2010 NYJC H2 P2 Q1 - PBworksjjphysics.pbworks.com/w/file/fetch/58627225/JJ H2 Physics Year 1...H1...

H1 Physics/ 2012/ Year 1/ Revision/ Measurement/ Solutions

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2010 RI H2 P2 Q1

1 (a) vernier callipers

zero error (do not accept parallax)

(b)

dV h

V d h

V d h

V

V

2

3

3

3

=π4

= 964.665 cm

Δ 2Δ Δ= +

2×0.01 0.1Δ = + ×964.665

8.50 17.0

= 8 cm

= (965 ±8) cm

2010 VJC H2 P3 Q1

2(a) Random errors are errors with different magnitudes and signs in repeated measurements, and can be

reduced by taking the average of many readings.

Systematic errors are errors of measurements that occur according to some 'fixed rule or pattern' such

that they yield a consistent over-estimation or under-estimation of the true value.

(b)

Hence k = (4.6 0.3) N m-1

(c) Advantages of many sets of readings and drawing graph: (state any 3)

1. By drawing best fit line, random errors are reduced.

2. Systematic error can be spotted eg. if the F-e graph does not pass

through origin.

3. Can spot erroneous points that are out of the trend.

4. Can check if the proportionality limit is exceeded eg. if the F-e

graph turns into a curve instead of a straight line.

k F

e

mg

(x2 x1)

0.020 x 9.81

(36.30 32.00)

4.563 N m1

k

km

me

e

1

200.1

4.3 0.073

k 0.073 x 4.563 = 0.3 N m-1 (to 1 sf)


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2010 ACJC H2 P2 Q1

2010 IJC H2 P2 Q1

4 (a) kg m s-1

= [mass] [velocity] = [momentum] or [impulse] [B1]

electric field strength =

-2

-1 -3

electric forceelectric field strength =

charge

electric forceelectric field strength =

charge

[mass] [acceleration] [E] =

[current] [time]

kg m s [E] =

A s

[E] = kg m A s [B1]

(b) (i) Estimate the density of plastic to be 2 times that of water = 2 g cm-3

Mass of protractor = (Volume) (Density)

Mass of protractor = ½ r 2 x

Mass of protractor = ½ (5.0)2 (0.1) (2)

Mass of protractor = 7.9 g

[M1] for logical reason. [A1] for acceptable range of 5 to 50 g

3(a)(i) 3V d6

, where d = 22 cm (accept values from18 cm to 24 cm)

=∴V 5.6 x 10-3

m3 (accept values from 3.0 to 7.5 x 10

-3 m

3 ); accept

corresponding values if alternative units are used (e.g. mm3 or m

3)

3(a)(ii) 2

2

1= mvKE , where v = 10 m s

-1 (accept values from 8 m s

-1 to 12 m s

-1) and

mass = 60 kg (accept values from 40 kg to 100 kg)

=KE∴ 3000 J (accept values from 1900 J to 7200 J); accept corresponding

values if alternative units are used (e.g. kJ)

(iii) Density is between 800 to 1100 kg m-3

(b)(i) The values are close to each other but very far from the expected value which should be less than density of water (1000 kg m

3).

Therefore, set B is precise but inaccurate.

(b)(ii) - mass of beaker is included in the measurement of mass

- balance calibrated incorrectly

- negligence in zero-ing the weighing balance

(either one of the above)


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(ii) Micrometer Screw Gauge [B1]

(iii) Repeat the measurement of x at different parts of the protractor and taking average. [B1]

2010 MI H2 P2 Q1

5a Scalar: Magnitude and no direction (mass)

Vector: Magnitude and direction (weight)

B1

B1

b

V vB

vA

vB

vA

x : Ax B

x 10sin45 5sin35 4.2

y : Ay B

y 10cos45 5cos35 11.2

V 4.22 11.22 11.9

a 11.9

30 0.40

tan1 4.2

11.2 20.6

Hence 0.40 m s-2

at bearing of 201

M1

M1

A1

2010 NJC H2 P2 Q1

6(a) A systematic error is one that occurs consistently more or consistently less than the actual reading. [1]

A random error is one that gives a scatter of readings about a mean value. [1]

(b) Method 1:

Volume = 2h = 40.46 cm

3

3 [1]

Method 2

Conclusion:

2010 CJC H2 P2 Q1

7 (a) Mass and temperature [1]

(b) (i) Base units of thermal energy = base units of energy

= kg ms-2

.m

= kg m2 s

-2

[2]

(ii) Base units of the constant c (heat capacity)

= base units of energy_____________

[2]


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(units of mass)(units of temperature)

= kg m2 s

-2

kg.K

= m2 s

-2 K

-1

2010 MJC H2 P2 Q6

8 (a) d/h = 394/450 = 0.876 ~ 0.90 [B1]

Since Chart No.49 has the correct fcu, fy and d/h, Chart No.49 is appropriate to

be used for this design. [B1]

(i)

(ii)

M/bh2 = 2.25 [M1]

N/bh = 27.3 [M1

From Chart No.49,

100Asc/bh = 2.6 [M2 - allow errors up to 0.05 i.e. 2.55 < x < 2.65]

(If 100Asc/bh = 2.7 or 2.5, allocate only one mark)

Therefore, Asc = 2340 mm2 [A1]

Provided four 32 mm reinforcement bars ~ 3220 mm2 > Asc, therefore the

provided reinforcement steel is sufficient. [B1]

(c)

N/bh > fcu [B1]

Column would be crushed/break/crack. [B1]

(d)

At 12 metres, the column is too slender with respect to its dimension. Might

twist and buckle under load. [B1, any of the underlined words]

(e)

From Chart No.39, corresponding N/bh = 21.0; allow 21.0 ± 0.5 [B1]

From Char No.49, corresponding N/bh = 25.5; allow 25.5 ± 0.5 [B1

Decrease in N = 3.48bh kN

Percentage decrease of N = 3.48/25.42 x100 = 13.7% [M1 and A1; ecf

allowed]


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2010 SRJC H2 P2 Q1

9) (a) (i)

= m8

(ii) percentage uncertainty introduces into the value of Q = 4 x 0.02 / 2.05 x 100% = 3.90 %

(b) (i) Peter’s acceleration downwards decreases (or velocity increases at a decreasing rate), as the weight remains the same but the air resistance increases with increasing velocity, resulting in the net force downwards decreasing with time, till terminal velocity is reached when the weight is equal to the air resistance.

(ii) (iii) The velocity is decreasing at a decreasing rate because the net force is upwards.

However, as the net force upwards is decreasing with time, the deceleration is decreasing. Hence the

gradient is getting less negative, until it is equal to 0 m s-2

where terminal velocity is reached.

END

H2 Physics/ 2012/ Year 1/ Revision/ Kinematics/ Solutions

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2010 NYJC H2 P3 Q1

1 (a)

(b)

2 2

2

1

1

2

0 2( 9.81)(4.75 1.0 0.25)

8.29

8.29sin75 8.58

y y y

y

y

o

v u a s

u

u m s

u m su

(ii)

2

2

1

2

11.0 (8.29) ( 9.81)

2

1.80 0.113 ( )

(8.58cos75 ) 1.80 4.0

y y y

x x

o

s u t a t

t t

t s or s rej

s u t

m

(iii)

(iv)

The acceleration of a body at any instant is defined as the time rate of change of

velocity of the body at that instant.

+ve

+ve

v/m s-1

t/ s

8.29

1.80 0.845

- 9.37

He should go closer to the durian. Directly below the durian so that he can exert

his force in the vertical direction and initial vertical velocity is at its greatest.


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2010 PJC H2 P2 Q1

2(a) Velocity is the rate of change of displacement.

Acceleration is the rate of change of velocity.

(b)(i) 6.0 s

(b)(ii)

Displacement = 100.20.62

1200.6

2

1

= −20 m

(b)(iii)

2010 RVHS H2 P2 Q1

3(a) Length of S$2 = 12.7 cm (10 ~ 15 cm),

Human reaction time = 0.3 s (0.1 ~ 0.5 s)

u = 0, s = 0.127 m, a = 9.81 m s2

s = ut + ½ at2 0.127 = 0 + 4.905 t

2 t = 0.17 s

Since human reaction is around 0.3 s, more than 0.17 s, hence student B

would not be able to catch the note.

(b) Distance travelled by lecturer in time t = vL t

Distance travelled by student in time t = vs (t – to)

vL t = vs (t to) + d

t = (d vs to) / (vL vs )

2010 TJC H2 P2 Q1

4 (a) Distance = Area under the graph = 601262

1.)( ]

= 5.4 m [1]

(b) Max height =Area under the graph = m .. 0510012

1 [1]

(c) a = Gradient= 2s m 106.0

612

[1]

s / m

t / s

2.0 4.0 6.0 8.0 10.0 12.0


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(d) From graph, time taken to reach max height on rebound = time taken to next reach ground = 1.0 s.

