FUNDAMENTALS OF PHYSICS

1
raft
Motor Boat
V
V
1.1. Method : 1 (Relative approach)
V = Relative speed of
motor boat w.r.t.
river which is constant
Observer on raft see that speed of motor boat is constant because duty of motor boat is constant.
Hence if motor boat take 1 hrs in down stream journey then to reach again at raft motor boat
will take 1 hrs in upstream journey. Hence Total time in complete journey = 2 hrs.
Motion of raft : u = speed of river. Then 2 u = 6  u = 3 km/hr. Ans.
Method : 2 (With frame of ground)
Motion of raft : u (1 + t1
) = 6 ––– (i)
Motion of motor boat : (v + u) × 1 – (v – u) t1
= 6  v + u – vt1
+ ut1
= 6
 v – vt1
+ u (1 + t1
) = 6; from (i)  v – vt1
+ 6 = 6  t1
= 1 hrs.
put t1
= 1 hr in (i) : u (1 + 1) = 6  u = 3km/hr Ans.
Q.1.2: Total distance travel by point is S. Then Time taken in first journey :
t1
=
0V
2
S
Time taken in second journey :
2
t
V
2
t
V
2
S 2
2
2
1   t2
=
21 VV
S

Mean velocity =
21 tt
S

=
210 VV
S
V
S/2
S


 Mean velocity =
021
210
2VVV
)V(V2V


Ans.
Q.1.3 Method : 1 (Graphical Approach) Given tan  = 
timeTotal
trapziumofArea
tan
2
t–
)t(
2
1
Time
ntDisplaceme







 





4
–1t Ans.
Fixed point
6 km u
distance = 2 × u
raft
1 hr.
t1
6 km
Part One
Physical Fundamenatls of Mechanics
1.1 Kinematics
V
t


t
 – t
2
 – t
2

Compact ISC Physics (XII)2
Method : 2 (Analytical)
Total displacement (S) =
22
2
t–
2
1
t
2
t–
2
t–
2
1







 







 







 

Time taken =  =






 





 





 



22
2
t–
2
1
t
2
t–
2
t–
2
1
S
t = 


4
–1 Ans.
Q.1.4 (a)
S
2m
20 s
t
average velocity =
20
2
= 0.1 m/s = 10 cm/s Ans.
(b) Velocitywill be maximum when slope of S (t) curve will be maximum.
V =
10–14
0.4–1.4
m/s = 25 cm/s Ans.
(c) Instanteneous velocity may be equal to mean velocity when
slope of line joining final and initial point will be same to
slope at point on curve. Ans. : to
= 16 s
Q.1.5 Velocity of B with respect to A : 12A–B V–VV


Position of B with respect to A : 12ABA–B r–rr–rr


A
B
VB – A
Particle B will be collide with A if velocity of B with respect A is directed toward observer A.
Then A–BA–BA–BA–B rˆVˆr||V 

Then |r–r|
r–r
|V–V|
V–V
12
12
12
12




 Ans.
Q.1.6
N
EW
S
V = 15 km/hrW – E
V = 30 km/hrS – E
60º
Y
X
S
16 s
approx
t
S
1.4
14
t
0.4
10
A
r1
v1
v2
r2
B
Y
X


3
E–SV

= 30 i; jˆ60sin15–iˆ60cos15V E–W 


E–WE–SS–W VV–jˆ60sin15–iˆ30)–60cos(15V


km/hr4060)sin(1530)–60cos(15|V| 22
S–W 

 19º
30–60cos15
60sin15–
tan  Ans
Method 2
30 km/hr
VW – S15 km/hr
60º 
we know E–SE–WS–W V–VV


60cosVV2–VV|V| E–SE–W
2
E–S
2
E–WS–W 

 40 km/hr Ans.
from figure : tan  =
60cos15–30
60sin15
    = 19º
Person : 1
Q.1.7
2 km/hr

d
2.5 km/hr
Time to cross the river : t =
cos2.5
d
––––– (i)
from figure : sin  =
5
4
2.5
2
  cos  =
5
3
 Put in (i) : t =
1.5
d
––––– (ii)
Person : B
2 km/hr

d
2.5 km/hr
x
Using trigonometry : x = d tan 1
Time to reach at destination point : t = t1
+ t2
Here t1
= Time to cross the river =
2.5
d
t2
= Time to walk on bank =
u
tand
u
x 
 And tan
=
5
4
2.5
2
  t =
5
4
u
d
2.5
d
 –––– (iii)
from (ii) and (iii) :
5u
4d
2.5
d
1.5
d
 u = 3 km/hr Ans.

Q.1.8 Boat A :
d
d
time to reach again at same point.
tA
=
– WV
d
WV
d


–––– (i) where V = Speed of boat w.r.t. river
W = Speed of water w.r.t. earth.
Boat B : d
W
V
Time to reach again at same point. tB
=
22
– WV
2d
––––– (ii)
Also given V =  ––––– (iii) now 1–
–
2d
–
dd
t
t
2
222
B
A









1–t
t
2B
A


 = 1.8s
1.9: Method : 1 
d Vm – r
x
Vr – E
A
Time to cross the river: t = cos|V|
d
mr

Then Drift (x) = (Vm – r
sin  – Vr – E
)
cosV
d
mr
Since Vm – r
< Vr – E
. Hence
this is not possible that drift will be zero. Hence we should have to minimize
drift (x). Hence
d
dx
= 0
d sec2
 –
mr
E–r
V
d)(V
sec tan  = 0  sin =
2
1
V
V
E–r
r–m
   = 30º
Angle made by boat with flow velocity of water = 30º + 90º = 120º Ans.
Method : 2 (Vector addition method)

Vr – E
Vm – r
E–rr–mE–m VVV

 Hence  will taken any value between
(0 – 180º) hence we can draw a semi circle of radius of length |V| r–m

. Then.
30º 90º

5
Vr – E
A
C4
C3
C2
C1
B
And resultant is given by C1
, C2
, C3
and C4
, ....... Cn
. But for minimum drift resultant must be
tangent at semicircle. Then cos  =
2
1
V
V
E–r
mr
   = 60º Then   = 180 – 60º = 120º
1.10: Relative acceleration of particle (1) w.r.t. (2) = g – g = 0; Relative velocity
of particle (1) w.r.t. (2) = V1 – 2
= )–(90cosV2V–VV 00
2
0
2
0 
= V0
)sin–(12  Where 90 –  is angle b/w two velocity..
Since there is no relative acceleration of particle (1) hence its relative velocity
does not change w.r.t time then. Distance b/w two particle at time t is : Distance = V0
t )sin–(12 
= 22 m Ans.
Q.1.11 : Method : 1 (Vector) + y
3 m/s 4 m/s x (+)
1
2
Initial velocity in y direction
of both particle zero. Hence vertical velocity of particles (1) at time t : Vy
= u + at  Vy
= g t; Velocity
of particle (1) at time t = : jˆtgiˆ3–V1 

; Velocity of particle (2) at time t : jˆtgiˆ4V2 


Since 0V.VVV 2121 

Then – 12 + g2
t2
= 0  t = 0.12 s. Hence distance between two
particle will be : Distance = Vrelative × t = (4 – (–3)) × 0.12 = 7 × 0.12  2.5 m Ans.
Method : 2 (Graphical) : Since 21 VV

   +  = 90
  = 90 –   tan  = tan (90 – )  tan  = cot 
tan  × tan  = 1 ––––– (i)
And tan  =
3
gt
 tan  =
4
gt
Vr – E

C
Vmr

V0
 = 60º
g g
1 2
V0
3 m/s 4 m/s
V1
gt gt
 




put in (i): 1
4
gt
3
gt












 t = 0.12
Distance = Vrel × time = 7 0.12  2.5 m Ans.
1.12 : Method : 1 (Velocity of approach) Since particle A heading to particle B and B to C and C
to A. Then position of all particle at t = dt is as figure 2.
Again position of all particle at t = 2 dt is as figure 3.
Again position at t = 3 dt and so on ........ Since at any time
all particle travel same distance then at each moment of time,
all particle will be at equilateral tringle. Then by symmetry you
can say that all particle will be met at centriod of tringle then
path of each particle will as :
Suppose at any instant of time, distance b/w particle A and
centriod P is r. Then, line joing particle A and P make 30º angle
with side of equilateral tringe then
dt
dr
will always constant and
equal to v cos 30º then –
dt
dr
= V cos 30º, (–)ive sign because
r is dicreasing function. Finally r = 0 while initial / 3r a
 
t
0
0
3
a
dt
30cosv
dr
–
 t =
3V
2a
 A P =
2
a
sec 30º =
3
a
Ans.
a
60º 60º
60º
A
B
C
a
V
V
V
a
Vdt
Vdt
Vdt
Vdt
Vdt
Vdt
Vdt
Vdt
Vdt

A
B
C
30º
P30º
A
P
A 30º
a
2/
/ 3
a

7
Method : 2 (Relative approach)
let distance b/w A and B at time t is r then. At any instant of time, rate of decreasement of
distance b/w two particle A and B will be constant as shown in figure.
Then
dt
dr
= – (V + V cos 60º) =
2
3V
–   
dt
0
0
a
dt
3V–
dr2
 t =
V
3a
Ans.
1.13: Suppose at time t distance b/w A and B is r. Then rate of decreasement of r is :
dt
dr–
=
– v + u cos    
o t
0
dtθ)cosu(–vdr–

