PHYS1131 Solutions Tut 5 11

8/3/2019 PHYS1131 Solutions Tut 5 11

1/13

PHYS1131 Higher Physics 1ASolutions Homework Problem Set 5

Q1.

Q2.

mtea = 0.520kg,mice = 0.520kg,Tice = 0oC, Tf =?,m =?

ctea = cw = 4186Jkg1K1, Lfus = 3.33 10

5Jkg1

The heat needed to melt the ice is given by mL = 0.5203.33105 = 173kJ

a)Heat lost by tea = Heat gained by iceQlostbytea = mteacw(Tf Ti) = 0.520 4186 (Tf 90.0)Qgainedbyice = mL + micecw(Tf Ti)Assume all the ice melts m = 0.520kgQgainedbyice = 0.520 3.33 10

5 + 0.520 4186 (Tf Ti:ice) 2177 (Tf 90) = 173160 + 2177 Tf

Tf =195930 173160

2 2177= 5.2oCall the ice is melted

b)Qlostbytea = 0.520 4186 (Tf 70.0)

the maximum value is when Tf = 0, Qlostbytea = 152kJThis is not enough to melt all the iceTf = 0.0

oC

mL = 152000m = 0.458So 0.520 0.458 = 62g of ice remain


2/13

Q3. Lf= 3.4.105J/kg, m =0.1kg

Q = Lf.m 3.4.104J, The internal energy of the gas is the same after the cycle

So, all work done goes in to melting the ice. i.e. W = Q

Q4. A = 1.8 m2, d = 1.0cm, Ts=330, T0=1.0

0C, k = 0.04 W/m.K.

a).

WAd

TTk

dt

dQ

Ad

Tk

dt

dQ

s230.

.

0=

=

=

As comparison, at rest the human body produces about 80W. The minimumamount of food needed is then 80 * 24hrs = 6900kJ = 1650 cals. Standard dietis 2300-3000 cals.Is 80W a lot or a little? 736W is one horse power. Running up stairs (80kg,

height change = 1ms-1) W

t

mgh800

1

1.10.80==

b). k = 0.60 W/m.K.

dt

dQ

dt

Qd

k

k1515 =

=

, i.e. heat flows 15 times faster.

Q5.Th > Tc

k1 k1

k2 k2

Th Tc

L L

k1 k2

k2 k1

TjTh Tc

L L

a) b)


3/13

Q6. See question sheet for p-V diagram.For path iaf: Q = 50J, W = -20JFor path ibf: Q = 36Ja). Q + W = E, E = 30J

JibfW 6)( =

b. (f i), Q + W = -E = -30 J as return to position iW = 13JQ = -43J


4/13

c) For Eint,i=10J, Eint.f= 10 + 30 = 40 Jd) ib: Q 6 = 22 10 = 12 Q = 18J

bf : Q = 18J since W = 0 along bf as V constant.

Q7.

Po = 1 atmA = 2.0 cm2

V = 0.30 cm/s

= 6 x 10-4 g cm-3

(a) We have steam = 6 x 10-4 g cm-3 constant

Hence rate of change of amount of steam m.

steam = V.

Where V.

= rate of change of volume

= Av v the speed of piston

m

.

steam = - Av

= -6 x 10-4 x 2 x 0.30 g/s

= -3.6 x 10-4 g/s

Hence this is rate of condensation steam to water.

(b) Rate of energy lost converting steam to water = Rate of loss of Heat

i.e. m.

LFusion = Q.

Q.

= 3.6 x 10-7 x 2.26 x 106 J/s

= 0.81 J/s

This answer is approximately correct. In fact the latent heat of fusion changes with pressure.

The pressure inside the piston is ~ 2 atm so L is ~2.20 x 106 J/s giving dQ/dt = 0.79 J/s

(c) From First Law =dE

dt=

dQ

dt+

dW

dt

Now dW = -PdV p consists of term from weight of piston plus atmosphere pressure

i.e P =mpistong

A

+ Po anddV

dt

= Av

sodW

dt= mpistong + poA( )v = -(2.0 x 9.8 + 1.013 x105 x 2 x 10-4) (-0.003) J/s

= + 0.120 J/s

sodE

dt=

dQ

dt+

dW

dt= -0.814 + 0.120 = - 0.69 J/s

PAST EXAM QUESTION

(a) External pressure on air in container ismg

A+ pA

within cylinder pV = nRT = constant we allow

temperature to equilibrate at T = Toso pV = const = pAVo

APo

v

m

A

pA

To

m

ho


5/13

Hence p =pAVo

V=

pAVo

Aho

Both p =mg

A+ pA Air = 0.029 g/mol

pA

VoAho=

mg

A+ pA

pA

VoAho=

mg+ pA

A

A

ho =pAVo

mg+ pAAi.e. no dependence on or To!

b)

Hence the system is stiffer for a rapid [adiabatic] than a slower (isothermal) change. To be

expected as a slower change will remain in thermal equilibrium, and heat will flow as it

occurs.


