2nd-Dispatch-DLPD_IIT-JEE_Class-XI_English_PC-(Physics).pdf

Physics Sheet Solutions 2nd Dispatch CLASS : XI

Contents Preface 1.

Newton's Laws of Motion Exercise

2.

96 - 113

Rigid Body Dynamics Exercise

8.

079 - 95

Centre of Mass Exercise

7.

059 - 078

Circular Motion Exercise

6.

044 - 058

Work, Power & Energy Exercise

5.

026 - 043

Gravitation Exercise

4.

001 - 025

Friction Exercise

3.

Page No.

114 - 156

Unit and Dimensions Exercise

157 - 161

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TOPIC : NEWTON'S LAWS OF MOTION EXERCISE-1 PART - I SECTION (A) A-1.

Gravitational, Electromagnetic, Nuclear.

A-4.

Newton's IIIrd Law

A-6.

Vertical wall does not exert force on sphere (N' = 0).

A-8.

action reaction pairs

(1) and (2) (3) and (4) (5) and (6) (7) and (8)

SECTION (B) B-1.  

N = F + mg N = mg + mg N = 2mg

[equilibrium]

B-3.

If is obvious that block can`t accelerate in y direction  N – mg cos  = 0  N = mg cos 

RESONANCE

SOLN_NEWTON'S LAWS OF MOTION - 1

B-5.

Due to symmetry normal reactions due to left and right wall are same in magnitude W – N cos 60 – N cos 60 = 0 [Equilibrium in vertical direction] W–

 B-7.

N1 cos300 N1

N N – =0 2 2 = 50 +

N 3 – 2 = 2 2

N1 sin300

=



N=W

N2 2

50 ................ (1)

N2 2

N1 = 2 N2 ..............................(2) Solving equation (1) & (2) N1 = 136.57 N N2 = 96.58 N

SECTION (C) C-1.   

Since string 2 is massless net force on it must be zero. T2 = F = 10 N T1 + mg = T2 [Equilibrium of block] T1 + 1 × 10 = 10 T1 = 0

C-3.

RESONANCE

SOLN_NEWTON'S LAWS OF MOTION - 2

TC – 10 = 0 TC = 10 N TB – TC – 5 = 0 TB – 10 – 5 = 0 TB = 15 N TA – TB – 5 = 0 TA = 20 N

    C-5.

[Equilibrium of block] [Equilibrium of 2]

[Equilibrium of 1]

For finding distance travelled we need to know the acceleration and initial velocity of block. m 2g – T = m 2a [Newton’s second law for m2] T – m 1g = m 1a [Newton’s second law for m1] m2g – m1g = (m2 + m1)a [adding both the equation] 

a=

(m2 – m1)g m2  m1

a=

6–3 ×g 63

a=

g 10 = m/s2 3 3

s=ut+ S=

1 at2 2

=0×2+

1 10 × × 22 2 3

20 m 3

T – m1g = m, a 

g





T = m1  g  3 

=3×

40 3

T = 40 N

Force exerted by clamp on pulley is 2T  2 × 40 = 80 N C-7.

VA = VP2 = 10m/s  For pulley

V1  V2 VP = 2

P P

VB – VC  VP2 = 2

VB  2  VB = 22 m/s  2

and

VP1 

VA  10 2

VP P2

 10 =

0=

P1

VP

V1 1 2 V2

VA  VP2

V A. A

Vp2 = 10m/s .

C  2m/s = VC

VB. B

2

 VA = 10 m/s 

SECTION (D) D-1.

Since string is inextensible length of string can’t change  rate of decreases of length of left string = rate of increase of length of right string 

V1 cos 1 = V2 cos 2

RESONANCE



V1 cos 2 V2 = cos 1

SOLN_NEWTON'S LAWS OF MOTION - 3

D-3.

Since rod is rigid, its length can’t increase.  velocity of approach of A and B point of rod is zero.  u sin  – v cos  = 0  v = u tan  at any angle  x and y coordinates of center of mass are

X

 cos  2

...............(i)

 sin  2 from (i) and (ii) Y

...............(ii)

2 4 equation of circle. X2  Y 2 

D-5.

