Weight - University of Sydneysydney.edu.au/science/physics/pdfs/current/jphys/exams/...93/11(solutions) Semester 1, 2007 Page 1 of 14 PHYS1901 Advanced Exam Solutions Semester 1, 2007

93/11(solutions) Semester 1, 2007 Page 1 of 14

PHYS1901 Advanced Exam Solutions Semester 1, 2007

SECTION A

Question 1 (common question F4, R1)

a)

(Drawing – 1 mark) No, they are not an action-reaction pair since

(i) they act on the same object (ii) different kinds (weight is gravitational while normal is electromagnetic)

(1 mark)

b) When Jack is simply standing on the table, the normal reaction force from the table need

only be strong enough to balance his weight. When he jumps, Jack is moving and so has momentum mv . The table must apply a

sufficient force over a time to balance his weight and provide enough impulse to bring Jack to rest. Through Newton’s third law, Jack applies the same force on the table. As the interaction time will be short, the required force can be large, overcoming the structural forces in the table and breaking it.

(2 marks) c) While the required impulse is the same, the mattress extends the time of the interaction and

hence reduces the required force. This reduces the force on the table and it does not break.

(1 mark)

(Total 5 marks)

Normal force

Weight


Question 2

(a) Work done W = F.5

xdx

(1 mark)

F =x 2

so W =x 25

xdx =

x

5

x

=5 x

(1 mark) (b) Using the Work – KE Theorem

W =1

2mv f

2 1

2mv1

2

(1 mark) closest is when Vf = 0

(1 mark)

5 x

=

1

2mvi

2

Substituting in the various values x = 2.82 10 10 m

(1 mark)

(Total 5 marks)


Question 3 a.) Use Newton’s second law and take upwards to be positive:

Hookes law gives F = - kx where x is the extension of the spring beyond its natural length Ignoring the signs which just indicate direction,

total length = 2L +3mg

k+

3ma

k

(1 mark) b) Once the rocket is turned off, a = -g total length = 2L

(1 mark) Rocket , astronaut and weights are now in “free fall”, although still going up. No force is required from the spring, so stretch = 0.

(1 mark)

(Total 5 marks)

For Mass A1 F1 F2 mg= ma Using expression for F2

F1 =2m g + a( )

so, x1 =2m

kg + a( )

For Mass B1 F2 mg= ma F2 = m g + a( )

x2 =m

kg + a( )

(1 mark)

(1 mark)

mg

F1

A

F2 mg

F2

B

A

B


Question 4 (Common question R4) a.) At freezing point. Since B shows the ice rising to the freezing point but not going above it,

as would happen if all the ice melted. (1 mark)

b.) No liquid freezes. Since liquid never drops below the freezing point.

(2 marks)

c.) Ice partly melts. In the interval when the ice is at 0 °C and the water is above 0 °C, energy transfers from ice to water, partly melting the ice.

(2 marks)

(Answer and reason required in each case)

(Total 5 marks)


Question 5 (Common Question R6)

a.) fL = fs

v + vL

v + vs

21,000 = 20,000 330 + 0

330 + vs

330 + vs

330 =

20

21

1+vs

330 =

20

21

vs = 20

211 330

= 15.7m.s1

(Correct use of equation - 2 marks)

(Answer – 1 mark) b.) The wall acts as a source with frequency 21 kHz. The bat hears this Doppler shifted to higher frequencies by another 1 kHz because of its

own motion, so f = 22 kHz. or

f2 = fs

v + v2

v + vs

= 21,000330 +15.7

330 + 0

= 22kHz

or Another way to think of this: there is a virtual image of the bat on the far side of the wall.

The bat and its image are flying towards each other, and their relative speed is twice the airspeed of the bat. Hence, the bat hears twice the frequency shift that the bird hears.

