FUNDAMENTALS OF PHYSICS

1

raftMotor 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 boatwill 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 = 0V2

S Time taken in second journey :

2t V

2t V

2S 2

22

1 t2 = 21 V V

S

Mean velocity = 21 t t

S =

210 V VS

VS/2

S

Mean velocity = 021

2102V V V

)V (V 2V

Ans.

Q.1.3 Method : 1 (Graphical Approach) Given tan =

timeTotal

trapziumof Area tan

2t– ) t (

21

Time

ntDisplaceme

4– 1 t Ans.

Fixed point

6 km u

distance = 2 × u

raft

1 hr.

t1

6 km

Part OnePhysical Fundamenatls of Mechanics

1.1 Kinematics

V

t

t

– t2

– t2

Compact ISC Physics (XII)2

Method : 2 (Analytical)

Total displacement (S) = 22

2t–

21 t

2t–

2t–

21

Time taken = =

22

2t–

21 t

2t–

2t–

21

St =

4– 1 Ans.

Q.1.4 (a)

S

2m

20 s t

average velocity = 202

= 0.1 m/s = 10 cm/s Ans.

(b) Velocity will be maximum when slope of S (t) curve will be maximum.

V = 10– 140.4– 1.4

m/s = 25 cm/s Ans.

(c) Instanteneous velocity may be equal to mean velocity whenslope of line joining final and initial point will be same toslope at point on curve. Ans. : to = 16 s

Q.1.5 Velocity of B with respect to A : 12A– B V– V V

Position of B with respect to A : 12ABA– B r– r r– r r

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 r V r || 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 sapprox

t

S

1.4

14t

0.4

10

A

r1

v1

v2

r2

B

Y

X

3

E– SV

= 30 i; j 60sin 15– i 60 cos 15 V E– W

E– W E– SS– W V V– j 60sin 15– i 30)– 60 cos (15 V

km/hr 40 60)sin (15 30)– 60 cos (15 |V | 22S– W

19º

30– 60 cos 1560sin 15– tan Ans

Method 2

30 km/hr

VW – S15 km/hr

60º

we know E– SE– W S– W V– V V

60 cos V V 2– V V |V| E– SE– W 2

E– S2

E– W S– W

40 km/hr Ans.

from figure : tan = 60 cos 15– 3060sin 15

= 19º

Person : 1

Q.1.72 km/hr

d

2.5 km/hr

Time to cross the river : t = cos 2.5

d ––––– (i)

from figure : sin = 54

2.52

cos = 53

Put in (i) : t = 1.5d

––––– (ii)

Person : B

2 km/hr

d2.5 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 = 2.5d

t2 = Time to walk on bank = u tan d

ux And tan = 5

4 2.52

t = 54

ud

2.5d

–––– (iii)

from (ii) and (iii) : 5u4d

2.5d

1.5d

u = 3 km/hr Ans.


Q.1.8 Boat A : dd

time to reach again at same point.

tA = – W Vd

W Vd

–––– (i) where V = Speed of boat w.r.t. river

W = Speed of water w.r.t. earth.

Boat B : dW

V Time to reach again at same point. tB =

22 – W V

2d ––––– (ii)

Also given V = ––––– (iii) now 1–

–

2d –

d

d

tt

2222

B

A

1–

tt

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) cos Vd

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 VV

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– r r– mE– m V V V

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 – EA

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 = 21

VV

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 = )– (90 cos V 2V– V V 0020

20

= V0 )sin – (1 2 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 – (1 2

= 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 = : j t g i 3– V1

; Velocity of particle (2) at time t : j t g i 4 V2

Since 0 V . V V V 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 V V

+ = 90

= 90 – tan = tan (90 – ) tan = cot

tan × tan = 1 ––––– (i)

And tan = 3gt

tan = 4gt

Vr – E

C

Vmr

V0

= 60º

g g

1 2V0

3 m/s 4 m/s

V1gt gt


put in (i): 1 4gt

3gt

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 Cto 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 timeall particle travel same distance then at each moment of time,all particle will be at equilateral tringle. Then by symmetry youcan say that all particle will be met at centriod of tringle thenpath of each particle will as :Suppose at any instant of time, distance b/w particle A andcentriod P is r. Then, line joing particle A and P make 30º angle

with side of equilateral tringe then dtdr

will always constant and

equal to v cos 30º then – dtdr

= V cos 30º, (–)ive sign because

r is dicreasing function. Finally r = 0 while initial / 3r a

t

0

0

3a

dt 30 cos v

dr– t = 3V

2a A P =

2a

sec 30º = 3a

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

P30ºA

P

A 30ºa2/

/ 3a

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 ofdistance b/w two particle A and B will be constant as shown in figure.

Then dtdr

= – (V + V cos 60º) = 2

3V– dt

0

0

a

dt 3V– dr 2 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 : dtdr–

=

– v + u cos o t

0dt θ) cosu (–v dr –

–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 dtdx

= – V cos

+ u dt u) cos V(– 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

Vut

t = 22 u– V V

1.14: With frame of train : With frame of train, train appear in restthen 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. Sinceevent (2) will be happen after time then. Distance travelledby headlight (A) = Distance travelled by head light (B) =

u1 t1 +

a t21 = wt() +

ww

t Then distance

b/ two events is =

2 t – w = 0.24 km. Ans.

AA

1

A11

B11

C11

C1

C

B1

B

V cos 6

0º

V cos 6

0º

V V

V

VV

V60º

60º 60ºV

V cos 60º

B

A

W

AB

event 2

B A event 1

w (t + /2)

A

VB 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 = 60km 24.0

= 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 = 2g2.4 .

= 0.288 Distance travelled = 2 × 0.288 + 0.7

1.3 m Ans.

1.16: Method : 1 (Relative velocity) from figure

tan = 1

2VV

cos = 22

21

1

V V

V

also tan = 1

2

1 VV y

y =

1

21 V

V BC = 1

212 V

V –

Shortest distance = CM = BC cos = 22

21

1

1

212

V V

V VV –

shaft

a=1.2 m/s2

2.7m

2.4m/sH

0.7m

Shortest distance

= 22

21

2112

V V

|V – V |

now t = 2

22

1

22122

21

122

21

22

21

V V V V

V V

tan CM sec V V

BM AB V V

AM

22

21

2211

V V V V

t

Ans.

