General Physics I (PHYS1414)
Supplementary Examples
with
Selected Answers
¾ ¿ Example (September 20, 2012)
A block of mass m slides down on the frictionless inclined surface of a wedge which is
fixed on a scale. If the mass of the wedge is also m, find the weight of the system
recorded by the scale.
m
Fixed
wedge
m
Slides down
Frictionless
surface

Scale
A.
mg + mg cos 
B.
mg + mg cos  sin 
C.
mg + mg cos2
D.
mg + mg sin2
E.
2 mg
Answer:
N sin

N
N
N cos
Frictionless
surface
g sin
mg
N
mg
The block slides down
along the inclined
R 
f
Fixed wedge
Answer:
Since the wedge is fixed on the scale, the block slides down the wedge with an
acceleration g sin . On the other hand, we notice that there is no acceleration in the
direction perpendicular to the inclined plane, hence the normal reaction, N given by the
wedge to the block is mg cos . The same force acts on the wedge by the block. As the
scale measures the reaction force R which is given by mg + N cos, that is
R  mg  mg cos 2  .
! " # " $ # $ % & ' ' $ ' # $
$ # & $ ¼
¼
¼
( ¾ ) * ¾ + # , - !$ # & $ ( ¾
)
¿ ¾
*
. ¿ + # ¼
t,
tldhod L.
Qtes
eyvl ftru(
v'ws
4
(o(L
ILre
is
.fi1t-f
r
{rA
e-
?
v
I
I
4v
Ma
th'l
e-L
*f
{L
??
l-' 3h
4 2lrL ? ,X I
(i^Q.
a
e
)
o
?
[v
-7
_2.1 v
(^=
I
tIv
v
---
v
l-i2
x
v
tso
asJ
t#
(i--
&r
{+
dT
-{,+
-v'
-{)
)
tv
vF;t
e)
2.
,-'= l!'
yrl + vci(f
--
{'rt
ru,ft
Wo
+h"t
i*f
qnr(
;
i
I
L:,
[r*.
t wr,r ))
+
v*^f
*
vtf)
."r'/
u'- +'xr"
t
d3
a.'=
h-o?e
0=
i=
t
+
/tt't
(
x
ot'f
-(t
0."
le;S
v'fai
I.-wst
v tn^'4
i T
l, lrt'f)
q r-_fi
f
.
LJ,tf t tu' f)L
L*f I - v t-f
V si-{
(t*
(t'#=
-r)
Example (September 20, 2012)
A man traveling East at 8 km h1 finds that the wind seems to blow directly from the
North. On doubling his speed he finds that it appears to come NE. Find the velocity of the
wind.
Answer:
Let the velocity of the wind be

w  x iˆ  y ˆj .
 8 iˆ
Then the velocity of the wind relative to the man is

w  8 iˆ  ( x  8) iˆ  y ˆj
Parallel to
 ĵ

w
But this is from the North, and is therefore parallel to  ĵ . Hence, we obtain x  8 = 0.
 16 iˆ
When the man doubles his speed, the velocity of
the wind relative to him is

w  16 iˆ  ( x  16) iˆ  y ˆj .
But this is from the NE and is therefore parallel to

w
 ( iˆ  ˆj ) . Hence y = x  16 = 8.
The velocity of the wind is 8 iˆ  8 ˆj , which is equivalent to 8 2 km h1 from NW.
Parallel to
 ( iˆ  ˆj )
Example (September 20, 2012)
A swimmer wishes to across a swift, straight river of width d. The speed of the swimmer
in still water is u and that of the water is v, where v > u. If the swimmer wants to cross the
swift with a minimum time, find the minimum time and the position that he reaches in the
opposite bank.
Answer:
s
A
B

ĵ
d
C

u

u

O

v
Q
iˆ

The minimum time of travel can be obtained if the swimmer directs in the direction of j .

The magnitude of u is contributed completely in the direction of crossing the swift. It is
the fastest way. Hence the minimum time t = d/u. The distance downstream s = vt =
(vd)/u.
Example (Challenging)
As an extension of the last example, keeping all information above, what is the direction
along which the swimmer should proceed such that the downstream distance he has
traveled when he reaches the opposite bank is the smallest possible? What is the
corresponding time required?
Answer:
s
A
B

ĵ
d
C

u

u


O

v
iˆ
Q
The downstream distance is smallest when OB is tangent to the circle, centered Q, radius
u.
u
d 1  ( )2
d cos
u
d v2  u2
v
sin   , and AB  d cot  
.


u
v
u
sin 
v
The time taken is t 
d

u cos
d
u
u 1  ( )2
v

dv
u v2  u 2
.
-)
F*t
J-
,)
F
+
4
ar>
ts+
'gtt
r hst
:
J
t
Fs's
t*,
(
-l
d
-)
Y:'
lrt
Fs'st
--
YtntS''
tJ
tvr(
' tqt
3
bf
h tss's + hlr ,a^t'
[.
lr€
[-.
5s
\{e lu're
4
\
t
*dF sln
(-'. d
Sis
$ro .
:
tbspv.rg
thr
x#fir'
)cree
?ff-"
( :-g- {t.9.- s'a^€ Nonittt^t 2-4
hnfi[io,,.,
.)
vrrs
inrufio,l 4t
^r,e
)nf nl' ,
h tsn^,
a^d
I
S
J
d
)
'wi*l^'
d
c,.'.sfa^(,
Yr-
S't m.vitt
-a
-i
Je
-l
5r
^s'
t
l
)
J
Example (Challenging, September 27, 2012)
A wedge with mass M rests on a frictionless horizontal tabletop. A block with mass m is
placed on the wedge. There is no friction between the block and the wedge. The system is
released from rest. Calculate 1) the acceleration of the wedge, 2) the horizontal and
vertical components of the acceleration of the block, check the limit when M  .
m
M

Answer:
Before doing this problem, we look at two remarks first.

If M is not moving then we have the trajectory
Trajectory
sketched in the right.