Thus, time taken to next reach ground = 1.0 + 1.6 = 2.6 s [1]

(e) [2]

2010 HCI H2 P2 Q1

5

(a) vy2 = uy

2 + 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]

2010 IJC H2 P2 Q2

6 (a) (i) Taking into consideration, the vertical component of the motion

Vertical component of initial velocity

oy uu 3020 sin)(sin 10.0 m s

-1

ghuv yy 222

0.6 2.6

t / s

s / m

0

5.4

0.4

1 m for correct shape, 1 m for correct values

+


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h).().( 81920100 2 [M1]

h = 5.10 m [A1]

(ii) The time taken to reach maximum height

gtuv yy

t).(. 8190100

t = 1.02 s [C1]

cos (20)cos30o

x xv u u 17.3 m s-1

The horizontal distance (or the range)

= ))(( flightoftimeux

= (17.3)(1.022)

= 35.3 m [A1]

(b)

(i) The maximum height and the horizontal distance will both be reduced. [B1]

Note: The path should be shorter in height and in range slightly. No mark for excessive reduction

in height and range as the effect of air resistance is slight. The path should be asymmetrical

as the net force on the clay disc is varied.

(ii) The maximum vertical height is smaller because the net force acting against the object moving up

is larger (weight plus air resistance). The deceleration on the object is greater, thus the maximum

vertical displacement is reduced. [B1]

The shorter time of flight due to the reduced maximum height causes the horizontal distance

covered to be reduced. Or the additional horizontal force due to air resistance causes the object to

be displaced less horizontally. [B1]

2010 JJC H2 P2 Q1

7 (a) 90 km h-1

= (90 x 1000) / 3600 = 25 m s-1

. Conversion [1]

OR 35 m s-1

= (35 x 3600) / 1000 = 126 km h-1

.

Since 35 m s-1

25 m s-1

, the motorcycle has exceeded the speed limit.

(b) 0 4.0 13.0 52v u at m s-1

Working [1]

(c)

smotorcycle = ut = (35)(20 + 13.0) = 1155 m value of smotorcycle [1]

scar = 12

at2 = 1

2(4)(13.0)

2 = 338 m value of scar [1] (1)

30o

20 m s-1

N

Fig. 2.1

A


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s = smotorcycle scar = 1155 338 = 817 m value of s [1] (1)

817 100 = 717 m working [1]

(d)

v = 52 35 = 17 m s-1

Working [1]

t =717

17= 42.2 s answer [1]

2010 MI H2 P2 Q2

8(a) Velocity changes by 10 m s-1

per second.

[Do not accept definition of acceleration.]

(b)(i)

t s

x

ux

300

900 0.333 s

(ii) No air resistance.

(iii)

sy u

yt

1

2gt 2

sy 0

1

29.81 0.3332

sy 0.544 m

(iv)

vx

2 ux

2 2as

0 9002 2a 4 a 101000 m s-2

Lower

2010 CJC H2 P2 Q2

9 (a) (i) Using “v2 = u

2 + 2 a s”

v2 = 0 + 2 x 9.81 x 80

v = 39.6 m s-1

i.e. 40 m s-1

.

[2]

(ii) Magnitude of resultant velocity:

V = ( 67 2 + 40

2 ) = 78 m s

-1 .

Direction of resultant velocity:

Let be the angle between resultant velocity and the vertical

Tan = 67 / 40 = 1.675

= 59

The resultant velocity is 78 m s-1

making 59 with the vertical.

(b) The horizontal component of the velocity will be decreased. This is due to the air resistance

which provides the negative acceleration to the horizontal component.

[1]

The vertical component of the velocity would continue to increase. However, due to the

increasing magnitude of the velocity, the drag force due to the air resistance will also increase

at a rate proportional to the squared-value of the speed. Thus, the vertical component will

[2]


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increase at a decreasing rate until terminal velocity is reached.

(c)

[2]

2010 MJC H2 P3 Q1

10 (a) Sx = ut + ½ at2

Sx = ½ (2.25) 0.502 [M1]

Sx = 0.281 m

Sy= ut + ½ at2

Sy= 0.5 x 9.81 x 0.502 [M1]

Sy= 1.23 m

Net displacement = 2 2x yS S = 1.26 m [A1]

Angle with respect to horizontal = 77.1 degree below horizontal [ A1]

(b) (i) Graph must be negative before any mark is awarded.

Graph: 1 mark for identifying that acceleration is zero at start and end of graph

Graph: 1 mark for identifying that acceleration is negative and constant at middle part of

graph

b ii) The acceleration between time t1 and t2 is negative value because the

negative slope of v-t graph.

The acceleration between time t1 and t2 is a constant maximum negative value

since it corresponds to the steepest gradient of the v-t graph.

END

Start of fall

Horizontal distance from the point of drop off

vertical distance from the point of drop off

(ii) (i)

H2 Physics/ 2012/ Year 1/ Revision/ Forces/ Solutions

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2010 NYJC H2 P2 Q1

1 (a) Resultant external force acting on the body is zero.

Resultant torque about any point is zero.

(b)

(c)

2010 PJC H2 P2 Q2

2(a) Upthrust 53.35316.381.91050.4800 4 N

(b)

At equilibrium,

kx + U = mg

x

Umgk

312

102.0

5316.381.91050.48000 4

k N m−1

(c)(i) When the sphere is lifted out of the liquid, the volume of liquid displaced is reduced. This

causes the upthrust acting on sphere to decrease. To maintain equilibrium, the tension will

increase and the string breaks when the tension exceeds the maximum allowable value.

(c)(ii) At breaking point,

mgUT

81.91050.4800081.98000.32 4 V

41023.4 V m3

2010 PJC H2 P3 Q1

3(a) The rate of change of momentum of a body is directly proportional to the resultant force that

acts on it and has the same direction as the resultant force.

Taking moments about the hinge,

0 0

0

0

0

cos30 0.50 sin60 0.75

0.50 12 9.81cos30

0.75sin6078 N

mg T

T

Let the force due to plane A be FA and the force due to plane B be FB. These 2

forces must be perpendicular to the sides of the respective planes.

Resolving vertically,

FB cos 300 + FA cos 70

0 = mg

Resolving horizontally,

FB sin 300 = FA sin 70

0

Solving simultaneously,

FA = 99.6 N

FB = 187 N


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(b)(i)

T – 1.0g = 1.0a ------- (1)

4.0g(sin45°) – T – 15 = 4.0a ------- (2)

Solving (1) and (2),

T ≈ 10.4 N

a ≈ 0.587 ms−2

(ii) Loss in GPE = Gain in KE + WD against friction

1545sin

0.15.00.4

2

10.4

2

10.10.4

22

vg

v ≈ 3.04358 ms−1

, v ≈ 3.04 ms−1

(iii) The total momentum of a system is constant, provided no external resultant force acts on it.

(iv)

By conservation of momentum,

4.0(1.6) = 4.0v1 + 6.0(0.7)

v1 = 0.55 ms−1

(v) Velocity of approach = 1.6 – 0 = 1.6 ms−1

Velocity of separation = 0.7 – 0.55 = 0.15 ms−1

Since velocity of approach is not equal to velocity of separation, collision is not elastic.

T

1.0g

T

15

4.0g

4.0gsin45° 4.0gcos45°

N

A A C C

1.6 ms−1

v1 0.7 ms

−1


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2010 RI H2 P2 Q2

4(a) U W T

ρVg

V

U

ρ

1 1

1

-3 3

2

2 -3

-3

= -

= 50 - 40

=10 N

10 =

10=

1000×9.81

=1.02×10 m

= 50 -34

=16 N

16=

1.02×10 ×9.81

=1600 kg m

(b) (i)

(ii)

s

s

Taking moments about O,

5.00×500+(70×9.81× )=1000sin53.0°×10.0

=7.99 m

(iii)

R T

R T

R

x

y

2 2

At equilibrium, the net vertical and horizontal forces must be zero.