1

2y

0V1

B

CM

V2

1

2V2

0V1

9

Method : 2 (Velocity of approach)

At shortest distance velocity of approach = 0. Then V1 cos = V2 sin tan = 2

1VV

In tringle B A O : tan = 2

1

1

2VV

tV– – t V

V22 t – l2 V2 = l1 V1 – V2

1 t t = 22

21

21

V VV V

Ans.

And Shortest distance is : )– t (V t)V– ( 2 2

21

Shortest distance = 22

21

21

V V

V V

Method :3

V t1

1 1– V t

V1

V t – 2 2

V2

)– t (V t)V– ( 2 2

21 ... (1). for shortest

distance dtdl

= 0 )– t (V t)V– (

V )– t (V 2 V t)V– ( 2– 2

22

1

2 211

= 0

t = 22

21

21

V VV V

Ans. put value of t in (1) : 2

221

21

V V

|V – V |

Ans.

1.17: Method : 1

Time to reach at point D : T = V/n sec

V tan – m

. For

minimum time : dθdT

= 0 Then – V

sec2 + Vn

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 A1

B1

2

B

90–

O

BCm– 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) mediumin speed(1) mediumin speed

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 = τR π

Ans.

(b) Mean velocity = τR 2

Ans. (c) We know ....(i)

And t2

2 2τ

2 ii from

(i) and (ii) :

2τ 2

τ 2 now V = RW then

V = R

τ 2

. Average accleration :

cf V– V = τ

0– τR 2

Avg. Accn = 2τR 2

Ans.

B C

medium (1)i

medium (2)

BR

A

Time taken =

WX

13

6

1m/s2

–1m/s2

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)– (1 t a r

a t– t a 2

(a) at 2– a dtrd V

a 2–

dtvd a cc

a V

(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 – 2t) t =

Then position B is :

r = a

4a – 1 . The Distance travelled in up and down journcy is:

2a

4a

4a

Ans.

1.21: (a) V = V0 (1 – t/) dt τt– 1 V dx

x

0

t

00

x = V0

2τt

– t 2

Position at time t is

x then : x = V0 t

2τt– 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

2τt– l Ans.

Case II : t > Total distance travelled = AB + BC

= V0

2τt

– l t V– 2τ

– l V 2τ

– l 00 =

2t – l t V–

2 V

2 V

000

Distance = V0 – V0 t

2τt– l when t > Ans.

1.22 : (a) v = 1

x x differentiate w. r. t. time : 21

21

21

xα x2α a

dtdx xα

21

dtdv ––

a = 2α 21

Since acceleration is constant; velocity will be : V = u + at V = 2α 21

t

C

AB

t =


(b) S = ut + 21

at2 S = 21

2α2

t2 t = α

s2 Mean velocity <V> = s2

s

ts

< V > = s

Ans.

1.23: Calculation of time : w = – a v ( – i ve sign is used to show deacceleration)

dtdv

= – a v ot

0

0

0v

dt a – v

dv t0 =

aV 2

1

0 Ans.Ans.

Calculation of distance : v a– v a– dxdv v

0

0

21

x

0

0

v

dx a– dv v x0 = 3aV 2

3

0

1.24 : (a) j bt– iat r 2 x = at y = – bt2 y = – b

2

ax

a2y = – bx2 Ans.

(b) j2bt – i a dtrd v

j2bt – i a v 222 tb 4 a |v|

j 2b– dtvd a

j 2b– 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 = 2bta

Ans.

(d)t

j bt– iat ts V

2

mean

jbt – i a Vmean

222mean tb a |V|

1.25: (a) x = at y = at (1 – t) y = x

ax – 1 Ans.

(b) vx = dtdx

= a ay = dt

dvy = a – 2 a t j t) 2a– (a i a v 22 t) 2a– (a a |v|

2t) 2– (1 1 a |v|

j 2a– dtvd 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 )(2a t) 2a– (a a

t) 2a– (a 2a– 2

1

t = 1

Method : 2

t 2a– aa 4– tan

– a = a – 2 a t t = Ans.

1.26 : x = a sin wt y = a (1 – cos wt)

1–

ay

= – cos wt ––– (i) ax

= sin wt ––– (ii)

1 1– ay

ax

22

x2 + (y – a)2 = a2 Equation of circle of radius a. j dtdy i

dtdx v

jsin wt aw i wt cos wa v wa |v|

= const. Uniform circular motion.

(a) Distance travelled in time is = (aw) Ans.

(b) Since motion is uniform circular motion. Hence only radial acceleration ispresent then Angle b/w velocity vector and accn will be

Ans.

1.27 : y = ax – bx2 differentiate w.r.t. time : dtdx2x b–

dtdx a

dtdy

Vy = a VVx – 2bx Vx –––– (i)

At x = 0 Vy = a Vx ––– (ii) differentiate equation (i) w.r.t. time : 2x

xxy 2bV– dt

dV2bx – dt

dV a dt

dV

ay = a ax – 2bx ax – 2b Vx2 At x = 0 Given ax = 0 and ay = w ay = | – 2b Vx

2 | w = 2b Vx2 –– (iii)

Speed at origin will be : V = 2y

2x V V = 2b

wa 2bw 2

V = 2bw

(1 + a2) Ans.

1.28 : (a) 2 ta 21 t u s

20 tg

21 t V s

g

V 0

Ans.

a

(0, a)


(b) ts v

tg 21 v v 0

Ans.

we know T = gsin u 2

u sin = gg . v– 0

g2

gg . v–

2g v v 0

0

20

0 gg . v g – v v

1.29 : (a) S = 0 in y direction 0 = (V0 sin ) T – 21

g T2 T = gsin V 2 0

(b) At maximum height final velocity in y direction = 0 Vy2 = uy

2 + 2 aH

02 = (V0 sin )2 + 2 (–g) H H = 2g sin V 22

0 Range R = V0 cos

gsin 2V0

R = g 2sin V2

0 when H = R g

cos sin 2V 2g

sin V 20

220

tan = 4 = tan– 1 (4)

(c) x = (v0 cos ) t y = (v0 sin ) t – 21

g t2 y = v0 sin 2

00 cos vx g

21–

cos vx

y = x tan – cos 2u

gx22

2

Ans.

