2012 JC1 H2 MYE (Solution)

1

2012 NYJC JC1 Mid-year Exam – 9646 H2 Physics [Turn over

SECTION A

1 The calculation of a physical quantity is given to be

( )( ) ( )( )1-21-9 m5.0C12mC3.2N100.3 ×

Which of the following is the correct S.I. derived unit for the answer?

A kg m2 s‒2 C‒1 B kg m2 s‒2

C kg m2 s‒3 C‒1 D kg m2 s‒3

2 Four physical quantities P, Q, R and S are

related by the equation P = Q – RS. Which

statement must be correct?

A P, Q, R and S are all scalar quantities.

B P, Q, R and S are all vector quantities.

C P, Q, R and S have the same unit.

D The unit of R is the same as the unit of

P / S.

3 Three forces of 3 N, 4 N and 5 N are in equilibrium. The angle between the 3 N and

5 N forces is

A 37° C 127°

B 53° D 143°

4 Which of the following pairs include a vector

and a scalar?

A displacement, weight

B pressure, power

C speed, kinetic energy

D work, acceleration

• For homogeneous equation, each of the 3 terms

MUST have the same units. The unit of RS must

equal that of P and Q. Hence R and S cannot

separately have the same unit as P and Q. (Not C)

• All 3 terms CAN be scalars but R,S can each be

vectors such that their product is a scalar e.g.

FORCE multiplied with DISPLACEMENT to give

WORK which is a scalar. Similarly all terms can be

vectors with R a scalar and S a vector such that RS

is a vector. (Not A nor B)

• [P] = [Q] = [RS] = [R][S] i.e. [R] = [P] / [S]

• 3,4,5 will form a right angled closed triangle.

• The angle between 2 vectors is the acute or

obtuse angle formed when the 2 vectors are

aligned to start at the same point. 3N

4N 5N

5N ?

Remember that work is scalar

and its associated quantities

such as all forms of energy ,

power are also scalar.

5 Four sets of measurements are made of the diameter d of a wire using a micrometre

screw gauge, as shown in the table.

If the actual diameter of the wire is 1.49 mm, which set of measurements could be

described as accurate but not precise?

Set of readings Readings, d / mm

A 1.49 1.46 1.52 1.50

B 1.48 1.58 1.51 1.40

C 1.35 1.37 1.42 1.42

D 1.32 1.37 1.41 1.50

Answer: B

Accurate � How close they are to the true value

Precise � How close the measured readings are from one another

Readings d/ mm Avg. Value

1.49 1.46 1.52 1.50 1.4925 Accurate. More precise

1.48 1.58 1.51 1.40 1.4925 Accurate. Less precise

1.35 1.37 1.42 1.42 1.39 Not accurate

1.32 1.37 1.41 1.50 1.4 Not accurate

6 A train accelerates from rest at (1.5 ± 0.5) m s‒2 over a distance of (400 ± 5) m. What

would its final velocity be at the end of the distance covered?

A (35 ± 12) m s‒1

B (30 ± 10) m s‒1

C (35 ± 6) m s‒1

D (34.6 ± 6.0) m s‒1

Answer: C

v2 = u2 + 2 a s = 2(1.5)(400) � v = 34.641

2

1 1 0.5 5( ) ( )(34.641) 5.99 6

2 2 1.5 400

35 6

v a s

v a s

a sv v

a s

v v

∆ ∆ ∆= +

∆ ∆∆ = + = + = =

∴ ± ∆ = ±

3


Ball released

from rest with

speed = 0

(taking down

as positive)

7 At time t = 0 s, a ball is released from rest above a floor. In the velocity-time graph, at

which point in time does the ball reach its maximum height after bouncing from the

floor?

8 A motorist travelling at 10 m s–1 can bring his car to rest in a braking distance of 10 m.

In what distance could he bring the car to rest from a speed of 30 m s–1 using the same

braking force?

A 17 m C 52 m

B 30 m D 90 m

9 A student throws a ball vertically upwards and it returns to him in 3.0 s. Then he throws

the ball at the same speed again, but at an angle of 30° to the horizontal. How long will

it take for the ball to return to its initial level? Neglect air resistance.