If M is moving to the left, then the actual trajectory is sketched as below, where
aM is the acceleration of the block relative to the wedge of mass M.
a1
a
aM

Actual trajectory
Now, we look at the force diagram of the wedge and the block.
For M:
N 2 sin   Ma1
(1)
For m:
mg  N 2 cos  ma M sin 
(2)
N 2 sin   m(a M cos  a1 )
(3)
N1
a1
N2


N2
a

mg
Mg
This is a set of simultaneous equations with 3 unknowns. After solving, we obtain
mg sin  cos

a1  M  m sin 2 

a  ( M  m) g sin 
 M
M  m sin 2 
(*)
Since the horizontal and vertical components of a are given by
a x  a M cos  a1 and a y  a M sin  ,
Mg sin  cos

a x  M  m sin 2 
hence we have 
2
a  ( M  m) g sin 
 y
M  m sin 2 
The reaction between the wedge and the block is given by the expression of a1 and
equation (1), i.e.
N2 
mMg cos
.
M  m sin 2 
Consider the equation set (*), when M is very large, e.g. M   , we have
a1  0 and a M  g sin  .
Æ Ì Ì ½ ½ ! ½
! " % " # "# $
" # "# $
#
½
& ½ ! " # ! $ ! % & % & % &
! ¼ " ! ¼ # ¼ ! $ !% & ! & ! &
! & ' ! Example (October 8, 2012)
Three particles A, B and C, of masses 4, 6 and 8 kg, respectively, lie at rest on a smooth
horizontal table. They are connected by taut light inextensible strings AB and BC and
 ABC  120 o . An impulse I is applied to C. If the magnitude of I is 88 Ns and it acts in
the direction BC, find the initial speeds of A, B and C.
Answer:
B
6 kg
I2
u
I1
u
A
4 kg
I2
120o
I1
V
v
C
8 kg
I = 88 Ns
Let I1 and I2 be the impulsive tensions in the strings BC and AB respectively. Since A is
acted on only by I2, its initial speed u will be in the direction of AB. Since the string AB is
taut, B must have a speed u in the direction AB. Also, let B have a speed v in a direction
perpendicular to AB [this is necessary because I1 acts in a different direction to I2].
Finally, since both I and I1, the impulses acting on C, have the same direction BC, let C
have an initial speed V in the direction BC.
Since BC is taut, the speeds of B and C in the direction BC are equal.

u cos 60  v cos 30  V

u / 2 v 3 / 2  V.
(i )
Considering the motion of A,
4u = I2.
(ii)
Considering the motion of B along and perpendicular to AB,
6u  I 1 cos 60  I 2 .

6u  I 1 / 2  I 2
(iii )
6v  I 1 cos 30

6v  I 1 3 / 2.
(iv)
Considering the motion of C,
8V = 88  I1.
(v)
Eliminating I2 from equations (ii) and (iii)
6u  I 1 / 2  4u

I 1  20u.
(a)
Substituting from equation (a) in equation (iv),
6v  20u 3 / 2
5
v
u
3

(b)
Substituting from equation (b) in equation (i),
u
5
3


u V
2
3 2
V  3u
(c )
Substituting for V and I1 [equations (a) and (c)] in equation (v),

8  3u  88  20u
44u  88

u  2.
From (b)
v  10 / 3 .
From (c)
V  6.

Speed of A is 2 m/s,
Speed of B is
u2  v2  4 
4
Speed of C is 6 m/s.
100
3
7
m/s.
3
Example (October 11, 2012)
Four particles each of mass m are connected by inextensible strings. They are in the form
of square as shown in the figure. If the particle A is acted by an impulse P, find the
velocities of A, B, C, and D immediately after the impulse.
B
P
C
A
D
Answer:
v3
B
v2
T2
v1
C
A
P
T1
T2
v2
D
v3
Let v1 be the velocity of A, v2 and v3 be the components of the velocities of B and D, v4 be
the velocity of C after the impulse P is applied. Let T1 and T2 be the impulsive tensions.
Impulse consideration:
For particle A: P  2T1 cos 45  mv1

P  2T1  mv1
For particle B: T1  mv 2
(1)
(2)
and
T2  mv3
(3)
For particle C: 2T2 cos 45  mv 4
2T2  mv4
and
(4)
Velocity consideration:
Along BA:

Along CB:

v1 cos 45  v 2
v1  2v 2
(5)
v3  v 4 cos 45
2v 3  v 4
(6)
Substitute (2) into (1) and substitute (3) into (4):
P  2mv2  mv1
and
(7)
 2mv3  mv 4
 2v 3  v 4
(8)
After solving (5), (6), (7) and (8) for the velocities, we get
v1 
P
,
2m
v2 
P
2 2m
The initial velocity of particle A is v1 
v2 
P
2 2m
v3  v 4  0 .
,
P
. The initial velocities of B and D are
2m
. The velocity of particle C is zero.
Example
A gun of mass M fires a shell of mass m and recoils horizontally. If the shell travels with
speed v relative to the barrel, find the speed with which the barrel begins to recoil if
(a)
the barrel is horizontal,
(b)
the barrel is inclined at an angle  to the horizontal.
Answer:
(a)
V
M
v
m
Let the barrel be recoiling with speed V. The speed of the shell as it leaves the
barrel is v  V. Before firing the shell, the gun is at rest and the total momentum is
zero. By the conservation of momentum,
m (v  V)  MV = 0.
Hence V = mv / (M + m).
(b)
When the gun is inclined at angle  . The shell leaves the barrel with a velocity
which is the resultant of two components, v and V. By the conservation of
momentum in the direction of recoil,
m (v cos   V)  MV = 0.
m
Hence V = mv cos  / (M + m).
V
M

v
Example (Challenging)
A particle of mass m slides down the smooth inclined face of a wedge of mass 2m, and
inclination , which is free to move in a smooth horizontal table. Use equations of
momentum and energy to obtain an expression for
m
the velocity of the particle relative to the wedge
when the particle has moved a relative distance s
2m

from rest down the inclined face of the wedge.
Answer:
Let the velocity of the wedge be V and that of the particle relative to the wedge be u. By
conservation of horizontal momentum, since there is no horizontal force acting on the
system,
2mV  m (u cos  V) = 0.