(d) Radius of currature = onaccelerati normal

V2 R0 = cos g

V20

gV cos 0

A

V0 RA = g cos V 22

0 Ans.

1.30 : w = g sin andw = g cos Here is first dicreasing and thenincreasing and projection of total accn on velocity vector

will be (–)ive then. wv = v. w1

= – g sin

– g –

V

y

x

V cos 0

V0V sin 0

g

V0

g wn

w

V

t

wV

w

g sin w = w

V

w = g cos n

15

1.31 :

Time to collide with incline T = g

2g 2

cos g

cos2gh 2

cos gu 2 y

Length MO = (velocity in MO) × T =

sin cos 2gh sin cos 2gh

Range OC = (MO) sin = cos

MO =

cos g

2gh 2 )sin cos 2gh 2 R = 8 h sin Ans.

1.32: We know R = g2θsin u 2

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 = gusin

T1 = 1031.5sin 240

= 24.3s = 0.41min Ans.

T2 = 1059.5sin 240

= 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

220

2

cos V 2 xg

.... (i)

Particle (ii): y = x tan 2 – 2

220

2

cos V 2 xg

.... (ii)

From (i) and (ii) : x tan 1

cos 2V xg –

122

0

2

= x tan 2

222

0 cos 2Vgt–

V = 2gh0

Just before collision

2gh2gh si

n

Just after collision

2gh s

in

2gh c

os

O

C

x

M

y

5.1×10 m3

240m/s

250 m/s = v0

250 m/s = v 0

60º45º=


x = 2

21

22121

20

cos– cos cos cos )– (sin

gV

.... (iii) Time for particle (1) : t1 =

10 θ cos Vx

TIme for particle (ii) : t2 = 20 θ cos V

x Then t =

210 cos

1–

cos1

Vx

.... (iv)

Put value of x on (iv) : t =

21

210θ cos θ cos)θ– (θsin

gV 2

= 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

210θ cos θ cos)θ– (θsin

g

2V= 11s Ans.

1.34 : (a) tan = ayV0

Time to reach at hight y : t = y/V0

Also dtdx

= ay dtdx

= a [V0 t] t

0

x

00 dt t aV dx

x =

0

02

0Vy

2V a

2 taV

x =

0V 2a

y2 Ans.

(b) ay = 0 ax = dtdy a

dtdVx = aVy = aV0 anet = aV0 Tangential acceleration = a

= aV0 cos a = 2

0

2

220

0

)(ay/V 1

y a (ay) V

(ay) V a

Ans.

Radial accn or normal acc

n : (an) = a V0 sin an =

00 V

y a V a Ans.

y

0 x x

v = ayx

v0

y

17

1.35: (a) tan = bxa x = at .... (i) Vy = (a t) b

y

0

t

0dt t ab dy

2 tab y

2

Then y =

ax

2ab

y =

2ab

x2 Ans.

(b) ax = 0 ay = dtdx

b dt

dVy = ab anet = ab Then normal accn : an = anet cos

anet = ab Then an = ab 22 (bx) a

a

R =

b a) xb (a

aV

2

3/2222

n

2

R =

/

axb 1

ba

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 smal l time dt particle displacementby 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 = v x

bx=V y

v

a na net

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 dsdv v

= a cos v dv = a (ds) cos = a dx

x

0

v

0dx a dv v

2v = ax v = xa Ans.

1.37: Angle travel by particle: = 2n. Speed of particle : v = Rw = at w =

t

0

n2

0

dt Rat ds

dtdθ

Rat

2n = 2Rat2

t2 = anR

t2 = anR

Now : Radial accn : ar = R ta R

V222

= a

Rn 4 Ra2

= 4na Tangential accn : at = dtdv

= a Total accn = 2t

2r a a = 22 na)(4 a

Total accn = a 2n)(4 1 = 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

0Rdt dV V–

RV

dtdv (–) sign because

V is decreasing R

tV 1

V V R

t V

VV

0

0

0

1–

(b) anet = 2r

2t a a = at at =

Rv2

= –v dsdv

–v dv = Rv2

dt

Where ds = distance travel by particle in time dt. –v dv = Rv2

ds

V

V

S

00Rds

vdv– – n

RS

VV

0 Then V = V0 e–S/R at =

Re V

Rv 2S/R–2

02

ds V

dx

a

R

v

R

at

ar

Givena = aat t = 0u = V

t r

0

19

anet = 2 R

V Re 2 V 22S/R–2

0 anet = 2 R

V e R

2 V 2

2S/R

20 Ans.

1.39: Method : 1

v = a s Squaring both side : v2 = as

Compare with : v2 = u2 + 2 accS u = 0 acc = 2a

Hence motion is constant magnitude tangential accn. Then.

At any time t : at = a/2 ar = R

V2 =

Ras

Since we know that velocity vector and tangential accn is parallel then.

tan R2S

a/2 Ras

aa

t

r Ans.

Method : 2

v2 = as differiate w.r.t time : 2v = dtds a

dtdv

= av 2a

dtdv

= at

ar = Ras R

V2 tan =

R2S

aa

tr Ans.

1.40 : (a) = a sin wt now = 0 t = 0

= ± a t = 23 ,2 , V = wtcos wa

dtd

a = sin wt wa dtdv 2 Then at = – aw2 sin wt

ar = R

wtcos wa R

V 2222 anet = 2

r2t a a

anet = aw2 wtcos Ra

wt sin 42

22 –––– (i)

at t = 0 = 0 at t = 23or 2 anet =

Rwa 22

anet = aw2 Ans.

at

ar

at

ar

x = a sin wt

y


(b) For minimum value of acceleration 0 dt

danet cos wt = a 2R put in equation(i)

amin = aw2

2aR– 1 =

a 2R– 1 a = 2

2

a 2R– 1 a Ans.Ans.

1.41: Speed of particle when distance is S. V2 = 2aS –––– (i) (Because a = const.)

S = 21

at2 t2 = a2S

Then wn = b 2

22

a4bS

a2S

Radius of curvature : R = 2

2

n

2

4bS2aSa

wV

R = 2bSa3

Ans.

Net accn : w = 2n

2T w a w =

2

2

22

a4bS a

Ans.

1.42: (a) y = ax2 diff. w.r.t. time : dtdx2ax

dtdy

Vy = 2ax VVx

Again diff. w.r.t. time : ay = 2ax ax + 2aVx2

At x = 0 Vy = 0 Then Vx = v ay = 2aV2

Since speed is const. its tangential acceleration will be zero. Then

ax = dtdV

= 0 anet = 2 a VV2 R = 2

2

n

2

2avV a

V R = 2a1

Ans.