 –l = – vt + u 
t
0
cos  dt.... (i)
Now since 
t
0
cos  dt is not known then to find this integration,
we use rate of decreasement of component: Then
dt
dx
= – V cos 
+ u dtu)cosV(–dx
t
0
0
0
  0 = ut – V  cos
t
0
dt 
 cos
t
0
dt =
V
ut
now put in (1) : –l = –vt + u 







V
ut
 t = 22
u–V
V 
1.14:With frame of train : With frame of train, train appear in rest
then distance b/ there two event is equal to l.
With frame of earth : When event (1) will happen. Velocity of train is V = u + at = wt. Since
event (2) will be happen after time  then. Distance travelled
by headlight (A) = Distance travelled by head light (B) =
u1
t1
+


a t2
1
= wt() +


w
w 







t Then distance
b/ two events is = 




  2t– w = 0.24 km. Ans.
A
A
1
A11
B
11
C11
C
1
C
B
1
B
V
cos60º
V
cos 60º
V V
V
V
V
V
60º
60º 60º
V
V cos 60º
B

A
W

AB
event 2
B A

event 1
w (t + /2) 
 A

V
B u
x
y x
r

 It we want that both event will be happen at same point then velocity of reference frame
will be Vreference frame
=
60
km24.0
= 4 m/s Ans.
1.15: (a) For observer inside lift at t = 2s. velocity of bolt = 0
Accn
of bolt = 10 + 1.2 = 11.2 m/s2
. Then assume t time is taken by
bolt to reach at floor. s =


at2
 2.7 =


× 11.2 t2
 t = 0.7s
(b) Velocity of bolt with respect to ground at t = 2 sec.
V =1.2×2 = 2.4 m/s Displacement then S= ut +


at2
 S = 2.4 (0.7) –


×10 (.7)2
 – 0.7 m
Distance : H = 2g
2.4. 
= 0.288 Distance travelled = 2 × 0.288 + 0.7
 1.3 m Ans.
1.16: Method : 1 (Relative velocity) from figure
tan  =
1
2
V
V
 cos = 2
2
2
1
1
VV
V

also tan  =
1
2
1 V
Vy


y =
1
2
1
V
V
  BC =
1
21
2
V
V
–


Shortest distance = CM = BC cos = 2
2
2
1
1
1
21
2
VV
V
V
V
–








 

shaft
a=1.2 m/s
2
2.7m
2.4m/s
H
0.7m
Shortest distance
= 2
2
2
1
2112
VV
|V–V|


now t = 2
2
2
1
221
2
2
2
1
1
2
2
2
1
2
2
2
1
VV
VV
VV
tanCMsec
VV
BMAB
VV
AM










 
2
2
2
1
2211
VV
VV
t




Ans.
1
2
y
0V1
B
C
M

V2
1
2
V2
0
V1

9
Method : 2 (Velocity of approach)
At shortest distance velocity of approach = 0. Then V1
cos  = V2
sin   tan  =
2
1
V
V
In tringle BA  O : tan  =
2
1
1
2
V
V
tV–
–tV





V2
2
t – l2
V2
= l1
V1
– V2
1
t  t = 2
2
2
1
21
VV
VV

  
Ans.
And Shortest distance is : )–t(Vt)V–( 2
2
2
1   
Shortest distance =
2
2
2
1
21
VV
VV

  
Method :3
V t1
1 1– V t
V1
V t –2 2
V2
)–t(Vt)V–( 2
2
2
1    ... (1). for shortest
distance
dt
dl
= 0 
)–t(Vt)V–(
V)–t(V2Vt)V–(2–
2
2
2
1
2211








= 0
t = 2
2
2
1
21
VV
VV

  
Ans. put value of t in (1) : 2
2
2
1
21
VV
|V–V|

  

Ans.
1.17: Method : 1
Time to reach at point D : T =
V/n
sec
V
tan–m 

 
. For
minimum time :
dθ
dT
= 0 Then –
V

sec2
 +
V
n 
sec
× tan  = 0  sec  = n tan   sin  =

1
. Then
distance BC = l tan   BC =

–

Ans.
1 1– V t
V1
V t2
V2
V t1
A A
1
B
1
2
B
90–
O
B
C
m– tan 
m
D
A


tan


Method: 2 (Help of light wave)
We know that light travel via that path in which time will be
less. Then.
(2)mediuminspeed
(1)mediuminspeed
θsin
isin
 =
V
V
sin
90sin 


 sin 


 length BC = l tan =
1–η2

Ans.
1.18:
1.19: (a) Mean velocity in irodov is misprint and it is mean speed then mean speed =
τ
Rπ
Ans.
(b) Mean velocity =
τ
R2
Ans. (c) We know 
  

       
   ....(i)
And 


t2
  


2
 
2
τ
2 
ii from
(i) and (ii) : 






 
2
τ
2
  
τ
2 
now V = RW then
V = R 




 
τ
2
. Average accleration :

cf V–V
=
τ
0–
τ
R2 
 Avg. Accn
= 2
τ
R2 
Ans.
B C
medium (1)
i



medium (2)
B
R
A
Time taken = 
WX
1
3
6
1m/s
2
–1m/s
2
x
Distance
t
t
t

11
1.20:This is one dimension motion be cause direction of position vector r
 is same as constant
vector a

. Then t)–(1tar 

at–ta 2 

(a) at2–a
dt
rd
V


  a2–
dt
vd
acc



  aV

 (1 – 2 t) Ans.
(b) At initial position t = 0  r

= 0 Now at final position r

= 0 then. a

t (1 – t) = 0
t = 0 ro t = 
 Ans.
Initial A B At point B velocity will be equal to zero so
that particle will be turn back. Then 0 = a

(1 – 2t)  t =


Then position B is :
r = a 
















4
a
–1 . The Distance travelled in up and down journcy is:




 2
a
4
a
4
a
Ans.
1.21:(a) V = V0
(1 – t/)  dt
τ
t
–1Vdx
x
0
t
0
0  





 x = V0 







2τ
t
–t
2
 Position at time t is
x then : x = V0
t 





2τ
t
–t Ans.
(b) Henc we see that velocity will change dirn
at t =  because When t >   v = 
And t <     v = – ive. Then Case I : t <  x VV0
t 





2τ
t
–l Ans.
Case II : t >  Total distance travelled = AB + BC
= V0
 
















 





 
2τ
t
–ltV–
2τ
–lV
2τ
–l 00 = 









2
t
–ltV–
2
V
2
V
0
00
Distance = V0
 – V0
t 





2τ
t
–l when t >  Ans.
1.22 : (a) v = 
1
xx differentiate w. r. t. time : 2
1
2
1
2
1
xαx
2
α
a
dt
dx
xα
2
1
dt
dv ––

a =
2
α
2
1
Since acceleration is constant; velocity will be : V = u + at  V =
2
α
2
1
t
C
A
B
t = 

(b) S = ut +
2
1
at2
 S =
2
1








2
α2
t2
 t =
α
s2
Mean velocity <V> =
s2
s
t
s 

< V > = s


Ans.
1.23:Calculation of time : w = – a v ( – i ve sign is used to show deacceleration)
dt
dv
= – a v   
ot
0
0
0v
dta–
v
dv
 t0
=
a
V 2
1
0
Ans.Ans.
Calculation of distance : va–  va–
dx
dv
v    
0
0
2
1
x
0
0
v
dxa–dvv x0
=
3a
V 2
3
0
1.24 : (a) jˆbt–iâtr 2


 x = at  y = – bt2
 y= – b
2
a
x






a2
y = – bx2
Ans.
(b) jˆ2bt–iâ
dt
rd
v 


 jˆ2bt–iâv 
 222
tb4a|v| 

 jˆ2b–
dt
vd
a 


 jˆ2b–a 

2b|a| 

(c) Since direction of acceleration is toward y dirn then angle made by velocity vector with y axis
is known as angle b/w two vectors then. tan  =
2bt
a
Ans.
(d)
t
jˆbt–iât
t
s
V
2
mean 

 jˆbt–iâVmean 

 222
mean tba|V| 

1.25: (a) x = at  y= at (1 – t)  y= x 








a
x
–1 Ans.
(b) vx
=
dt
dx
= a  ay
=
dt
dvy
= a – 2 a  t  jˆt)2a–(aiâv 

 22
t)2a–(aa|v| 

 2
t)2–(11a|v| 

 jˆ2a–
dt
vd
a 


  2a|a|

Ans.