1
V  u cos  .
3
By conservation of energy,
mgs sin  

1
1
(2m)V 2  m (u cos   V ) 2  (u sin  ) 2
2
2
1
2
1
 gs sin   u 2 cos 2  u 2 cos 2   u 2 sin 2  ,
9
9
2
1
 6 gs sin   2
u
 .
2
 2  sin  

V
m
V
u

2m
! "
% ! ¼ ¾
! #
! $ ! & % ' ( ! ) ! % % 2.
n6ls
ca)
^{ tdl
+.r*
d!*ra+t
ci."
mfu^*4 r%
sw"ll
v--
J Yw)o d^
d-
f,
"r..'
I
3
f"
Jo--o
cot
Y
t [o ra
Y-so,,t
0
"
"Lr
)
rt..dlt"
S-2n r
|\'3
I
e,^O
rd.V
l'^S
o
dz?
J
0
e;o f, s.^e
['
)u
tT:-
f-
t z
5,4,
,f
OI'
o
Y
Y=
)
t\
JL<.c
WL
z.
ut--o
f
Cd",-
=1o
r4rr**tv2)
J
kftt
fx,y) = (r
-.?,,
o)
v
,
el
r'u'7'-
(t)
(L{r,.,
"frl cr[r,,.0,t
e
?\ >nx-dx
=
[.r-t( -, f1
(
,tl*
widlq d x
1t ftu
I + ,Iun
{:
-,3
;r\
>\
r
d^
v"lr^. fu^i&t,I
r.v.,i
'i[u'i'''"-l
-'l
4
?
('x
)->
=\I:
=
i
q.a
z3 l- l)
=T
)1 \
'-\
X
-
rr
->t)
-:-.r
z-v
I d dx
I
t
0
q
-)i-
7I
-\
+
z/Y
3
-*)
I
t:.^",-
9 =o
)-_
v
C
J-{
I
(X
\/)
tn) :
(.\
\
u,
o'*
,t
{--
>y
^*.+ry\ t,.t i^r."'r
sk"t\
¾ ! " ! #
$ % &
& &
&
! &
& & & " & ! ' & ( &
& ) ! pir fra*.
I
Thg ot^n^ l4novcS f,nnn tl^rdy
X*'\
T
ilb,
?-
( dr.gp.d
v3 "xY
lryh'{
A
while
[zsfi','tce
h S11d
*'"kes
At
?,'*t
Ty
r@
raevi.rS cb,era
-- -( E! v
{,+
F ? -(v'
S"*.
Vl,tnc
-$= -T
r wc
(f )
+
Whc,r^
+
= ( r' -- Z(3y
hove
uUt+
Rs 3(XJ *
R=
$tuc
6t.
$* F "tI oh tl^t
:". nir, +b r+1 it
F
L
x)
t +tt^J rh +\* fin.- -Tbrl
:^tl,rrr.f*
Tr^r
ay
'J
n
@o
) (3
t
-3^I
P.-3Cf"L,
U=
LC€J
i
t2r,
l7*r-rtol -(a
tvtl
$,nn
a?
R
-e
t =r
LI
T--t-
hf")'(uP lJ
LLii - tuvi MaLL
i
jtre
WLrrr
ftt
R'-
(:-.. U' l- )
,
LtrV1.
3LU
&r"tl ir
*lro(a
Mrry
^j*
L i', n2
l/r totte )4 - ( = M - ly r-
a
4t +i.rt t
nt t=o
j*t
f, svll
c^
*
fl* t"tb
.)
)
nU?
&t,
fcI
o*t
i*,#
fr+ t-- o
k',M rt")
| +ve
-Q +
c)
(T?
-R+
(ru.
!'
lrvt
dhr,^
'- n^(o) - rJ-o) (i
*iel
=
u,
R ?- <I1-t (i'
haxe
+hL
"rT
(z cxt*
*"nI
Ju.q
4vt, ? 3c17
j-4 ,f"(c,n V2 ,tr" !'
pz g&1 q, gt'19
(
v
At f,"ra f
(.51
c M. il('trd
J
+ rY 4t (L-v)
hI
=
f,n
gt
7"nn
!-qt r-tL!)
:
(L'y
rA
?-
C
L
1r
+
I'y
vl0r,4 -
tJ
frt t-- o
At fi^af
- L'rJ
a
0
€)
)
r
I
i
\
:
3.
l*t
t
I""
ri)
ii-t[t'*
"l=I,, ='tY
_
$r'z
Y
tr@tt
x-
aa
ir=
lc*^
.
Yrtl'
F
Y"n
=
x-
1O
YoA'-
T
(l- -tL)
v,
(o^--
9,nn
(;i^
(lt-R?
R -- lL*R=
{-
li"'
({F sr-')']
eu
R'- )(.t ?
{
W[,r^
T=
(
(t)
R=
t
3t7L = 3l4qtr#
Z,r
ly
(v
= voo oo.rfar,f )
,
/{"1u'{
,, ('"X ?e
ftnet
*fP- Efri*r
t ,tve
H
(- rw ?
('hI
fitqt)
? nfi+ Ai"
l"
t ,^ct*t z)y
t
a3
(c
=)
76av?!
.,.itl,q sgeed Vo
.i
Vo
Thr claai^
(v, = (.,or^tt*tf)
?o
Fi"d F
LW+ )r"
l)
(:
,*
^QJ+Y
.
fuo ftve
V=
Fub
v
tol
Vo
F-ry--(-IAt
o"2Vo
4v
.
=i.\
Ut',
Tt
!,
:vv.tQ
dta-
-
Vg
.C
G,
lt. Lcc,.*r9 F- trII--
2i
&t
--
7ro
r"'(o) -- C-vJ h'o
+
F = A'1l.t 2v7
4
e -- 2( ll
t V,'),.
+
(0,=
-*tta"(
)
f +ve
qte' (?
--*
-_
*(ay))
- [rt
\
(-
^x1
4
e- aYf
(-
a T
^W
-e LrJtr
-/
ai
9.^a
3=
A
u
j-l/t
}.srvt
{'t
('
?
+ L3 Tu c-.ki*uaov.er
uf w ft c'o"rta"^t
r'
1 -^t
rlSJ* 1"3
C^raQzr,q'rl4 4.
F,*dF.
o- ,r^ff*t
^1
x,JIt 71nt l'n1
=)
??
VY1-
4
(,
L,1
+ 3tnT
Qt 3") *
! " ! ! #
!
" " # ¼ ! " # ¼ ! ! ! $ # "! ! # % !
! # $ ¼ ! " ¼ ! " # ! ! " & ¼ eu
uf
n hviau?lav lwwtw(
\
ptodiarr
l{.dreA
t/
a
c,i4,
e\
\
'\
Mqli6|^
fqD=
ioD
1qethe
t
[i,.^i lan,t"
,\
Q?'- t
tf
c'-
c-(t'|, 4 (0 roqoh
nG
J r,r|
( ---h,4
I
L n1 -T -- vtz1,
et,f
3t
) J= L\\>?
,lir
4?=
trz
I rre+
t
lul
r Y["
\Y
t
Itn, nn?
h. t6*
Qn.*^">
