2010 H2 Physics Ans - HCI

2010 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 H2 PHYSICS

1

2010 HCI H2 Physics Preliminary Examinations Suggested solution for Paper 2 Q1 (a) vy

2 = uy2 + 2aysy

0 = (usin80.0°) 2 + 2 (- 9.81) (1.75) [M1] u = 5.95 m s-1 [C1] (b) vy = uu + ayt 0 = 5.95 sin80.0o – 9.81(T/2),

where T is time of flight and is twice the time to reach maximum height, T = 1.1946 s [C1]

Range Sx= 5.95 cos80.0° (1.1946) = 1.23 m. [A1]

(c) 1. Large angle gives a longer time of flight for a given projection speed and hence

juggler has more time to manipulate balls. [B1] 2. Large angle keeps (horizontal) range small and of the order of the natural distance

between the juggler’s hands. [B1] (d) The left hand has 1.19/2 = 0.597 seconds before ball 1 arrives. [A1] (e) The minimum distance between the two hands must be at least greater than the diameter

of the balls. [B1] Otherwise, an upward moving ball will collide with a downward moving ball. [B1]

Q2 (a) Decrease in GPE = Increase in EPE + Work done against drag force [M1]

m g H= ½ k e2 + F H 80.0(10)H = ½ (100)(H – 20.0)2 + 300 (H) [C1] 800 H = 50( H2 – 40H +400) + 300 H 16 H = H2 – 40H + 400 + 6 H 0 = H2 – 50H +400 0 = (H – 40) (H – 10) H = 40 m , 10 m (rejected) [A1]

(b) T = ke, where e is the extension of the rope = 100 (40.0 – 20.0) = 2000 N [A1] (c) T – mg = ma [M1]

2260 – 80.0(10) = 80.0a a = 15 m s-2 [A1] Direction: Vertically upwards [B1]

+

www.erwintuition.com



2

(d)

[B1] 3(a)

[B1] mark each for each clearly labeled force X 3 (b) Normal horizontal contact force of wall on cube provides the centripetal force for cube

to spiral. [B1] Component of the weight down the slope causes the acceleration of cube down slope.

[B1] (ci) Since the horizontal normal contact force is always perpendicular to the direction of

motion, work done is zero. [B1]

(ci) Rate of increase of kinetic energy = rate of decrease of gravitational potential energy. [M1]

= m g v sinθ. [A1] 4(a) Gravitational potential at a point in a gravitational field is the work done per unit mass by external force to bring a mass from infinity to that point [B1] without a change in kinetic energy. [B1]

4(bi) Gm

Vr

= − [A1]

4(bii) Distance of any point on the ring to P = 2 2

R h+ [M1]

Potential at P, Vp = 22

4

hR

Gm

+− [A1]

Horizontal Normal Contact force from wall into the paper

Weight of cube

Normal contact force from ground

Figure 3b

The direction of the normal contact force from the wall should be drawn properly into the page using the right symbol. In the event where it is really hard to draw, a clearly written statement to explain the direction is required.

X

Tension in rope

Height Daniel falls through 0

H 20.0




3

4(biii) Work done W by external force at constant kinetic energy = the change in gravitational potential energy of the system. [A1] W required from h to h1

= (∆Vp)M =

+−

+

−222

1

2

114

hRhRMmG

As h1 gets smaller, W gets more negative (or W is negative but increase in

magnitude). [A1]

5 (a) (i)

V = 1.398 – 0.281 = 1.117 V I = 0.31x10-3 A R = V/I = 3600 Ω

[A1]

(ii) R = ρl/A, thus ρ = RA/l = 3600 x 0.800 x 0.210 / 0.900 = 670 Ω (2 s.f.)

[M1] [A1]

(iii) Any one of these:

• Among all the readings given, the least significant or most imprecise is the current reading (only 2 s.f.). The current reading will be subject to significant random errors. EITHER: Increase the area of the copper plates in the soil. This will decrease the resistance of the sample of soil to be measured and increase the current readings for the same voltage applied. OR use a higher voltage supply so that current will be greater.

[B1]

• Use a variable voltage supply in place of the battery. Plot the graph of the voltmeter reading V against the current I. The gradient equals the resistance R of the soil and the y-intercept the emf due to the copper plate interacting with the soil. Calculate the resistivity using the value of resistance R obtained from the gradient of the graph.

• Vary the area of the copper plates in the soil (use different depths), find the corresponding R of the soil between the plates using V and I, and plot the graph of the R against A-1. The gradient is ρl where l = x. Calculate resistivity ρ as gradient divided by x.