2bt
x
y
a

13
Method:1
(c) Velocity=
a (–2a t)
a
acceleration = 2a
cos  = |v||a|
v.a


 222
)(2at)2a–(aa
t)2a–(a2a–
2
1



t =

1
Method:2
t2a–a
a
4
– tan

  – a = a – 2 a t  t = 
 Ans.
1.26 : x = a sin wt y= a (1 – cos wt)  







1–
a
y
= – cos wt ––– (i) 
a
x
= sin wt ––– (ii)
11–
a
y
a
x
22
















 x2
+ (y – a)2
= a2
Equation of circle of radius a. jˆ
dt
dy
iˆ
dt
dx
v 

jˆsin wtawiˆwtcoswav 

 wa|v| 

= const.  Uniformcircular motion.
(a) Distance travelled in time  is = (aw)  Ans.
(b) Since motion is uniform circular motion. Hence only radial acceleration is
present then Angle b/w velocity vector and accn will be 
 Ans.
1.27 : y = ax – bx2
differentiate w.r.t. time :
dt
dx
2xb–
dt
dx
a
dt
dy
 Vy
= a VVx
– 2bx Vx
–––– (i)
At x= 0 Vy
=a Vx
––– (ii)differentiateequation (i)w.r.t.time :
2
x
xxy
2bV–
dt
dV
2bx–
dt
dV
a
dt
dV

ay
= a ax
– 2bx ax
– 2b Vx
2
At x = 0 Given ax
= 0 and ay
=w ay
= |– 2b Vx
2
|w =2b Vx
2
–– (iii)
Speed at origin will be : V =
2
y
2
x VV  =
2b
wa
2b
w 2
 V =
2b
w
(1 + a2
) Ans.
1.28 : (a)
2
ta
2
1
tus

 
2
0 tg
2
1
tVs


g

V
0
Ans.
a
(0, a)

(b)
t
s
v


  tg
2
1
vv 0

 Ans.
we know T = g
sinu2 
 u sin  = g
g.v– 0


g
2
g
g.v–
2
g
vv 0
0 










 2
0
0
g
g.v
g–vv

 

1.29 : (a) S = 0 in y direction  0 = (V0
sin ) T –
2
1
g T2
T = g
sinV2 0 
(b) At maximum height final velocityin ydirection = 0 Vy
2
= uy
2
+2 aH
02
= (V0
sin )2
+ 2 (–g) H H =
2g
sinV 22
0 
Range R = V0
cos  






 
g
sin2V0

R =
g
2sinV2
0 
when H= R 
g
cossin2V
2g
sinV 2
0
22
0 


tan  = 4  = tan– 1
(4)
(c) x = (v0
cos ) t  y = (v0
sin ) t –
2
1
g t2
 y = v0
sin 
2
00 cosv
x
g
2
1
–
cosv
x

















y = x tan  –
cos2u
gx
22
2
Ans.
(d) Radius of currature =
onacceleratinormal
V2
 R0
=
cosg
V2
0

g
V cos0 
A
V0
RA
=
g
cosV 22
0 
Ans.
1.30 : w
= g sin  andw
= g cos Here  is first dicreasing and then
increasing and projection of total accn
on velocity vector
will be (–)ive then. wv
= vˆ.w1

= – g sin 
– g
 –
V
y

x
V cos0 
V0V sin0 


g
V0

g
wn
w
V

t
wV
w
g sinw
= w
V

w = g cosn 

15
1.31 :
Time to collide with incline T =
g
2g2
cosg
cos2gh2
cosg
u2 y





Length MO = (velocityin MO) × T = 




  sincos2ghsincos2gh
Range OC = (MO) sin  =
cos
MO
=





 
cosg
2gh2)sincos2gh2
R = 8 h sin  Ans.
1.32: We know R =
g
2θsinu2
5.1 × 103 =
g
sin)( 2
 
 1
= 32.5º. Also we know that range
will be same for 2
= 90 – 31.5 = 59.5º. Then time of flight will be: T1
= g
usin 
T1
=
10
31.5sin240
= 24.3s = 0.41min Ans.
T2
=
10
59.5sin240
= 42.3s = 0.69 min Ans.
1.33: Method: 1
  º   
  º. For both particles x and y co-ordinate must be same then.
Particle (1): y = x tan 1
–
1
22
0
2
cosV2
xg

.... (i)
Particle (ii): y = x tan 2
–
2
22
0
2
cosV2
xg

.... (ii)
From (i) and (ii) : x tan 1
cos2V
xg–
1
22
0
2

= x tan 2
2
22
0 cos2V
gt–


V = 2gh0

Just before collision
2gh
2gh sin 


Just after collision
2gh
sin

2gh
cos

O
C
x
M
y
5.1×10 m
3
240m/s

250 m/s = v0
250 m/s = v0
60º
45º= 

x =
2
2
1
2
2121
2
0
cos–cos
coscos)–(sin
g
V


.... (iii)  Time for particle (1) : t1
=
10 θcosV
x
TIme for particle (ii) : t2
=
20 θcosV
x
Then t = 





 210 cos
1
–
cos
1
V
x
.... (iv)
Put value of x on (iv) :t = 





 21
210
θcosθcos
)θ–(θsin
g
V2
= 11s Ans.
Method: 2
Particle (1) : x = V0
cos
(t + t) .... (i)  y = V0
sin
(t + t) –


g (t + t)2
.... (ii)
Particle (2) : x = V0
cos
(t).... (iii)  y = V0
sin
(t) –


gt2
.... (iv)  From (i), (ii),
(iii) and (iv): t = 





 21
210
θcosθcos
)θ–(θsin
g
2V
= 11s Ans.
1.34 : (a) tan = ay
V0
 Time to reach at hight y : t = y/V0
Also
dt
dx
= ay 
dt
dx
= a [V0
t]   
t
0
x
0
0 dttaVdx
x =








0
0
2
0
V
y
2
Va
2
taV
x = 







0V2
a
y2
Ans.
(b) ay
= 0 ax
=
dt
dya
dt
dVx  = aVy
= aV0
 anet
= aV0
Tangential acceleration = a
= aV0
cos  a
=
2
0
2
22
0
0
)(ay/V1
ya
(ay)V
(ay)Va



Ans.
Radial acc
n
or normal acc
n
: (an
) = a V0
sin  an
=







0
0 V
yaVa Ans.
y
0 x x
v = ayx
v0

y

17
1.35: (a) tan  =
bx
a
 x = at .... (i) Vy
= (a t) b
 
y
0
t
0
dttabdy 
2
tab
y
2

Then y =







a
x
2
ab
 y = 





2a
b
x2
Ans.
(b) ax
= 0 ay
=
dt
dx
b
dt
dVy
 = ab  anet
= ab  Then normal accn
: an
= anet
cos
anet
= ab Then an
= ab
22
(bx)a
a

 R =
ba
)xb(a
a
V
2
3/2222
n
2

 
R =

















/
a
xb
1
b
a
Ans.
1.36: Method: 1 (Work-Energy)
Tangential accn
is given by : 

.a|w| = a cos 
Suppose at time t particle at posetion A and in small time dt particle displacement
by ds.
Work done : dw = Ft
ds = (ma cos) ds = ma (ds cos)  dw = ma (ds cos) = ma
(ds) cos  ma  ds cos
Using work energy theorem : W = max 


mV2
= max  V = xa Ans.
y
x
a = vx
bx=Vy

v
an
anet
0

 a 
ds
x

Method : 2 (Kinematics)
Tangential accn
: at = 

.a = a cos. Also since we know
that tangential accn is rate of change of speed then.
dt
|v|d
= a cos 
ds
dvv
= a cos  v dv = a (ds) cos = a dx
 
x
0
v
0
dxadvv 

2
v
= ax  v = xa Ans.
1.37: Angle travel by particle:  = 2n. Speed of particle : v = Rw = at w =



t
0
n2
0
dt
R
at
ds
dt
dθ
R
at
2n =
2R
at2
 t2
=
a
nR

t2
=
a
nR
Now : Radial accn
: ar
=
R
ta
R
V
222
 =
a
Rn4
R
a2

= 4na Tangential accn
: at
=
dt
dv
= a  Total accn
= 2
t
2
r aa  = 22
na)(4a 
Total accn
= a 2
n)(41  = 0.8 m/s2
Ans.
1.38: (a) At any time t : at
= ar
=
R
V2
Since at
is rate of change of speed then.
–  
V
V
t
0
2–
2
0
R
dt
dVV–
R
V
dt
dv
(–) sign because
V is decreasing
R
tV
1
V
VR
t
V
V
V
0
0
0
1–


(b) anet
= 2
r
2
t aa  = at  at =
R
v2
= –v
ds
dv
 –v dv =
R
v2
dt
Where ds = distance travel by particle in time dt. –v dv =
R
v2
ds
 
V
V
S
00
R
ds
v
dv
–
–  n
R
S
V
V
0
 Then V = V0
e–S/R
 at
=
R
eV
R
v 2S/R–2
0
2


ds V
dx
a
R
v
R
at
ar
Given
a = a
at t = 0
u = V
t r
0

19
anet
= 2
R
V
R
e2V 22S/R–2
0
  anet
= 2
R
V
eR
2V 2
2S/R
2
0
 Ans.
1.39: Method : 1
v = a s Squaring both side : v2
= as
Compare with : v2
= u2
+ 2 acc
S  u = 0 acc
=
2
a
Hence motion is constant magnitude tangential accn
. Then.
At any time t : at
= a/2  ar
=
R
V2
=
R
as
Since we know that velocity vector and tangential accn
is parallel then.
tan   R
2S
a/2R
as
a
a
t
r
 Ans.
Method : 2
v2
= as  differiate w.r.t time : 2v =
dt
ds
a
dt
dv
 = av  2
a
dt
dv
 = at
ar
=
R
as
R
V2
  tan  =
R
2S
a
a
t
r  Ans.
1.40 : (a)  = a sin wt now  = 0    t = 0
 = ± a   t = 2
3,
2
 , V = wtcoswa
dt
d


a = sin wtwa
dt
dv 2
 Then at
= – aw2
sin wt
ar
=
R
wtcoswa
R
V 2222
  anet
= 2
r
2
t aa 
anet
= aw2 wtcos
R
a
wtsin 4
2
2
2
 –––– (i)
at t = 0   = 0  at t = 2
3or
2
  anet
=
R
wa 22
 anet
= aw2
Ans.
at
ar

at
ar

x
 = a sin wt
y

(b) For minimum value of acceleration  0
dt
danet   cos wt = a2
R
put in equation(i)
amin
= aw2