e tr = Lwr,f h.-) 4.*
t
6*^ )y- -+Y'^+
Vt.f
hA
=
y.
z
fi^,
(t", ,+ wtr.) oo^
+wt.
1("^;-rm'f
- (l-,.+rn/
(o.,tryx) e*c,r.4
- =';
("-
r
f*u J( L4- 4
c.rNr
t hly.,= ctr,tua,\X*n
h,X, + hrX L= (n,.t r4>) io,
- \re + f}1>4 -- (n,*r"t r) a.^
l,r,r,Fr
tt
.l
)
a
+
9
a
+
(nnr-
A3
x^t )
a.
e- (y^l.f
Y?ttb1 z
;
w?- vh(
|l!t--lvl ,
(rnr-h,)
Ft, t
4'cH
A"x
q -- v./l.t !&{ r-
a-=
hr)
t{t4
a
1
?
( m.
-?ra
(Urrt.
t
t)
rmz
f
lr = J rrF*^
a -- _l
t'-e
&u =
&n7
Roe7
tf
Q;
t,
,=
fnns'a40
r; ttt
'- lj; f
u
Jo
' *,.,_,e
(j "r^
ds
{rit
frn,t'o
l,'
s t^*'):"'
q 7"
6
tZ
)r= A."so
l1= Ro.,so&2
t
-
R"*^o)
fyh"a I
.,
1L
,
I
fil"sruA
T,, I*+ TL
ff
TI=
e) Tr=
!nu
kI f')*n o,
-+Tv
[r'^ fft^
?I* --"Ly t T^ f
r
t
Yrf
fAY
L
-fe
t-
{hletw
$ r
+
t
Lo*ri
.))
;
t-
;
c
+
ro
Lwt r;
2^;
:)
4
n
,rr1;
V.' =
O
Ñ
Ý
ܾ
Ý
Ý! Ü! " # Þ ! Ñ
Ö
$ % % # & % % " # ' % # Å Ê " # & Å Ê '
d7,,i,\*
''r^)
-- Qnn)'"
i,)
J,^ --' I zn
7 ,r,('
r &r
rftr)
Lr
l/r1"
)
n
{--
nr'
tiir
nf-
)
L
'ltt-
ol*,utfw
T
Jlia*,fiW -
f
d'^ r
7-
.L
r
1,rt*u
t *,i L'(,
.l
--
-{Z
Il-)
-7
*vn
n<-'-t
rrt'- l')
("
;\-------___\
Z\ I
r a
trL- L ) oL
dtta=(t
'7-
. Y') d'7
-9.
dun= (u(o'-?- )aL
t-.
-z t
(r*_.{rf j )+}
tc
-+
7
= otik
z-
^r'
+
f
tI**= 4l ,- -{r'' { ]"
P*
.P
4.Gu')u
= Lrfurou)
-L L \-I-/
3'
la ( nr - i Y
{5
/\
g \+'d /
r-t-
,r
cii)
T
N,te
{J,t
4t a-.
r
l
At^
Lrn,-'ur)
R,
*""*
fi*d llorro,
!-
Ro-
+a
is ^ 1,,*i-i-;.a ca*e
IV
|
'h^[('w
w?*a
)*"ok+.t-rs
i
( l+t[ri - -l ]
(n -- zY-l--l,.-\
?"t-u-r;)
El
t'>)
t,n')
1
I
-a4
I.
- M- K' ]
i-Lr,R,'
J.^
K(
Kz
-(
= I r +-
3 Lt t(
({n rzj, U:
o))n'/ r\r
J
I('\
-L-.1
((?)(1)
"t
ff"-;r.,.
9ol
t5
- ei
Z
R. R- f
{2. +
t- h" tt"tr
--1
3 Mp."
!t#
+7(
')
n,?_ sJ
J
R,+
R.'+ n, 1P-t Ar,
l,
(: -r. l-- iu
Rt"e,L\'))
rv
t[1
-
=
(.r
l
P,E-I,Y7
?
3M
r
ul- u; J
q-Rl
3
Ri
re,
1L
ni + e'
! " " # $ " $ %
¾ $ '' &
( ! " ) ' ' ' *
$ " ' ! '
' ! " " # ( ! '! + ( , ' ) " ¾ % ¾ &
' - ( " . ' ' - )
, " ' $ %
¾ $ '' &
/
' 0
' & " " ' - $ , ¼ ' ( " ' ! - $ ' ¾ z.(n)"
T--
[')
(. (/
AlL
#
#vLlr-
ad
M,IL
v
v
t?
u
,M1 L
\r,
il{?
Le
R
=-
0L
:1
d
a,=
IR
+L
LL(a)-J
\=t /
4?
=NGP
r=uy
kflL
tr
r\ :- lnaa
+ ?*
tb)
(t)
lf,D
r'* T*
a
X"u)
.'z
3
ciil
h
/s\
t
,ft
*
! ! ! ! " ! !
! ! ! # ¼ ! $ ¼ !
! ! ! !
! ! !" Æ! ! ! !" !
! ! % ! ! !
! &! ' % ! ! ! ! ! ! ! ! !" () ! !" !
! ) * !" # + ! ! ! # ! + ! ! ! !
! !" ! ! ! ! ! ! ! ! ! , ! ! &!
%
! !" ! "! ! !
! ! ! ! !" ! - ! " ! * ) &! % ! ! . / 0 ! 1 !" ! !
!"
! ! ¼ ¼ ! !
! ) 23!4 ! ! * ! ! " ! !" " ! 5
¼
¼
O
Jiredio,,r
"f
LoTc@
Jir.4 ? 3 o
J'
rN
(=rxry
J__rJ
G
-
-
I
"I
J
"T
But,
-J
T
II JL
-f
Further Example (November 22, 2012)
1. A thin rod of mass m and length l is held vertical on the top of a horizontal floor, where the rod’s lower end
is hinged to a fixed joint on the floor. The rod is released and it hits the floor when it’s angular speed is ω.
Obtain ω by the following methods.
(a) The Newton’s second law of motion for rotation.
(b) The work-energy theorem for a torque.
(c) The conservation of mechanical energy.
1
176n2lc-
1ne lo0
C
40r* rQl = o
= mA cAaTq
oQs =
'aco "Te
oQe
t\, r?
.l 1Cs
tn4 = bg
Huq-
co4o
&_
CA
aI[
cB
Q,rn
cc
t "ri[n =o
<-l
yr&
luarc - oTy
lIe
%
I o1o= l-Toto
{.-
,..
a(a
--
T+-Ts'o
=) _oTe
-'-(T+-Tr)= I-To
tT.
r-
9
-k
q
10.3 Gas Laws