(Answer – 1 mark)

(Correct argument - 1 mark)

(Total 5 marks)


Question 6 a) It has four degrees of freedom:

i) The position of string relative to the pulleys, measured via the height of M or the

distance of m from the pulley (but not both, since the string does not stretch)

ii) The angular position of m iii) The linear velocity of the string, measured via one of the following: string relative to

pulleys, velocity of M, etc.

iv) The angular velocity of m Note: 1 & 2 could be replaced by the two-dimensional position of m, while 3 & 4 could be replaced by the two-dimensional velocity of m.

(1 mark for having 4 degrees of freedom and 0.5 for defining each one - 3 marks) b) The nonlinearity arises because sin appears in the equation of motion where is the angle

of the swinging mass from the vertical. For large swing amplitudes, sin cannot be approximated by .

(1 mark for sin and 1 for the need to have large swing amplitudes – 2 marks)

(Total 5 marks)


Question 7 (Common question R8) a) P.E. = mg h = 5.00 9.80 3.00 = 147 J

(2 marks)

b) Apply conservation of mechanical energy to path of the mass since the tension from the chain acting on the mass is always at right angles to the path and therefore does no work on the mass.

(1 mark) PE + KE = 0

mgh +1

2mv2 = 0

v = 2gh mv = m 2gh

= 5.00 2 9.8 3.00( )1

2

= 38.4 kg.m.s 1

(2 marks) c) Angular momentum = mvl

at lowest position = 38.4 8.00 L = 307 kg.m 2.s 1

(2 marks) d.) After breakage the horizontal motion continues unchanged, the vertical motion accelerates

under gravity. The path is a parabola.

(2 marks) e.) There is no change to the angular-momentum in the instant after the chain breaks. Circular

or linear motion are not distinguished L = 307 kg.m 2.s 1 Later, gravity causes a change in the result.

(1 mark)

(Total 10 marks)


Question 8 (a)

Set up x-y coordinates as shown. Coordinates of CM: yCM = 0 by symmetry

xCM =

M

30 +

2M

3d

M

3+

2M

3

=2d

3

( 2 marks) (b) Work done by gravity = KE( ) = GPE( )

=1

2M 2

+1

2 12

+1

2 22

(1 mark)

Now = R

and 1 =1

2M1 R2

=1

2

M

3R2

and 2 =1

2

2M

3R2

Same radius same for both rollers

Moments of inertia

R

CM

F1 N1

N2 F2

Mg

x y M 2

M1

2M

3


So Work done =1

2M 2

+1

22 1

6MR2

+1

3MR2

=1

2M 2

+1

4M 2 , since = R

=3

4M 2

But Work done =M

3+

2M

3gh

= 4gh 3

(Valid derivation 2 marks) (c) a = acceleration in x direction

Use s (down the hill) = h

sin

(negative x direction) and v2=u2

+2as

4gh

3= 0 + 2a

h

sin

a2g

3sin

or, for a bit more effort:

a =d

dt

=d

dt

4g

3h

1

2

but h = xsin if = angle of slope of hill

a =4g

3

d

dtx sin

=4gsin

3

1

2 x

dx

dt

Usedx

dt= = 4gh 3

then a =g sin

3

1

h sin2

gh

3

=2g

3sin

- ve sign downhill

(2 marks)


(d) Consider y components of forces. N1 + N2 Mg cos = 0 (no acceleration in the y direction) Consider x components:

F1 + F2 Mg sin = 0

F1 + F2 = Mg sin

At point of slippage. (say = max) .

F1 = μsN1 and F2 = μsN2

μs N1 + N2( ) = Mg sin max

and from above N1 + N2 = Mg cos max Hence

tan max = μs

max = tan 1 μs

27

(working - 2 marks)

(result – 1 mark)

(Total 10 marks)

Coefficient of static friction


Question 9 a)

T =H

kA=

0.03m 280W

0.2W .m-1.K 1 2.0 m2

= 21K

(2 marks) b) The actual temperature differential is only a few degrees. Therefore the temperature

gradient in most of the body is not sufficient to generate enough heat flow.