= cos53.0° = 602 N

= 500+(70×9.81) - sin53.0° = 388 N

= 602 +388

= 716 N

2010 RI H2 P3 Q5

5 (a) N = W cos

= (3.00 x 9.81) cos 30.0o= 25.5 N

(b) a// = g sin

= 9.81 sin 30.0o = 4.91 m s-2

(c) h/s// = sin

s// = h / sin

=0.500 / sin 30.0o = 1.00 m

weight of beam

force by worker on beam

tension reaction force


4 | P a g e

v//2 = u//

2 + 2 a// s//

= 02 + 2 x 4.91 x 1.00

= 9.82 or 9.81

v// = 9.82 = 3.13 (shown)

(d) sy = uy t + ½ ay t2

2.00 = (3.13 sin 30.0o) t + ½ (9.81) t2

4.905 t2 + 1.565 t – 2.00 = 0

t = 0.499 s

sx = ux t + ½ ax t2

= (3.13 cos 30.0o) (0.499) + 0 = 1.35 m

(e) (i) Impulse = area under F-t graph

= ½ (0.200) (360) = 36.0 Ns

(ii) vy = uy + ay t

= (3.13 sin 30.0o) + (9.81) (0.499 = 6.46 m s-1

py = m [vy – uy]

36.0 = 3.00 [vy – (–6.46)]

vy = 5.54 m s-1

(iii) Since the KE or speed after collision is smaller, the collision is inelastic. During collision, the KE of the ball is converted into sound energy, thermal energy and/or elastic PE as the ball deforms.

2010 RVHS H2 P3 Q2

6(a) The torque (or the moment) of a force F about an axis is the product of that

force and the perpendicular distance from the line of action of the force to the

axis.

(b) (i) Taking moments about the elbow,

Ty x 3.5 = 65 x 10 + 20 x 35

Ty = 385.7 N

T = Ty / cos 20 = 410 N

(ii) Tx = 385.7 / tan 20 = 140.4 N

Fx = 0

Tx = Rx = 140.4 N

Fy = 0

Ty = Ry + 65 + 20

Ry = 385.7 -65 – 20 = 300.7 N

Resultant force acting at elbow (pivot)

= 2 2

x yR R = 332 N


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State direction

(c) N2L: rate of change of momentum is proportionally to the force applied

Larger change in momentum (higher speed)

Shorter time interval

N3L: force exerted on hand equal in magnitude and opposite in direction to the

force on wood

2010 ACJC H2 P3 Q4

7 (a) F spring= kx

5 x 9.81 = k(0.05)

k=.981 N m-1

(No not penalize for inconsistent –ve sign that appears) – Pls annotate BOD

(b) (i) At a height 40cm, the contraction of spring is 10 cm.

Energy stored in spring = ½ (981)(0.1)2

= 4.905 J

Energy gain = Energy lost

½ mv2= mgh + 4.905

½ (5)v2= (5)(9.81)(0.1)+4.905 = 9.81

v= 1.98 m s-1

(ii) By conservation of energy,

Let final extension be x (measured from natural length)

Energy at natural length= Energy at lowest point

9.81 = - (5)(9.81)x + ½ (981)x2

Initial energies = final energies

4.905 =-5 (9.81)(x+0.1) + ½ (981)x2

490.5 x2 – 49.05 x – 9.81 = 0

50 x2 – 5x – 1 = 0

Hence x = -0.10 (initial released point) or + 0.20 (lowest point)

Hence lowest point is (0.30 – 0.20) = 0.10 m above the ground.

(c) In reality, it is larger. There will be some energy lost due presence of air resistance/friction in spring and thus

less energy will be available for conversion to elastic potential energy.

2010 MI H2 P2 Q4

8(a) Boat has greater volume compared to solid block (more empty space

within)

Hence displaces more water and therefore achieves sufficient upthrust

to support its weight.

(b)(i) T W mg 0.200 9.81 1.96 N

(ii)

V m

0.200

8000 2.5 105 m3


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(b)(iii)

T U W

T W U

T 1.96 2.5 105 800 9.81 1.76 N

(c)

2010 NJC H2 P2 Q2

9 (a) (i)

1 mistake : minus 0.5 mark

2 mistakes : zero mark

[1]

(ii) The gutter is in rotational equilibrium.

Taking moment about the right side of the gutter,

Tcos (1.5) = Mgcos(1.5/2) where is the angle of inclination.

T = M/2(9.81)

Since T = weight of empty bucket, hence mass of empty bucket = M/2 =

25.0 g

[2]

(b) (i) Dirty water flows into the bucket. The heavier bucket now provides a

clockwise moment about the pivot that is greater than that by the weight of

the gutter.

The gutter is lifted and subsequent clean water flows along the gutter into

the tank.

[1]

(ii) The bucket will only gradually start to move downwards as it fills with

water and hence the motion is less jerky.

[1]

(c) (i) Considering the vertical motion:

vf2 = 2gh + vi

2

velocity of water just before hitting bottom of bucket vf = 2.80 ms-1

, where h

[3]


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= 30+10 cm and vi = 0.50sin, sin-1 = 5/150

Resultant force on water, assuming all water comes to rest upon impact

= (2.80 – 0) dt

dm

= (2.80 – 0) dt

dV

50x10-6

(1000)(2.80-0) = 0.140 N

(ii) The assumption made in (c)i is that the final speed of the water is zero.

Otherwise, the water will rebound. The change in momentum of the water

and hence the force on the bucket will be larger.

[2]

2010 CJC H2 P2 Q3

11 (a) (i) P = Fv , v is a constant

F = 900x103 / (20x1000/60x60) = 1.62 x 10

5 N

[2]

(ii) Additional power needed

= mgh / t

= mh xsinӨ / t

= mg v sin sinӨ

= 2.2x105 x 9.81x (20x 1000/60x60) (1/60) x= 2.0 x10

5 W

[2]

(iii) T sin Ө = ma

Tcos Ө = m

tan Ө = ma / mg

Ө = tan-1

(a/g) = tan-1

(0.667 / 9.81) = 3.89o

[3]

(b) (i) There is an upthrust acting on the object. Hence tension is now reduced and reading will be

decreased. [1]

(ii) Upthrust = volume of object x density of water x acceleration due to gravity

Upthrust = initial reading – final reading

volume of object = ( initial reading – final reading ) / density of water x acceleration due

to gravity

Knowing the values of tensions, density of water and acceleration due gravity, volume can be

found

[3]

2010 AJC H2 P2 Q1

12(a) When an object is floating in equilibrium, the upthrust (which is equal to the weight

of the fluid displaced) is equal to the weight of the object. OR A floating object in a fluid displaces a weight of fluid equal to its own weight.

motion

acceleration

Ө

ma

mg


8 | P a g e

(b)(i)

θ = angle of inclination Take moments about A, U 0.75L cos θ = W 0.5L cos θ (Note: cos θ optional

0.75 U = 0.5 W = 0.5 x 60 = 30 N Because ⊥ length to lines of

U = 40 N action of forces proportional

to 0.5L, 0.75L (similar Δ))

(ii) Upthrust = weight of water displaced = Volume x density x g

40 = Volume x 1.0 x103 x 9.81 Volume of water displaced = 4.0775 x 10-3 m3 Volume of rod = 2 x 4.0775 x 10-3 = 8.155 x 10-3 m3 Density = mass/volume = (60/9.81)/ 8.155 x 10-3= 750 kg m-3

(iii) Since there is now no tension, the unpthrust has to increase until it is equal to

the weight of the object. hence more volume of the rod must be submerged to displace more water for arger upthrust.

2010 AJC H2 P3 Q5

13 a (i) Linear momentum is the product of mass and velocity of an object (ii) N2L: the rate of change of momentum of a body is proportional to the external net

force acting on it and occurs in the direction of the force. (iii) From (i) momentum P = mass m x velocity v

From (ii) force F = dP/dt = d (mv)/dt = m d(v)/dt dv/dt = a hence F = m a

(iv) When objects of a system interact, the total momentum before impact is equal to the

total momentum after impact, if no net external force acts on the system. OR total linear momentum of an isolated system of bodies is constant.

(b) (i) Momentum = m v cos 60 = 0.30 x 4.0 = 1.2 kg m s-1 to the right (ii) Total momentum before = total momentum after

1.2 = PA + PB = 0 + PB PB = 1.2 kgms-1 to the right


9 | P a g e

(iii) From (ii), velocity of B = PB/m =1.2/0.1 = 12 m s-1

Initial KE = ½ m (v cos 60)2 = ½ × 0.30 × 4.02 = 2.4 J

Final KE = ½ mBvB2 + 0 = ½ × 0.1 × 122 = 7.2 J

Additional KE = 7.2 – 2.4 = 4.8 J (iv)

(v) Time to reach from P to O vertically = time to reach from O to P vertically

0 = v sin 60 – gt t = 8.0 sin 60/9.81 = 0.706 s

(vi) At P, both A and B have the same vertical velocity (0 ms-1).

Both experience the same downward vertical acceleration, g. Both fall the same vertical distance ( the height of P) above the ground. Hence they both take the same time to reach the ground.

(vii) From (b) (iii) and (v)

t = 0.706 s distance = vB x t = 12 x 0.706 = 8.47 m

(c) (i) It will reach a point lower that P.

For the upward motion, air resistance is acting downward against the velocity, the deceleration of the motion is more than g. It will slow down faster and reach zero vertical velocity earlier and hence the highest point lower than P.