A 1.5 s C 3.0 s

B 2.1 s D 6.0 s

velocity v

time t A

B

D C 0

Ball falling

down with

increasing

velocity at

the constant

rate of g

(slope of

line )

At time A,

Ball makes

contact

with

ground

with a

large

downward

speed

(positive)

At time B,

Ball is still

in contact

with

ground and

its speed

reduces to

zero.

At time C,

Ball loses

contact

with

ground and

rebounds

from

ground

with a large

upward

speed

(negative).

Between C and D, Ball

moves away from

ground on its upward

journey with

decreasing (negative)

velocity. It decelerates

at the rate g (positive

slope) until its speed

reduces to zero at D

when it reaches its

maximum height.

10 The figure below shows the path of a ball leaving the ground at point X, reaching a

maximum height at point Y, and returning to the ground at point Z.

If air resistance is negligible, the vertical component of

A velocity is maximum at Y.

B acceleration at Y is zero.

C velocity has the same magnitude at X and Z.

D acceleration is lesser at Y than at Z.

11 Two blocks X and Y, of masses m and 3m respectively, are connected to block Z of

mass 4m by a light rod R. The blocks are accelerated along a smooth horizontal

surface by a force F as shown in the diagram below.

What is the magnitude of the force exerted by block Y on block X during the

acceleration?

A F / 8 C F / 2

B 3F / 8 D 7F / 8

θ

Initial velocity

X

Y

Z Ground

[C]

net Force F on all blocks = (1 + 3 + 4) m a ⇒⇒⇒⇒ a = F/8m

net force on Z = FRZ = (4m) a = 4m (F/8m) = F/2

FRZ = FRY (same compression in the same light rod)

net force on Y = FXY - FRY = FXY - F/2 = (3m) a

FXY = F/2 + (3m) a = F/2 + 3m (F/8m)

= 7F/8 = FYX = force by Y on X

5


12 A stationary body explodes into two fragments of masses m and 2m. The fragments

gain kinetic energies X and Y respectively.

What is the value of the ratio X

Y?

A 0.25 C 2.0

B 0.50 D 4.0

Ans: C

From Conservation of Momentum,

Initial momentum of system = Final momentum of system

0 = px + py

py = – px

Use 2p

KE =m2

to establish the ratio,

= = =

xx

y x

pp

X m m

pY p

m m

22

2 2

2( ) 2( )2.0

2(2 ) 2(2 )

13 A small boat is about to be towed up from the sea on rollers, via a pulling cable

connected to a pulley system as shown in the diagram. Which is the correct free body

diagram for the boat, paying attention to the points of application of all forces?

U: upthrust

T: pull by cable

W: weight of small boat

N: normal reaction force.

Answer: B

14 A hinged uniform rod of length 1.0 m is held in the horizontal position by a cable as

shown. The cable is attached to the rod at a point 0.25 m away from the hinge.

Three forces act on the rod. The forces are the weight W of the door, the tension T in

the cable and the force H at the hinge. The diagram is not drawn to scale.

Which list gives the three forces in increasing order of magnitude?

A W, H, T C H, T, W

B W, T, H D T, H, W

T T

T T

U

U

U

U

W W

W W

N N

N N

A B

C D

7


15 Two blocks X and Y, of masses 1.50 kg and 0.50 kg respectively, are connected by a

light string passing through a smooth pulley as shown.

If Y is released from its rest position 0.80 m above the ground, what is the speed with

which it will hit the ground?

A 1.4 m s‒1 C 4.0 m s‒1

B 2.0 m s‒1 D 16 m s‒1

16 A load is slowly lowered down a smooth slope using a rope as shown.

The forces acting on the load are its weight W, the pull of the rope T, and the normal

reaction force from the slope N.

Which of the following correctly describes the work done by W, T, and N?

W T N

A positive negative zero

B negative positive zero

C negative zero positive

D zero negative positive

Answers:

1 2 3 4 5 6 7 8

A D C D B C D D

9 10 11 12 13 14 15 16

A C D C B A B A

pulley X

Y

0.80 m

smooth surface

load

SECTION B – Answer ALL the questions.

