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© JJC 2014 8866/ JC1 Promo P2/ 2014 [Turn Over

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JURONG JUNIOR COLLEGE JC1 Promotional Examination 2014

Name Class 15S

PHYSICS Higher 1

8866/02

Paper 2 Structured Questions Candidates answer on the Question Paper No Additional Materials are required

24 Sept 2014 1 hour 30 minutes

READ THESE INSTRUCTIONS FIRST Do not open this booklet until you are told to do so. Write your name and class in the spaces provided at the top of this page. There are five questions in this paper. Answer all the questions in the spaces provided. Write in dark blue or black pen. You may use a soft pencil for any diagrams, graphs or rough working. Do not use highlighters, glue or correction fluid. The number of marks is given in brackets [ ] at the end of each question or part question.

For Examiner’s Use

1 /10

2 /12

3 /12

4

/12

5

/14

Total /60

(This question paper consists of 16 printed pages)


2

Data

speed of light in free space, c = 3.00 x 108 m s-1

elementary charge, e = 1.60 x 10-19 C

the Planck constant, h = 6.63 x 10-34 J s

unified atomic mass constant, u = 1.66 x 10-27 kg

rest mass of electron, me = 9.11 x 10-31 kg

rest mass of proton, mp = 1.67 x 10-27 kg

acceleration of free fall, g = 9.81 m s-2

Formulae

uniformly accelerated motion, s = ut + ½at2

v2 = u2+ 2as

work done on/by a gas, W = pV

hydrostatic pressure, p = gh

resistors in series, R = R1 + R2 + …

resistors in parallel, 1/R = 1/R1 + 1/R2 + …


3

Answer all the questions in the spaces provided.

1. (a) (i) State the S.I. unit for torque.

[1]

(ii) Express the joule in terms of S.I. base units. [2]

(iii) It can be shown that the S.I. base unit for torque is the same as the joule, J. Suggest why it is inappropriate to express the unit of torque as J.

[1]

(b) A student attempting to determine the density of a solid cylinder obtains the following readings:

diameter of cylinder = 10.0 0.1 mm

height of cylinder = 9.0 0.1 mm

mass of cylinder = 55.3 0.1 g….(13.6 0.1 g)

The density of the cylinder was calculated as 78273…..19280 kg m-3.

(i) Calculate the absolute uncertainty in the value of the density.


4

absolute uncertainty = kg m-3 [3]

(ii) State and explain which measurement gives rise to the least uncertainty.

[2]

(iii) Express the density of the cylinder with its associated uncertainty. [1]


5

2. (a) Define speed of an object.

[1]

(b) Fig. 2.1 shows the variation with time t of the velocity v for an object.

Fig. 2.1

(i) State the time at which the object is at maximum displacement from the starting point.

time = s [1]

(ii) Calculate the displacement of the object at t = 12.0 s.

displacement = m [2]


6

(iii) On Fig. 2.2, sketch a graph to show the variation with time t of the displacement s for the object. (You are not expected to label values of the displacement.) [2]

Fig. 2.2

(c) A car, on a steep incline of 37.0° below the horizontal, rolls down from rest with a constant

acceleration of 4.00 m s2 along the incline. It travels 50.0 m down to the edge of a vertical cliff, which is 30.0 m above the ocean surface as shown in Fig. 2.3.

Fig. 2.3

50 m

30 m

37.0o


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(i) Determine the speed of the car when it reaches the edge of the cliff,

speed = m s-1 [2]

(ii) 1. Determine the horizontal component of the car’s velocity just before it lands in the ocean.

velocity = m s-1 [1]

2. Show that the vertical component of the car’s velocity just before it lands in the ocean is 27.1 m s-1. [1]

3. Hence determine the final velocity, v of the car just before it lands in the ocean.

velocity = m s-1 [1]

direction = [1]


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3 (a) (i) Explain what is meant by moment of a force.

[1]

(ii) A uniform rigid rod of weight W1 of 400 N is attached to a vertical beam by a hinge as shown in Fig. 3.1. The other end of the rod is fastened to a support cable. The structure is used to support a load W2 of 2000 N.

Fig. 3.1

1. Show that the tension T in the support cable is 4400 N. [2]

rod

hinge

load

Support cable

60o

vertical beam


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2. Determine the magnitude and direction of the resultant force acting on the rod by the hinge.

magnitude = N [3]

direction = [1]

(b) (i) State Hooke’s law.

[2]


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(ii) A sphere of mass 3.00 kg rests on a frictionless slope, supported by a spring as shown in Fig. 3.2.

Fig.3.2

The spring obeys Hooke’s Law. The spring constant is 500 N m-1.

1. Calculate the component of weight along the slope.

component of weight = N [2]

2. Hence, calculate the compression of the spring.

compression = m [1]

spring


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4 (a) State Newton’s Second Law of motion.