V1 =

10 – 20 2

[constrained relation of P1 ] V1 = – 5 m/s 10 =

–5  V2 2

V2 = 25 m/s VC = V2 = 25 m/s upwards VP1 = V1 = – 5 m/s

 VP = 5 m/s downward [because we have assumed upward direction as +ve for V1]

SECTION (E) E-1.

Since point A is massless net force on it must be zero other wise it will have  acceleration.  F1 – 60 cos 45 = 0 F1 = 30 2 N F2 – 60 cos 45 = 0



F2 = 30 2 N W – 60 sin 45 = 0 W = 30 

E-3.

2 N 

F = ma



ˆ ˆ a = ax i + a y j

=

d2 x dt 2

d2 y ˆi + ˆ dt 2 i

= (10) ˆi + (18 t) ˆj

at t = 2 sec t = 2 sec 

ˆ a = 10 ˆi + 36 j



ˆ ˆ F = 3 (10 i  36 j )

= 30 ˆi  108 ˆj 

F =

302  108 2 = 112.08 N

RESONANCE

SOLN_NEWTON'S LAWS OF MOTION - 4

E-5.

R4 – mg = ma

R4 – 1 = 0.1 × 2

R4 = 1.2 N

R3 – mg – R4 = ma R3 – 1 – 1.2 = 0.1 × 2

R3 = 2.4 N



Similarly R2 = 3.6 N R1 = 4.8 N F =6N Fnet = ma = 0.1 × 2 = 0.2 N

m

B m

> T

2a

B

2x 2a

> >

(a) When the block m is pulled 2x towards left the pully rises vertically up by x amount.  aB = 2aA F.B.D. of blocks

T x

T

T = m2a ............. (1)

2T

>

E-7.

A 2m a

F.B.D.

2T FBD of A

FBD

A 2m a

2mg – 2 T = 2ma

2mg mg – T = ma ................(2) (1) + (2)  mg = 3ma a = g/3  aB = 2g/3

RESONANCE

SOLN_NEWTON'S LAWS OF MOTION - 5

(b)  = xB + 3xA 2

+3

aB

d2 x A

dt dt  0 = – aB + 3aA

B m

2

xB

T T

 aB = 3aA ........... (1) For B

^

0=

d2 x B

A

^

^ T

^

2T^ 3m

aA

T = maB ....................... (2) 3mg

For A

3mg – 3T = 3maA ............... (3) mg – T = maA

By (1) , (2) & (3)  aB = 3g/4 Ans.

SECTION (F) F-1.

Reading of weighing machine is equal to the normal reaction Normal reaction is not affected by velocity of lift, it is only affected by acceleration of lift. For I, II and III a = 0 N – mg = 0 [Equilibrium of man] N = mg = 600 N For IV, VI and VII a = +2 m/s2 N – mg = ma [Newtons II law] N = 60 × 2 + 60 × 10 = 720 N For V and VIII a = – 2 m/s2 N – mg = ma [Newtons II law] N = 60 × (–2) + 60 × 10 = 480 N

F-3.

(a) For A For B 

(b)



T AB = 2mg, T BC = mg 2mg + mg = maA  T AB – mg – T BC = maB 2mg – mg – mg = maB  T BC – mg = mac  T AB = 2mg T AB – mg = maB 2mg – mg = maB aB = g () aA = 0 & aC = g().

TAB

aA = 3g A m

maB = aB = 0 ac = 0.

TAB

m B m

Cm

mg

mg mg

TAB

B m

TBC

TBC

aB

mg

SECTION (G) G-1.

If we take both A and B as a system then there is no external force on system.  mAaA + mBaB = 0 [Newton’s II law for system ] 60 aA + 75 × 3 = 0

aA =

–15 m/s2 4

–ve sign means that acceleration is in direction opposite to the assumed direction.

RESONANCE

SOLN_NEWTON'S LAWS OF MOTION - 6

G-3.

4F – (M + m)g = a =

G-5.

(M + m)a

4F 4F – (M  m)g = –g Mm Mm aA

tan 45º = a B

(wedge constrained relation)

N sin 45 = ma For Rod A  mg – N cos 45 = ma From equation (1) & (2) a =

...........(i) ...........(ii)

g 2

SECTION (H) H-1.