17 The volume of water V flowing through a pipe in time t is given by

η

π

L

ppr

t

V

8

)( 21

4−

=

where p1 and p2 are the respective pressures at each end of the pipe, r is the pipe’s

radius, L is the pipe’s length, and η is the viscosity of water. In order to determine the

viscosity of water, the following measurements are taken.

V = (10.0 ± 0.1) cm3

r = (0.43 ± 0.01) mm

t = (4.0 ± 0.1) s

L = (15.5 ± 0.1) cm

p1 = (2.150 ± 0.005) × 105 Pa

p2 = (1.000 ± 0.005) × 105 Pa

(a) Use the equation to find the S.I. base units of η.

S.I. base units of η = ......................... [2]

(b) Verify that the value of η is 3.98 × 10‒3.

[1]

4

1 2

4

1 2

24

24

3

-1 1

( )

8

( )

8

kg m s(m ) (s)

m[ ] [ ] [ ][ ]

[ ] [ ] (m) (m )

kg m s

r p pV

t l

r p p t

l V

r p t

l V

−

−

π −=

η

π −η =

η = =

=

4

1 2

3 4 5

-2 -6

-3

( )

8

(0.43 10 ) (2.150 1.000 ) 10 4.0

8 (15.5 10 ) (10.0 10 )

3.98 10

−

−=

× − × ×=

× ×

= ×

r p p t

l V

πη

π

9


(c) Calculate the associated uncertainty for the viscosity of water, ∆η.

uncertainty ∆η = ......................... [4]

18 A small rubber block is released from rest and slides down a smooth ramp of length

0.500 m inclined at angle of 35°. It then rebounds with the same speed upon collision

with the wall at the bottom of the ramp.

(a) (i) Calculate the block’s acceleration along the ramp. [2]

(ii) Calculate the time taken by the block to reach the bottom of the ramp. [1]

(iii) Find the speed of the block when it is at the bottom of the ramp. [1]

(b) (i) On Fig. 18.1, sketch the variation with time of the rubber block’s velocity,

from the moment of release to the moment it attains maximum height.

Label with appropriate values.

0.500 m

35°

∑F// → mg sin θ = ma

a = g sin θ = 9.81 sin 35°

= 5.63 m s‒2

s = ut + ½ a t2

t = √( 2s / g sin θ)

= √(2 × 0.500/ 9.81 sin 35°)

= 0.42 s

v = u + at = g sin θ × t

= 9.81 sin 35° × 0.42

= 2.37 m s‒1

( )( )

4

1 2

1 2

1 2

3 -3

( )

8

4

0.01 0.005 0.005 0.1 0.1 0.14

0.43 2.150 1.000 4.0 15.5 10.0

0.14317

0.14317 3.984 10 0.6 10−

−=

∆ + ∆∆ ∆ ∆ ∆ ∆= + + + +

−

+ = + + + +

−

=

∴∆ = × = ×

r p p t

l V

p pr t l V

r p p t l V

πη

η

η

η

[2]

(ii) On Fig. 18.2, sketch the variation with time of the acceleration of the rubber

block, from the moment of release to the moment it attains maximum

height. Label with appropriate values.

[2]

19 A dive bomber, diving with constant speed u at

an angle of 53° to the vertical, releases a bomb

at an altitude of 730 m. The bomb hits the ground

5.0 s after release.

(a) Calculate the speed, u, of the bomber. [2]

Taking downwards as positive,

sy = uy t + ½ ay t2

730 = uy (5.0) + ½ (9.81) (5.0)2 uy = 121 m s-1 u cos 53 = 121 u = 200 m s-1

(b) Calculate the horizontal distance travelled by the bomb during its flight. [1]

sx = ux t = (200 sin 530) (5.0) = 800 m

a/m s‒2

t/s

Fig. 18.2

5.63

0.42

0.84

v/m s‒1

t/s

Fig. 18.1

2.37

‒2.37

0.42

0.84

53°

11


(c) Sketch a graph showing how

(i) vx, the horizontal component of the velocity of the bomb, and

(ii) vy, the vertical component of the velocity of the bomb

vary with time, from the time the bomb is released to the time the bomb hits the

ground.