(ii) Sketch on Fig. 5.1 the path of the firework when it fails to explode where

1. air resistance is negligible. Label the path as R.

⇒ R path is symmetrical with the path OP

2. air resistance is not negligible. Label the path as S.

⇒ S path is not symmetrical, lower than P and shorter horizontal distance, reaches

peak earlier than P.


10 | P a g e

2010 SRJC H2 P2 Q2

14 (a) (i)

(ii) Taking moments about B,

mg x 0.5 = FA x 1.5

Fa = 130.8 N

(iii) Let the distance of the pivot on the left from the centre of gravity be SA

Let the distance of the pivot on the right from the centre of gravity be SB

Fa + Fb = (40.0) (9.81) = 392.4 N

Fa = 392.4 - Fb

Taking moments about the centre of gravity,

Fa x Sa = Fb x Sb

As the support at point A is shifted right, Sa decreases. So 1+Sa / Sb increases, and Fb decreases.

(b) Impulse = Change in momentum of object = Area under F-t graph

Impulse = (0.5 x 10.0 x 100.0) = 500 N s

Δp = m Δv

Δv = Δp / m = 500 / 200= 2.5 m s-1

Δv = vf - vi

vf = Δv + vi = 2.5 + 15.0 = 17.5 m s-1


11 | P a g e

(c)

Let the frictional force between X and the table be FX,T.

Let the frictional force between X and Z be FX,Z.

From the free-body diagram for Z,

T + mzg sin20° - 11 = mza

Since velocity of Z is constant,

T = 11 - mzg sin20°

= 11 – (2)(9.81)sin20°

= 4.290 N

From the free-body diagram for X,

100 – (T + 11 + mxg sin20° + FX,T ) = mxa

Since velocity of X is constant,

100 – (T + 11 + (20)(9.81) sin20° + FX,T ) = 0

100 – [4.290 + 11 + (20)(9.81) sin20° + FX,T ] = 0

100 – 4.290 – 11 – (20)(9.81) sin20° – FX,T = 0

FX,T = 17.6 N

END

H2 Physics/ 2012/ Year 1/ Revision/ Dynamics/ Solutions

1 | P a g e

2010 NYJC H2 P3 Q6

1(a) Linear momentum is defined as the product of a body’s mass and its linear velocity

(b)(i) Rate of change of momentum of air = (mv-mu)/t

= m/t(v) = pr2v(v) = pr

2v

2

By N2L, force acting on air by propeller = pr2v

2 backward

N3L, force acting on propeller by air = pr2v

2 forward. (shown)

(ii) F= ma

pr2v

2 = ma

a = (1.2)()(0.72)(20

2)/420 = 1.76 m s

-2

(c)(i) Rate of rate of transfer of KE = 1/2mv2 /t = ½()(r

2)(v) (v

2)

= (0.5)(1.2)()(0.702)(20

3) = 7400 W

(ii) Since speed is constant, a = 0

F= 0, F – f = 0

Hence f = 740 N

By conservation of energy,

Power = rate of transfer of KE to air + rate of work done against f

16 000 = 7400 + (740)(u)

u = 12 m s-1

(d)(i)

(ii) When the front part is wider, it can displace more water in front, causing the point of application of U to be

closer to the front. This will help to create a restoring torque to stabilize the boat.

(iii) The external forces acting on the system will be weight, upthrust and drag force. Since mg is counter by U,

hence the drag force will results in a net external force acting on the system. Therefore, total momentum is not

conserved.

(e)(i) Since the propeller is above water, there is no concern about whether the propeller will get struck in the river

bed.

(ii) Problem 1: A bigger propeller will results in an greater uneven distribution of the weight affecting the

stability of the boat.

Problem2: When the diameter of the air column increase, air turbulence may come which will greatly reduce

the efficiency of the propeller.

Problem 3: The perpendicular distance between the thrust and drag force will increase. The increase in torque

will affect the stability of boat.

2010 RVHS H2 P2 Q2

2(a) Lsin + Lsin = D

(40.0 sin 50.0 + 40.0 sin ) = 50.0

= 28.9 (Shown)

(b) Vertical height Jane is above Tarzan, h = 40.0 cos 40.0 cos 50.0 = 9.307 m


1

2MJvJ

2 + MJgh = F D (work done against friction), Therefore, vJ = 6.12 m s

1

Force on propeller U, upthrust

water level

Total weight

of airboat

Drag

force


2 | P a g e

(c) By conservation of energy,

1

2(MJ + MT)v

2 + F D = (MJ + MT)gh

Therefore, v = 9.90 m s1

2010 SAJC H2 P2 Q1

3 Graph A : [ 1 mark per point, max. 3 marks]

- The direction in which the object is travelling is suddenly reversed. This occurs when something hits a

vertical wall

- very large acceleration is required

- example : ball bouncing against wall / ball hit by racquet

Graph B : [ 1 mark per point]

- acceleration going from zero to a high value.

- example : letting go of a ball, dropping an egg

- acceleration changes suddenly as the force holding the object in place is removed

Graph C : [ 1 mark per point]

- constant force being applied drops to zero suddenly.

- example taking foot off car accelerator/ceasing to pedal bicycle

- so zero subsequent acceleration

Generally,

for explaining what is happening, for sensible example, or relating to sketch graph

2010 SAJC H2 P3 Q5

4(a)(i) Newton’s 1st law: Every object continues in its state of rest or uniform motion in a straight line unless a net

(external) force acts on it to change that state.

The law implies that a “force” is that quantity which is required to change velocity.

(ii) Mathematically, resultant force, F dt

mvd )( = k

dt

mvd )( (where the value of the proportionality const k

would depend on the definition of the unit of force.)

Since the definition of the newton was chosen as “the force which causes a mass of 1 kg to have an

acceleration of 1 m s-2

,

thus, 1 N = k x 1 kg x 1 m s-2

k = 1

Thus, F = dt

mvd )(

= m

dv

dt (since m = const)

= ma

(b) As the stone accelerates towards Earth, its downward momentum is increasing from an initial value of zero

momentum.

Since there is no net force acting on the system of stone & Earth, the total momentum of the stone-Earth

system must remain unchanged, ie = 0.Thus, the Earth must accelerate upwards towards the falling stone.

{Assume air resistance is negligible}

(c)(i) (ii) Given KEα = 9.2 x 10

-13 J

Thus ½ mαvα2 = 9.2 x 10

-13 J, where mα = 4u = 4x1.66x10

-27 kg

→ Vα = 27

13

1066.14

102.92

xx

xx

= 1.67 x 107 = 1.7 x 10

7 m s

-1 (proved)

HeRnRa 4

2

220

86

224

88


3 | P a g e

(iii) Since no net force is involved in radioactive decay, & since total momentum before decay = zero (as Ra is

initially stationary),

by principle of conservation of momentum, mαvα = mRnvRn [ 1 ]

Thus, vRn = mαvα

mRn =

(4u)(1.7 x 107)

220u

= 3.1 x 105 m s

-1 [ 1 ]

(d)(i) Centre of gravity of an object is defined as that point through which the entire weight of the object may be considered to

act

(ii) Principle of Moments:For a body to be in equilibrium, the sum of all the anticlockwise moments about any point

must be equal to the sum of all the clockwise moments about that same point.

(iii)

Moment exerted by the weight about the base

= mg x (2.3cos450)

= 180 x 9.81 x 2.3cos450

= 2.9 x 103 N m (shown)

(iv) Applying the principle of moments about the base,

T x (4.0 sin350) = 2.9 x 10

3

→ T = 1.26 x 103 = 1.3 x 10

3 N

2010 TJC H2 P3 Q1

5 (a) head-on : collision takes place along the line joining the centres of the colliding bodies. [1]

elastic collision: both the momentum and the kinetic energy are conserved. [1]

(b)

speed of separation = speed of approach

(1)---

0

211

112

vvuor

uvv

[1]

(c) By conservation of linear momentum,

u1 v1

v2

Before collision After collision


4 | P a g e

.13

11 ratio Thus

13

11

13

2- (1) from

13

2132 (2)(1)

(2)---12

12

1

1

111121

1221

211

211

u

v

uuuuvv

uvvu

vvu

mvmvmu

28.013

111

1

2

12

1

fraction required (d)

2

2

1

1

2

1

2

1

2

1

u

v

mu

vum

(e) This is a head-on elastic collision between two equal masses. After collision, the incident neutron will stop while

the target neutron moves off with the velocity of the incident neutron.

The ratio of the final to the initial speed of the incident neutron would be zero

The fraction of the kinetic energy of the incident neutron that is transferred to the target neutron would be 1 or

100 %.

2010 VJC H2 P3 Q3

6 (a) In a system of interacting bodies, the total momentum of the system remains constant provided no external

force acts on the system.