Pseudo force depends on mass of object and acceleration of observer (frame) which is zero in this problem.  Pseudo force is zero.

H-3.

F.B.D. in frame of lift It is obevious that block can accelerate only in x direction. ma is Pseudo force.  mg sin  + ma sin  = max [Newton`s II law for block in x direction]  ax = (g + a) sin 

PART - II SECTION (A) A-1.

Experimental fact.

A-3.

Force exerted by string is always along the string and of pull type. When there is a contact between a point and a surface the normal reaction is perpendicular to the surface and of push type.

SECTION (B) B-1.

F – N = 2 ma, N = ma1

[Newton`s II law for block A] [Newton`s II law for block B]

F 3



N=



N = 2 ma2 [Newtons II law for block A] F – N = m 2a [Newtons II law for block B] N = 2F/3 so the ratio is 1 : 2

RESONANCE

SOLN_NEWTON'S LAWS OF MOTION - 7

B-3.

F – N = Ma N – N’ = ma N’ = M’a 

N’ = M’

[Newtons law for block of mass M] [Newtons law for block of mass m] [Newtons law for block of mass M’]

F M  m  M'

N = (m + M’)

F M  m  M'



N > N’

SECTION (C) C-1.

Point A is mass less so net force on it most be zero otherwise it will have  acceleration.  F – Tsin = 0 [Equilibrium of A in horizontal direction] F

 C-3.

T = sin 

10 – T2 = 1 a [ Newton’s II law for A ] T2 + 30 – T1 = 3 a [ Newton’s II law for B ] T1 – 30 = 3a [ Newton’s II law for C ] 

a=



T2 =

C-5.

g 7 6g 7

Mg – T = Ma [ Newton’s II law for M] T – mg = ma [ Newton’s II law for m] 

T=

2 m Mg mM

If m 8r For any point outside, the shells acts as point situated at centre. Distance from centre of hollow shell = (y – 4r) Field due to hollow shell = –

4GM ( y – 4r )

Distance from centre of solid spere = (y – r) Field due to solid spere = –

GM ( y – r )2

 4GM GM  ˆ  (– j ) Total field =  y – 4r   ( y – r )2 

RESONANCE

SOLN_GRAVITATION - 50

x

3.

(a)

Force will be due to the mass of the sphere upto the radius r In case (i) 0 < r < b ; Mass M = 0, therefore F(r) = 0 In case (ii) b < r < a ; Mass M =

(iii) a < r <  ; Mass M =

r2

(b)



Uf – Ui = –

 a3  b3 4 4  (a3 – b3), therefore F(r) = Gpm 2 3 3  r

   

   

 Fc .dr

r1

(i) 0 < r < b ;

u(r) = - 2 Gm(a2 — b2)

(ii) b < r < a ; u(r) = (iii) a < r <  ; u (r) = 5.

 b3 4 4  (r3 – b3), therefore F(r) = Gpm r  2 3 3 r 

2Gm (3ra2 - 2b3 - r3) 3r 4Gm 3 (a  b 3 ) 3r

(a)

The gravitation field is uniform inside the cavity and is directed along OO´ . Hence the particle will strike at A.

(b)

The gravitational field at any point P inside cavity.

4  |E | = G 3

4 4 2 G = Gy OO´ = GR 3 3 3   Total workdone = m | E | . S = m . 2 GR . R 3 2 Applying work - energy theorem Workdone by all force = Change in kinetic energy

–

1 m . 2 GR . R = mv2 2 3 2

6.

v =

2GR 2 3

(a)

GMsm 1 mv2 = 2 R

(b)

1 mve2 – 2

Ans.

G m =0 2R

GmM s 1 m (V + Ve)2 = 2 R V =

 2 – 1

or

V=

2G S R

or

Ve =

G R

or

V + Ve =

2GMs R

GMs R

RESONANCE

SOLN_GRAVITATION - 51

7.