Boyle’s law
Keeping the temperature constant, the volume of a gas varies inversely with the pressure,
which is equivalent to saying that pressure times volume is constant.
V

1
P
or
PV = constant
Charles’s law
When the pressure is kept constant, the volume of a given amount of any gas varies
directly with the temperature (in Kelvin).
V  T or

V
 constant
T
Gay-Lussac’s law
When the volume is kept constant, the absolute pressure of a given amount of any gas
varies directly with the temperature (in Kelvin).
PT

or
P
 constant
T
The ideal gas law
As the pressure increases with the number of moles of gas molecules linearly, we have
P  n , where n is the number of moles of gas molecules. From the Boyle’s law and the
Gay-Lussac’s law, we have P 
we conclude that P 
1
and P  T respectively. Summing up the relations,
V
nT
PV
, i.e.
 n  constant . The proportional constant is nearly
V
T
the same for all gases at low pressures and is referred as the universal gas constant R,
where R = 8.31451 J/(mol K). The ideal gas law is given by
PV  n RT .
The Avogadro’s number NA = 6.022  1023 is the number of molecules per mole, and so
nNA equals N molecules present in a sample of gas. Now, the ideal gas law can be
rewritten as
PV  n RT 
N
R
RT  N (
)T  N k B T ,
NA
NA
where kB is the Boltzmann’s constant and kB = 1.38066  1023 J/K.
Example
A cylindrical flask of cross-sectional area A is fitted with an airtight piston that is free to
slide up and down. Contained within the flask is an ideal gas. Initially the pressure
applied by the piston is 130 kPa and the height of the piston above the base of the flask is
25 cm. When additional mass is added to the piston, the pressure increases to 170 kPa.
Assuming the system is always at the temperature 290 K, find the new height of the
piston.
Piston
Piston
h1
P1
h2
P2
Answer:
The system is basically isotherms, that is, the temperature is kept constant. The Boyle’s
law is applied here as PV = constant, or P1V1 = P2V2.
(130103)(0.25 A) = (170103)(h2 A)
gives h2 = 19 cm.
10.4 Kinetic theory
Pressure and temperature of a gas are macroscopic quantities. However, the macroscopic
quantities connect the microscopic quantities, such as the position or velocity of an
individual molecule. The kinetic theory of gas states the connection. We imagine a gas to
be made up of a collection of molecules moving about inside a container of volume V. In
particular, we assume the following:

The container holds a very large number N of identical molecules. Each molecule
has a mass m, and behaves as point particle.

The molecules move about the container in a random manner. They obey
Newton’s laws of motion at all times.

When molecules hit the walls of the container or collide with one another, they
bounce elastically. Other than these collisions, the molecules have no interactions.
(a)
The origin of pressure
Imagine a container that is a cube of length L on a side. Its volume, then, is V = L3. In
addition, consider a given molecule of mass m that moves in the negative x direction
toward a wall. If its velocity is vx and then vx after rebound from the container’s wall.
The change of momentum of the molecule is given by
p x  p f , x  pi , x  mv x  ( mv x )  2mv x .
As the time required for this round trip of length 2L is
L
 vx
t  2 L / v x ,
vx
the average force exerted by the wall on the molecule is
F
mv x2
p 2mv x


.
t 2 L / v x
L
The average pressure exerted by the wall is P 
Round trip = 2L
F F mv x2 / L mv x2 mv x2