(b) An elastic collision is a collision in which the kinetic energy of the system is conserved.

(c) (i) Strain energy stored = . Jkx 40.0100.2100.22

1

2

1 2232

(ii) By Conservation of Energy,

KE of block after collision = Strain energy stored in spring

1

22

894.0

40.0)0.1(2

140.0

2

1

smv

vmv

(iii) By conservation of momentum, 22112211 vmvmumum

211 )0.1()2.0(0)2.0( vvu

For elastic collision, 1221 vvuu

121

121 0

uvv

vvu

1

121

2121

68.2)894.0(4.0

2.12.14.0

)(2.02.0

smuvu

vuvu

Speed of ball = 2.68 m s-1

(iv) By Conservation of Energy,

Gain in KE of ball = Loss in GPE of ball

COM [1] Working [1] Answer [1]

Equation [1] Working [1] Answer [1]


5 | P a g e

7.50

)0.1)(81.9(2

68.2

2)cos1(

)cos(02

1

22

2

gl

v

llmgmv

2010 YJC H2 P3 Q1

7(a) X is the resultant of the weights of all parts of the car

Z is the resultant of the normal force on the car and frictional force on the tyres.

(b) 1. Zx 2000 = 1200 (8) Zx = 11600 N

Zy = 1200 (9.81) = 11772 N

Thus, Z = (116002 + 11772

2) = 16.5 kN

2. Angle = tan1

(11772/11600) = 45.4

2010 TPJC H2 P2 Q1

8 (a) From Newton’s Second Law, Fnet = ma

Since air resistance is negligible, the only force acting on each object is weight. Thus,

ma = mg [M1]

a = g

=> same acceleration for both objects [A1], so strike the ground at the same time. This is

despite both objects having different masses.

Comments:

The majority were unable to score full marks for this part. Despite the question requesting

for the use of Newton’s Second Law, many students used kinematics instead. Moreover,

these students claimed that acceleration for both objects is the same without accounting for

the difference in masses. There were also a number of students who left their answers with

just the definition of Newton’s Second Law.

[2]

(b) .

[B2] for diagram – must include upthrust otherwise minus 1 mark.

[B1] for stating fluid resistance or viscous force increases when velocity increases.

[B1] for net force = 0 when weight = viscous force+ upthrust

Comments:

This question was also not answered well. At least 40 % of students were given zero credit

when they state that upthrust increases when velocity increases. These students failed to

realize that upthrust and viscous force are two different forces. The better students were

able to state that upthrust is constant while viscous force is dependent on velocity.

Another area of concern is the drawing of free body diagrams. It is disappointing to see that

many were poorly drawn or poorly labeled at this point in time.

(c) Fnet = R – mg [M1]

ma = kv2 – mg

(80)a = (35)(5.0)2 – (80)(9.81)

a = 1.1 ms-2

[A1]

Direction: upwards [A1]

Weight

R

U


6 | P a g e

Comments:

The majority were able to score full marks for this part. However, there were some students

who could not visualize the two forces (air resistance and weight) acting on the parachutist,

and hence omitted one of the forces in the calculations.

There were also a few cases of pure carelessness where the “square” for velocity v in R =

kv2 was omitted.

2010 TPJC H2 P2 Q2

9 (a) The principle of conservation of momentum states that the total momentum of a closed

system of colliding objects remains constant (or total momentum before and total

momentum after is equal) if no net external force act on that system.

Comments:

The condition “no resultant or no net external force acting on the system” must be

present in order for credit to be given.

(b) Δp = area under graph = ½ (40 10–3

) (150) = 3.0 kg m s–1

[C1]

Comments:

Many students did not realise that the time axis was given in milliseconds and thus

were unable to gain any credit.

(c)

[2]

(d) Δp = pf – pi

- 3.0 = (2.0)(vP) – (2.0)(2.0)

vP = 0.50 m s-1

to the right [C1]

3.0 = (1.0)(vQ) – (1.0)( - 1.0)

vQ = 2.0 m s-1

to the right [C1]

Comments:

The majority did not get any credits for this part. Those who got it wrong either

assumed that the collision was elastic or that it was completely inelastic. The majority

also failed to see the link for this part with parts b) and c).

(e) Initial total KE = ½ (2.0) (2.0)2 + ½ (1.0) (1.0)

2 = 4.5 J

final total KE = ½ (2.0) (0.50)2 + ½ (1.0) (2.0)

2 = 2.25 J

Percentage loss in KE

= [4.5 – 2.25 / 4.5] x 100% [C1]

= 50 % [A1]

Comments:

F / N

0 t / ms

-150

20 40

[M1] for correct graph (Newton’s 3rd

Law)

[B1] for correct label of numerical values

(at least 1 value on each axis).


7 | P a g e

v

The majority was given full credit with error carried forward from the previous part.

It was a pity that there were a few students who did not complete this question upon

realizing their part d) was wrong.

2010 HCI H2 P3 Q1

10 (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]

NF 51067.1 [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]

2010 IJC H2 P3 Q5

11

(a) (i) Newton’s First Law of Motion states that every body continues in its state of rest or uniform motion in a

straight line unless a net external force acts on it to change that state. [B1]

Hence, if we observe an object initially in a state at rest but which suddenly moves, or an object moving at

uniform speed in a straight line, but which suddenly reduces or increases in speed or changes in direction, we

say that there must be a net external force acting on it. [B1]

(ii)

1.

[B1] for diagram showing 3 arrows.

Description: An object that is projected vertically upwards in the earth’s gravitational field. [B1]

F

a


8 | P a g e

v

u

Fws

Fsw

2.

B1] for diagram showing 3 arrows.

Description: An object that is undergoing circular motion. [B1]

(b)(i)

mu

2 7 2

6

1 1Initial KE (1.2 10 )(0.50)

2 2

1.5 10 J [A1]

(ii)

2 2

2

2

27

5

5

Using 2

0=(0.5) 2 (15)

-(0.5) = [M1]

2(15)

-(0.5)net force = (1.2 10 )[ ] [M1]

2(15)

= -1.0 10 N

magnitude of retarding force is 1.0 10

v u as

a

a

ma

N [A1]

(iii)

2

Using ,

(0.50)0 0.50 [ ] [M1]

2 15

60 s [A1]

v u at

t

t

(iv) Viscous force D

F v where v is the instantaneous velocity [B1]

Viscous force will be greatest initially as the initial speed is greatest, and it will reduce to zero when

the final speed is zero.

The viscous force will provide a greater retardation force [B1], and the time taken to stop the ship will

be shorter [B1].

(v)

m v u 7

6

p = ( - ) 1.2 10 (0 0.50) [M1]

p = - 6.0 10 N s [A1]

Fws = force exerted by water on the ship

Fsw = force exerted by ship on the water

a

F


9 | P a g e

(Diagram - B1)

By considering the ship and the water as an isolated system and with no external forces acting on the ship

and the water [B1], the momentum lost by the ship is equal to the momentum gained by the water [B1].

2010 JJC H2 P3 Q6

12 (a) The rate of change of momentum of an object is directly proportional to the resultant force

acting on it. The change in momentum takes place in the direction of that force. [2 or 0]

(b) (i)

[1] two correct labeled forces

(ii)

[1] two correct labeled forces

-1 per missing, unlabelled or incorrect force

(iii) Due to friction, B exerts a force of 40 N on A towards the left.

Using F ma

-2

40 40

1.0 m s

A

A

a

a

[1] value

Mass A accelerates towards the left at 1.0 m s-2

[1] direction

(iv) Normal contact forces between A and B , FAB and FBA . [1]

Frictional forces between A and B, f and f’ . [1]

(We only accept these two pairs “action-reaction forces between A and B only.”

(c) The total momentum of a system before and after an interaction is constant if there are no

external forces acting on the system. [2 or 0]

(d) (i) Mass of man and rifle minus the mass of 10 bullets.