[2]

(d) Draw a vector diagram to illustrate the concept of the change of velocity, ∆v, of

the bomb from the time of release to the time it hits the ground.

[2]

20 (a) State Newton’s second law of motion.

.................................................................................................................................

............................................................................................................................ [1]

(b) A lorry of mass m = 1000 kg pulls horizontally on a trailer of mass M = 1500 kg.

When they accelerate at 3.0 m s−2, the total resistance to motion, R, is constant

at 5000 N. One quarter of this resistance is acting on the trailer.

lorry trailer

v

t

vx

vy

The rate of change of momentum is directly proportional to the net force acting on a

body and the change in momentum takes place in the direction of the net force.

∆v v

-u

(i) Calculate the driving force on the lorry. [2]

(ii) On the diagram of the trailer below, draw all the horizontal and vertical

forces acting on it. Label all forces clearly.

[2]

(iii) Calculate the force exerted by the lorry on the trailer. [1]

21 Fig. 21.1 shows a man standing on top of a stationary wind-surfing board. The

sailboat, consisting of the surfing board and sail, floats on the sea.

Weight

Normal reaction from ground Tension T or pull

by lorry on trailer

through the

connecting rod

¼ R

Fig. 21.1

Net force on lorry and trailer = (m + M) a

= driving force F (on lorry) ‒ total resistance on both lorry and trailer

F - R = (1000 + 1500) × 3 = 7500 N

F = 7500 + 5000 = 12500 N

Net force on trailer = Ma = 1500 × 3 = 4500 N

T ‒ ¼ R = 4500

T = 4500 + ¼(5000)

= 4500 + 1250 = 5750 N

13


(a) State two conditions required for the sailboat to be in a state of equilibrium.

.................................................................................................................................

.................................................................................................................................

............................................................................................................................ [2]

(b) (i) The total mass of the sailboat, including the man standing on it, is 90 kg.

Determine the total upthrust acting on the sailboat.

upthrust = ......................... N [2]

(ii) Hence, given that the density of seawater is 1020 kg m‒3, calculate the

volume of seawater displaced by the sailboat.

volume = ......................... m3 [2]

Resultant force acting on a sailboat is zero.

Resultant torque acting on a sailboat is zero.

Comments:

Some students did not write in words for the condition making it difficult

to understand the conditions.

Common mistakes:

• Resultant moment is zero.(‘about any point’ is missing)

• Upthrust equal to weight of boat.(too specific, may have other

forces).

• Torque acting on sailboat is zero.(‘net’ is missing)

∑∑∑∑Fy = 0 U – mg = 0

U = (90)(9.81) = 880 N

Comments:

The correct answer was calculated by the majority of candidates.

However, a significant number of candidates did not explain clearly why

U=mg. In this case, the sailboat is at equilibrium and hence ΣΣΣΣF=0.

U = weight of fluid displaced = Vpg

∴∴∴∴V = U/pg = 880/(1020)(9.81) = 0.088 m3

The vast majority of candidates correctly calculated the volume

displaced.

(c) Moments later, the sailboat is cruising at constant speed. Fig. 21.2 (next page)

shows some of the forces acting on the mast of the sailboat with the sail

attached to it.

The base of the mast is connected to the surfing board by a universal hinge. The

wind exerts a force of 200 N on the sail, perpendicular to the mast. The mast is

4.3 m long, and its mass is uniformly distributed.

(i) Show that the man pulls on the sail with a force of 460 N.[1]

Pulling force of man, T

Fig. 21.2

Force by wind = 200 N

Universal hinge

1.5 m

Weight of mast and sail = 50 N 3.5 m

75°

70°

∑∑∑∑Mhinge = 0 (200)(3.5) – 50(2.15cos70o) – Tsin75o(1.5) = 0

T = 460 N (shown)

Comments:

The vast majority of candidates realised they need to apply the condition

that resultant moment about any point = 0. However, only a minority are

able to mathematically worked out the correct answer.

15


(ii) Hence, determine the magnitude of the force, R, exerted by the universal

hinge.

force exerted by hinge = ......................... N [3]

(d) Fig. 21.3 (next page) shows the top view of the sailboat, with some of the forces

that act on it.