2a
R
–1   =







a2
R
–1a   = 2
2
a2
R
–1a Ans.Ans.
1.41: Speed of particle when distance is S. V2
= 2aS –––– (i) (Because a
= const.)
S =
2
1
at2
 t2
=
a
2S
Then wn
= b 2
22
a
4bS
a
2S






Radius of curvature : R = 2
2
n
2
4bS
2aSa
w
V
  R =
2bS
a3
Ans.
Net accn : w = 2
n
2
T wa   w =
2
2
2
2
a
4bS
a








 Ans.
1.42: (a) y = ax2
diff. w.r.t. time :
dt
dx
2ax
dt
dy
 Vy
= 2ax VVx
Again diff. w.r.t. time : ay
= 2ax ax
+ 2aVx
2
At x = 0 Vy
= 0 Then Vx
= v  ay
= 2aV2
Since speed is const. its tangential acceleration will be zero. Then
ax
=
dt
dV
= 0  anet
= 2 a VV2
 R = 2
2
n
2
2av
V
a
V   R =
2a
1
Ans.
(b) 2
2
2
2
b
y
a
x
 = 1 diff. w.r.t. time : 0
b
V2y
a
V2x
2
y
2
x

Again diff. w.r.t. time : 0
b
2V
b
a2y
a
V2
a
a2x
2
2
y
2
y
2
2
x
2
x 
At x = 0 and y = b : 0
b
V2
b
a2
a
V2
2
2
yy
2
2
x 
a = a

w = bt
n
4
x
y
V0

21
As shown in figure : Vy
= 0 and Vx
= V0
 0
b
a2
a
V2 y
2
2
0
  ay
=
2
02
V
a
b
Since speed is Const. tangential accn = ax
= 0 at; t = 0  anet
=
2
02
V
a
b
Radius of curvature : R =
n
2
0
a
V
R = 2
02
2
0
V
a
b
V
 R =
b
a2
Ans.
1.43 : Given
dt
d
= w = const. Then
dt
d
2
dt
)d(2 


= 2w = const.
Hence angular velocity of point A w.r.t. point P is
constnat then. WP
= 2W and velocity will be perpendicular
to position of A w.r.t. P then V = R WP
= 2RW Ans.
Since WP
= const. 0
dt
dWP  Then  Tangential accn
= 0
Radial accn
= R WP
2
= 4 RW2
= anet
 Direction is toward the centre. Ans.
1.44 :  = at2

dt
d
= w = 2at 
dt
dw
=  = 2a
Tangential acceleration : at
= R = 2aR Radial acceleration
: ar
= RW2
= R (2at)2
= 4 a2
t2
R
And. v = RW = R 2at 2at = R
v –––– (i)
Now anet
= 2
r
2
t aa  = 2aR 22
)(2at1
from (i) : anet
=
22
)(2at1
t
v
 = 0.7 m/s Ans.
1.45 : Since acceleration is constant. l =
2
1
at2
and
V = at Then l =
2
Vt
–––––– (i) and also. angular acceleration is const. then
 =
2
1
 t2
and w = at then  =
2
1
wt and since particle taken n turn in its journey..
V
O
A



P
at

ar
l
Va

 = 2  n Now 2 =
2
1
wt –––– (ii)
(II)
(I)
:
W
V
2



 W =

v2
Ans.
1.46 :  = at – bt3
 w =
dt
d
= a – 3bt2
when body is in rest then w = 0  a – 3bt2
= 0
t =
3b
a
 in
= 0 and final
= t (a – bt2
) = 





3
2a
3b
a
Wavg
=
t
– infinal


=
 
3
2a
3b
a
3
2a
3b
a
 Ans.
Win
= a and Wfinal
= a – 3b 





3b
a
= 0  avg
= 3ab
3b
a
a
t
– WW infinal
 Ans.
And.  =
dt
dw
= – 6bt = – 6b
3b
a
  = 2 3ab Ans.
1.47 : Given  =  = at To find tangential accn : (at
) : at
= R = Rat
To find radial accn : (ar
) :  =
dt
dw
 ar
= Rw2
 
t
0
w
0
dtdw  w =
2
at
dtat
2t
0
  ar
=
4
taR 42
Then
at

ar
tan  =
t
r
a
a
 tan  =
taR4
taR 42
 t = s7tan
a
4
3  Ans.
1.48 : Given   W    = angular accn.   = k W  k = const.
 Wk
d
dW
– W 

 (–)ive because W is decreasing. –



0
0
W
2
1
dkdwW
0
at
ar
y
x
R

23

k3
W2 2
3
0 ––– (i)  = Angular displacement.  Also  = k W
– dW = k W dt   
t
0
0
w
2
1–
dtkdwW
0
 t
k
2W 2
–1
0  ––– (ii)
Avg. angular velocity : Wavg
= <W> =
t

=
3
W
k
W2
k3
W2 0
2
1
0
2
3
0











Ans.
1.49: (a) Given W = W0
– a –––– (i). At t = 0   = 0  W = W0
Also
dt
d
= W0
– a   

 t
00 0
dt
a–W
d
 t)a–(Wn
a
1–
0
0 


 at–
W
a–W
n
0
0


  = )e–(1
a
W at–0
Ans.
(b) Put value of  in equation (i) : W = W0
– a 





)e–(1
a
W at–0
 W = WW0
e–at
Ans.
1.50: Given  = 0
cos    = W
d
dW
= 0
cos  



0
0
w
0
dcosdww 
2
W2
= 0
sin  
W = ±  sin2 0 Ans.
1.51: (a) Instanteneous axis of rotation is passing through that point
which velocity is zero always. If we observe carefully its
point must be at line joining AB. Then. at time t :
VP – 0
= RW = Rt = yt

y
x
B
A
V
y

–0
0
0
2–
 sin2–w 0
 sin2w 0
 cos0

x = vt ––– (i) Velocity of point P :
VP
= 0  v = ty x = v 





y
v
t = y
v
 put in (i)
y =
x
v2

Ans.
(b) Velocity of point O = 0
VO – P
= Rw VP
= u + at = wt
Then yw = wt  t =
w
wy
x =
2
1
at2
=
2
1
w
2
w
wy






 x =
w
yw
2
1 22
Ans.
1.52 : Since there is no slipping at ground. VC
= 0  V = RW where VC
= velocity of contact point.
To find distance, we have to find speed of a particle of rim then.  = wt  Time to one complets
journey = 2/w VP
= )–(cosvv2vv 22
 = 2 v sin /2 = 2V sin (wt)/2  Again
dt
ds
= 2 v sin wt/2  


w2
0
s
0
dtsin wt/2v2ds S =
w
v8
= 8 R Ans.
y
xO
P (x, y)
Vy t
y
x
O
P
(x, y)
yw = RW
y
x
C
w
V
A
W
O

R
C
B
P 
v
v

1.53: (a) Acceleration of point C : aC
= w  Velocity of point
C at time t : VC
= u + at = wt
Angular accn
of ball about its centre:
 =
R
w
R
w
R
aC   Angular velocity of ball at
time t : w = w0
t =
R
w
t  Velocity of point AA
w.r.t. centre C : VA – C
= RW = w t iˆ VC – E
= w t iˆ  VA – E
= 2 w t Ans.

25
Point B :
A
wt

wtC
B
y
x
VB–C
= Rw = wt (– j) VC–E
= wt i  VB–E
= wt (i – j) VB–E
= wt Ans.
Point O : VO–C
= – wt i  VC–E
= wt i  VO–E
= O Ans.
(b) Acceleration Calculations :
PointA:
A
ar
C
at
y
x
aC–E
= wi .... (i)  aA–C
= at
i – ar
j  Where at
= tangential acceleration. ar
= radial acceleration.
at
= R = w  ar
=
R
tw
R
2(wt)
R
v
222
  aA–C
= wi –
R
tw 22
j .... (ii)
(i) + (ii) : aA–E
= 2 wi –
R
tw 22
j  aA–E
= 2w










2R
wt2
Ans.
wt
wtC
0
A
wt
wt
y
x

Point B : a

C–E
= wi .... (i)  a

B–C
= (–R) jˆ + 







R
v2
iˆ =
– wjˆ –
R
tw 22
iˆ .... (ii)
(i) + (ii) : a

B
= 







R
tw
–w
22
iˆ + w jˆ  aB
= w










2R
wt2
Ans.
Point O: a

C–E
= w iˆ .... (i)  a

O–C
= (–R) iˆ +








R
v2
jˆ =
–w iˆ –
R
tw 22
jˆ .... (ii)
(i) + (ii) : a

O–E
= –
R
tw 22
j  aO–E
=
R
tw 22
Ans.
1.54 : Velocity of point A = 2v  Velocity of point B = v   accn
of point A =
R
v2
 accn
of point B =
R
v2
Again Radius of curvature =
onacceleratinormal
(speed)2
 RA
=
/RV
(2V)
2
2
= 4R
Again an
=
R
v2
cos 45º =
R2
v2
 RB
= R22
R2
v
)2(V
2


1.55 : Method : 1 (axis is rotating): w

1–0
= w1 iˆ  w

1–0
= w2 jˆ  w

1–2
= w1 iˆ – w2 jˆ
w1–2
= 2
2
2
1 ww   

1–2
=
dt
wd 2–1

= w1
dt
iˆd
– w2 dt
jd
direction of
dt
iˆd
is toward y axis and
this is rate of change of dirn
of x axis.
y
x

RC
B
y
x
C
0
A
2V
B v
v
V
x
B
an
v
2
R
V
45º
V
V 2
v
2
A
2V
R

27
Method : 2 (axis is not rotating)
w

1–C
= w1
cos  iˆ + w1
sin  kˆ   w

C–0
= w2
j  w

1–0
= w1
cos  iˆ + w1
sin  kˆ + w2 jˆ  2
2
2
10–1 ww|w| 


dt
wd 0–1


 = – w1
sin  






 
dt
d
iˆ + w1
cos  






 
dt
d
kˆ
And
dt
d
= w2
 

= – w1
w2
sin  iˆ + w1
w2
cos  kˆ
|

| = w1
w2
. Here x axis is directly attached with observer. Then
dt
iˆd
= – w2
kˆ  dθ1|iˆd|  
dt
dθ
dt
|iˆd|
 = w2
But
dt
jˆd
= 0 Because with frame of this observed dirn
of axis does not rotating then. 