 3 
.
A L2
V
L2
L
Since a gas consists of molecules of different velocities, the relation above should be
replaced by P 
m(v x2 ) av
, where ‘av’ represents the average value. Now, the gas
V
molecules moves in a three dimensional space, we have
v av2  (v x2 ) av  (v y2 ) av  (v z2 ) av  3(v x2 ) av which yields P 
1
1 mv av2
) or P   v av2 ,
N(
3
3
V
where N is the number of molecules in the container and  is the density of the gas.
Rearrange the equation, we finally obtain an equation in kinetic theory of gas
PV 
1
N mv av2 .
3
The L.H.S. of the above equation can be expressed in terms of the kinetic energy of
individual molecule, e.g. PV 
1
2 mv 2
2
N mv av2  N ( av )  N K av . But, as we know that
3
3
2
3
PV  N k B T , we then have a most important result
K av 
(b)
3
k BT .
2
RMS speed of a gas molecule
The square root of (v 2 ) av is given a special name – the root mean square (rms) speed. As
K av 
3k T
1
3
m(v 2 ) av  k B T , we obtain (v 2 ) av  B , hence
2
2
m
v rms 
3k B T
.
m
The rms speed of the gas molecules can be expressed in terms of the molecular mass M,
3RT
, since kB = R/NA.
M
e.g. v rms 
(c)
The speed distribution of molecules
The distribution of speeds of molecules is governed by the Maxwell speed distribution.
The following figure shows the distribution of speeds of oxygen gas at the temperatures T
= 300 K, and T = 1100 K. Note that the most probable speed increases with increasing
temperature.
10.5 The internal energy of an ideal gas
The internal energy U of a substance is the sum of its potential energy and kinetic energy.
In an ideal gas, as there are no interactions between gas molecules, the potential energy is
then zero. As a result, the internal energy U of an ideal gas is represented by its kinetic
energy.
U
Since U 
3
N k BT .
2
3
3
N k B T  (n N A )k B T and R  N A k B , we obtain a similar relation, but this
2
2
time in terms of moles U 
3
nRT .
2
Example
Two identical containers A and B are connected by a tap S that is initially closed. A
contains an ideal gas at a pressure P1 and temperature T1. B contains the same gas at a
pressure P2 and a temperature T2. The taps is then opened. If the temperatures of
containers A and B are maintained and remain unchanged, find the molar ratio of the gas
in the two containers. Find also the pressure of the gas mixture.
P1
A
S
T1
P2
B
T2
Answer:
Before opening S, the number of moles of gas in containers A and B:
n1 
P1V
PV
and n 2  2 ,
RT1
RT2
where V is the volume of containers A and B.
After opening S, the number of moles of gas in containers A (or B):
n1 ' 
P 'V
P 'V
and n2 ' 
.
RT1
RT2
P 'V
n ' RT1 T2
Therefore, 1 
 .
n2 ' P'V T1
RT2
As n1  n 2  n1 ' n2 ' , we have
P1V P2V P 'V P 'V
, where P’ is the common



RT1 RT2 RT1 RT2
pressure in the containers.
After simplification, we obtain
P1 P2 P ' P '

  . Rearrange the expression
T1 T2 T1 T2
P1 P2

T1 T2
P T  P2T1
.
P' 
 1 2
1 1
T1  T2

T1 T2
10.6 Important terms used in thermodynamics

Heat: Heat is the energy transferred between objects because of a temperature
difference

Thermal equilibrium: Objects that are in thermal contact, but have no heat
exchange between them, are said to be in thermal equilibrium.

Temperature: Temperature is the quantity that determines whether or not two
objects will be in thermal equilibrium.
10.7 The zeroth law of thermodynamics
If object A is in thermal equilibrium with object C, and object B is separately in thermal
equilibrium with object C, then objects A and B will be in thermal equilibrium if they are
placed in thermal contact.
10.8 The first law of thermodynamics
The change in a system’s internal energy, U, is related to the heat Q and the work W as
follows:
U  Q  W , where W    PdV .
Q is positive
System gains heat
Q is negative
System loses heat
W is positive
Work done on system
W is negative
Work done by system
It is noted that if work is done on an insulated system (i.e. Q = 0) its internal energy
increases. The work done W has a positive value when the external world does work on
the system (e.g. compression), and it has a negative value when work is done by the
system (e.g. expansion).
U  U f  U i  W
On the other hand, if the system does work, and / or heat is removed, its internal energy
decreases.
10.9 Thermal Processes
All processes discussed in this section are assumed to be quasi-static, which means the
processes occur so slowly that at any given time the system and its surroundings are
essentially in equilibrium. Thus, in a quasi-static process, the pressure and temperature
are always uniform throughout the system. If friction or dissipative forces are neglected,
the process is considered to be reversible. For a reversible process, it must be possible to
return both the system and its surroundings to exactly the same states they were in before
the process began.
(a)
Isothermal processes
Constant-temperature
heat bath