(b) (i) The labels 60 W and 120 V indicate that the normal operating conditions

for the bulb. If a DC voltage of 120 V is applied across the bulb, then the power dissipated is 60 W. If an alternating voltage is applied, then the RMS value of the voltage should be 120 V so that the mean power dissipated is 60 W.

Vrms = 170/√2 = 120 V Mean P = IrmsVrms Irms = Mean P/V rms = 60/ 120 = 0.500 A

[M1] [A1]




4

(ii) Compare the given equation with V=V0 sin (2πf t).

Therefore 100 = 2f, and f = 50 Hz.

[B2]

6 (a) The gap between the STM probe tip and the sample surface acts like

a potential barrier to the electrons.

[B1]

When a pd is applied between the probe and the sample, there is a finite probability that electrons can tunnel through this potential barrier due to the wave nature of the electrons even though the electron does not have sufficient kinetic energy.

[B1]

The probability of tunneling is kdeT 2−∝ where d is the width of the

potential barrier. Thus tunneling current varies exponentially with the distance of gap d between the probe and the sample surface. As probe scans across the surface, the distance d changes and the variation in tunneling current can be detected and used to plot the topography of the sample surface.

[B1]

1st mark: identifying potential barrier 2nd mark: identifying electrons as the particles tunneling the barrier 3rd mark: the tunneling current varies exponentially with the width of the potential barrier which is identified as the distance of the gap.

(b) (i) Work function energy refers to how much additional energy needs to be provided to the electron for it to cross the potential barrier classically. This equals to U-E in the equation for k. Substituting the values,

( )

( )

2 31 192

2 234

8 9.11 10 (4.0 1.6 10 )8 ( )

6.63 10

m U Ek

h

ππ− −

−

× × ×−= =

×

101.02 10= ×

[M1]

[A1]

120

60

t / s 0

P / W

T = 1/f

= 1/50

= 0.020 s

P=V2/R = (V02/R)sin2(2πf t)

[1] – Correct sine-squared curve (note P=0 when t=0) [1] – All three values labelled (120W, 60W, T = 0.020 s)

Mean power

Peak power




5

(ii)

0

2

2

20

kd

kd

kd

T e

T e

T e

−

−

−

∝

=

2 ( )

0

ok d dTe

T

− −=

( )10 102 1.02 10 ( 1 10 )0.0002

0.0001

de

−− × − ×

=

d = 6.60 x 10-11 m

[C1]

[A1]

(c) DNA complexes are poor electrical conductors. As a result, electrons that have tunneled through to the molecule accumulates. This results in a strong electrostatic field that will disrupt the tunneling process. The conducting film helps to channel the tunneling current away.

[A1]

7 (a) Units of E = Units of

Ae

Fl= Units of

A

F

= kg m s-2 m-2 = kg m-1 s-2

[M1] [A1]

(b)(i) The extension of the material is proportional to the applied load if the limit of proportionality is not exceeded.

[B1]

(ii) Mark X on (2.0, 12.4) [B1] (iii) Material B.

Glass is brittle and is unlikely to undergo extended stretching/plastic deformation after it reaches its limit of proportionality before its breaking point.

[A1] [B1]

(c)(i)

)106.1)(1001.1(

)60.1)(0.6(37 −−

××==

Ae

FlE

= 5.89 x 1010 kg m-1 s-2

[C1] [A1]

(ii) Work = Area under the force-extension graph

= )0148.0(8.14)0036.0)(4.1412(2

1)002.0)(12(

2

1+++

= 0.278 J Working must be clearly shown to be given full credit. Accepted range (0.264, 0.291)

[C1] [A1]

(d) Rod Z will break first. Each rod X and Y only experiences half of the force on Z. The maximum force that rods X and Y can take is more than half of Z, hence Z will reach breaking point first.

[A1] [B1]




6

Question 8 Mark Aim

B1 The aim of this experiment is to investigate how the attractive force varies with the distance x.

(Some students used another quantity instead of force, eg. Extension of spring. In such cases, students need to elaborate on how this quantity is related to the attractive force and justify in order to score this mark.)

Preliminary work

D1

The attractive force can be measured by the formula F = ke where e is the extension of the spring, which is given by e = L – L0, where L0 is the initial length of the spring when it is not stretched and L is the length of the spring when it is stretched. The spring constant k can be measured by

D2 i) suspending the spring vertically from a retort stand. Its length is measured using a meter rule Lo. ii) attach a mass of 50g to the bottom of the spring and measure its new length L. iii) calculate the spring constant k from k = F/e = (0.050 )(9.81)/(L-L0)

(Calibration curves are acceptable but elaborations on how to get the data for this curve and how to use it are necessary. Spring balance is also acceptable as a means to measure the force directly, however, as with scale instruments, need elaboration on zero error).