1–2
= –
w1
w2
kˆ  1–2
= w1
w2
1.56 : w

= at iˆ + b t2 jˆ 
dt
dw


= a iˆ + 2b t jˆ (a)







 t
a
b
1at|w|


(b)








a
t2ab
1at||

 

w = w cos   cos  


w
.w

cos  
2224222
322
t4batbta
t2bta


   = 17º Ans
y
x0
w2
w1
y
x0
w2
y
x0
w2
w1
w2z

C
1
y
z
passing
y axis
1 unit
1 unit
d
j
d i
i

1.57: Method : 1 (axis are moving): It we see carefully
then x axis is moving while ydirection is not rotating.
Angular velocity of disc with respect to centre M
is : iˆ
R
V
W O–M 

.... (i) Angular velocity of centre
M w.r.t. origen. jˆ
OM
V
W O–M 

 OM = R cot 
 jˆtan
R
V
W O–M 

. Since angular velocity is vector
quantityit followvectoradditionlaw ]jˆtaniˆ[
R
V
W O–D 




cosR
V
tan1
R
V
|W| 2
O–D

Ans.
Again Angular acceleration () is :
dt
wd
β

 









dt
jˆd
tan
dt
iˆd
R
V
β

And since ydirn
isconstant
dt
jˆd
=0  But
dt
iˆd
 |iˆd| = d dt
d
dt
iˆd 
 = wMO
 kˆtan
R
V
–
dt
iˆd

kˆtan
R
V
R
V
β 

















 kˆtan
R
V
–β 2
2


 2
2
R
V
|β| 

tan 
Method : 2 (axis are not rotating)
angular velocity of disc w.r.t. M : WD–M
=
R
V
cos  iˆ +
R
V
sin  jˆ angular velocity of M w.r.f. O
: jˆtan
R
V
jˆ
cotR
V
W O–M 



angular velocity of disc
wrt O: W

D–O
= W

D–M
+ W

M–0
 W

D–O
=
R
V
cos  iˆ

R
V
sin  jˆ  
R
V
tan  jˆ 





















cosR
V
tan
R
V
sin
R
V
cos
R
V
|W| O–M

Ans.
y
x
R0
v
M
p


z
y axis
y axis
d
j
d i
i
y
x0
w =0
D
z


V
R
V
M

29
0jˆ
dt
dθ
θcos
R
V
iˆ
dt
dθ
θsin
R
V–
dt
wd
α O–D
O–D 














and
dt
dθ
is angular velocity of M w.r.t.
0 : d/dt =
R
V
tan   2
2
O–D
R
V–
α 

sin  tan  iˆ + 2
2
R
V
cos  tan  jˆ | O–Dα

| = 2
2
R
V
tan  Ans.
1.58 : Method : 1 : (Direction of Co-ordinate axis is fixed)
At time t line AB rotate by  angle then A–PW

=
w0
cos  iˆ + w0
sin  kˆ   
 t jˆ
| A–PW

| = 22
0
22
0
22
0 tsinwcosw  
| A–PW

| = w0










0
0
w
t
Now
jˆkˆ
dt
d
cosW
dt
d
)iˆ(sin– W
dt
Wd
000
A–P







 





= –W0
0
t sin  iˆ + W0
0
t cos  kˆ + 0
jˆ  2
0
22
0
2
0 tw|| 


22
00 tw1|| 

Ans.
Method : 2 (x axis is moving) : Here we take line AB along
x dirn
. jˆtiˆwW 00A–P 

 2
0
2
0A–P
w
t
1w|W|





jˆ
dt
jˆd
t
dt
iˆd
w
dt
wd
000
A–P 


 Here 0
dt
jˆd

But
dt
iˆd
= 
t kˆ  Then kˆtkˆ
dt
ds
dt
di
0 

= w0

t kˆ +  jˆ   2
1
20
20 tw1|| 

Ans.
y
xA
0
w0
w0
z

y
x
A
0
w0
z
P


1.2 The FundamentalsEquation of Dynamcis
1.59: Where F is force due to air which will be constant then. mg – F
= mw .... (1). When m mass is taken then F – (m – m) g + (m
– m) w .... (ii) From (i) and (ii) : m =
w
mw2

Ans.
1.60 : Net pulling force = (  m) a  m0
g – k m1
g – k m2
g = (m1
+ m2
+ m0
) a  a = 









021
210
mmm
)m(mk–m
g Ans.
F. B. D of m2
:
m2
a
k m g2
T
T – k m2
g = m2
a. Put value of a then T =
210
0
mmm
mk)(1


m2
g Ans.
1.61 : (a) Pulling force F = m1
g sin  + m2
g sin  – k1
m1
g cos  – k2
m2
g cos . Then acceleration
a is : a =
21
2211
mm
)mkm(kcosg–m2)(m1sing


.... (i)
F. B. D of m1 : N = Normal force between two blocks. m2
g sin  + N + k1
m1
g cos  = m1
a .... (ii) put value of (a) in (ii) : N =
21
2121
mm
αcosmm)k–(k
 Ans.
F
mg
m
F
(m – m) g
m – m w
km g2
m2m1
a
km g1
m0
m g0
(b) For minimum value of acceleration will be zero and friction force will at maxm
value then
0 = [g sin (m1
+ m2
) – g cos (k1
m1
+ k2
m2
)] / m1
+m2
 tan  =
21
2211
mm
mkmk


a
m
g
sin
1


k
m
g cos
2
2

k
m
g
cos
1
1

m
g
sin
2

m 2
m
1
N
m
g sin
1


k m
g cos
1
1

m1

31
1.62 : Upward Journey : We know S = Vt –
2
1
at2
where V = final velocity in this situation Vf
= 0
Then  =
2
1
(g sin  + µg cos ) t2
–––– (i)
t = Time taken in upward journey.
Downward journey : We know S = ut +
2
1
at2
u = initial
velocity. Then  =
2
1
(g sin  – µg cos ) (t)2
–– (ii)
and u = 0  now : (ii)
(i)
 = 2
1
cosug–sing
cosugsing









  µ = 









1
1–
2
2
tan  Ans.
1.63 : (a) Starts coming down. m2
g > m1
g sin  + fmax
m2
g > m1
g sin  + k m1
g cos  
1
2
m
m
> sin  + k cos  (b) m1
g sin  > m2
g + k m1
g cos   g sin  >
1
2
m
m
+ k cos  
1
2
m
m
< g sin  – k cos  (c) At rest : Friction will be static : g sin  – k cos  <
1
2
m
m
<
sin  + k cos 



a
v
v
=
0
f
a =
(g sin
+
µ
g cos
)



a
v
u = 0
a = (g sin
– µg cos )


m
1

m2
m g2
m
g
sin
1

f
=
k
m
gcos
m
ax
1


m g2
m
g
sin
1

a
fmax
m g2
m
g
sin
1

a
fmax

1.64: To find tendency of sliding check value of m2
g
and m1
g sin  = m1
g sin 30º = m1
g/2  0.5 m1
g
 Here m2
g = m1
 g =
3
2
m1
g  0.66 m1
g Then
m2
g > m1
g sin Block m2
has tendency to
move down ward.
Then.Pullingforce =m2
g– m1
gsin – k m1
gcos acceleration a=
21
112
mm
cosgmk–singm–gm


a =
1
]cosk–sin–[g


Ans.
1.65 : (a) Before no sliding b/w m1
and m2
:
F.B.D. of System : Acceleration of both block will be same. w1
= w2
= w =
2121 mm
at
mm
F



But friction b/w m1
and m2
will be static then. m1
fr
fr = m1
[w1
] =
21
1
mm
atm
 < k m2
g  t <
1
212
ma
)m(mgmk 
 w1
= w2
=
21 mm
at
 ––– (i)
If t > k m2
g
1
21
ma
)m(m 
. Sleeping b/w two block will be start then F.B.D. of m2
:
m2 F = at
km g2
w2
 w2
=
2
2
m
gkm–at
–––(ii) Ans.
F.B.D. of m1
: m1
k m g2
 w1
=
1
2
m
gmk
––––– (iii) Ans.
Assume to
=
1
212
ma
)m(mgmk 
 t > to
: Slope of w2
>
Slope of w1
= w2
because of Slope of w2
= 