Applying angular momentum conservation : mv0 = mvd v0= vd .......... (i)

1 mv02 + 0 2

Intial energy =

GMs 1 mv2 – d 2 Applying energy conservation ,

Final energy =

GMsm 1 1 mv02 = mv2 – d 2 2 2GM s .......... d From equation (i) and (ii) :

v02 = v2 –

v 022

v02 =

d2 +

d

2

2GMs v0

2

(ii)

2GM s d

–

d – 2 = 0

Solving this quadratic

d = –

GMs v0

2

2   2 GMs  1   v 0  – 1   =  GM  v 02      

2

 GMs   v 2  0

+

   2  

Ans.

PART - II 1.

Gravitational field at ‘m’ due to hollowed - out lead sphere = { Field due to solid spere } – { Field due to mass that was removed } Field due to solid sphere =

GM d

Field due to removed mass =

2

= E1 =

GM' x2

3 M 4 R M’ = 4 ×   3 2 R 3 3

And

E2 =

4R 2

= E2

=

M 8

R 2

x = d–

So ,

GM

GM R  8 d –  2 

=

2

GM = 2 18 R 2 3 R  8   2 

GM

Enet = E1 – E2 =

GM  1 1   –  R 2  4 18 

Fnet = mEnet

RESONANCE

=

=

7GMm 36 R 2

7GM 36 R 2

Ans.

SOLN_GRAVITATION - 52

4.

5.

Gm1 3 r1 = 4 Gm 2 r2



m1 + m2 = m



4R



V = KVe = K

Ve =

m1

m

2GM R

m2

=

4 r12

4 r22

m1

=

2

Initial total energy =

1 2GMm mv2 – 2 R

Final total energy =

1 2GMm m02 – 2 x

or

4 r12

=

5 Gm = Ans. 3 Gm1 R r1

2GM R

1 2GM 2GMm m.K2 – 2 R R

=

Applying energy conservation

1 2GM 2GMm mx2. – = 2 R R 1 1 x2 = – x R R 9.

Fg =

x =

0–

2GMm x

R 1– k2

Ans.

GMmr R3

pressing force

= Fg cos  =

GMm

= 2 R2

GMmr cos  R3

= constant

a=

Fg sin  GMr sin  = m R3

a=

GMy R3

10.*

In elliptical orbit sun is at one of the foci hence the distance between the planet and sun changes as planet revolves hence linear speed, kinetic energy and potential energy of planet donot remain constant

11.*

S =

2 , 1 .5

E =

2 24

1  1 – west to east = 2    1.5 24 

T west to east =

2

= 1.6 hours

 west to east

Similarly 1   1  east to west = 2   1 . 5 24  

T east to west =

24 hours 17

RESONANCE

SOLN_GRAVITATION - 53

EXERCISE-3 1.

P.E. = –

GMm r

Total energy = 

T.E. = 0

K.E. =



GMm 1 + mV2 r 2

if 

GMm 1  mV 2  0  v = r 2

2GM T.E. is – ve r

For v <

1 mV2 2

2GM r

2GM , T.E. is + ve r

for v >

GM i.e. equal to orbital velocity, path is circular.. r If T.E. is negative, path is elliptical. If T.E. is zero, path is parabolic. If T.E. is positive, path is hyperbolic. If V is

2.

5.

(A) (B) (C) (D)

At centre of thin spherical shell V  0, E = 0. At centre of solid sphere V  0 , E = 0. At centre of spherical cavity inside solid sphere V  0, E  0. At centre of two point masses V  0, E=0.

 4 2  3   T =  GM  R    y = mx + c

 GM  R=  2  4 

2

(3) Slope

=m=

log R =



2 1 log T + log 3 3

 GM   2  4 



 20   10 11  M  3   = 18 log 4  10

M = 6 × 1029 Kg T 2  R3

 RA   RB

3

 R  =  A   RB

2

t=

2

   =  B   A

rel = 80 – 0 = 70

6.

T2 / 3

2 3

 GM  1 intercept c = log  2  = 6 3  4 

(4) (5)

1/ 3

  



 rel = (rel) t

 R     4R  

3

2

 =  B  A

  

   B  A

  1  =   8 

2 = (T0) t

2 T 0

Let M and R be the mass and radius of the earth respectively. If m be the mass of satellite, then escape velocity from earth e =

( 2 Rg)

Velocity of satellite s =

e = 2

Further ]s =



 GM    =  r 

(2 R g) / 2

......... (1)

 R 2g    R h  

R 2g Rh h = R = 6400 km

 2s =

RESONANCE

SOLN_GRAVITATION - 54

7.

