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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.

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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



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.


∴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



∴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. 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.



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:



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.


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.



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


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.



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.


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.



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


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 )



x=10 ln (2t ln 5+1 )ln5


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.



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


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



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 )


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 )



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.


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.



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


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



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 )


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...

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Transcript of 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...