If the sailboat is moving forward on the sea at constant speed, indicate with an

arrow on Fig. 21.3 the likely direction of the force due to the wind, Fw.

Explain your answer clearly.

Drag force, FD Fig. 21.3

Sideward

force, F

Forward

motion

Fw

∑∑∑∑Fy= 0 Ry + (200sin20o) – 50 – 460(sin5o) = 0 Ry = 21.7 N

∑∑∑∑Fx= 0 Rx - (200cos20o) + 460(cos5o) = 0 Ry = 270 N

∴∴∴∴R=√[(Rx)2 + (Ry)2] = 271 N

Comments:

Answers here were disappointing as the vast majority of candidates did

not realised they need to apply the condition that ΣΣΣΣFy = 0 and ΣΣΣΣFX = 0.

Hence very few get full credit for this part.

.................................................................................................................................

............................................................................................................................ [2]

(e) Fig. 21.4 shows the side view of the sail boat with all the forces acting on it.

(i) T

he sideward force F from Fig. 21.3 is the same force exerted on the fin, F,

shown in Fig. 21.4. This force arises when the wind exerts a force against

the sail. With reference to Fig. 21.3 and 21.4, explain how this force

exerted on the fin comes about.

Weight of man fin

Force exerted

on sail by wind

Weight of sailboat

Fig. 21.4

Upthrust

Force exerted on

fin by seawater, F

Since the sailboat is moving with a constant speed, no acceleration. By N2L, resultant force = 0. Hence FW has to act in a direction shown to ensure that.

Comments:

Answers here were disappointing as there are significant number of

candidates need to explain that Fnet =0 when they indicate the direction of

Fw.

Common mistakes:

• Consider forward motion as a force.

• A forward net force is required to move forward at constant

speed.(Based on Newton’s 1st law, Fnet = 0).

• Draw Fw directly opposite to sideward force.

17


.......................................................................................................................

.................................................................................................................. [1]

(ii) Suggest why the surface area of the fin can be much smaller than the area

of the sail.

.......................................................................................................................

.................................................................................................................. [2]

(iii) With reference to Fig. 21.4, explain why it is important for the man to lean

back to keep the sail boat stable.

.......................................................................................................................

.......................................................................................................................

.................................................................................................................. [2]

The force exerted on the sail result in a tendency of motion between the sailboat and the water. Hence the sea water will exert a force on the fin to resist it as it is contact with the fin.

Comments:

Many candidates had no appreciation of the situation. Some just assume the force on fin will just appear from nowhere to ensure equilibrium without focusing on the condition for it to arise. In this case, there must be a relative motion between the fin and seawater.

Since the density of water is much larger than air, the effective area of the fin can be smaller than the sail to give a force of same magnitude.

Comments:

Many candidates had no appreciation of the situation as density of fluid

affect the rate of change of momentum of fluid and hence the force acting

on it. Many also did not realize the two force must have the same

magnitude.

Based on the Fig, the man has to learn back to create an clockwise moment about the CG to counter the clockwise moment produced by upthrust and the force by the wind. This to ensure the system can be in rotational equilibrium.

The force by wind and force on fin will create a anti-clockwise torque .

Hence the man need to lean back to create increase the clockwise

torque to ensure no resultant torque and the sailboat can be stable.

Comments:

Many candidates show good appreciation of the situation but the answer

given is too vague and lack specific detail to be given full credit. A vast

majority did not realize that they need to take moment about the C.G. of

the boat.

(f) The surf board is designed with the foot straps at the rear part of the board, as

shown in Fig. 21.5, rather than the centre part.

(i) Such a design allows the surf board to move across the sea surface

inclined at an angle to the surface. Explain how this inclined angle reduces

the volume of seawater displaced by the sailboat, compared to your

answer in (b)(ii).

.......................................................................................................................

.......................................................................................................................

.................................................................................................................. [2]

(ii) Hence explain how having foot straps at the rear part rather than the centre

part will allow the surfer to cruise at a higher speed.

.......................................................................................................................

.......................................................................................................................

.................................................................................................................. [1]

Foot straps

Fig. 21.5

When the board is inclined and moving across the sea, there will be water continuously hitting the board. This will resultis an additional upward force. Since U + F = mg, the upthrust will decrease. Hence the volume of seawater displaced will also decrease.