(b) 2

2

2

2

by

ax

= 1 diff. w.r.t. time : 0 b

V2y

aV2x

2y

2x

Again diff. w.r.t. time : 0 b

2V

b

a2y

aV 2

a

a2x 2

2y

2y

2

2x

2x

At x = 0 and y = b : 0 b

V 2

ba 2

aV 2

2

2yy

2

2x

a = a

w = btn

4

x

y

V0

21

As shown in figure : Vy = 0 and Vx = V0 0 ba 2

aV 2 y2

20 ay =

202 V

ab

Since speed is Const. tangential accn = ax = 0 at; t = 0 anet = 202 V

ab

Radius of curvature : R = n

20

aV R = 2

02

20

V a

bV

R = b

a2 Ans.

1.43 : Given dtd

= w = const. Then dtd 2

dt)d(2

= 2w = const.

Hence angular velocity of point A w.r.t. point P isconstnat then. WP = 2W and velocity will be perpendicularto 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 WP2 = 4 RW2 = anet Direction is toward the centre. Ans.

1.44 : = at2 dtd

= w = 2at dtdw

= = 2a

Tangential acceleration : at = R = 2aR Radial acceleration: ar = RW2 = R (2at)2 = 4 a2 t2 R

And. v = RW = R 2at 2at = Rv –––– (i)

Now anet = 2r

2t a a = 2aR 22 )(2at 1

from (i) : anet = 22 )(2at 1 tv

= 0.7 m/s Ans.

1.45 : Since acceleration is constant. l = 21

at2 and

V = at Then l = 2

Vt –––––– (i) and also. angular acceleration is const. then

= 21 t2 and w = at then =

21

wt and since particle taken n turn in its journey..

V

O

A

P

at

ar

l

Va


= 2 n Now 2 = 21

wt –––– (ii)

(II)(I) : W

V

2

W = v2

Ans.

1.46 : = at – bt3 w = dtd

= a – 3bt2 when body is in rest then w = 0 a – 3bt2 = 0

t = 3ba

in = 0 and final = t (a – bt2) =

32a 3b

a

Wavg = t– infinal

=

32a

3ba

32a 3b

a Ans.

Win = a and Wfinal = a – 3b

3ba

= 0 avg = 3ab

3baa

t– W W infinal Ans.

And. = dtdw

= – 6bt = – 6b 3ba

= 2 3ab Ans.

1.47 : Given = = at To find tangential accn : (at) : at = R = Rat

To find radial accn : (ar) : = dtdw

ar = Rw2

t

0

w

0 dt dw w = 2

at dtat 2 t

0 ar =

4 ta R 42

Then

at

ar

tan = t

r

aa

tan = ta R 4 ta R 42

t = s 7 tan a4 3 Ans.

1.48 : Given W = angular accn. = k W k = const.

Wk ddW– W

(–)ive because W is decreasing. –

0

0

W

21

dk dw W0

at

ar

y

x

R

23

k 3

W2 230 ––– (i) = Angular displacement. Also = k W

– dW = k W dt t

0

0

w2

1– dtk dw W

0 t

k2W 2

–10 ––– (ii)

Avg. angular velocity : Wavg = <W> = t

= 3

W

k W2

k 3

W2 0

210

230

Ans.

1.49: (a) Given W = W0 – a –––– (i). At t = 0 = 0 W = W0

Also dtd

= W0 – a

t

0

0 0dt

a– Wd

t )a– (Wn a1–

0 0

at– W

a– Wn

0

0 = )e– (1

aW 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 ddW

= 0 cos

0 0 w

0 d cos dw w

2W2

= 0 sin

W = ± sin 2 0 Ans.

1.51: (a) Instanteneous axis of rotation is passing through that pointwhich velocity is zero always. If we observe carefully itspoint must be at line joining AB. Then. at time t :VP – 0 = RW = Rt = yt

y

x

B

A

V

y

–0

0

0

2– sin 2– w 0

sin 2 w 0

cos 0


x = vt ––– (i) Velocity of point P :

VP = 0 v = ty x = v

yv

t = yv put in (i)

y = xv2

Ans.

(b) Velocity of point O = 0

VO – P = Rw VP = u + at = wt

Then yw = wt t = wwy

x = 21

at2 = 21

w 2

wwy

x = wyw

21 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 = )– ( cos v v2 v v 22 = 2 v sin /2 = 2V sin (wt)/2 Again

dtds

= 2 v sin wt/2

w2

0

s

0 dt sin wt/2 v2 ds 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

RC

B

P

v

v

1.53: (a) Acceleration of point C : aC = w Velocity of pointC at time t : VC = u + at = wt

Angular accn of ball about its centre:

= Rw

Rw

RaC Angular velocity of ball at

time t : w = w0 t = Rw

t Velocity of point AA

w.r.t. centre C : VA – C = RW = w t i VC – E = w t i 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 :

Point A:

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) Rv

222 aA–C = wi –

R tw 22

j .... (ii)

(i) + (ii) : aA–E = 2 wi – R tw 22

j aA–E = 2w

2Rwt

2 Ans.

wt

wtC

0

A

wt

wt

y

x


Point B : a

C–E = wi .... (i) a

B–C = (–R) j +

Rv2

i =

– w j– R tw 22

i .... (ii)

(i) + (ii) : a

B =

R tw– w

22

i + w j aB = w

2Rwt

2 Ans.

Point O: a

C–E = w i .... (i) a

O–C = (–R) i +

Rv2

j =

–w i – R tw 22

j .... (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 = Rv2

accn of point B = Rv2

Again Radius of curvature = onaccelerati normal

(speed)2 RA =

/RV(2V)

2

2 = 4R

Again an = Rv2

cos 45º = R 2

v2

RB = R 2 2

R 2v

)2(V2

1.55 : Method : 1 (axis is rotating): w

1–0 = w1 i w

1–0 = w2 j w

1–2 = w1 i – w2 j

w1–2 = 22

21 w w

1–2 = dt

wd 2–1

= w1 dti d

– w2 dtj d

direction of dt

i d 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

v2

RV

45º

V V 2

v2

A2V

R

27

Method : 2 (axis is not rotating)

w

1–C = w1 cos i + w1 sin k w

C–0 = w2 j w

1–0 = w1

cos i + w1 sin k + w2 j 22

210–1 w w |w|

dtw d 0–1

= – w1 sin

dtd

i + w1 cos

dtd

k

And dtd

= w2 = – w1 w2 sin i + w1 w2 cos k

| | = w1 w2. Here x axis is directly attached with observer. Then

dti d

= – w2 k dθ 1|i d| dtdθ

dt|i d| = w2

But dt

j d = 0 Because with frame of this observed dirn of axis does not rotating then.