Blood flow close to the surface of some parts of the body allows larger heat flow at those location, but still far short of what is required.

(2 marks)

c)

Hout = e AT24

= 1( ) 5.67 10 8 W.m-2.K-4( ) 2.0m2( ) 307K( )4

= 1007W

Hin = e AT14

= 1( ) 5.67 10 8 W.m-2.K-4( ) 2.0m2( ) 296K( )4

= 871W

(2 marks) Net rate of heat transfer = 1007 – 871 = 136W

(1 mark) d) Based on these results, conduction is not the dominant cooling mechanism

• Radiation is a significant cooling mechanism, although not sufficient by itself Other factors

• Perspiration can be significant (although < radiation), especially with a fan to move the air over the skin

(1 mark of each valid point – maximum 3 marks)

(Total 10 marks)

[Reference: hyperphysics.phy_astr.gsu.edu/hbase/thermo/coobod.html]


Question 10 (Common Question R10) a) i)

e = 1QC

QH

= 0.20QC

QH

= 0.80

(1 mark) Reduce heat loss by half

QC

QH

= 0.4 e = 0.6

(1 mark)

(Alternatively: 20% of energy goes into doing work on the wheels. It is half of this which may be recovered. Any reasonable estimate of energy lost in braking followed by decent working was given full marks.)

ii) This requires making QC = 0 From the second law of thermodynamics, it is impossible for any process to convert heat QH( ) completely into mechanical work. (Otherwise it would be possible to

construct a perpetual motion machine, with e = 1, or 100% efficiency).

(2 marks) b) i.) T varies for the water, so

Swater = dQ

T=

mcwaterdT

T1

2

1

2= mcwaterln

T2

T1

= 0.250 4190 ln 338

293

= 150 J.K 1

(2 marks)

ii.) T in constant for the heating element, so

Selement = Q

T=

mcwater T

T=

0.250 4190 45

393= 120 J.K-1

(2 marks) iii.) Stotal = Swater + Selement = 30 J.K 1

(1 mark) iv.) This process is irreversible because Stot > 0.

(1 mark)

(Total 10 marks)


Question 11 Assume pendulum mass is m and falling weight mass is M.

a) T = 2l

g = 1.004 s 1.0 s (2 marks)

b) Change in height = L * (1 - cos (15°) ) = 0.00852m energy = mgh = 0.5 * 9.8 * 0.00852 = 4.2 10 2 J (3 marks) c) Energy lost = Mgh = 0.300 * 9.8 * 2.0 = 5.88 J 5.9 J

(2 marks) d) In one second the energy lost it Mgh/ (24*3600) = 6.8 10 5 J Assuming the energy loss is constant, the time to expend the available energy 4.17 10 2 J / 6.8 10 5 J = 613 s

(3 marks) So it will keep swinging for about 600 seconds = 10 minutes (in fact it will be more, since the amplitude will decay exponentially)

(Total 10 marks)


Question 12

a) =F

μ=

20 N

50 10 3kg.m-1= 20ms 1

T =2

=2 3.14 rad

31.4 rad. s 1= 0.20 s

=f

= 2 = 4.0 m

(3 marks) b) y(x,t) = Asin(kx t)

= ym sin20

x t

(1 mark) c) y x,t( ) = 0.005( )sin 1.57x 31.4 t( )

d) (2 marks)

(2 marks)

e) y(x,t) = (0.005)sin[(1.57)x (31.4)t]

v =dy(x,t)

dt= (0.157)sin[(1.57)x (31.4)t]

vmax = 0.157m.s 1

vwave

vtransverse

=20 m.s 1

0.157m.s 1

= 127.4

130

(2 marks)

(Total 10 marks)

Weight - University of Sydneysydney.edu.au/science/physics/pdfs/current/jphys/exams/...93/11(solutions) Semester 1, 2007 Page 1 of 14 PHYS1901 Advanced Exam Solutions Semester 1, 2007

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