2100 (10 4 10 )

FAB WA

Contact

force f ’

FBA

WB

100 N

f


10 | P a g e

99.6 kg [1] ans

(ii) By conservation of linear momentum,

2 20 10 4 10 800 100 (10 4 10 ) manv

-13.21 m smanv [1] ans

(iii) Method 1 (consider bullet)

Using ( )F t m v u

2(10) 10 4 10 (800 0)bulletaverageF [1] sub

32.0 NbulletaverageF [1] ans

By Newton’s 3rd

Law, force exerted on bullets by rifleman = force exerted on

rifleman by bullets

32.0 NManAverageF

Method 2 (consider man)

Using 'F t m v mu

(10) 99.6(3.21) 100(0)

32 N

ManAverage

ManAverage

F

F

(iv) - we are only considering the average force in our calculation. [1]

- in reality the duration of each shot is much shorter[1] and hence the impact force is

greater [1] (BOD)

(v) With the rubber pad, the impact duration becomes longer [1]. As such the recoil force

when firing the rifle becomes smaller [1]. OR some energy is absorbed by material

and less energy transferred to man [2]

(Recoil is weaker because of longer impact duration)

2010 MI H2 P3 Q1

13(a) W = 80 (9.81) = 780 N

(b)(i) Weight acting downwards from centre of object

Normal force acting upwards from base of object

(ii)

a F

m

N W

m

2.5 N 780

80

N 980 N

(b)(iii)

a F

m

N W

m

a 1.5W W

m

a 0.5g 4.9 m s-2


11 | P a g e

(c)

(d)(i)

P W

t

Fs

t

P 8000 1103 250

60

P 37.5 kW

(ii)

Eo

Ei

0.75

37.5

Ei

0.75

Ei 50.0 kW

2010 AJC H2 P3 Q1

1 (a) A body remains at rest or continue to move in a straight line with uniform velocity unless a net ext force acts on

it. From this law, it shows that a force is needed to change the state of motion of the object.

If a body changes from rest to motion or from uniform velocity to non-uniform velocity, a

force must be acting on the object.

(b) (i) Taking the two gliders as a system, the resultant force acting it = 10000 – ( 2000 +2000)

= 6000 N

Using F = ma ,

6000 = (1200 + 1200) a

a = 6000/2400 = 2.5 m s-2

(ii) using v2 = u

2 + 2as, s = 40

2/ [2(2.5)]= 320 m

OR

Time to reach 40 m s-1

= t, Using v = u + a t, 40 = 0 + 2.5 t ⇒ t = 16 s

using s = ut + 1/2 a t2, s = 0 + 1/2 (2.5) 16

2 = 320 m

OR

work done = kinetic energy gain, (ma)s = 1/2 m v2, s = 402/[2(2.5)] = 320m

(iii) If one glider is towed, it will move with greater acceleration as it has less mass and

friction and therefore greater resultant force acting on it.

from v2 = u

2 + 2as greater acceleration, smaller distance required to reach the speed

END

H2 Physics/ 2012/ Year 1/ Revision/ WEP/ Solutions

`1 | P a g e

2010 NYJC H2 P2 Q2

1(a) Extension 30.2- 25 = 5.2 m

Total EPE = 2 × ½ kx2 = (19000)(5.2)

2 = 510 kJ (proven)

(b)

Total elastic

potential energy /kJ

Gravitational

potential energy of

capsule /kJ

Kinetic energy of

capsule /kJ

Ground level 510 0 0

30 m above the

ground 0 88 422

Highest point 336 174 0

(c) When the cord is unstretched,

Ht above the ground = 30 – (252 – 4

2) = 10.8 m above the ground.


gain in KE + gain in GPE = loss in EPE

½(300)v2 + (300)(9.81)(10.8) = 510 000 v = 56.5 m s

-1

(d) The apparent weight of the passenger will be the greatest at the lowest point immediately after the

capsule is released. The EPE is the largest, this implies that extension is the largest which will results in the

largest force.

2010 SAJC H2 P2 Q2

2 (a) GPE ( ) ( )A A B Bm g h m g h

2.00 9.81 0.500sin40 5.00 9.81 ( 0.500sin50 ) [ 1 ]

12.481 J 12.5 J (3 sig fig) [ 1 ]

(b) Loss in GPE = Gain in KE + Work done against friction [ 1 ]

2112.481 0.500 3.00 2

2mv [ 1 ] for correct W.D. against friction

21(2.00 5.00) 9.481

2v

-11.65 m s (3 sig fig)v [ 1 ]

Or

ΣF = ma 5gsin50° - 2gsin40° - 3 x 2 = (5 + 2)a

a = 2.709 m s-2

v2 = u

2 + 2as v

2 = 0

2 + 2(2.709)s v = 1.65 m s

-1

(c)

Energy / J

Time 0

Ep [ 1 ]

Wf [ 1 ]

Ek [ 1 ]

12.5

9.5

3.0


`2 | P a g e

2010 ACJC H2 P3 Q4

2010 CJC H2 P3 Q5

4 (a) The direction of the velocity is always changing although the speed is constant

velocity change acceleration since acceleration = rate of change of velocity

Acceleration is directed towards the center of the circle.

(b) (i) Principle of conservation of momentum states that the total momentum of a system of

interacting particles is constant if there is no external resultant force acting on it.

(ii) Completely inelastic collision:

The two particles stick together

By principle of conservation of linear momentum mv = 2mV V = v/2 both move off

at a speed of v/2

(iii) Bullet embedded in wood:

Apply conservation of momentum

( 0.025) 150 = ( 2 + 0.025) v

v = 1.852 ms-1

Bullet and wood swing up:

Apply conservation of energy

½ m v2 = m g h

½ (1.8522) = (9.81) h h = 0.17 m

Hence x = -0.10 (initial released point) or + 0.20 (lowest point)

Hence lowest point is (0.30 – 0.20) = 0.10 m above the ground.

4c In reality, it is larger. There will be some energy lost due presence of air

resistance/friction in spring and thus less energy will be available for

conversion to elastic potential energy.

4a) F spring= kx

5 x 9.81 = k(0.05)

k=.981 N m-1

(No not penalize for inconsistent –ve sign that appears) – Pls annotate BOD

4b

(i)


At a height 40cm, the contraction of spring is 10 cm.

Energy stored in spring = ½ (981)(0.1)2

= 4.905 J

Energy gain = Energy lost

½ mv2= mgh + 4.905

½ (5)v2= (5)(9.81)(0.1)+4.905 = 9.81

v= 1.98 m s-1

(ii) By conservation of energy,

Let final extension be x (measured from natural length)

Energy at natural length= Energy at

lowest point

9.81 = - (5)(9.81)x + ½ (981)x2

Initial energies = final energies

4.905 =-5 (9.81)(x+0.1) + ½

(981)x2

490.5 x2 – 49.05 x – 9.81 = 0

50 x2 – 5x – 1 = 0


`3 | P a g e

(c) (i) Total kinetic energy before impact = ½ x 22000 x 32 + ½ x 66000 x 1

2

= 99000 + 33000

= 132 000 J

Total kinetic energy after impact = 0 + ½ x 66000 x 22 = 132 000 J

Since total kinetic energy of the system is conserved, collision is elastic.

OR

speed of approach = I 3 - 1 I = 2 ms-1

speed of separation = I 2 – 0I = 2 ms-1

since masses are the same and speed of approach = speed of separation

elastic collision

(ii) kinetic energy half-way through the collision

= ½ x 22000 x 1.52 + ½ x 66000 x 1.5

2

= 24750 + 74250 = 99 000 J < total kinetic energy after the collision.

Change in total kinetic energy = 33 000 J .

Reason: difference in energy is stored as elastic potential energy in the springs

(iii) Impulse = change in momentum

= 22 000 x 3 - 0

= 66 000 N s

(iv) Based on the principle of conservation of momentum, change in momentum cannot be

altered

Since impulse = change in momentum impulse would not be altered.

2010 HCI H2 P2 Q2

5(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]

(d)

[B1]

Tension in rope

Height Daniel falls through 0

H 20.0


`4 | P a g e

2010 HCI H2 P2 Q3

6(a)

(b) Normal horizontal contact force of wall on cube provides the centripetal force for cube to spiral.

Component of the weight down the slope causes the acceleration of cube down slope

(ci) Since the horizontal normal contact force is always perpendicular to the direction of motion, work

done is zero.

(ci) Rate of increase of kinetic energy = rate of decrease of gravitational potential energy

= m g v sin.

2010 SRJC H2 P2 Q8

7(a)(i) power = 0.40 kW

(ii) Using P = Fv (where F = resistive force)

0.40 x 103 = F x 2.00 F = 200 N

(b) (i) From graph,

Power for 350 kg boat to travel at 2.50 m s-1

, P = 1.90 kW

Energy required = Pt = (1.90 x 103) x (30 x 60) = 3.42 MJ

(ii)

(iii) The estimate was made based on the assumption that the mass of the boat is constant at 350 kg. The

amount of fuel expended is 0.190 kg, which is a very small fraction of 350 kg. The estimate should be a

reasonable one.

(c) (i) Since the distance travelled by boat A and B are equal, the area under both graphs should be equal.

V = 3.00 m s-1

(ii) Boat B is travelling at a constant speed, the energy output from the engine is used to do work against

resistive forces. [1] Boat A is travelling with a constant acceleration. The energy output from the engine is

used to increase the kinetic energy of the boat as well as do work against the resistive forces. [1]

Therefore, Boat A will expend more petrol. [1]

END

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

H2 Physics/ 2012/ Year 1/ Revision/ Thermal Physics/ Solutions

1 | P a g e

2010 NYJC H2 P3 Q7

1(a)

(b)

(c) (i)

(ii)

(d)

(e)

(f)

(g) (i)

(ii)

D to A 850 -750 100

It is the sum of a random distribution [1] of kinetic and potential energies [1]

associated with the molecules of a system.