1–2 = –

w1 w2 k 1–2 = w1 w2

1.56 : w = at i + b t2 j dtdw

= a i + 2b t j (a)

t

ab 1at |w|

(b)

a t2ab 1at ||

w = w cos cos

w . w

cos 2224222

322

t4b a tb ta

t2b t a

= 17º Ans

y

x0

w2

w1

y

x0

w2

y

x0

w2

w1

w2z

C1

y

z passing

y axis 1 unit

1 unitd

jd ii


1.57: Method : 1 (axis are moving): It we see carefullythen x axis is moving while y direction is not rotating.Angular velocity of disc with respect to centre M

is : i RV W O–M

.... (i) Angular velocity of centre

M w.r.t. origen. j OMV W O–M

OM = R cot

j tan RV

W O–M

. Since angular velocity is vector

quantity it follow vector addition law ] j tan i[ RV W O–D

cos R

V tan 1 RV |W| 2

O–D

Ans.

Again Angular acceleration () is : dtwd β

dtj d

tan dt

i d

RV

β

And since y dirn is constant dt

j d = 0 But

dti d | id | = d dt

d dt

id = wMO k tan RV–

dtid

k tan RV

RV

β

k tan

RV– β 2

2

2

2

RV |β|

tan

Method : 2 (axis are not rotating)

angular velocity of disc w.r.t. M : WD–M = RV

cos i +

RV

sin j angular velocity of M w.r.f. O

: j tan RV j

cot RV W O–M

angular velocity of disc

wrt O: W

D–O = W

D–M + W

M–0 W

D–O = RV

cos i

RV

sin j RV

tan j

cos RV

tan RV

sin RV

cos RV

|W| O–M

Ans.

y

xR0

v

M

p

z

y axis

y axisd

jd ii

y

x0

w =0

D

z

VR

VM

29

0 j dtdθ θ cos

RV i

dtdθ θsin

RV–

dtwd α O–D

O–D

and dt

dθ is angular velocity of M w.r.t.

0 : d/dt = RV

tan 2

2O–D

RV– α

sin tan i + 2

2

RV

cos tan j | O–Dα

| = 2

2

RV

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 i + w0 sin k t j

| A–PW

| = 220

220

220 t sin w cos w

| A–PW

| = w0

0

0w

t Now

j k dtd cos W

dtd ) i ( sin – W

dtW d 000

A–P

= –W0 0t sin i + W0 0 t cos k + 0j 2

022

020 t w ||

2200 t w 1 ||

Ans.

Method : 2 (x axis is moving) : Here we take line AB along

x dirn. jt i w W 00A–P

20

2

0A–Pw

t 1 w |W|

j dt

j d t dt

i d w dt

wd 000

A–P

Here 0 dt

j d

But dt

i d = t k Then k t k

dtds

dtdi

0 = w0 t k + j 2

12020 t w 1 ||

Ans.

y

xA

0

w0

w0

z

y

xA

0w0

z

P


1.2 The Fundamentals Equation 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 mw 2

Ans.

1.60 : Net pulling force = ( m) a m0g – k m1 g – k m2

g = (m1 + m2 + m0) a a =

021

210m m m

)m (mk – mg 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

0m m m

m k) (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

2211m m

)m k m (k cos g– m2) (m1 sin g

.... (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

2121m m

α cos m m )k–(k Ans.

F

mg

m

F

(m – m) g

m – m w

km g2

m2m1

a

km g1m0

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

2211m m

m k m k

a

mg s

in

1

k mg c

os

22

k mg co

s

11

mg s

in

2

m 2

m 1

N

m g sin

1

k m g cos

11

m 1

31

1.62 : Upward Journey : We know S = Vt – 21

at2where V = final velocity in this situation Vf = 0

Then = 21

(g sin + µg cos ) t2 –––– (i)

t = Time taken in upward journey.

Downward journey : We know S = ut + 21

at2 u = initial

velocity. Then = 21

(g sin – µg cos ) (t)2 –– (ii)

and u = 0 now : (ii)(i)

= 21

cos ug– sin g cos ug sin g

µ =

1 1–

2

2

tan Ans.

1.63 : (a) Starts coming down. m2g > m1 g sin + fmax m2g > m1 g sin + k m1 g cos 1

2mm

> sin + k cos (b) m1 g sin > m2 g + k m1 g cos g sin > 1

2mm

+ k cos

1

2mm

< g sin – k cos (c) At rest : Friction will be static : g sin – k cos < 1

2mm

<

sin + k cos

a

v

v = 0

f

a = (g

sin + µ g cos

)

a

vu = 0

a = (g sin – µg cos )

m 1

m2

m g2

m g

sin

1

f

= k m

g co

s

max

1

m g2

m g

sin

1

a

f max m g2

m g

sin

1

a

f max


1.64: To find tendency of sliding check value of m2gand m1 g sin = m1 g sin 30º = m1 g/2 0.5 m1g

Here m2 g = m1 g = 32

m1 g 0.66 m1g Then

m2 g > m1 g sin Block m2 has tendency tomove down ward.

Then. Pulling force = m2g – m1 g sin – k m1 g cos acceleration a = 21

112m m

cos g mk – sin g m– g m

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 m m

at

m mF

But friction b/w m1 and m2 will be static then. m1

fr

fr = m1 [w1] = 21

1m mat m

< k m2 g t < 1

212m a

)m (m g mk w1 = w2 =

21 m mat ––– (i)

If t > k m2 g 1

21m a

)m (m . Sleeping b/w two block will be start then F.B.D. of m2 :

m2 F = atkm g2 w2 w2 =

2

2m

g km– at –––(ii) Ans.

F.B.D. of m1 : m1

k m g2 w1 = 1

2m

g mk ––––– (iii) Ans.

Assume to = 1

212m a

)m (m g mk t > to : Slope of w2 >

Slope of w1 = w2 because of Slope of w2 =

2ma

from

(ii) [After sliding] Slope of w1 =

21 m ma

from (i) [Before sliding]

Slope w1 = 0 from (iii) [After sliding]

m 1

m2

m g2

m g

sin

1

f

= k m

g cos

max

1

k

Given mm

1

2.

m2

m1

fr = k m g2 F = at

m2

m1

w F = at

w

t0t

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 + 21

at2 sec = 21

(g sin – kg cos ) t2

A

R

k = coefficient of friction mg sin

k mg cos

t2 = ] cosk – cos [sin g

2 ) cosk – (sin g

sec 22

t = ] cosk – cos [sin g

22

–– (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 = – k1 Ans.