4 2 3 x Gm Hence time period of revolution T is T2 =

 8.

T = 2

x3 Gm

T = 2

8R g

(Put x = 2R)

Now total energy at height h = total energy at earth's surface (from principle of conservation of energy)

m 1 m = m2 – GM Rh 2 R



0–GM

or

1 GM m GMm m2 = – 2 R 2R

v=

gR

( h = R)

9 to 11 Let the angular speed of revolution of both stars be  about the common centre , that is, centre of mass of system. The centripetal force on star of mass m is m2

2d Gm(2m) 4 2 3 d = . Solving we get T= 3 d2 3Gm

The ratio of angular momentum is simply the ratio of moment of inertia about center of mass of system. 2

Lm LM

 2d  m  Im 3    2  2 I M d 2m  3

Similarly, The ratio of kinetic energy is simply the ratio of moment of inertia about center of mass of system. 2

+

Km KM

 2d  1 I m 2 m  3  2   2  2 1 I M 2 2m d  2 3

12.

Till the particle reaches the centre of planet, force on both bodies are in direction of their respective velocities, hence kinetic energies of both keep on increasing . After the particle crosses the centre of planet, forces on both are retarding in nature. Hence as the particle passes through the centre of the planet, sum of kinetic energies of both the bodies is maximum. Therefore statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

13._

It is minimum  –ve (iii) Energy density =

15. 14.

for closed paths (circular or elliptical) the total mechanical energy is negative. (i) g’ = g – R cos 2  At equator = 0  g’ = g – R  0 = g – R 

RESONANCE

B2 and B increases by a large factor.. 2 0  r

SOLN_GRAVITATION - 55

 =



(ii)

g = R

9.8

= 1.24 × 10 –3 rad/s

6400  103

dA L = = constant because angular momentum of planet (L) about the centre of sun is constant. dt 2m Thus, this law comes from law of conservation of angular momentum.

(iii)

T r 3 / 2



T2  r2    T1 =  r1 

3/2

 r2  T2 =    r1 

or

3/2



 3.5 R   T1 =   7R 

3/2

(24) h = 8.48 h

EXERCISE-4 PART - I 1.

Time period of a satellite very close to earth’s surface is 84.6 minutes. Time period increases as the distance of the satellite from the surface of earth increase. So, time period of spy satellite orbiting a few hundred km, above the earth’s surface should be slightly greater than 84.6 minutes. Therefore, the most appropriate option is (C) or 2 hrs.

2.

(A) Gravitational field is a conservative force field. In a conservative force field work done is path independent.  W I = W II = W III

3.

speed of particle at A VA = escape velocity on the surface of moon =

2GM R

At highest point B, VB = 0 From energy conservation.  UB U A  1 –  mVA2 = VB – VA = m  m   m 2

or

VA2  UB UA  UA – GM  –  , also  [3R2 – r2] 2 m m m 2R 3  

2 GM – GM  – GM  R    2     1 . 5 R – 0 . 5 R – –   R R  h  R3  100        2

or or

1 3 1  99  1 –1  – –   R R  h 2R 2  100  R h = 99.5 R  99R Ans rA C A

4.

mA

rB

com

Gm A m B



(rA  rB ) 2 m A rA



TA2



B mB

4 2 = mArA

TA2

4 2 = m B rB

TB2

mB rB TB2

As C is com  mArA = mBrB hence TA = TB 5.

(A) It is similar equation as v = ù a 2  x 2 in SHM. (B) Particle on positive x-axis move towards origin with speed decreasing as x decreasing. (C) It is spring mass system performing SHM. (D) Object moves away from Earth so its speed will decrease, since its speed is greater than escape velocity so it will never return back.

RESONANCE

SOLN_GRAVITATION - 56

6.

 = 0 =0 Case I r < R FC = mg

rR

mV 2 r

r mV 2  (g = acceleration due to gravity at surface of sphere) R r g r R

V= Case II

for r < R

r>R GMm

mV 2 r

=

r2

GM  r

V=

g R r

So

7.

