 4mg4mg+ku2 )
m⋅ yud yud yu
=mg+k yu2
d yud yu
=mg+k yu
2
m yud yud yu
=m yu
mg+k yu2
∫1d yu=m∫yu
mg+k yu2 d yu
© NSW Department of Education, 2019 27
yu=m2kln (mg+k yu2 )+c
When yu=0 , yu=u2
0= m2kln(mg+ k u24 )+c
c=−m2k
ln( 4mg+ku24 )yu=
m2kln (mg+k yu2 )− m
2kln( 4mg+ku24 )
yu=m2kln( 4mg+4k yu
2
4mg+k u2 )The maximum height, yu=H reached is when yu=0
H= m2kln ( 4mg+4k ×024mg+k u2 )
H= m2kln( 4mg4mg+ku2 )
28 MEX-M1 Sample questions