Comments:

Many candidates had great difficulty accounting why upthrust decrease.

They lack good appreciation of the situation and only one candidate was

able to score full marks for this part. Some even mentioned that the

weight can decrease and the board is inclined.

Since volume of seawater displaced decrease, the surface area in contact with the seawater may decrease. Hence the drag force will also decrease. Since there will be less work done against the drag force, the sailboat can move at higher speed.

Comments:

Many candidates did not do well in this part. They did not realize that the

surface area in contact with seawater decrease was due to a decrease of

volume displaced rather than the inclined angle.

19


22 (a) The graph of Force vs. Extension shows the tensile behaviour of a bungee

cord.

With reference to the above graph, state and justify whether the bungee cord

obeys Hooke’s Law.

.................................................................................................................................

.................................................................................................................................

............................................................................................................................ [2]

(b) The figure below shows man doing a bungee jump off a platform 80.0 m above a

lake, using the bungee cord from (a).

The man has a mass of 800 N. He falls through a distance of 32.5 m before the

cord becomes taut and begins to exert a force on him.

80.0 m

extension / m

Fo

rce

/ k

N

0 5 10 15 20 25 30

1.0

2.0

3.0

4.0

Yes. The force-extension graph is a straight-line graph that passes through the

origin, which shows that the cord’s extension is proportional to the force exerted

on it.

(i) State the man’s acceleration at the point when he has fallen a distance of

32.5 m.

acceleration = ......................... m s‒2, direction ......................... [1]

The man falls through a total distance of 61.0 m before he stops for the first time.

(ii) Deduce the extension of the cord when he stops for the first time.

extension = ......................... m [1]

(iii) Deduce the acceleration of the man at that point in (ii).

acceleration = ......................... m s‒2, direction = ......................... [3]

(iv) Show that the potential energy stored in the cord at this point is 49 kJ.

[1]

(c) (i) The gravitational potential energy at the surface of the lake is taken to be

zero. Complete the table below with values of elastic potential energy of

the cord, as well as the gravitational potential energy and kinetic energy of

the man at the three points stated, giving your answers to two significant

figures. [4]

At the platform

32.5 m below

the platform

61.0 m below

the platform

Gravitational

Potential Energy / kJ 64 38 15

Elastic Potential

Energy / kJ zero zero 49

Kinetic Energy / kJ zero 26 zero

9.81 downward

Extension = 61.0 ‒ 32.5 = 28.5 m

Force from cord = 3.40 kN (from graph)

Net force on man = 3400 + (‒800) = 2600 N = m a

Solving, a = 2600/(800/9.81) = 31.9 m s‒2 upward

Potential energy stored = Area under graph (from x=0 to x=28.5 m)

= ½(3400)(28.5) = 4.85 × 104 J

or by C.O.E., Potential energy stored = G.P.E. lost

= (800) (61) = 4.88 × 104 J

21


(ii) The sketch graph below (not to scale) shows the variation of the man’s

kinetic energy with the distance from the platform as he falls.

On the same axes above, sketch two clearly labelled graphs to show how

the gravitational potential energy of the man and the elastic potential

energy of the cord vary with the man’s distance from the platform. [2]

(iii) Using the graph provided in (a), show that the man has maximum kinetic

energy at a distance of 39 m from the platform. [2]

(iii) Calculate the maximum kinetic energy during the fall. [3]

61.0 32.5 0.0

Energ

y

Distance from

platform / m

Kinetic

Gravitational

Elastic

Max

KE

When falling with greatest speed, acceleration = 0

Net force on man = T ‒ W = 0

Thus T = 800 N

From graph, extension of cord = 6.5 (to 6.75) m

Distance from platform = 32.5 + 6.5 = 39.0 m

Total energy = 64 kJ

GPE = 800 (80 ‒ 39) J = 33 kJ

EPE = ½ (6.5)(800) = 2.6 kJ

Kinetic energy = 64 ‒ 33 ‒ 2.6 = 28 kJ

2012 JC1 H2 MYE (Solution)

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Transcript of 2012 JC1 H2 MYE (Solution)