2m
a
from
(ii) [After sliding] Slope of w1
= 







 21 mm
a
from (i) [Before sliding]
Slope w1
= 0 from (iii) [After sliding]
m
1

m2
m g2
m
gsin
1

f
=
k
m
g
cos
m
ax
1

k
Given 
m
m
1
2
.
m2
m1
fr = k m g2 F = at
m2
m1
w F = at
w
t0
t
w1
w = w1 2
w2

33
1.66 : F.B.D. of block : ma = mg sin  – k mg cos   a = g sin  – kg cos 
Time to reach at bottom : s = ut +
2
1
at2
 sec  =
2
1
(g sin  – kg cos ) t2
A
R


k = coefficient of friction mg sin 

k mg cos 
t2
=
]cosk–cos[sing
2
)cosk–(sing
sec2
2



 
t =
]cosk–cos[sing
2
2


–– (i)
To minimise t  (sin  cos  – k cos2
) will be minimum also. Assume. x = sin  cos
 – k cos2
  Now
d
dx
= 0 – sin  sin  + cos2
 + 2 k cos  sin  = 0
cos 2 = – k sin 2 tan 2 = – k
1 Ans.
Put value of k :  = 49º Put value of  in equation (i) :
tmin
=
49]cos0.14–49ºcos49º[sin10
2.102
2

 1.0 s. Ans.
1.67 : When block pull up Direction of friction will be downward.
F.B.D. of block : N = mg cos  – T sin   fr = k (mg cos  – T sin ). At just sliding:
T cos  = mg sin  + k (mg cos  – T sin ) T =


sinkcos
cosmgksinmg
 Tmin
 

m k
T


m
T
mg sin 
mg
mg cos 
µN
T
cos T
sin 
N
(cos +k sin ) min  minimumvalueof cos +k sin = 2
k1  Tmin
=
2
k1
)cosk(sinmg



1.68 : (a) At time of breaking off the plane vertical component of
F

must be equal to weight mg. Then F sin  = mg = at sin
 t = sina
mg . Motion equation of block : a1
= Accelration
of block. F cos  = m a1
 a1
=
dt
dv
m
cosat


 

1t
0
V
0
dtt
cosa
dVm

2
t
αcosa
mV 2
1 


sina
gm
2
1
αcosa
mV
22
22
 V ==


sina2
cosgm
2
2
Ans.
(b)  

t
0
V
0
dtt
cosa
dVm

2
t
αcosa
mV 2

1.69 : Motion equation :  = aS 
3
mg
cos  = m a1
a1
=
acceleration of block of mass m. a1
=
m3
mg
cos  =
3
g
cos
  a1
=
3
g
cos as   
s
0
v
0
dsascos
3
g
dvv 
s
0
v
0
2
a
assin
3
g
2
v
  v2
=
a3
g2
sin as  v = assin
a3
g2
 v =
m2
cosa 
t2

2
tcos
m2
a
dt
ds
  


t
0
2
s
0
dtt
m2
cosa
ds  s = 3
3
x
t
m2
cosa 
 s =


sina6
cosgm
32
32
Ans.Ans.
1.70 : Method : 1
Motion of mass 2 m : T – k2mg = 2 mw  T = 2
mw + 2kmg
Motion of motor : T – kmg = ma  2 mw + 2 k mg – k mg = ma  a = 2 w – kg
Acceleration of 2 m w.r.t. m : a2m – m
= 2w – kg + w = 3w – kg  Suppose in time t both
will be met then.  =
2
1
(3w – kg) t2
 t = kg–3w
2 
Ans.
m
F
=
at

µ = 0
m
F
=
m
g/3

2m w
k2mg
m
a
k mgT T
kk

35
Method : 2
On system : Tension will be internal force hence k2 mg – k mg = – 2 mw + ma 
a = 2w – kg  a2m – m
= 3 w – kg then  =
2
1
(3w – kg) t2
 t = kg–3w
2 
Ans.
1.71: For observer inside elevator : a1–2
=
21
010212
mm
wm–wmgm–gm


=
21
012
mm
)w(g)m–(m


But in form of vector : a1–car
=
21
012
mm
)w–g()m–(m


Ans.
Acceleration of m1
w. r. t. shaft : a1–shaft
= a1–car
+ acar
=
21
012
mm
)w–(g)m–(m

+ w0
   a1–shaft
=
21
021112
mm
wmmgm)m–(m


Ans.
Tension in string : T – m1
g = m1
a1–shft
  T =
21
021
mm
)w(gmm2


Force applied by pulley on ceiling = 2 T =
21
021
mm
)w(gmm4


 as vector form:
2T = )w–g(
mm
mm4
0
21
21 
 Ans.
1.72: Motion of body (2) : mg – T/2 =  ma  2 mg – T
= 2 ma .... (i)
Motion of body (1): T – mg sin  = m a/2 .... (ii) from (1)
and (ii) :
]1[
]sin–[22g


 Ans.
1.73 : Equationof motion : T = m0





 
2
aa 21
.... (i) m1
g
= T/2 = m1
a1
.... (ii) m2
g – T/2 = m2
a2
.... (iii)
from(i), (ii) and(iii): a1
= )m(mmmm
g).m–(mmmm4
21021
21021


Ans.
m g1
m1
T
T T
m2
m g2
2T
shaft
w0

m
mg sin 
a/2
T
T/2
a
m
mg

T/2
T/2
m2
m1
a1
mg
m g2
m0
T
a + a1 2
a2
2

1.74: Motion equation on system: Mg – mg = M a1
+ m a2
.... (i) Motion equation
of m : fr – mg = ma2
.... (ii) Since length of rod is  then.
2
1
 (a1
+ a2
)
t2
.... (iii). From (i), (ii) and (iii) : fr
= 2
tm)–(M
mm 
Ans.
1.75:  = 100 cm – T + mg = m a1
.... (i) + 2T mg = m
2
a1
.... (ii) from
(i) and (ii) : a1
=
4
g2)(–


arel
=
2
a3 1
. Suppose t time is taken then
 =
2
1
arel
t2
 t = g)–(2
4)(


Ans.
1.76 : Motion equations T – mg = ma .... (i)  mg – 2T =  m a/2 ....
(ii) from (i) and (ii) : a =
4
2)–(2g


. When body
(1) travel h distance then in same time body (2) travel 2h distance
in upward dirn
using constraint relation. Nowvelocity of body (2) Just
before string slack.
V2
= 2a (2h). It body (2) travel x distance again then V2
= 2a (2h) = 2gx
x = g
ah2
. Total hight from ground : 4 = 4
h6
h
g
ah2


 Ans.
1.77: F. B. D. of N sin  = m a2
.... (i) F. B. D. of rod : mg – N cos  = m a1
.... (ii) Constraints:
a2
sin = a1
cos  .... (iii) From (i), (ii) and (iii) : a1
=
 cot1
g
2 a2
=
 cottan
g
T
2T
mg
nm
T
a1

m
nmg
a
2
1
mg
Mg
Ma1
m
a2
mg
2
mg
T a
2T
h
a
2
N
mg
a1

a1

a2


m a2
a1

N a2


37
1.78: F. B. D. of m : mg – KN – T = m a2
.... (i) N = m a1
.... (ii) F. B. D. of wedge : a1
= a2
....
(iv) From (i), (ii), (iii) and (iv) : a1
=
m
M
k2
g
kmM2m
mg



Ans.
mg
T
m
KN
M
a1
a2
1.79: F.B.D of bodies on frame of wedge: Since system is stationary on frame of wedge hence :
mg = kma + ma + kmg  a =
k1
k)–(1g

Ans.
a2
a1
mg
m
N
TKN
1.80: F.B.D. of block (2) with frame of wedge: At maximum accn w : fr1
will be maxm
then fr1
= k [mg
cos  + mw sin ]. Since block is under rest with frame of wedge, then equilibrium equation
of block along incline. k (mg cos  + mw sin ) + mg sin  = mw cos   w =
k–cot
)cotk(1g


a
m
ma
mg
2 m N
1 m
ma
m
ma
kmg kma
mg
 mg
mw


fr1

2
mg
mw
w
m
1
fr
1.81 : Method : 1
F.B.D. of System a2
= accn
of bar w.r.t. incline. Since no force on system in horizontal direction
then O = ma1
+ m [a1
– a2
cos ] –––– (1) F.B.D. of bar w.r.t. wedge : m2
a1
cos  + m2
g
sin  = m2
a2
–––– (ii) from (1) and (2) : a1
=


sinmm
cossingm
2
21
2
Ans.