Piston
Piston
Q
Consider the following reversible processes at constant temperature, the piston moves
down without friction. Applying the first law of thermodynamics, U  Q  W , and
making use the property of ideal gas which has its internal energy depends only on the
temperature, we have U  0 and Q  W , since the process is carried out at constant
temperature. The work done, W, is positive as work is done on the system. Hence we
have a negative Q, that is, heat is given out by the system to the external world. The
reversed process in the figure below performs the same idea, but this time the work W is
negative, as work is done by the system. Since U  0 again, the heat Q is then positive
or we conclude that heat is gained by the system from the external world.
Constant-temperature
heat bath
Piston
Piston
Q
Remarks: PV plots
The PV plot of one mole of an ideal gas at the temperatures 300 K, 500 K, 700 K and 900
K. Notice that each isotherm has the shape of a hyperbola. As the temperature is
increased, however, the isotherms move farther from the origin. Thus, the pressure
corresponding to a given volume increases with temperature, as one would expect.
In an isothermal expansion from the volume Vi to the volume Vf, the absolute value of the
work done is equal to the shaded area. For n moles of an ideal gas at the temperature T,
the work done on the gas is
Vf
Vf
Vi
Vi
W    P dV   
V f dV
nRT
  nRT ln V
dV  nRT 
Vi V
V
Vf
Vi
 Vf 
  nRT ln  
 Vi 
Example
A cylinder contains 0.50 mol of an ideal gas at a temperature of 310 K. As the gas
expands isothermally from an initial volume of 0.31 m3 to a final volume of 0.45 m3, find
the amount of heat that must be added to the gas in order to maintain a constant
temperature.
Vf
Vi
T = constant
Q
Answer:
Since the process is isothermal, the internal energy of an ideal gas has no change, i.e.
U  0 . The relation U  Q  W gives Q = W, where W is negative (work is done by
the system) . Now, we know that the gas gains heat from the surrounding (positive Q). As
Vf
the work done W is given by   PdV , we obtain W as
Vi
Vf
Vf
Vi
Vi
W    PdV   
Vf
V f dV
nRT
V
dV  nRT 
 nRT ln V V f  nRT ln( ) .
Vi V
i
Vi
V
Substituting n = 0.50 mol, R = 8.31 J/(mol K), Vf = 0.45 m3 and Vi = 0.31 m3, we obtain
Q = W = 480 J.
(b)
Adiabatic processes
A process is said to be adiabatic if there are no heat flows into or out of the system. We
can have such process if the system is well insulated. The figure below is an adiabatic
compression, the gas heats up and its pressure increases.
Insulation
Piston
P increases
T increases
Similarly, for an adiabatic expansion, the gas cools down and the pressure decrease.
Example
When a rod that fits snugly into a cylinder is pushed
downward rapidly, the temperature of the gas increases
before there is time for heat to flow out of the system. Thus,
the process is essentially adiabatic. As a result, the
temperature of the gas can increase enough to ignite bits of
paper in the cylinder. In a diesel engine the same principle is
used to ignite an air-gasoline mixture.
Example
An amount of gas is compressed from volume A isothermally and the PV plot is obtained
as shown in curve (ii). If the same gas is compressed adiabatically, which curve would
show the correct curve? Curve (i) or curve (iii)?
Answer:
In an adiabatic compression, no heat change occurs in the process. According to the first
law of thermodynamics, U  Q  W  W , U is positive as W is positive during
compression. The increasing temperature gives an increasing pressure for a given volume
of gas. Hence, curve (iii) is the correct curve.
10.10 Specific heats for an ideal gas
(a)
Constant volume
A heat Q flows into a constant-volume container, which is filled with n mole ideal gas,
the temperature of the gas rises by the amount T, and its pressure increases as well. The
heat Q is given by QV = n CV T, where CV is the molar specific heat of a gas at constant
volume. In other words, we can write CV 
(b)
QV
.
nT
Constant pressure
Similarly, the whole process can be carried out at constant pressure, but this time with the
temperature and volume varies. If a heat Q is added to the gas then its temperature and
volume increase. The molar specific heat of a gas at constant pressure is related by QP =
n CP T, where CP is the molar specific heat of a gas at constant pressure. And, we have
CP 
QP
.
nT
According to the first law of thermodynamics U  Q  W , we have in process (a) the
internal energy U  Q , since W = PV = P(0) = 0. But, U 
3
nRT , hence we
2
3
nRT
QV
3
 R.
have CV 
 2
nT
2
nT
For process (b), it is a constant-pressure case, as heat is added to the container, the
increasing volume involves a work done W by the gas, where W   P V   nR T .
Making using the relation U  Q  W again and we obtain
Q  U  W 
3
5
nRT  nR T  nRT .
2
2
5
nRT
QP
5
2
 R.
Plug it into the expression of CP, we obtain C P 