An absolute scale means that the scale is theoretical and does not depend on the

property of any particular substance.

273.15 20 293.15 K

293 K

T

5

( )

0.90 990 160 [1]

1.4 10 J [1]

Q mc

Let mi be the initial mass of the air.

5

5

1.0 10 0.10 (180 273.15) [1]

1.0 10 0.10 0.030 [1]

(180 273.15)

i

r

i

pV nRT

mR

M

mR

Let mf be the final mass of the air.

5

5

1.0 10 0.10 (25 273.15)

1.0 10 0.10 0.030 [1]

(25 273.15)

f

r

f

pV nRT

mR

M

mR

m = mf – mi = 0.041 kg [1]

22 2 2

22

2 2

2 2

2 180 25180

180

25

,3 31 3

[1],2 2

28 (180 273.15) 1.15

32 (25 273.15)

NO O ONO

O N

N O

kT kTm v kT v v

m m

v m T

v m T

This is to ensure that the rate of heat lost from the apparatus to the

environment remains unchanged for both of the experiments.

Q/t = (m/t) c (T) + heat loss

0.1525 ( ) (19 15) (1) [1]

600.23

37 ( ) (19 15) (2) [1]60

loss

loss

c P

c P

Using equations (1) & (2),

Ploss = 2.5 W [1]


2 | P a g e

2010 PJC H2 P3 Q6

2(a) It means that the absolute scale does not depend on the

property of a particular substance.

[1]

(b)(i) An ideal gas is one that obeys the equation of state for an

ideal gas, pV = nRT for all values of pressure, volume and

temperature.

[1]

(b)(ii) Absolute zero on Kelvin scale is the temperature at which all

substances have a minimal internal energy.

[1] minimal

internal energy

Do not accept ‘zero

internal energy’

(b)(iii) Internal energy of a gas is the sum of all the microscopic

kinetic and potential energies of the molecules or atoms in

the gas.

[1]

[1]

(c)(i)

53.91 2.76 10 0.0160

8.31 (27 273.15)

0.737698 mol

0.738 mol

pV nRT

pVn

RT

[1] finding p

[1] substitution

[1] final ans

(c)(ii)

5

no. of moles of air in supply of air

11.65 10 0.0117

8.31 300.15

5.464777 mol

pV

RT

No. of moles of gas needed for 4 tyres 4 0.737698 2.950792 mol

there is sufficient air in the supply to fill up 4 tyres.

Pressure in supply after filling 4 tyres

5 5

(5.464777 2.950792) 8.31 300.15

0.0117

5.359 10 Pa 4.00 10 Pa

[1] finding amount

of air in supply

[1] find amount of

gas for 4 tyres

[1] for showing

pressure does not

fall below

4.00×105 Pa


3 | P a g e

(c)(iii)

23

21

3

2

31.38 10 300.15

2

6.21 10 J

U KT

[1] substitution

[1] final ans

(c)(iv)

21 23

3

2

6.21 10 6.02 10

3740 J

U NKT

[1] for correct

substitution and

answer

(d)(i) Using nRTpV , product of p and V at a point is

proportional to T.

At pt. A, 250105.05 2 AAVp kgm2s

−2

At pt. B, 4801042.1 2 BBVp kgm2s

−2

Hence temperature at B is higher; temperature of air

increases when the air is compressed

[1]

[1]

[1]

(d)ii)

[1] correct shape of

lines

[1] direction of

processes shown

[1] correct value at

p ~ 2.1 Pa for point

2010 RI H2 P2 Q4

3 (a) It is the sum of the kinetic energies and potential energies of the particles in the gas.

(b) (i) pV nRTUsing ,

pVT

nR

5 31.0 10 5.0 10

0.20 8.31

300 K

V / 10–3 m3

p / 105 Pa

Fig. 6.2

A

B

5.0 4.0 3.0 2.0 1.0 0

1.0

2.0

3.0

4.0


4 | P a g e

(ii)

V/ 103 m

3

0

p/ 105 Pa

(1)

(3)

(2)

2.5 5.0

1.0

2.0

2010 RVHS H2 P2 Q3

4a) (i) To find gradient at start of graph

Range 0.07 – 0.18 K s–1

or 4.4 – 11.0 K min–1

1

1

(ii) Formula ∆Q/∆t = mc∆T/∆t used

Value for rate within acceptable range 18 – 50 W

1

1

or 1100 –

3000 J min–1

(b) (rate of) energy lost to the surroundings OR due to

evaporation[do not credit boiling] approaches (rate of )

energy supply OR increases with temperature difference.

1

(c) (i) Curve of reducing gradient starting at 20 °C, 0 s initially

below given graph.

1

Reference of need to heat mug

Hence reduced rate of temperature rise

Reference to insulating properties of mug

2 Any 2 points

2010 SAJC H2 P3 Q6

5 (a) (i) internal energy = sum of p.e. + k.e. of molecules [ 1 ]

no p.e. as no force of attraction [ 1 ]

mean k.e. of molecules α absolute temperature [ 1 ]

(ii) minimum k.e. of molecules is 0 [ 1 ]

minimum absolute temperature is 0 (or cannot be negative) [ 1 ]

(b) (i) AB: k.e. increases with temperature [ 1 ]

separation, and thus p.e. remains constant or increase very slightly

[ 1 ]

(ii) BC: k.e. remains constant as temperature constant [ 1 ]

separation, and thus p.e increases greatly; as change of phase/state occurs

[ 1 ]

(iii) CD: k.e. increases with temperature [ 1 ]

p.e. separation remains constant [ 1 ]


5 | P a g e

(c) (i) n = m

M ; ρ =

m

V ; p =

nRT

V =

(m

V )RT

M [ 1 ]

= ρRT

M [ 1 ]

(ii) suitable test, i.e. ratio test, half-height, etc [ 2 ]

test carried out successfully [ 1 ]

(iii) p at 8 km = 3.5 ± 0.3 x 104 Pa [ 1 ]

p

po =

ρ

ρo

ρ = 0.35 x 1.3 = 0.455 (kg m-3

) [ 1 ]

(iv) p

ρT = constant

10

5

1.3 x 293 =

3 x 104

ρ x 250 [ 1 ]

ρ = 0.457 kg m-3

[ 1 ]

2010 TJC H2 P2 Q4

6 (a) (i) The increase in internal energy of a system is equal to the sum of the heat absorbed by the system minus

the work done by the system.

(b) (i)

atm

5

5

P=P +hρg

=1.0×10 + 14 1000 9.81

=2.4×10 Pa

(ii)

1 1 2 2

35 -2

2 5

-5 3

2

P V =P V

4(2.4×10 ) π 1.50×10

3V =

1.0×10

V =3.4×10 m

(iii) [1]

T = 298K

P/Pa

V/m3

1.4 x 10-5 3.4 x 10-5

2.4 x 105

1.0 x 105

[1]

[1] [1] [1]

1 m correct labeling 1 m correct line and arrow


6 | P a g e

(iv) At constant temperature, U = 0. Since volume of air bubble increases, work done by air bubble is

positive. From relation stated in (i), Q is positive. Hence heat is absorbed by the air bubble as its rises.

[2]

OR

At constant temperature, U = 0. Since volume of air bubble increases, work done on air bubble is

negative. From first law of thermodynamics, Q is positive. Hence heat is absorbed by the air bubble

as its rises.

2010 YJC H2 P3 Q2b

7bi)

7bii)

2010 SRJC H2 P3 Q7

8 (a) (i)

(ii)

(b)

(c) (i)

The change in volume/potential energy when ethanol vaporizes is much greater than the change in volume/potential energy during melting. OR There is additional work done against the atmospheric pressure during vaporization.