Put value of k : = 49º Put value of in equation (i) :

tmin = 49] cos 0.14– 49º cos 49º[sin 102.10 2

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 =

sin k cos cos mgk sin mg

Tmin

m k

T

m

T

mg sin

mgmg cos µN

T cos T sin

N

(cos + k sin ) min minimum value of cos + k sin = 2k 1 Tmin = 2k 1

) cosk (sin mg


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 = sin amg . Motion equation of block : a1 = Accelration

of block. F cos = m a1 a1 = dtdv

m cosat

1t

0

V

0

dt t cos a

dV m 2t

α cos amV 2

1

sin a

g m 21

α cos amV

22

22 V = =

sin a 2 cos g m

2

2 Ans.

(b)

t

0

V

0

dt t cos a

dV m

2t

α cos amV 2

1.69 : Motion equation : = aS 3mg

cos = m a1 a1 =

acceleration of block of mass m. a1 = m 3mg

cos = 3g

cos

a1 = 3g

cos as s

0

v

0

ds as cos 3g

dv v

s

0

v

0

2

aassin

3g

2v

v2 = a 3g 2

sin as v = assin a 3g 2

v = m 2

cos a t2

2 t cos m 2a

dtds

t

0

2s

0

dt t m 2

cos a ds s = 3

3

xt

m 2 cos a

s =

sin a 6 cos g m

32

32 Ans.Ans.

1.70 : Method : 1

Motion of mass 2 m : T – k2mg = 2 mw T = 2mw + 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. = 21

(3w – kg) t2 t = kg– 3w 2

Ans.

m

F = at

µ = 0

m

F = mg/3

2m wk2mg

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 = 21

(3w – kg) t2 t = kg– 3w 2

Ans.

1.71: For observer inside elevator : a1–2 = 21

010212m m

wm– wm g m– g m

=

21

012m m

) w (g )m– (m

But in form of vector : a1–car = 21

012m m

)w – g( )m– (m

Ans.

Acceleration of m1 w. r. t. shaft : a1–shaft = a1–car + acar = 21

012m m

) w– (g )m– (m

+ w0 a1–shaft = 21

021112m m

wm m g m )m– (m

Ans.

Tension in string : T – m1 g = m1 a1–shft T = 21

021m m

) w (g m m 2

Force applied by pulley on ceiling = 2 T = 21

021m m

) w (g m m 4

as vector form:

2T = )w– g( m mm m 4

021

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 – [2 2g

Ans.

1.73 : Equationof motion : T = m0

2

a a 21 .... (i) m1g

= T/2 = m1 a1 .... (ii) m2 g – T/2 = m2 a2 .... (iii)

from (i), (ii) and (iii) : a1 = )m (m m m m g ).m– (m m m m 4

21021

21021

Ans.

m g1

m1

TT T

m2

m g2

2T

shaft

w0

m

mg sin

a/2T

T/2a

mmg

T/2T/2

m2m1

a1

mg

m g2

m0T

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. 21

(a1 + a2)

t2 .... (iii). From (i), (ii) and (iii) : fr = 2 tm)– (M m m

Ans.

1.75: = 100 cm – T + mg = m a1 .... (i) + 2T mg = m 2a1 .... (ii) from

(i) and (ii) : a1 = 4 g 2) (–

arel = 2a 3 1 . Suppose t time is taken then

= 21

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) Justbefore string slack.

V2 = 2a (2h). It body (2) travel x distance again then V2 = 2a (2h) = 2gx

x = gah 2

. Total hight from ground : 4 = 4 h 6

h gah 2

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 = cot 1

g2 a2 = cot tan

g

T

2T

mg

nm

Ta1

m

nmg

a 2

1

mgMg

Ma1

ma2

mg

2mgT a

2T

h

a2

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 =

mM k 2

g km M 2m

mg

Ans.

mg

T

m

KN

M

a1a2

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 = k 1k)– (1 g

Ans.

a2

a1

mg

mN

TKN

1.80: F.B.D. of block (2) with frame of wedge: At maximum accn w : fr1 will be maxm then fr1 = k [mgcos + 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 )cot k (1 g

am

ma

mg

2 m N

1 mma

m

ma

kmg kma

mg

mg

mw

fr1

2

mg

mw

w

m

1fr

1.81 : Method : 1

F.B.D. of System a2 = accn of bar w.r.t. incline. Since no force on system in horizontal directionthen 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 =

sin m m cos sin g m

221

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 = m2g cos –––– (i) F.B.D. of wedge : N sin = m1 a1 –––– (ii) from (i) and

(ii) : a1 =

sin m m

sin cos g m2

21

2

2m 2

1

m 1

a2

m 2

m1 a1

m a2 1

m 2

m g2

Method : 3

F.B.D. of bar with frame of ground observes : Motionequation in perpendicular dirn of incline then : m2g cos –N = m2 a1 sin ––– (i)

1.84: v = 360 km/hr = sm 100

36001000 360

At point A : N – mg = R

mv2 N = mg +

Rmv2

= 70 g + 500[100] 70 2

= 70

g + 500100 100 70

= 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, Rin (ii) N = 0.7 kN Ans.

a2

m g2

N

a1

vC

v

BR

Av

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

= 21

mv2 v = cos g 2 ar = 2v = 2 g cos

anet = cos 4 sin g a a 22rt 22 anet = g cos 3 1 2

Now : T – mg cos =

2mv = 2 mg cos T = 3 mg cos Ans.

(b) Component of velocity in y dirn: vy = v sin

vy = cos g 2 sin v2y = 2 g cos sin2

For vy maximum v2y will be maximum. Then x = cos sin2 will

be maximum. ddx = 0 = 2 cos2 sin – sin3 = 0 2 cos2

= sin2 tan = 2 sin = 32 cos = 3

1 v2

y

= 2 g 32

3

1 3 3

g 4 vmax

y

T = 3 mg cos = 3 mg

31

= mg 3 Ans.

(c) If no acceliration in y direction then. ar cos = at sin 2 g cos2

= g sin2 tan = 2 cos = 31

Ans.