Control Variable B2 The current in the coil is to be kept constant. This can be done by connecting

an ammeter to the coil as shown in the following diagram. (Good answers include how a rheostat is necessary to allow for adjustments

to the resistance in the circuit as a means of ensuring constant current. Some students forgot to include a cell in the circuit diagram)

A4

A5 The rheostat is first used to vary the resistance in the circuit and hence the

current. The current should be fixed at a value such that when x = 6.0 cm, the extension in the spring is more than 1.0 cm. This is to ensure that values of e recorded in the experiment are large enough such that it can be measured with acceptable percentage error with a meter rule.

(students should elaborate that an acceptable range is one whereby variations in data are not just realistic, but also measurable, according to the resolution of the measuring instruments)

B3 More batteries can be connected in series to increase the current until the resulting change in e is large enough.

A4

A

A




7

Hypothesis D3 Suppose the relationship between the attractive force F and the distance x is

expressed in power law, F = cxn where c and n are constants to be determined. lg F = n lg x + lg c If we plot lgF against lgx, then a straight line will indicate that the relationship is valid and the gradient is n and the y-intercept is lg c.

(some students neglect the interpretation of the gradient and vertical intercept of the graph)

Procedure

A1 a) Setup the apparatus as shown in Fig 1.1. b) Use a meter rule to measure the length of the spring when it is

unstretched. Record this as L0.

D4 c) Switch on the circuit to allow current to flow through the coil. Monitor the ammeter to ensure that current is constant. Ensure that an attractive force is exerted on the bar magnet by the coil (this occurs when the spring is stretched). If there is a repulsive force instead, switch the polarity of the battery in the circuit to reverse the direction of current flow.

(Its important to include this step in verifying that an attractive force is present and if not,elaborate on how one can rectify the problem)

A2 d) Measure the distance between the magnet and the coil x. Also measure the length of the stretched spring and record it as L.

A3 e) Shift the magnet, together with the spring, corkboard and brick, for at least 5 different values of x. For each x, record the new stretched length of the spring L. Tabulate the data in the table below,

x /m L /m F =k(L-Lo)/N lg (x/m) lg (F/N)

f) Plot a graph of lgF against lgx

A1-3 Basic Procedure A4 Labeled Diagrams A5 Actions taken to set range readings B1 Correct Dependent and Independent Variables B2 Valid choice of control variable B3 Action to improve accuracy of readings D1 Theory on how to calculate the attractive force D2 Pre experiment work necessary to measure spring constant D3 Linearization D4 Action to ensure correct current flow (12 marks in total)



This booklet consists of 7 printed pages, inclusive of this page.

2010 HCI H2 Physics Preliminary Examinations Suggested solution for Paper 3 SECTION A Q1 (a) AB: Car travels at constant speed. Tom's reaction time. BC: Car decelerates uniformly. Tom applies car's brakes. CD: Car's velocity changes from one direction to the opposite direction. Actual collision of the two

vehicles. DE: Both vehicles stuck together after the collision and slow down to a rest. Friction of road on

the vehicles causes them to stop eventually. [B4]

(b)

( )

×

×−−=

==

36001.0

1000)]20(30[1200

)(

F

dt

mvd

dt

dpF

[M1]

NF51067.1 ×= [A1]

(c) N3L: CVVC FF =

t

)vv(m

t

)vv(m 1V2VV1C2CC −=

− [M1]

30

50

vv

vv

m

m

1V2V

1C2C

C

V=

−

−=

kg2000)1200(3

5mV == [A1]

(d) Inelastic. [B1] (EITHER)

The 2 vehicles stuck together and move as one after the collision, as observed from the velocity-time graph. (OR) The total KE of the 2 vehicles after the collision is less than the total KE of the 2 vehicles before the collision. [B1]

Q2

(a)(i) The increase in the internal energy of the system, ∆U, is the sum of the heat supplied to the

system, Q, and the work done on the system, W. [B2] (a)(ii) Internal energy U of a system is the sum of all microscopic kinetic energies of the particles and their potential energies. [B2] (b)(i) Work done by gas = (0.070 – 0.020) x (11 x 105) [M1] = 55 kJ [A1]




Page 2 of 7

(b)(ii)

Q3

(a) x = 2 a sin θ [B1]

(b)(i) x = n λ [M1]

