2012 Nyjc Ph h1 p2 Promo Soln

8/17/2019 2012 Nyjc Ph h1 p2 Promo Soln

1/13

This document consists of 13 printed pages. [Turn over

NANYANG JUNIOR COLLEGE

JC 1 PROMOTIONAL EXAMINATION

Higher 1

CANDIDATE

NAME

CLASSTUTOR’S

NAME

PHYSICS 8866/02

Paper 2 Structured Questions 5 October 2012

1 hour 30 minutes

Candidates answer on the Question Paper.

No Additional Materials are required

READ THESE INSTRUCTIONS FIRST

Write your name and class on all the work you hand in.

Write in dark blue or black pen on both sides of the paper.

You may use a soft pencil for any diagrams, graphs or rough working.

Do not use staples, paper clips, highlighters, glue or correction fluid.

Answer all questions.

The number of marks is given in brackets [ ] at the end of each question or

part question.

For Examiner’s Use

1[8]

2[8]

3[8]

4[16]

5[20]

Total


2/13

2

3.00 108

1

µ 4π 10 7

1

ε 8.85 10 12

1

(1/(36π)) 10

1

1.60 10 1

6.63 10 34

1.66 10 27

.11 10 31

1.67 10 27

6.67 10 11

2

2

.81 2

+

+ 2

/ Δ

ρ


3/13

3

[Turn over

1)(a)(i) Explain how random error may arise when using a micrometer screw gauge to

measure the diameter of a pencil.

…………………………………………………………………………………………………

…………………………………………………………………………………………………

………………………………………………………………………………………………[1]

Suggested answers:

Non-uniformity of the pencil

Parallax error when reading the micrometer screw gauge

(ii) Suggest how this error can be minimised.

………………………………………………………………………………………………..

…………………………………………………………………………………………………

…………………………………………………………………………………………………

………………………………………………………………………………………………[2]

This random error can be reduced by taking a large number of individual measurements

and calculating the average. The average value will comprise a minimized random error

(b) To determine the volume of glass in a length of glass tubing, an experimenter obtains the

following measurements:

length L of tubing: 400 ± 5 mm

external diameter D of tubing: 15.0 ± 0.2 mm

internal diameter d of tubing: 10.0 ± 0.2 mm

The volume V of the glass is calculated using the formula

)(4

1 22d D LV −= π

(i) Show that the absolute uncertainty for )( 22d D −∆ is 10 mm2. [2]

Solution:

2 2 2 2

2 2 2

2 2

( )

2 2(15.0)(0.2) 6

4

D d D d

DD D mm

D

d mm

∆ − = ∆ + ∆

∆ ∆ = × = =

∆ =


4/13

4

(ii) Hence calculate the percentage uncertainty in the value of V using these readings.

Solution:

%909.01015

10

400

5

)(

)(2222

22

==−

+=−

−∆+

∆=

∆

d D

d D

L

L

V

V

Percentage uncertainty = ……………………………. [2]

2 A motorcycle helmet consists of a hard outer shell which is lined with a plastic foam material

which squashes relatively slowly when the helmet strikes a hard surface such as the road or

a car.

(a) Explain why the process of squashing 'relatively slowly ' is helpful in protecting a

motorcyclist wearing the helmet.

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………..

……………………………………………………………………………………………………[2]

(b) In one safety test for a motorcycle helmet, the helmet is loaded with a dummy head and

dropped so that it reaches a speed of 6.4 m s-1 before hitting a flat metal block. The totalmass of the dummy head and helmet in one test is 5.0 kg and it rebounds at a speed of

3.1 m s-1. The time of contact between the helmet and the metal block is approximately

4.0 ms.

(i) Calculate the magnitude of the change in momentum of the helmet.

change in momentum = ……………………… kg m s-1 [2]

(ii) The helmet 'passes' the safety test if the instrumentation connected to the dummy head

registers an average deceleration of less than 300 g where g is the gravitational

acceleration. Deduce with justification whether or not this helmet passes the test.

Squashing 'relatively slowly' increases the time for the motorcyclist head to come to

a stop.

This longer time reduces the impact force which is change in momentum

divided by time for the momentum change. A smaller force will cause less injury to

the motorcyclist.