1.86 : Extreme position : Only tangential acceleration is prerent at extremeposition. at = g sin

Lowest Position : Energy conservation : mg [ – cos ] = 21

m v2

v2 = 2 g (1 – cos ) ar =

2v = 2 g (1 – cos )

A/C condition : ar = at g sin = 2 g (1 – cos ) sin + 2

cos = 2 cos = 53 = 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 ) = 21

mv2 mgR (1 – cos ) = 21

mRg cos

1 – cos = 2

cos cos = 3

2

2 cos 3

= 1 v2 =

Rg

32

v = 3Rg 2

Ans.

1.88: Top view : Spring force = F = N = normal reaction on sleeve. Since no accelerationin tangential direction. N sin = cos ––– (i). Equation in radial direction: N cos

+ sin = m r w2 = m ( + ) cosec w2 from (i):

sin

cos x (cos ) + sin

= m ( + ) cosec (w2) x = m + m w2 = 2mw– x m

Ans.

A

A

v

mg

RR

N

w

r

v

l

w

1.89: k = k0 =

Rr– 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 rmv

mg Rr

– 12

v2

= k0 g

Rr

– r 2

. To maximum value of v : x = Rr– r

2 will be maxm. Then dr

dx = 0

0 R2r

– 1 r = 2R 4

R g k 4R–

2R g k V 0

02max

Vmax = gR k

21

0 Ans.Ans.

O

R

r kmgmv /r2

41

1.90: Tangential and radial both acceleration is only provided by friction becausefriction 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

d w2

Rv2

at = Tangential acceleration = w

anet = 2

22

2

R4d 1 w w

Rd 2w

Fnet = m w 2

2

R4d

1 = k mg

Squaring :

2

22

R4d 1 w = k2 g2 d = 1–

wkg

2R

2

Ans.

1.91 : y = a sin αx k = friction coefficient. 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 will be minimum. And we know : speedis constant also we are secing that radius of curvature will be

minimum at maximum point of curve then. ay = 2

2

2

2 a v dt

y d

sin αx . At maximum value of curve : sin αx = 1

ay = 2

2

αa v

R = a vα v

av

2

22

y

2 R = a

2 . Then v2 g aαk

2 v

akg 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 2 T

R

v /R2

w

y

x

y

x

v

ay

mg

N

w

T T

(dm)

N

gT

T

= very small angle.sin =


– R 2m

(R 2 ) g cot = (R 2 ) R 2m

Rw2 T –

2 wR m

2cot g m 2

T = 2

m [Rw2

+ g cot ] T = 2g m

gRw

cot 2

Ans.

1.93 :1

2mm

= 0 T1 – m1 g = m1 a ––– (i) m2 g – T2 = m2 a ––– (ii)

Relation between T1 and T2 : (T + dT) sin 2d + 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 2d – T cos 2

d = dfr

dT = dfr ––– (i) dT = µ T d

0

T

T

d µ TdT

2

1

n (T2 – T1) = µ

T2 = T1 eµ ––– (iii)

from (i) and (ii) : a) (g ma)– (g m

TT

1

2

1

2

from (iii) : µ

1

2 e TT

Then eµ =

a g

a– g a ga– g

mm

1

2 ––– (iv). Since pulley is fixed : Before skiding a =

0 = 0 eµ = 0 µ = 1

n 0

(b) from equaiton (iv) : en0 =

a g

a– g a g

a– g 0

a = g 0

0n

]– [n

Ans.

T1T2

m1m2

m g1m g2

a

dT + dT

dN

dfr

T

y

x

43

1.94: Hence velocity along y axis is not responsible for circularmotion only velocity along Z-axis i s responsible.

Then N = RV m 2

Z VZ = V0 cos N = R

cos V m 220

1.95: x = a sin wt ax = 2

2

dtxd

= – aw2 sin wt

y = b cos wt ay = 2

2

dtyd

= – bw2 cos wt jay i a a xnet

= j wt cos wb – isin wt wa– 22 ]j wt cos b isin wt [a mw– F 2

j wt cos b isin wt a jy i x r 2 wr m– F

where r

= position vector of particle. F = mw2 22 wt)cos (b sin wt) (a

F = mw2 22 y x Ans.

1.96 : Method : 1 (Impluse equation)

(a) We know P

= Impluse in time t = t

0

dt F = t

0

dt mg– P

= – mg t Ans.

v0

xR

Z

O

y

V0

w

x

y

gV0

m

(b) T = gsin V 2 0

g . V0

= V0 g cos (90 + ) gsin V 2

mg P 0

– g . V0

= V0

g sin = 2 m V0 sin V0 sin = gg . V– 0

gg . V

2m P 0

Ans.

Method : 2 (Kinematic)

(a) j sin V i cos V V 000

j gt)– sin (V i cos V V 00f

where fV

= final veloci ty vecto r then 0f V m– V m P

j t g m– P

Ans.

V0

y

x

V0

90+

g


(b) T = gsin V 2 0

j gsin V 2

g m– P 0

P

= – 2 m V0 sin j

From method : 1 : V0 sin = gg . V– 0

j g

g . V m 2 P 0

Ans.

1.97 : m

– t) (T t a F

(a) w m F

w = linear acceleration. ]– tt [

ma

w dtVd 2

t

0

2 t

0

V

0

dt t– dt t ma Vd

3–

2 t

ma

V32

f

3P

6 a V m

Ans.

(b) ]– t t [ ma

w 2

t

0

2t

0

V

0

dt t– dt t ma Vd

3t

– 2t

ma

V32

0

3

0

2S

0

dt 3t – dt

2t

ma dS

12–

6 .

ma

S43

12

ma S

4 S =

m 12 a 4

Ans.

1.98: sin wt F F 0

a = dtdV sin wt

mF0

t

0

0V

0

dtsin wt mF

dV V = t

0

0 wt cos mwF

–

V = wt]cos– [1 mwF0

t

0

0S

0

dt wt]cos– [1 mwF

dS S =

sin wt

w1– t

mwF0

S = 20

mwF

[tw – sin wt] distance

distance will be increasing function w.r.t. time then

dtdS

= (+)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 mF a

dtdv 0 cos wt

t

0

00

0

dt wt cos mF

dV 0 = wmF0 sin wt

sin wt = 0 wt = t = wπ

Ans.

Now : t

0

00

0

wt cos mF

dV V = wmF0 sin wt ... . (i )

t

0

0s

0

dt sin wt wm

F ds

S = wt)cos– (1 wmF–

20

at t = wπ s = 2

0

wm2F

velocity will be maximum when sin wt = 1

Vmax = wmF0 Ans.