[1]

(b) A water droplet in a cloud falls through air and reaches terminal velocity eventually.

Use Newton’s Second Law to explain the situation. (Assume upthrust is negligible.)

[3]

(c) A man pushes horizontally on a heavy rock resting on the ground but it does not move.

(i) Use Newton’s Law(s) to explain the situation.

[2]


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(ii) A student explained it as “the pushing force being balanced by the reaction of this

force.”

Use Newton’s Third Law to comment on the validity of this explanation.

[1]

(d) A steel ball A of mass 0.20 kg, moving with a velocity of 4.0 m s-1, collides elastically and head-

on with another steel ball B of mass 0.40 kg and moving with a velocity of 0.50 m s-1 as shown in Fig. 4.

Fig. 4

(i) Explain the significance of the word “head-on”.

[1]

(ii) After the collision, the two balls separate.

1. Apply the law of conservation of momentum to write a mathematical equation to represent this collision. [1]

A B

4.0 m s-1 0.50 m s-1


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2. Determine the magnitude of the final velocities of balls A and B.

magnitude of the final velocity of A = m s-1

magnitude of the final velocity of B = m s-1 [3]


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5 (a) (i) Define work.

[1]

(ii) A constant force F displaces an object, moving at a constant velocity v, by a distance s over a time interval t, and that F, v and s all point along the same line as shown in Fig. 5.1.

Derive an expression for the power P, in terms of the force F and the velocity v of the object. Explain your working clearly. [3]

Fig. 5.1

t

F F

s

v v


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(b) A car is travelling along a horizontal road with speed v. The power P, required to overcome external forces opposing the motion is given by the expression

P = cv + kv3

where c and k are constants.

(i) The first term cv in the given expression is related to the rate of work done to overcome a constant force. Give an example of the force related to the second term kv3 in the expression.

[1]

(ii) For one particular car, the numerical values of c and of k are 240 kg m s-2 and 0.98 kg m-1 respectively. Calculate the power required to enable the car to travel along a horizontal road at 100 km h-1.

power = W [3]

(c) The car in (b) has mass 720 kg. Using your answer to (b)(ii) where appropriate, calculate, for the car travelling at 100 km h-1,

(i) its kinetic energy,

kinetic energy = J [2]


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THE END

(ii) the magnitude of the external force opposing the motion of the car,

external force = N [3]

(iii) the work done in overcoming the force during a time of 5.0 minutes.

work done = J [1]


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JURONG JUNIOR COLLEGE PHYSICS DEPARTMENT

2014 JC1 Promo 8866 H1 Physics Paper 2

Suggested Solutions

Qn Suggested solution Remarks

1(a)(i) N m [1]

(ii) Starting from E = mgh Base S I unit = kg m s-2 m = kg m2 s-2

any correct working [1] final ans [1]

(iii) Torque(moment) is a physical quantity which measures the turning effect of a force. It is a vector. Energy is associated with work done and is a scalar. Expressing the unit of torque as J would therefore be misleading.

[1]

(b)(i) x100%

h

Δh

d

Δd2

m

Δmx100%

ρ

Δρ

x100%9.0

0.1

10.0

0.12

55.3

0.1

= 3.3%

3Δρ 0.033 x 19280 636 600 kg m

correct expression [1] correct substituition[1] final ans [1]

(ii) The mass gives rise to the least uncertainty. The fractional uncertainty is the lowest.

statement [1] explanation [1]

(iii) 19300 600 kg m-3 expression [1]

2(a) Speed is the rate of change of distance travelled by the object. [1]

(b)(i) From Fig. 2.1, time at which object has maximum displacement = 6.0 s (Occurs at point where velocity values are changing sign from negative to positive. Change in sign of velocity value denotes that the object is moving in the opposite direction.)

[1]

(ii) Recall that area under v-t graph gives displacement. At t = 12.0 s, displacement of the object = ½ (-20 x 6.0) + ½ (2.0 + 6.0)(10) = -20 m If answer is stated at +20m, award max 1 mark unless additional sketch is given.

[1] - subst [1] – ans


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(iii) [1] – correct shape (there must be a negative turning point at 6.0s) [1] – zero gradient at t = 0s, 6.0s and 12.0s. 1 mark if graph is correct except that s values are all positive, i.e. graph is drawn in positive y-region (unless answer in (b)(ii) was +20m, then award full marks)

(c)(i) v2 = u2 + 2as = 0 + (2)(4.00)(50.0) v = 20.0 m s-1

[1] - subs [1] – ans

(ii)1. In horizontal direction: ux = vx = 20.0cos37° = 16.0 ms-1

[1] - ans

2. In vertical direction, taking downwards as positive: uy = 20.0sin37° vy

2 =uy2 + 2aysy

= (20.0sin37°)2 + 2(9.81)(30.0) vy = 27.1 ms-1 (shown)

[1] - subst

3.

v = 16.02 + 27.12 = 31.5 ms-1 θ = tan-1 (vy/vx) = tan-1 (27.1/16.0) = 59.5° Velocity of car is 31.5 ms-1 at an angle of 59.5° below the horizontal.