Change in momentum = mv - mu = (5.0 x( - 3.1)) - (5.0 x( 6.4))

= 5.0 x (- 9.5 ) = - 47.5 kg m s-1

Magnitude of the change in momentum = 47.5 kg m s-1


5/13

5

[Turn over

[2]

(iii) The efficiency of the plastic foam material is defined as the percentage of the kinetic energy

that is dissipated during the test. Calculate the efficiency of the material.

efficiency = …………………… % [2]

3 A ball of mass 2.0 kg is dropped from a height 1.5 m above a spring with spring constant

25 N m-1. The subsequent positions of the ball and spring are as shown in Fig 3.

(a) Show that the compression of the spring when the ball is in equilibrium is 0.78 m. [2]

( )

78.0

25

81.0.2

=

=

=

=

Decelerating force = Change in momentum / time = 47.5 / 4 x 10-

= 11880 N = m a

Decelerating acceleration a = F / m = 11880 / 5.0 = 2375 m s-2

= 242 g < 300 g . It passed the test!!

Efficiency = (KE dissipated / initial KE) x 100

= [( ½ m v2 - ½ m u2 )/ ½ m u2 ] x 100 = [(v2 - u2 )/ u2 ] x 100

= [ | ( 3.12 - 6.42 ) | / 6.42 ] x 100 = (31.35 / 40.96) x 100

= 77 %

1.5 m

At the lowest

position

At start of

release of

ball

At start of

compression

of spring

At equilibrium

compression of

spring

Fig. 3


6/13

6

(b) By applying the principle of conservation of energy, calculate the speed of the ball

when it is at its equilibrium position. [4]

Loss in GPE = Gain in KE + Gain IN EPE

( )

( )

2 2

2 2

-1

1 1

2 2

1 1

2 2

37.1

6.1 m s

mg h x mv kx

mv mg h x kx

v

+ = +

= + −

=

=

(c) Calculate y the compression of the spring when the ball is at its lowest position. [2]

=↑↓

( )

( )

( ) 5.2 4.0

022

2

2

1

2

2

2

=−=

=−−

=+

=+

(d) State an assumption made in your calculation.

No air resistance.

………………………………………………………………………………………………..

…………………………………………………………………………………………………

………………………………………………………………………………………………[1]


7/13

7

[Turn over

4 In an attempt to verify the relationship d = ½ a t 2, a student sets up the apparatus as shown

in Fig 4.1 where a is the linear acceleration.

Fig 4.1

(a) Define acceleration.

Acceleration is the rate of change of velocity (with respect to time). [1]

The time taken for the car to move from rest through a distance d is t. The values of d and t

obtained are shown in Table 4.2.

d / cm t / s t 2 / s 2

1 100 3.1 9.6

2 80 2.9 8.4

3 60 2.5 6.0

4 40 2.2 4.8

5 20 1.4 2.0

Table 4.2

(b) Complete the table by filling in the missing t2 values. [2]


8/13

8

(c) On the grid below, plot the 2 missing points for the graph of d vs t2 using the valuesin Table 4.2 and draw a best fit straight line. [3]

Fig 4.3

(d) i) Calculate the gradient of the best fit straight line obtained in Fig 4.3.

The coordinates are (0.6,0), ( 8.0, 78 )

Gradient of the graph = (78 - 0) / (8.0-0.6) = 10.5

gradient of the graph = ………………….….[2]

(ii) By making use of the relationship d = ½ a t2 and your answer in (d)(i), deduce the

acceleration of the toy car as it travels down the slope.

For d = ½ a t2

Gradient = ½ a = 10.5

a = 2(10.5) = 21.0 cm s-2 = 0.210 m s-2

acceleration = ………………….….m s-2 [2]

d / cm

t / s0 2 4 6 8 10

100

80

60

40

20


9/13

[Turn over

(e) The values of t and d are measured using a stopwatch and a metre rule respectively.

For each of the following types of experimental error, describe the feature of thegraph in Fig 4.3 that indicates its presence and suggest a possible source of thaterror in the experiment.

i) random error

Feature : The readings are scattered above and below the best fit line.

Possible source: The reaction time of the student is not consistent, resulting in

readings that is an underestimate or overestimate from the true value. [2]

ii) systematic error

Feature: The graph did not cut the origin.

Possible source: The ruler is placed incorrectly such the measured distance d is

always an overestimate from the true value. [2]

(f) The experiment is repeated with another toy car of identical shape and size but made

of denser material. Deduce and explain how the acceleration of this toy car will differ,

if at all, from the answer in (d)(ii).

The acceleration will remain constant since the denser material will increase the

mass, but acceleration is independent of the acceleration, a larger mass will

experience a larger force but the same acceleration. [2]

5 (a) (i) Define the moment of a force.