6/13

Q8

(a) The displacement at t = 2 seconds is obtained by simply substituting t = 2 into the

given equation, that is:

x = 6.0cos(3 2+

3

)

x = 6.0cos(19

3)

x = 6.0 1

2

So x = 3m

(b) To calculate the velocity, we must first differentiate x with respect to time:

dx

dt= v =

18

sin(3t+

3)ms-1

Then we substitute t = 2 into the equation to give v = 9 3 ms-1

(c) Acceleration is calculated in a similar fashion, this time by differentiating the velocity

equation:

d2x

dt2 =

a = 542cos(3t+

3) ms

-2

And substituting t = 2 again gives a= 272

ms-2

(d) Since the wave equation is already given in the form:

x = Acos(t+ ) metres

We can simply read off the phase of the wave. So =

3

is the phase constant.

(e) The frequency v is related to the angular frequency by the equation:

=

2

Hz

From part (d) we know that = 3 , so v = 1.5Hz.

(f) The period of motion is given by T=

1

f . We calculated the frequency in part (e), so the

period is simply the reciprocal of this, ie. T =2

3seconds.


7/13

Q 9

Q10

(a) The piston moves with simple harmonic motion, therefore:x = Acost (displacement)dx

dt= Asint (velocity)

d2x

dt2=

2A cost

d2x

dt2=

2x

(acceleration)

The block and piston will separate when the acceleration of the piston is greater than

the acceleration of the block, ie. when a = g.

g=2x

9.8 =2

1.0

2

x (using T =2

)

Therefore x = 0.25m is the amplitude

(b) We use the same equation as previously, but this time solve for :g=

2x

=

g

x

(rearranging the equation)

=

9.8

510

2

x

y

t =0, x=2


8/13

This gives = 14 rad s-1. Converting to seconds by using the equation f =

2

gives

f= 2.2Hz

Q11

Note: Tension is uniform along the spring.

x = x1+ x

2

F= kx

= kx1 kx

2

We also know that:

F= k1x1

and F = -k2x2

Therefore, setting both values for F equal to each other yields:

k1x1 = kx1 kx2

k1 =k ( x1+ x2 )

x1

=

k x1 +x1

n

x1

= k 1+1

n

= kn +1

n

And doing similarly for k2 gives:

k2x2 = kx

k2 =k(x1 + x2)

x2

=

knx2 + x2

x1

x1

= k(1+ n)

1 2

x

= n2


9/13

Q12

The total energy is

E =1

2kA

2

When x = A/3, The potential energy is:

2

2

2

18

1

)3

(2

1

2

1

kA

Ak

kxPE

=

=

=

And PE + KE = E, so

EkAKE

kAKEkA

9

8

9

4

2

1

18

1

2

22

==

=+

Q13

By conservation of momentum, the momentum of the bullet before it strikes the block will be

equal to the momentum of the combined bullet-block system after the strike. So:

mv= (M+m)dx

dt

For SHM,

x = Acost

dxdt

= Asint

Substituting this into the original conservation of momentum equation gives us:

A sint (M+m )= mv

The maximum amplitude for this system will occur when sint=1, thus:

A =mv

(m + M)and using =

k

m + M

,

Mk

m

v


10/13

A =

m + M

k

mv

m + M

=

mv

k(m +M )

Alternative: Momentum mV = (m + M)u

Energy conservation1

2kxmax

2=

1

2(M +m)u

2and substitute u to yield the same result.


11/13

Q14.

(a) Energy stored in spring is given by:E

spring

=

kx2

2

=

(294)(0.239)2

2

= 8.40J

=U

the kinetic energy is:

K= 1

2I

2

rotation

+ 1

2Mv

2

translation

For a cylinder rolling without slipping, v = r, this gives

=

1

2

MR2

2

v

R

2

+

Mv2

2

=

Mv2

4+

Mv2

2

=

3Mv2

4

The proportion of total energy used as translational energy is given by:

Translational EnergyTotal Energy

=

Mv2

23Mv

2

4

=

2

3

Therefore the total translational energy is 2/3 x 8.40 = 5.60J.

(b) The proportion of total energy that is rotational kinetic energy is given byRotational Kinetic Energy

Total Energy=

Mv2

4

Mv2

4+

Mv2

2

= 13

Therefore the rotational energy is 1/3 x 8.40 = 2.80J

(c) When extension isx then, applying N2L

M x

..

= - kx - Ffr

where FfrFrictional ForceFor the angular acceleration, , we have


12/13

Torque

= I

i.e. FfrR =

1

2MR

2

But =d

dt

=1

R

dv

dt

=x

..

R

So Ffr

=1

2M x

..

,

i.e. M x..

= kxM x

..

2

Hence x..

= 2x with 2 =

2k

3M

Thus SHM with period T =2

= 2

3M

2k

Q16

The system shown undergoes SHM in

a vertical direction. Find the equation

of motion and the frequency for the

system.

Solution:

l1

l2

T1

T2

x

m

Vertical SHMk1

k2


13/13

Q16

(a) Standard form is;

y = ym sin2(x

ft)

so that

ym=

0.1mf =1.0Hz

= 2.00m

= f=1.0Hz*2.00m = 2ms1

(b)

)00.2cos(2.0

)00.2cos(1.0*00.2

tx

txdx

dyy

=

==

maximum when cos=11

63.02.0

== msy

Q17

Linear mass density 103.00.2

060.0

== kgmm

kg

Wave velocity =F

=

500N

0.03=130ms

1

2m

F

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PHYS1131 Solutions Tut 5 11