2
nT
nT
Now, we conclude that CP is greater than CV, as extra work is required for expansion
while increasing the temperature.
C P  CV  R .
G"li"X C"
D*t'
8= nC'aT
oTJ
=
W
f
= - lf
(lu'--d1'a'^ice)
r"-"f
Q*
Q ..6
4v -- o
aU=d.
It
|n Raf
a_
,,' naT
1--t
4
?
Cv
-
=
(4^efi)
hAT
Fr"a,*
de.
= nCe dT
?V= xRT
'.' ll= -?&/ = - n R.dl
e&vt
laE _ = Q.+W
2
Q=
dE..t
+
;. ^C?d( =
ce
t(
+
Cf= C"+R
fr(E"
c(
aR df"
Rran F Jv -- ^ 4T
nKtf
e)d7
dQ= nCndT +
a-
z
ft'o
'- W
^G"
va7
wL lrft"o
fa
!
7ok
4n-
cv
r'--
Cv3
,
lr
{n
cr-- 3k
I
cy
tr?t
, tr= +
/ Y't,
laocnetomic
flt
Ji*touL
,dt
P"iat'ci"
lnt
ftA;rfrl+* ehrz"CI + W
AE* 3
A.= o
EE int
JE*
Thur
=
"
CrdT
nC'af
-[E,*=
lW
(lv =
d'(^RT)
=
9-""
n([f
, pdurvtf-c) Vrf.-7lv+ xRdT
tf
aR
t'Tr
Vrf
=
I
"C,
d'Qv)
Jtr..
w
lv
-r
ll ? t/
(
fkIL= -r
-) Jt.
)v
['+ tu a)
L--
?
a
tuv!
v
, rrv{
-Y
cv
-._
?4v
.! .f= crt R-
Cp
"%tT
-n Cr 4'f
V
nCpJT
I^
I
?i
)
"
e-
-rh9y)
r{'=
,r('
-.' P vr= o^stnf
frlhl.t:c (ro*qs
w. - l"r,, = - Qv'.[,,'+
v.
-r.t
.
f, Vr"
p (4Y,^rt
a
-?+v{
?
lnf
I
lu.
-f+t
=
itrvt,
(+ -V;
- ftvr)
=)
vl =
^
-ftt
1
/
t,v{=
t\'Tr)= nCnaf
lr,9
{v"
L .r-e, "K
hur'
d.rfat,ufth
Flqn,il-
ftN@)
N@
dv
.'
Nuh[cv
,f
wole'lcs heviaT s7""ls
v l, vtdv
N: {:
-
ilcv )dv
.d
^;toj
\
f-;3rcft,
sr0
:
0 po6
u|-A*^--
ffi"v
(e.ts|
Mcd4v
7"d tuk)
dfv
{,v
-=o
Al
z
u*Y
Vt, '
7.
t
v
lrj
e(ra4
t["
Va,
z nth," u^etl
;
le ef- llro-e;lt sferl
/
,ft4
n"
Vr,ry
. (r') ? fr v" NcDdV g
f"*
^,
vr-,
8RT
v f,lculdv = IgEL
-
r14
:
,KT
tvl
zRT.
=
n
W, (rr.(on, 7
Lrl,
:
vt
Vq = (tir')
f t.
G$f"4ive Yo(t"nne I'
L+t- "o((?sio,'t occ'^ttj'.
(,t )
1e
At
AF
a
i
"
V
vtO-
L,
evtl
=
N"f,
ie
3e
Nad vt
-
N
.6ar,t
Vr
.:-i
'' a
,
N'1, = c#l vol,
-
Nn
w -llis-,oq
I'
outl.,Ce t'h'r
n
1-V
'1'- Nnd"
\}
.v@
(Arc, o1
LT
no cpt I isbvr
o,^fr.de nit
fincz (V'"RT
\i'te
c clkfur.e betUre^ ." (ft ,bD
trfu,*t,
L=
V
JTnNd
(e*V".(*S' 'J? )
\{ha
t zoe
^t
a
1r'o*'tl\u:
Nz
A
Xo-g
-- ?e,a nuA
.
Further Examples (December 3, 2012)
1. (a) 2.0 mol of monatomic gas A initially has 5000 J of thermal energy. It interacts with 3.0 mol of monatomic
gas B, which initially has 8000 J of thermal energy.
i. Which gas has the higher initial temperature?
ii. What are the final thermal energies of each gas?
(b) 4.0 mol of monatomic gas A initially has 9000 J of thermal energy. It interacts with 3.0 mol of monatomic
gas B, which initially has 5000 J of thermal energy. How much heat energy is transferred between the
systems, and in which direction, as they come to thermal equilibrium?
2. A insulated rectangular container has a thin barrier inside it such that the container is divided into two volumes,
A and B. Gases in A and B can interact thermally through this thin barrier. Suppose A contains n1 moles of a
monatomic gas and initial temperature T1i , and B contains n2 moles of a diatomic gas and initial temperature
T2i .
(a) Show that the equilibrium thermal energies are
E1f =
E2f =
3n1
(E1i + E2i ) ,
3n1 + 5n2
5n2
(E1i + E2i ).
3n1 + 5n2
(b) Show that the equilibrium temperature is
Tf =
3n1 T1i + 5n2 T2i
.
3n1 + 5n2
1
Rrvc,rsihtc
(+)
(r,u$
(o, wwd, l4.onie
o+4
\
I
t
\a
@_-
----€
\\
s
t\
A(4u'
Ve("" e
f/olecnh
Co lt;slo
(b)
g*+wo*d
r,ov'<L
rtr :
c()
Wqrsrbb
Cr)
\*xt(,f
glal;itd,z
o_
a
a
\@
?g'e'e
t\-
?+
"o
?
(orr*nd, WltS&t+l-v
F4orc t
VorJ+-art4 sPuie
8"6'.
:
A$te" ,
YfuIYI:I
FfooiUZt
EIT*V't
A)
60"(a
).
I
=
A5 -- n?L^Y
v.
K
A$
nCv!,^+
a.
Sfinte-
or^
slates
. I/,^dt,;y.,A*fr
As?-g
T
4S=
'"[ora
%tAry)
i
a*'l +,
a( {t^ (vocfii
(
{=
As --+
T
q5 L
@^+-p,S
. D.(wrds
K
t
(w-te,rs\bte)
ftua,nelrG,
cs'ns1ar^,f
.
["'11^-^"{
)
(\vvzvtYsibb)
\v'n vg
4
t?,)?q
e= !n^ = -fr, = Ietv
nRI^t
--lr#
V;
= ncT/^Jg
V;
r,..
)
-)
l(J = d11t tu/
lA. = 4U- dh/
d.Q.= nfr&f + ?[V
&&= r. G aT t
=)
dd.
-(
=
+
Ag=
entwgl
:
LJ, (cin,
pV"
nRt)
"Crt* ""+
drS
+ a$= JT
a
n?.T
(Ji.,. dtr,l=-ldy)
__
^cu
f++
xR
l{ tv
J.v
I
V+
"crb\T: + ^R{.,^ V;
A )rtbte Vlfir*
Not telu"Ao.t ow hau {l,c t yt-t
or*ioel,'e7t tl^^rt sfa,ta
.
9,r^/,
W "( ftru
WU d""^f,
occutv w'uflt,^
( dwd U1*fen iG
U*r" g,1 oitrtrN iv,oreAtz, ( ,frn' lrrute/rsi b (s
Pcetil or
lncessrs )
yev^cJylr
, (r(
crntrffi (.f*
wutw
dccre a,,wg J
yLvevg',L(L
Further Examples (December 6, 2012)
1. As shown in the figure, which, if any, of the heat engines violate
(a) the first law of thermodynamics or
(b) the second law of thermodynamics?
Explain your answers.
2. The temperature of 3.15 mol of an ideal polyatomic gas (CV = 3R) is raised 52.0 K by each of three different
processes:
• at constant volume,
• at constant pressure, and
• by an adiabatic compression.
Neglect the vibrational modes of the gas. Complete a table, showing for each process
(a) the heat added,
(b) the work done on the gas,
(c) the change in internal energy of the gas, and
(d) the change in total translational kinetic energy of the gas molecules.
3. A Carnot engine has an efficiency of 22%. It operates between heat reservoirs differing in temperature by 75o C.
Find the temperatures of the reserviors.
4. A Carnot refrigerator operating between −20o C and +20o C extracts heat from the cold reservoir at the rate
200 J/s.
(a) What is the coefficient of performance of this refrigerator?
(b) What is the rate at which work is done on the refrigerator?
(c) What is the rate at which heat is exhausted to the hot side?
1
l,o,)
(_ro .,l;
\"
T-
)K
K
::
'v
/ t)
- -',
O
!"o. W,^= urwr
Lc)
G," = ai.
t \,,..1;^
t'5[,
=
=)
'o"Tfst
=) e," -- ,jz rf s -- 232
QH
#