…………………………………………………………………………………………..…

……………………………………………………………………………………… [1]

(ii) State the two conditions necessary for a body to be in static equilibrium.

…………………………………………………………………………………………..…

…………………………………………………………………………………………..…

…………………………………………………………………………………………..…

……………………………………………………………………………………… [2]

The moment of a force about a point is the product of the force andthe perpendicular distance of from the line of action of the force to thepoint.

Resultant force acting on a body is zero.

Resultant torque acting on a body is zero.


10/13

10

Fig 5.1 shows a disc and spring system which is mounted on a wall. It is used to move

two cargo boxes of equal weight towards each other on a rough floor by applying a force

at the wrench to turn the disc.

(b) Fig 5.2 shows two of the forces acting on the box A when it is about to slide.

(i) Complete the free body diagram by drawing two additional forces acting on Box A

on Fig 2.2. Label the forces clearly. [2]

Box A Box B

disc

wrench

Box A

Weight of Box A

Normal

reaction force

2.0 m

0.80 m

1.5 m

Rough floor

Fig 5.2

Fig 5.1

x


11/13

11

[Turn over

(ii) Box A has a mass of 500 kg. The maximum static frictional force between Box A

and the rough floor is 2200 N. Calculate the tension in the spring when Box A

begins to slide.

minimum tension = N [1]

(iii) Calculate x, the separation between the weight of Box A and the normal reaction

force.

separation, x = m [2]

(iv) Box B is identical to Box A except that it is attached to the spring and disc system at

a higher point. Deduce and explain whether the separation between the weight of

Box B and its normal reaction force is smaller, equal or greater than x.

…………………………………………………………………………………………..…

…………………………………………………………………………………………..…

…………………………………………………………………………………………..…

……………………………………………………………………………………… [2]

(v) The system may not be able to move the boxes keeping it upright if it is attached to

the boxes too high above the ground. By considering the rotational equilibrium of

the boxes, suggest why this is so.

…………………………………………………………………………………………..…

…………………………………………………………………………………………..…

……………………………………………………………………………………… [2]

ΣΣΣΣF = 0

T – f = 0

When the block is about to slide, f = max limiting value.

T = 2 200 N.

ΣΣΣΣT = 0

2 200(1.5) – 500(9.81)(x) = 0

x = 0.67 m.

Separation will be greater than x since point of connection is higherresulting in a greater torque.

If the connection point is too high, the perpendicular distance between Tand f will be large. This will results in a greater toppling moment/torque.The block may topple before it start to slide.


12/13

12

(c) The disc is free to rotate about its axle which is normal to its plane and passes throughthe centre C of the disc, as shown in Fig 5.3.

A wrench of length 40 cm is attached to the disc. When a force F is applied at rightangles to the lever at its end, equal forces are produced in the two springs.

As the wrench is turned from its initial position, the tension in each spring increases from

zero. When the wrench has been turned through an angle of θ, the boxes begin to slide.

(i) The spring constant of each spring is 2.0 × 104 N m-1. Using your answer in (b)(ii),calculate the extension in each spring when the boxes begin to slide.

extension = m [1]

(ii) Hence calculate the angle θ.

F

20 cm

40 cm

θ

Fig 5.3

wrench

e = r θθθθ

(0.11)= 0.1(θθθθ)

θθθθ = 1.1 rad

C

disc

F = ke

2200 = 2000 x e

e = 0.11 m


13/13

13

[Turn over

angle, θ = rad [2]

(iii) Calculate the torque produce by the forces in the springs.

torque produced = N m[1]

(iv) Hence calculate the magnitude of F, the force applied to the wrench.

magnitude of F = N [1]

(d) (i) As the boxes move towards each other at a slow constant speed, the kinetic

frictional force acting on each cargo is 1500 N. By using the principle ofconservation of energy, calculate the energy required to turn the disc by one

revolution to maintain the motion of the boxes.

energy required = J [2]

(ii) The system is modified to have a longer wrench so that the applied forcerequired to move the boxes at the same speed as in (i) may be reduced.

Deduce with explanation how this modification affects, if at all, the energy

required to move the boxes by the same distance.

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

T = 2200( 0.20) = 440 Nm

F(0.4) – 440 = 0

F = 1100 N

Energy supplied = work done against friction

=2 x 1500 x( 2ππππr)

= 1880 J

Energy required will remain the same since the work done against

friction is the same.

Documents

2012 Nyjc Ph h1 p2 Promo Soln