2 a sin θ = n λ a = (1)(0.165 x 10-9)/ 2 sin(23.5o) [M1] a = 2.07 x 10-10 m [A1] (b)(ii)

[M1]

(c)(i) 2 a sin θ = n λ

λ < 2a sinθ / n

λ is maximum when θ = 90o and n = 1 [M1]

λ < 4.14 x 10-10 m [A1]

(c)(ii) λ < 4.14 x 10-10 h/p < 4.14 x 10-10 [M1] h / (me vmin ) < 4.14 x 10-10 vmin > h / (me)(4.14 x 10-10) [M1] vmin > 1.76 x 106 m s-1 [A1]

4(a)(i) A nuclide with 3 protons and 4 neutrons. [B1] (a) (ii) Applying conservation of energy:

Gain in electric potential energy = loss in kinetic energy

4000004

106.13

4

0

19

0

=××

=

−

x

qVx

Qq

πε

πε

Solving, x = 1.08 x 10-14 m

[M1] [A1]

Process W / kJ Q / kJ ∆U / kJ

A 201 0 201

B -55 74 19

C -185 0 -185

D 0 -35 -35

1st order

2nd

order

AQ + CQ correct [1] DW correct [1] BW correct [1] (for –ve) ABC ∆U correct [1]




Page 3 of 7

(a)(ii) With quantum tunneling, the proton can still penetrate the barrier despite

having an energy that is less than the potential barrier.

[B1]

(b)(i) Energy released = (7.0138 + 1.0073 – 4.0015 x 2)(1.66 x 10-27)(3 x 108)2

= 2.70414 x 10-12 J Energy of each alpha = ½ (2.70414 x 10-12) = 1.3521 x 10-12 J

[M1] [A1] [A1]

(b)(ii)1.

Length = 1.3521×10−12

(5.0 ×103)(5.2 ×10−18)= 52 mm

[A1]

(b)(ii)2.

[A1]

Neutrons have no charge and are therefore able to penetrate more deeply into the positively charged nucleus, resulting in higher probability of nuclear reactions.

B1 (c)

SECTION B Q5 (a)(i) Advantage: precise (can accurately target 1 specific vehicle) or longer range because laser

is unidirectional/small divergence [B1]

Disadvantage: need to aim properly, need to ensure beam is reflected back to the gun [B1] (a)(ii) 1. Acceleration always directed towards a fixed point [B1]

and proportional to displacement from that point [B1]

(a)(ii) 2. SHM defining equation Cxx −=

Let )cos(BtAx = , )sin(BtABx −= [M1]

xBBtABx 22 )cos( −=−= , where 2

BC = [A1]

(a)(ii) 3. From table, maximum distance is 5.100 m, hence amplitude = 0.100 m. [B1]

Let t = 0 s when time = 1004 ms (laser speed gun time), hence )2

cos(100.0 tT

xπ

=

[M1]

When time = 1005 ms, t = 0.001 s, x = 0.095 m, [M1]

s 020.00198.0)001.02

cos(100.0095.0 ≈=⇒=⇒ TT

π [A1]

(a)(ii) 4. Using )2

cos(100.0 tT

xπ

= and t = 0.005 s, 0016.0)005.00198.0

2cos(100.0 −==

πx m [M1]

Hence Z = x + 5.000 = 4.998 m. [A1]




Page 4 of 7

(a)(ii) 5. 1000/1s 001.0 =∆=⇒=∆ tft samples per second [B1]

Laser gun computes the average speed = ∆x / ∆t [M1] ∆t should as small as possible so that it approximates instantaneous speed. [A1]

(b)(i) Excitation of electrons/atoms from lower to higher energy levels [B1]

with light of correct frequency f, such that the energy difference between the 2 levels is equal to hf. [B1]

(b)(ii) When using optical pumping for a 2 levels laser, any incoming photon can cause simulated

absorption as well as stimulated emission. [C1] Initially most of the atoms are in the ground state, the incoming photons will cause more stimulated absorption than stimulated emission. [M1] As more atoms become excited, the rate of stimulated emission will eventually increase until it at most equals that of stimulated absorption, when the number of excited atoms is equal to the number of ground state atoms. Thus it is very difficult to achieve population inversion. [A1]

6a) The magnetic flux density is defined as the force per unit length per unit current [B1] acting on an infinitely long current carrying conductor placed perpendicularly to the magnetic field. [B1]

6b) (i) d

IB

d

IB oo

π

µ

π

µ

22

2

2

1

1 == [C1]

LIBFLIBF 12122121 ==

L

FIB

d

IIIB

L

FF o 12

1221

2121

2=====

π

µ [A1]

d

6b)(ii) 1.