1.100 : (a) F = – rV V m–r

dtdV

t

0

V

V

dt m

r– VdV

0 t

mr– Vn V

V0

V = V0 e–r/m t V will be zero when t Ans.

(b) a = – mr

V = dsdv v

s

0

V

V

ds vmr– dV v

0V – V0 =

mr– S V = V0 m

r– S

Total distance travel by particle is S1 then final velocity = 0 0 = V0 mr– S1

S1 = rV m 0 Ans.

(c) 0V

= VV0 – mr

S2 S2 =

1– rV m 0

–––– (i) Also 0V

= VV0 2 tmr–

e

t2 = n rm –––– (i) <V> =

n

1)– ( V

tS 0

2

2 Ans.

1.101 : F v2 F = – kv2 a = mkv– 2

t

0

V

V2 dt

mk–

vdv

0

mkt–

V

1– V

V 0

mkt

V1

V1

– 0

t km

V V

V– V

0

0

––– (i)

V0 V

h


Also a = dxdv v

mkv– 2

v dv = m

dx kv– 2

h

0

V

V

dx k – Vdv m

0 m n

0VV

= – kh

k = hm

n 0V

V put in (i) : t =

VVn

h V V

V– V00

0

Ans.

1.102: Suppose at time t distance travel is x then F = mg sin – axmg 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 dxdv v

= g sin – a x g cos

mx

0

0

0

dx α) cos axg– αsin (g dv v 0 = (g sin ) x – 2

cos g xa 2 x =

cos g a

sin g 2 x = a

2

tan . Velocity will be maximum when dtdv = 0 = acceleration g sin = ax g cos

x = a1

tan now

tan ay

0

maxv

0

dx α) cos g x a– αsin (g dv v

maxV

= g sin

tan a1

2 cos g 2– tan

a1

VVmax =

ag

α cosα sin

ag 2

sin tan Ans.

1.103: K = friction coefficient Block start sliding when.

F = at1 = k m g t1 = ag mk . 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

= ma t

2t

0

v

0

t 2ma V t)( dt (

ma dV

t)( d t 2ma ds

t

0

2s

0

S = 6ma

(t)3 = 6ma

(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 = Kv2V0

Fnet = mg + kv2 a = g + m

kv 2

mkv mg

dsdv v–

2 mv dv = (mg – kv2) ds

0

v

h

02

0

ds – kv mgdv mv h = mg

mg KVn

2km 2

0 .... (i)

Downward Journey : Fnet = mg – kv2 mg

k v2

v a = m

kv– mg 2 = v dv/ds

mgmg KV

n 2km

h ds kv– mgdv mv

20

v

0

h

02

2

20

kv– mgmgn

2km

mgmg KVn

2km

v =

mgKV 1

v20

0

Ans.

1.105 :(a) Position vector of particle : r

= r cos i + r sin j

At time t : = wt )j sin i (cos F r F F

]j sin i [cos mF a

t

0

t

0

V

0

dt (sin wt) mF idt wt)(cos

mF v d

]j wt)cos– (1 i[sin wt

mwF V

speed = 22 wt)cos– (1 (sin wt) mwF |V|

Speed = 2

wtsin mw2F

Ans.

(b) Distance is calculated by speed. Then 2wtsin

mw2F V

dtds

dt 2wtsin

mw2F

ds t

0

s

0

S = Distance S = – t

02 2

wt cos 2 mw

2F S =

2wt cos– 1

mwF 4

2 .... (ii) velocity of particle

rw

m

F

0


will be zero : 0 = 0 2wtsin 2

wtsin mw2F

put value of t = w2π in (ii)

S = 2mwF 4

(1 – cos ) S = 2mwF 8

Average speed = /w2

8F/mw Time

Distance 2

Average speed = mw 4F

Ans.

1.106: K = tan = friction coefficient When = W = nettangent ial accelerat ion/ Wx = Acceleration

in x direction. Now W = m

cos mgk – cos sin mg

= g (sin cos – K cos ) = g [sin cos – sin ] W = g sin (cos – 1)... (i)

Acceleration in x dirn: Wx = m

cos cos mgk – sin mg WWx = g sin [1 – cos ] .... (ii)

From (i) and (ii) : W = –Wx dtV d –

dtV d 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 = cos 1V0

Ans.

1.107: Tangential force on system is F then F sin g (dm) = sin g )d R ( = Rg

t

0

d sin = R f in = 0 = – Rg f cos

f = R

f = RF =

Rg R cos– 1 a = m

g R R cos– 1 = x

g R R cos– 1 a =

R g R cos– 1

v

wmg sin

K mg cos

x

f

y

Rd

dm= R d

x

(dm)g

F

ddm

49

1.108: At time of break off. normal reaction will be zero. Where mw0

= Psuedo force because observer at sphere. R

mV20 = mg cos

– m w0 sin V20 = Rg cos – R w0 sin .... (i)

Energy equation: Wall forces = Kf – Ki Wpsuedo + Wmg = Kf –

Ki m w0 R sin + mg [R – R cos ] = 21

m V20 – 0 V2

0 = 2w0

R sin + 2 g R [1 – cos ].... (ii) From (i) and (ii) : V20 = 2R

RV– 2

0

+ 2 g R V20 = 2 g 3

R V0 = 3g R cos

gw

1 3

gw g 5 3

w

0

00

Ans.

1.109 : Given F nr1

F = nrK

Where K = Constannt.

A particle is said to steady if we displace particle away from origin, particle want to regainits position and also it we displaced particle toward origin, particle want to regain its original

position. Then. At steady state (mean position) : F = rV m 2

0 Then rV m

rK 2

0n .... (i)

It we increases r then. F – rV m 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

r1

rVm

r1

r

Kn – 20

n

> 0 .... (ii) From (i) and (ii) n < 1 Ans.

w = acceleration0

R

R00

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 wRg

. Case (i) : If Rw2 > g then cos 2 wRg

is defined and

only one equilibrium position will be exist and will be steady.

Case (ii) : If Rw2 < g then only 2 wRg

= 0º will be equilibrium

position because tangential fore along arch of ring due to mgwill be greater than that of pseudo force and object will comeat lower position of ring.

RN

Y

mg

m (R sin ) w2

Y

90–

FUNDAMENTALS OF PHYSICS

Engineering

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