[1] – ans for magnitude of v [1] – direction of v

3 (a)(i) Moment of a force on the body about any point is the product of that force and the perpendicular distance from that point to the line of action of the force.

[1]

s / m

t / s

2.0 4.0 6.0 8.0 10.0 12.0

vx

vy

v

θ


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(ii)1.

oLL T L

T

Taking moments about the hinge,

Clockwise Moments = Counter Clockwise Moments

400 2000 sin302

4400N (3sf)

[1] for correct clockwise moments [1] for correct anti-clockwise moments

2. y

y

y

x

x

F

F T

F

F

F T

o

0

2 2

0

sin30 400 2000

200N (upwards)

0

cos60 3810N (right)

Resultant force, R = 200 3810 3815.2 3820N

tan . o2003 00

3810(above horizontal)

[1] for correct vertical force [1] for correct horizontal force [1] for correct resultant force [1] for any correct direction

(b)(i) Hooke’s Law states that the force F required to stretch (or compress) an object (e.g. spring, wire) is directly proportional to the amount of extension (or compression) x, if the limit of proportionality is not exceeded.

[2]

(ii)1. sin

. sin

. .

o

o

F mg

30

3 9 81 30

14 715 14 7N

[1] for correct equation [1]

2. ., . .

FF kx x

k

14 715Using 0 02943 0 0294m

500

[1]

4(a) The rate of change of momentum is directly proportional to the resultant force acting on it and the change occurs in the direction of that resultant force.

[1]

(b) Initially, the weight of the droplet is greater than the resistive force. This downward net force results in the acceleration of the droplet, based on Newton’s Second Law. As the raindrop accelerates, the resistive force increases till it is equal to its weight. The raindrop experiences no net force and zero acceleration based on Newton’s Second Law. Hence it falls at terminal velocity.

[1] [1] [1] No mark


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(c)(i) Using Newton’s First Law, the rock remains in its state of rest because there is no resultant force. OR Using Newton’s Second Law, the rate of change of the rock’s momentum is zero because there is no resultant force. The horizontal pushing force is balanced by the frictional force between the ground and rock.

[1] [1]

(ii) Explanation is not valid because the pushing force and reaction force must act on different objects according to Newton’s Third Law.

[1]

(d)(i) The directions of the velocities of the two balls after collision will be along the same line as the directions of their velocities before the collision.

[1]

(d)(ii)1. Applying Conservation of momentum, (0.20)(4.0) + (0.40)(–0.50) = (0.40)VB + (0.20)(–VA) OR 0.40 VB – 0.20 VA = 0.60

[1] for either equation

2. Let final velocities of balls A and B be VA and VB respectively, Applying relative velocity equation for elastic collision, Relative velocity of separation = Relative velocity of approach VB – (VA) = 4.0 – (– 0.5) OR Applying conservation of KE, ½ mA(4.0)

2 + ½ mB(–0.5)

2 = ½ mA(VA)

2 + ½ mB(VB)

2

Solving the 2 equations simultaneously from (d)(i)1. and (d)(ii)2.,

VA = – 2.0 m s-1 VB = 2.5 m s-1

magnitude of final velocity of A = 2.0 m s-1 magnitude of final velocity of B = 2.5 m s-1

[1] for either equation [1] [1]

5(a) Work done by a force is the product of the force and the displacement in the direction of the force.

[1] – Ans

(b)

Power P = rate of work done by the force F (or work done by F per unit time)

Fs

Pt

However s

t = v, velocity of the object

Hence P = Fv

[1] define power [1] expression [1] expression

(b)(i) An example of such a force is air resistance. The second term kv3 is related to rate of work done to overcome a force that varies with the velocity (or a non-constant force).

[1] ans


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(ii)

P

1 1

3

4

100 1000100 km h = = 27.8 m s

60 60

power = (240 27.8) + [0.98 (27.8) ]

= 2.8 10 W

[1] conversion [1] sub [1] ans

(c)(i) 2 51

k.e. = x 720 x 27.8 = 2.78 x 10 J2

[1]- sub [1] - ans

MC Practice ecf if student use speed at 100 kmh-1. (ii) p f v

pf

v

4

power, = force, x velocity,

2.8 10 Engine force, = = = 1007 N

27.8

Hence the magnitude of the external force opposing the motion of the car

= Magnitude of the engine force (Since there is no resultant force on the car)

= 1007 N

[1] - sub [1] show understanding of no resultant force [1] - ans

MC Practice ecf if student use speed at 100 kmh-1 and wrong power calculated in b(ii) .

(iii) 4 6 work done = power time = 2.8 10 5 60 = 8.4 10 J [1] – ans

MC Practice ecf from b(ii)

The End

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