Method : 2
F.B.D. of bar : with frame of wedge, bar has zero accn
in perpendicular to incline then. N + m2
a1
sin  = m2
g cos  –––– (i) F.B.D. of wedge : N sin  = m1
a1
–––– (ii) from (i) and
(ii) : a1
=


sinmm
sincosgm
2
21
2

2
m2
1
m1

a2
m2
m1
a1
m a2 1
m2
m g2
Method : 3
F.B.D. of bar with frame of ground observes : Motion
equation in perpendicular dirn
of incline then : m2
g cos  –
N = m2
a1
sin  ––– (i)
1.84: v = 360 km/hr = s
m100
3600
1000360


At point A : N – mg =
R
mv2
N = mg +
R
mv2
= 70 g +
500
[100]70 2
= 70
g +
500
10010070 
= 70 g + 140 g = 210 g = 2.1 kN Ans.
At point B : N =
R
mv2
––– (i) put value of m, V and R N = 1.5 kN Ans.
At point C : N + mg = mv2
/R N = mv2
/R – mg ––– (ii) put value of V, m, R
in (ii) N = 0.7 kN Ans.
a2
m g2
N


a1
v
C
v
B
R
A
v
N
mg
N
m a2 1
m g2
N


 N
m1
a190 – 
N
mg

39
1.85: Tangential acceleration (at
) : mg sin  = m at
 at
= g sin 
Radial acceleration (ar
) : Energy conservation : mg  cos 
=
2
1
mv2
 v = cosg2   ar
=

2
v = 2 g cos  
anet
=  cos4singaa 22
rt 22 anet
=g  cos31 2
Now : T – mg cos  =

2
mv
= 2 mg cos   T = 3 mg cos  Ans.
(b) Component of velocity in y dirn: vy
= v sin 
vy
= cosg2  sin   v2
y
= 2 g  cos  sin2

For vy
maximum v2
y
will be maximum. Then x = cos  sin2
 will
be maximum. d
dx = 0 = 2 cos2
 sin  – sin3
 = 0  2 cos2

= sin2
  tan  = 2  sin  =
3
2
cos  =
3
1
 v2
y
= 2 g  3
2
3
1
 
33
g4
vmax
y

  T = 3 mg cos  = 3 mg 





3
1
= mg 3 Ans.
(c) If no acceliration in y direction then. ar
cos  = at
sin  2 g cos2
 = g sin2
  tan  = 2  cos  =
3
1
Ans.
1.86 : Extreme position : Only tangential acceleration is prerent at extreme
position. at
= g sin 
Lowest Position : Energy conservation : mg [ –  cos ] =
2
1
m v2
 v2
= 2 g  (1 – cos )  ar
=

2
v
= 2 g (1 – cos )
A/C condition : ar
= at
 g sin  = 2 g (1 – cos ) sin  + 2
cos  = 2  cos  = 5
3   = 53º Ans.
m
T

mg
v

v
y dirn.

at

ar


mgv

1.87: Particle will break off sphere when normal reaction will be zero.
mg cos  =
R
mv2
 v2
= Rg cos . Energy conservation : mgR
(1 – cos ) =
2
1
mv2
mgR (1 – cos ) =
2
1
mRg cos  
1 – cos  =
2
cos 
 cos  =
3
2

2
cos3 
= 1  v2
=
Rg 





3
2
 v =
3
Rg2
Ans.
1.88: Top view : Spring force = F =  N = normal reaction on sleeve. Since no acceleration
in tangential direction. N sin  =    cos  ––– (i). Equation in radial direction: N cos 
+   sin  = m r w2
= m ( + ) cosec  w2
from (i):


sin
cosx 
(cos ) +   sin 
= m ( + ) cosec  (w2
)  x   = m  + m   w2
  = 2
mw–x
m 
Ans.
A
A
v
mg
R
R
 
N
w

r
v


l

w

1.89:k = k0
= 





R
r
–1 friction coefficient. Suppose cyclist at radius of r. then friction provide centripital
force to motion on circular path. Then where kmg = friction force. k0 r
mv
mg
R
r
–1
2






 v2
= k0
g








R
r
–r
2
. To maximum value of v : x =
R
r
–r
2
will be maxm. Then
dr
dx
= 0 
0
R
2r
–1   r = 2
R  4
Rgk
4
R
–
2
R
gkV 0
0
2
max 





  Vmax
= gRk
2
1
0 Ans.Ans.
O
R
r kmg
mv /r2

41
1.90: Tangential and radial both acceleration is only provided by friction because
friction is acting as external force. Then maximum value of friction = k mg.
Velocity of car after d distance travel. v2
= 2 w
d. Then ar
= Radial
acceleration =
R
dw2
R
v2

   at
= Tangential acceleration = w 

anet
= 2
2
2
2
R
4d
1ww
R
d2w








 Fnet
= m w 2
2
R
4d
1 = k mg
Squaring : 







 2
2
2
R
4d
1w = k2
g2
 d = 1–
w
kg
2
R
2









Ans.
1.91 : y = a sin  α
x  k = frictioncoefficient. Centripital force
=
R
vm 2
and centripital force will be provided by friction.
At limiting condition : mgk
R
vm 2
 v2  k Rg.
For v maximum  R willbe minimum.And weknow : speed
is constant also we are secing that radius of curvature will be
minimum at maximum point of curve then. ay
= 2
2
2
2
av
dt
yd


sin  α
x . At maximum value of curve : sin α
x = 1 
ay
= 2
2
α
av
R =
av
αv
a
v
2
22
y
2
 R =
a
2
 . Then v2 g
a
α
k
2
   v
a
kg
 Ans.Ans.
1.92: F.B.D. of differential element of length dl. Equation of motion : 2T sin  – N cos  = (dm) Rw2
––– (i)  N sin  = (dm) g put value of N in (i) : 2 T  – (dm) g cot  = (dm) Rw2
2T
R
v /R2
w
y
x
y
x
v
ay
mg
N
w


 
T T
(dm)
N

g
T
T
 = very small angle.
sin = 

 –
R2
m

(R 2 ) g cot  = (R 2 )
R2
m

Rw2
 T –




2
wRm
2
cotgm 2
T =
2
m
[Rw2
+ g cot ]  T =
2
gm









g
Rw
cot
2
Ans.
1.93 :
1
2
m
m
= 0
  T1
– m1
g = m1
a ––– (i)  m2
g – T2
= m2
a ––– (ii)
Relation between T1
and T2
: (T + dT) sin 2
d + T sin 2
d = dN
Td = dN  dfr = µ dN = µ Td ––––– (i)
Since  is very small. sin d  d  cos d  1
In horizontal direction : (T + dT) cos 2
d – T cos 2
d = dfr
dT = dfr ––– (i)  dT = µ T d  


0
T
T
dµ
T
dT
2
1
  n (T2
– T1
) = µ 
T2
= T1
eµ
––– (iii)
from (i) and (ii) : a)(gm
a)–(gm
T
T
1
2
1
2


from (iii) :

 µ
1
2
e
T
T

Then eµ
= 












 ag
a–g
ag
a–g
m
m
1
2
––– (iv). Since pulley is fixed : Before skiding a =
0   = 0
 eµ
= 0
 µ =

1
n 0
(b) from equaiton (iv) : en0 =  





 ag
a–g
 ag
a–g0




 a = g
0
0
n
]–[n


Ans.
T1T2
m1m2
m g1
m g2
a
d
T + dT
dN
dfr
T
y
x

43
1.94: Hence velocity along yaxis is not responsible for circular
motion only velocity along Z-axis is responsible.
Then N =
R
Vm 2
Z
VZ
= V0
cos  N =
R
cosVm 22
0 
1.95: x = a sin wt  ax
= 2
2
dt
xd
= – aw2
sin wt 
y = b cos wt  ay = 2
2
dt
yd
= – bw2
cos wt  jˆayiˆaa xnet 

= jˆwtcoswb–iˆsin wtwa– 22
 ]jˆwtcosbiˆsin wt[amw–F 2



jˆwtcosbiˆsin wtajˆyiˆxr 

 2
wrm–F


where r

=positionvectorofparticle.F = mw2 22
wt)cos(bsin wt)(a 
 F = mw2
 22
yx  Ans.
1.96 : Method : 1 (Impluse equation)
(a) We know P

 = Impluse in time t = 
t
0
dtF = 
t
0
dtmg– P

 = – mg t Ans.

v0
x
R
Z
O
y
V0
w 
x
y
g
V0
m
(b) T = g
sinV2 0 
 g.V0

= V0
g cos (90 + )  g
sinV2
mgP 0 


– g.V0

= V0
g sin   = 2 m V0
sin   V0
sin  =
g
g.V– 0


g
g.V
2mP 0


 Ans.
Method : 2 (Kinematic)
(a) jˆsinViˆcosVV 000 

 jˆgt)–sin(ViˆcosVV 00f 

where fV

= final velocity vector then 0f Vm–VmP

 
jˆtgm–P 

Ans.
V0

y
x
V0

90+
g

(b) T = g
sinV2 0 
 jˆ
g
sinV2
gm–P 0 


 P

 = – 2 m V0
sin  jˆ
From method : 1 : V0
sin  =
g
g.V– 0

 jˆ
g
g.Vm2
P 0


 Ans.
1.97 : m
– t)(TtaF


(a) wmF

  w

=linearacceleration. ]– tt[
m
a
w
dt
Vd 2













 
 t
0
2
t
0
V
0
dtt–dtt
m
a
Vd









 

3
–
2
t
m
a
V
32
 f
3
P
6
a
Vm



 Ans.
(b) ]– tt[
m
a
w 2


 







 
t
0
2
t
0
V
0
dtt–dtt
m
a
Vd












3
t
–
2
t
m
a
V
32









 

0
3
0
2S
0
dt
3
t
–dt
2
t
m
a
dS









 