Q = thermal energy to raise temp to boiling point + thermal energy required to vaporize the ethanol

Q = mc∆ + m Lv

= ρ V (c ∆ + Lv) = (0.79 g cm-3) (1.0 cm3) [(2.4 Jg-1 K-1) (78-20 K) + 840 Jg-1] = 774 J


7 | P a g e

(ii) Describe and explain how the accuracy of c can be improved with the same setup, with accompanying

(d) (i)

(ii)

(iii)

2010 TPJC H2 P3 Q1

9(a) At B, pV = nRT

(4.3 x 106 x 1.2 x 10

-3) = 3.0 x 8.31 x TB [M1]

TB = 207 K


8 | P a g e

(b) At C, (2.2 x 106 x 3.8 x 10

-3) = 3.0 x 8.31 x TC

TC = 335 K [A1]

For monatomic ideal gas, internal energy U = total KE of all molecules as P.E = 0

U = 3nRT/2

Change in internal energy, U = 3nR T/2

From B to C, U = 3 x 3.0 x 8.31 x (335 – 207)/2

= 4790 J [A1]

(c) Work done from B to C = W = area under the graph from B to C

= 0.5(4.3 +2.2) x 106 x (3.8 – 1.2) x 10

-3 = 8450 J

(d) By the First Law of Thermodynamics,

U = q + W

4790 = q + (-8450)

Heat absorbed by gas in process BC, q = 1.32 x 104 J

2010 ACJC H2 P3 Q2

2010 CJC H2 P3 Q2

11 (a) The internal energy of a system is defined as the sum of the kinetic energy due to the random

motion of its molecules and potential energy due to intermolecular forces of attraction.

(b) (i) PV = nRT

(250 x 103)(7.5 x 10

-3) = n (8.31)(623)

n = 0.362 mol

10a An elastic collision is a collision in which the (total) kinetic energy of system is

conserved,

total linear momentum and total energy of the system are all conserved

b(i)

For an elastic collision with a stationary piston, the speed of the gas remains the

same when it bounce off in the opp direction

With a piston moving down, using the fact that relative speed of approach = rel

speed of separation, its rebound speed will be greater than that earlier

Since for ideal gas, T is dependent on ave KE of molecule alone,

Higher ave speed implies higher ave KE, hence Temperature rises

(ii) Apply ∆U=Q+W where

system

mentions that Q=0 and work done is done on gas(or W is +ve), hence higher

internal energy,

For ideal gas, Internal energy depends on temp alone

thus higher temperature

m

M

After collision v

m

M

Before Collision

u

V

v


9 | P a g e

no. of molecules = 0.362 x 6.02 x 1023

= = 2.2 x 1023

OR

PV = NkT

(250 x 103)(7.5 x 10

-3) = N (1.38 x 10

-23)(623)

N = 2.2 x 1023

(ii)

C

CC

A

AA

T

VP

T

VP

50

)V10(101

623

)10)(7.510 (250 C333

33C m101.49V

(iii) Net work done by gas = area of ΔABC

= 0.5 (7.5 x 10-3

– 1.490 x 10-3

)(250 – 101) x 103

= 450 J

(iv) Total change in internal energy is zero as there is no change in state of the gas or no change

in temperature.

2010 HCI H2 P3 Q2

12(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)

= 55 kJ

(b)(ii)

2010 IJC H2 P3 Q2

13 (a) (i) Internal energy of a system is the sum of all the microscopic kinetic and potential energies of the

atoms/molecules within the system. [B1] The kinetic energy is associated with the random motion of the

atoms/molecules while the potential energy is associated with the intermolecular forces between

molecules. [B1]

(ii) ΔU = Q + W [B1]

ΔU : increase in internal energy of the system

Q : thermal energy (heat) supplied to the system

W : work done on system [B1]

(b) (i) Since there is no change in temperature during boiling, the kinetic energy of the water molecules remains

constant [B1]. However, when the volume of the water expands during boiling, the separation between

water molecules increases, the potential energy of the atoms increases [B1]. Since internal energy of

water is the sum of potential and kinetic energy of the water molecules, there is a change (increase) in

internal energy.

Process W / kJ Q / kJ ΔU / kJ

A 201 0 201

B -55 74 19

C -185 0 -185

D 0 -35 -35

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


10 | P a g e

(ii) The volume occupied by the water molecules increases [B1] on vaporisation. Hence the water molecules

has to do work, pushing back the atmosphere [B1].

(iii) During boiling, the internal energy of the water molecules increases [M1] (ΔU: +ve). According to first

law of thermodynamics, the thermal energy supplied must be positive since work done on the boiling

water is negative [A1],

2010 JJC H2 P3 Q2

14 (a) Area enclosed by polygon

= 0.5(4+1)(4)(106)(10

3)(10

-6)

= 10000 J

net work done by gas = - 10000 J

(b) It is because the internal energy of an ideal gas is directly proportional to its

thermodynamic temperature and

in a complete cycle, the ideal gas returns to its original state; there is no change in its

temperature.

(c) 1 1

2 2

2

2

6 600

2

200 K

P T

P T

T

T

(d) Refrigerator.

From 1st law of thermodynamics,

0 ( )

U Q W

Q positive

Q negative

Heat is removed from the system.

2010 MI H2 P2 Q3

15(a)i) During melting, molecules move further apart from lattice

structure to clusters of molecules

Energy input is used to gain PE and not change KE, hence T

constant

B1

B1

(a)ii) During boiling, molecules escape from liquid and escape into

surrounding atmosphere.

Energy input required to gain PE and also to overcome

atmospheric pressure

B1

B1

(b)

V T

V1

V2

T

1

T2

0.00825

0.00424

T1

273

T1 531 K

M1

A1

(c)i) T 273.15 (Do not accept other values) B1


11 | P a g e

(ii) False. Need to measure on absolute scale, ie. thermodynamic

scale. In that case, difference is not 2 times.

B1

2010 NJC H2 P3 Q2

16 (a)(i) Show calculation to determine value of PV at C, B and one other point on the

curve BC. Conclude that since PV is constant for fixed mass of gas, the process BC is

isothermal.

Mass of ideal gas is fixed during the process.

(ii) Internal energy of an ideal gas is only made up of the sum of microscopic kinetic

energy due to random motion of its molecules

(b) Using Pc/Tc = PA/TA , TA = 192.5 K = 192.5 – 273.15 = - 80.7 oC

(c) Method 1

1224.12210)58.9(107.12

3)(

2

3 45 xxVVPU ABAB

Method 2

)5.192385)(31.8(2

3

2

3 nTnRU ABAB

But 053136.0)(31.8

c

cc

T

Vpn

Hence 1285.127 ABU

W is negative when the gas expands during the change AB.

At the same time, its internal energy has increased since its PV value increased during

the expansion.

Using the First Law of Thermodynamics, U = q + W, this implies that more thermal

energy must be supplied to the gas than its work done (i.e. q is positive and > W).

(d) Pressure of gas is due to collision of molecules with the container walls.

During contraction at constant temperature, the molecules move at and collide

with same speed. But due to contraction in volume, the frequency of

collision increases. Or more collisions per unit time increases.

This implies rate of change of momentum due to the collision of molecules

and hence pressure increases.

Note: A larger change in momentum from each collision (ie molecules move faster) or

higher frequency of collision (i.e. higher temperature or smaller volume) will

lead to larger pressure.

2010 CJC H2 P2 Q4

17 (a) (i) 4 minutes [1]

(ii) 1 Heater is switched off from t=0 to t = 4 min,

duration is 4 min,

mass of ice melted at 4 min = 0.011 kg (read off from graph at t=4 min)

mass of ice melted in 1 min

m1 = 0.011/ 4 min = 2.75 x 10-3

kg

[1]

2 Heater is swtiched on from t = 8 to t = 13 min

duration is 5 min,


12 | P a g e

mass of ice melted = 0.105 – 0.038 = 0.067 kg

mass of ice melted in 1 min

m2 = 0.067/ 5 = 1.34 x 10-2

kg

[1]

(b) Amount of ice melted in 1 minute due to heater alone

= 1.34 x 10-2

– 2.75 x 10-3

= 1.07 x 10-2

kg

Power of heater P = IV = 5.2 x 11.5 = 59.8 W

Amount of heat absorbed by ice in 1 minute

Q = 59.8 W x 60 s = 3588 J

Specific latent heat of fusion of ice

Q = mlf

lf = Q/ m

= 3588/ 1.07 x 10-2

= 3.37 x 105 J kg

-1

[3]

(c) The calculated value for the specific latent heat of fusion will be higher than the true value.

Ice taken from the freezer is below the melting point of ice. Part of the heater’s energy is

used to raise the ice’s temperature to its melting point before it starts melting and so more

heat has to be absorbed to melt the ice.

[2]

2010 MJC H2 P3 Q5

18 (a) The internal energy of a system is the sum of the potential energies and kinetic energies of

the molecules in the system B2

(b) (i) n is the number density or the number of molecules per unit volume B1

(ii)

A

Rk

N B1

(c) (i) Isothermal process/ Phase change. B1

Melting or boiling B1

(ii) Adiabatic process B1

Releasing gas from a balloon rapidly or pumping air into the bicycle’s tyre

B1

2010 AJC H2 P2 Q3 19 (a) PV = nRT

(c) Air conditioner or refrigerator.

END

2010 NYJC H2 P2 Q1 - PBworksjjphysics.pbworks.com/w/file/fetch/58627225/JJ H2 Physics Year 1...H1...

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