b)(ii) 2. The current in each turn of the coil of the spring produces a magnetic field that is perpendicular to the current in the adjacent coil. [B1]

Since the current in the adjacent spring coils is flowing in the same direction, by Fleming’s left hand rule, an attractive force will be exerted on the coils towards each other. [B1]

The spring coils will move towards each other and the length (vertical) will shorten. The change in the length of the spring is therefore a compression. [B1]

b)(ii) 3. )2(22

22

rd

IF

d

I

L

FlengthunitperForce oo π

π

µ

π

µ=⇒== where r is the radius of the

spring coil. [B1]

Since the spring obeys Hooke’s law, F = kx where k is spring constant.

A




Page 5 of 7

dk

rIxr

d

Ikx oo

22

)2(2

µπ

π

µ=⇒= [B1]

b)(ii) 4. Compression. [A1 6c)(i) According to the Faraday’s law of Electromagnetic Induction, when the metal axle Y of the

truck first enters the magnetic field entry point P with an initial velocity, there will be an induced electromotive force (emf) across the two metal wheels of Y whose magnitude is directly proportional to the rate of change of magnetic flux linkage. The wheel on the left (at the top according to figure) will be at a higher potential. [B1]

Since the axle Y, the two wheels and the railing form a complete loop, induced current will hence flow. [B1] Using Fleming’s left hand rule, a magnetic force opposite to the direction of motion would be produced due to this current flow. This opposing force on the wheels will cause braking to occur. [B1]

c)(ii) At any instant when axle Y enters magnetic field,

Induced emf E = IR = Bwv

Induced current R

BwvI = [B1]

Since FB = - BIw (using the direction of motion as +ve direction)

wR

BwvBaM )(−= [B1]

vvRM

wBa λ−=−= )(

22

∴a ∝ v ----------------(1) [B1]

c)(iii) (If the truck’s initial speed is high, the magnetic braking force experienced upon entering and leaving the magnetic field may not be able to bring the truck to rest completely.)

To improve, we can either have several regions of PQ placed close to each other for consecutive braking effects [B2] OR Have a stronger vertical magnetic field B. [B1] Any reasonable explanation that leads to a larger acceleration hence braking force eg. with reference to (1), would be awarded the second mark. [B1]

7(a) Similarities: 1. Both fields exert forces on moving charged particles. [B1] 2. They are non-contact forces/action-at-a-distance force [B1] Alternative answers 3. The forces they exert long range forces. 4. They are conservative fields Differences 1. Electric field will change the magnitude of the speed of charged particles whereas magnetic

field can be used to change the direction of motion of charged particles. [B1] 2. If the paths of charged particles are not straight line path, it will be parabolic and circular

paths in electric and magnetic field respectively. [B1]




Page 6 of 7

Alternative answers 3. The electric force is either parallel or anti-parallel to the Electric field whereas the magnetic

force is always mutually perpendicular to the magnetic field and the motion of charged particle.

4. There will not be any magnetic force on stationary charged particle whilst there will be electric force on charged particle regardless of its state of motion.

b(i) Using Fleming’s Left Hand Rule, the magnetic force will be mutually perpendicular to the velocity and the magnetic field. [B1] It provides the necessary centripetal force for uniform circular motion. [B1]

(ii)

]1[1C

]1[1C

2

2

r

vmvqB

r

vmFB

=⇒

=

∴ Bq

mvr =

(iii) ω

π2=T

Since Bq

mvr =

m

qB

r

v==⇒ω C1 [1]

qB

mT

π2=∴ C1 [1]

In the above derivation for the period, period is found to be dependent on mass and charge of deuteron and B only

(iv) [2]

(v)

1. Gain in KE crossing each gap = e V [C1] Hence gain in KE per revolution = 2 e V = 2 (100) ( 1.6 x 10-19) = 3.2 x 10-17 J [C1]

2. ]1[12

2C

m

qBf

qB

mT

π

π=⇒=∵

]1[1)1034.3(2

)106.1(400.027

19

C−

−

×

×=

π

K.E

Point 2 Point 3 Point 4

distance

B1 B1




Page 7 of 7

= 3.05 × 106 Hz [A1]

3. Number of revolutions to gain 1.00 MeV = 17

196

102.3

106.11000.1−

−

×

×××

= 5000 [C1]

Time taken = 5000 x )1060.1)(400.0(

)1034.3(219

27

−

−

×

×π [C1]

= 1.63 x 10-3 s [A1]

(Relativistic effects negligible at these speeds)



2010 H2 Physics Ans - HCI

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