12
–
6
.
m
a
S
43

12m
a
S
4

  S =
m12
a 4

Ans.
1.98: sin wtFF 0

  a =
dt
dV
sin wt
m
F0    
t
0
0
V
0
dtsin wt
m
F
dV  V =
t
0
0
wtcos
mw
F
–
V = wt]cos–[1
mw
F0
  
t
0
0
S
0
dtwt]cos–[1
mw
F
dS  S = 





sin wt
w
1
–t
mw
F0

S = 2
0
mw
F
[tw – sin wt] distance
distance will be increasing function w.r.t. time then
dt
dS
= (+)ive or > 0  w – w cos wt > 0 
cos wt < 1 0 < wt <  2 < wt < 3 
4 < wt < 5

wt
  

45
1.99: F = F0
cos wt 
m
F
a
dt
dv 0 cos wt   
t
0
0
0
0
dtwtcos
m
F
dV 0 =
wm
F0
sin wt 
sin wt = 0  wt =   t =
w
π
Ans.
Now :  
t
0
0
0
0
wtcos
m
F
dV  V =
wm
F0
sin wt .... (i)   
t
0
0
s
0
dtsin wt
wm
F
ds 
S = wt)cos–(1
wm
F–
2
0
at t = w
π s = 2
0
wm
2F
velocity will be maximum when sin wt = 1
 Vmax
=
wm
F0
Ans.
1.100 : (a) F = – rV  V
m
–r
dt
dV
   
t
0
V
V
dt
m
r–
V
dV
0
 t
m
r
–Vn
V
V0
  
V = V0
e–r/m
t  V will be zero when t  Ans.
(b) a = –
m
r
V =
ds
dv
v   
s
0
V
V
dsv
m
r
–dVv
0
V – V0
=
m
r
– S  V = V0
m
r
– S
Total distance travel by particle is S1
then final velocity = 0  0 = V0 m
r– S1

S1
=
r
Vm 0
Ans.
(c)

0V
= VV0
–
m
r
S2
 S2
= 






 1–
r
Vm 0
–––– (i)  Also

0V
= VV0
2tm
r–
e
t2
= n
r
m
 –––– (i)  <V> =



n
1)–(V
t
S 0
2
2

Ans.
1.101 : F  v2
 F = – kv2
 a =
m
kv– 2
  
t
0
V
V
2
dt
m
k–
v
dv
0

m
kt–
V
1–
V
V0
  m
kt
V
1
V
1
–
0
  t
k
m
VV
V–V
0
0








––– (i)
V0 V
h

Also a =
dx
dv
v
m
kv– 2
   v dv =
m
dxkv– 2
  
h
0
V
V
dxk–
V
dvm
0
 m n
0V
V
= – kh
k =
h
m
n
0V
V
put in (i) : t =
V
V
n
h
VV
V–V
00
0







Ans.
1.102: Suppose at time t distance travel is x then F = mg sin  – ax
mg cos   mw = mg sin  – ax mg cos  where w =
acceleration of block. w = g sin  – ax g cos  .... (i) v = g
sin  – ax g cos  
dx
dvv
= g sin  – a x g cos  
 
mx
0
0
0
dxα)cosaxg–αsin(gdvv  0 = (g sin ) x –
2
cosgxa 2

x = 

cosga
sing2
x =
a
2
tan . Velocity will be maximum when dt
dv = 0 = acceleration  g sin  = ax g cos  
x =
a
1
tan   now 


tanay
0
maxv
0
dxα)cosgxa–αsin(gdvv 


maxV
= g sin 
tan
a
1
2
cosg2
–tan
a
1















  VVmax
=
a
g
αcos
αsin
a
g 2
 sin  tan  Ans.
1.103: K = friction coefficient  Block start sliding when.
F = at1
= k m g  t1
= a
gmk . Assuming over time start from block start sliding then.
Suppose acceleration of block is w then. mw = at – kmg = a (t1
+ t) – kmg = a t  w
=
m
a
 t 
2
t
0
v
0
t
2m
a
Vt)(dt(
m
a
dV  

 t)(dt
2m
a
ds
t
0
2
s
0




S =
6m
a
(t)3
=
6m
a
(t – t1
)3
Ans.

mg sin 
K
mg cos 
k= ax
t = 0 t = t1
t = t – t1
t = t
mK F = at

47
1.104 : h
v = 0
V0
Upward journey :
mg
F = Kv
2
V0
Fnet
= mg + kv2
 a = g +
m
kv 2
 


m
kvmg
ds
dvv
–
2
mv dv = (mg – kv2
) ds
 

0
v
h
0
2
0
ds–
kvmg
dvmv
 h =
mg
mgKV
n
2k
m 2
0 
 .... (i)
Downward Journey : Fnet
= mg – kv2
mg
k v
2
v  a =
m
kv–mg 2
= v dv/ds
 mg
mgKV
n
2k
m
hds
kv–mg
dvmv 2
0
v
0
h
0
2

    2
2
0
kv–mg
mg
n
2k
m
mg
mgKV
n
2k
m
 

 v =
mg
KV
1
v
2
0
0

Ans.
1.105 :(a) Position vector of particle : r

= r cos  iˆ + r sin  jˆ
 At time t :  = wt  )jˆsiniˆ(cosFrˆFF 

 ]jˆsiniˆ[cos
m
F
a 


 
t
0
t
0
V
0
dt(sin wt)
m
F
iˆdtwt)(cos
m
F
vd


 ]jˆwt)cos–(1iˆ[sin wt
mw
F
V 


speed =
22
wt)cos–(1(sin wt)
mw
F
|V| 

 Speed = 2
wtsin
mw
2F
Ans.
(b) Distance is calculated by speed. Then 2
wtsin
mw
2F
V
dt
ds
  dt2
wtsin
mw
2F
ds
t
0
s
0
 
S = Distance S = –
t
0
2 2
wt
cos2
mw
2F
 S = 





2
wt
cos–1
mw
F4
2 .... (ii) velocity of particle

r
w
m
F
0

will be zero : 0 = 0
2
wtsin
2
wtsin
mw
2F
  put value of t = w
2π in (ii)
S = 2
mw
F4
(1 – cos )  S = 2
mw
F8
 Average speed =
/w2
8F/mw
Time
Distance 2


Average speed = mw
4F
 Ans.
1.106: K = tan  = friction coefficient  When = W
= net
tangential acceleration/ Wx
= Acceleration
in x direction. Now W
=
m
cosmgk–cossinmg 
= g (sin  cos  – K cos ) = g [sin  cos  – sin ]  W
= g sin  (cos  – 1)... (i)
Acceleration in x dirn: Wx
=
m
coscosmgk–sinmg 
 WWx
= g sin  [1 – cos ] .... (ii)
From (i) and (ii) : W
= –Wx

dt
Vd
–
dt
Vd xτ
   V
= –Vx
+ cos  at t = 0 V
= V0
and Vx
= 0  V
= – Vx
+ V0
.... (iii) Then const = V0
 Also Vx
= V
cos  ....
(iv) From (iii) and (iv) : V
=
 cos1
V0
Ans.
1.107: Tangential force on system is F then  F
   sing(dm) =   sing)dR( = Rg



t
0
dsin   = R f
  in
= 0 = –  Rg
f
cos


 f
= R
  f
=
R

F
=
Rg   R
cos–1   a
=
m
gR
  Rcos–1  =
x
gR
  R
cos–1   a
=

Rg
  R
cos–1 

v
w
mg sin 
K mg cos 


x
f

y

R
d
dm= R d 
x
(dm)g
F


d
dm


49
1.108: At time of break off. normal reaction will be zero. Where mw0
= Psuedo force because observer at sphere.
R
mV2
0
= mg cos 
– m w0
sin 
 V2
0
= Rg cos 
– R w0
sin 
.... (i)
Energy equation: Wall
forces = Kf
– Ki
 Wpsuedo
+ Wmg
= Kf
–
Ki
m w0
R sin 
+ mg [R – R cos 
] =
2
1
m V2
0
– 0  V2
0
= 2w0
R sin 
+ 2 g R [1 – cos 
].... (ii) From (i) and (ii) : V2
0
= 2R








R
V– 2
0
+ 2 g R V2
0
= 2 g 3
R  V0
=
3
gR
 cos 






















g
w
13
g
wg5
3
w
0
00
Ans.
1.109 : Given F n
r
1
   F = n
r
K
Where K = Constannt.
A particle is said to steady if we displace particle away from origin, particle want to regain
its position and also it we displaced particle toward origin, particle want to regain its original
position. Then. At steady state (mean position) : F =
r
Vm 2
0
Then
r
Vm
r
K 2
0
n
 .... (i)
It we increases r then. F –
r
Vm 2
> 0  K r–n
– mV2
r–1
> 0  Differentiate both side
with respect to r for small increase of r: –n k r–n–1
dr + mV2
r–2
dr > 0 and
r
1
r
Vm
r
1
r
Kn– 2
0
n




















> 0 .... (ii)  From (i) and (ii) n < 1 Ans.
w = acceleration0
R
R0
0
mw0
V0
mg
R
O F
v0
r
r
0
v

1.110: At steady state: mg sin  = m (R sin ) w2
cos   cos 
= 2
wR
g
. Case (i) : If Rw2
> g then cos 2
wR
g
is defined and
only one equilibrium position will be exist and will be steady.
Case (ii) : If Rw2
< g then only 2
wR
g
= 0º will be equilibrium
position because tangential fore along arch of ring due to mg
will be greater than that of pseudo force and object will come
at lower position of ring.
R
N
Y
mg
m (R sin ) w
2
Y
90–




FUNDAMENTALS OF PHYSICS

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