If only gravitational force acts on astronaut (that is in state of free fall), he shall feel weightless. Thus statement-2 is correct explanation of statement-1.

8.

W ext = U – UP W ext = 0 –



 –  –

Gdm  .1 x 

M W ext = G 2GM

=

7R 2

W ext = W ext =

9.

   7R 

P

2rdr 2

16R 2  r 2

=

x

rdr

2GM



7R 2

16R 2  r 2 3R

2GM zdz = [Z] z 7R 2

4R

r

dr

4R 2GM  2 2 16 R  r  3R 7R 2 

2GM 7R

2

4

2R – 5R



W ext =



2GM 7R 2

c.m.

A 2.2 Ms

5d 6

4



2 –5 .

B d 6

11 Ms

 5d   5d   dd ( 2.2Ms )      (11 Ms )      Total angular momentum about c.m.  6  6   66 = 6. Angular momentum of B about c.m. =  dd (11 Ms )       66

10.

g=

GM R2

4  (G) R 3  3  = 2 R

;

g R

g'  '   R'   2   R'  6        g =     R  =  3   R  = 11 Given, 

Ve  R 

RESONANCE

R' 3 6 = R 22

Ve =

GM = R

 4  (G) ()) R 3   3   R

; Ve = 3 km/hr. SOLN_GRAVITATION - 57

11.

12.

Ve =

2v 0



2GM = R

Ves =

KE =

1 1 mv 2e = m 2 v 0 2 2



4 2.G. R 3 3 = R

2



= mv02

4G  R 3

Ves  R Sarface area of P = A = 4RP2 Surface area of Q = 4A = 4 RQ2 mass R is MR = MP + MQ

 RQ = 2Rp

4 4 4 3 3 3 R R =  R P +  RQ  RR3 = RP3 + RQ3 3 3 3 = 9RP3 RR = 91/3 RP  RR > RQ > RP 

VR 1/3 and VP = 9

Therefore VR > VQ > VP

VP 1 VQ = 2

PART - II 1.

Electric charge on the moon = electric charge on the earth

2.

V=

3.

2GMe  10 2GMe = 10 = 110 k m/s Re Re 10 Acceleration due to gravity at leight h from earth surface.

2GM = R

g

g' = 1  h 

=

2

(r  x )

2

r – x = 2x x=



5.

Gm 2 (2R )2

Gm 2 4R 3

6.



1 2 = x rx



3x =





r 3

r 3

Gm G( 4m)  r /3 2r / 3

=

g g  2 9  h  1    h = 2R  R

G( 4m)

Gm x



R



4.

2

3Gm 6Gm 9Gm  = r r r

Ans.

= m2R

= 2 Gm 4R

3

v = R 

v=

Gm 4R

3

×R =

Gm 4R

 GMm  GMm  W = 0 –  R  R 

m = mgR = 1000 × 10 × 6400 × 103 R = 64 × 109 J = 6.4 × 1010 = gR2 ×

RESONANCE

SOLN_GRAVITATION - 58

TOPIC : WORK, POWER AND ENERGY EXERCISE-1 PART - I SECTION (A) A 1.

f = mg = F Displacement = vt (a) W mg = mg × vt cos90º = 0 (b) W N = N × vt cos90º = 0 (c) W f = –mgvt (d) W F = Fvt = mgvt.

A 6.

m = 500 g =

1 kg 2

mg sin  = fk 4  1 Wfk = (mg sin ) (2) =   (10) ×2=8J 5 2  

A 7.

W 1 = (mg sin )4 = (20 × 10 ×

4 ) (4) = 640 J 5

WF2 + WGrav = K = 0

WF2 – (mg sin ) (4) = 0

W 2 = WF = 4 mg sin  = 640 J 2

SECTION (B) B 2.

B 4.

W = Area under given graph from x = 0 to x = 35m =

1 1 × (20 + 40) × 10 – ×5×5 2 2

=

575 J. 2

F at any moment = mg

(  x )   mg( 

 =

W= 

0

 x)

dx



mg . 2

SECTION (C) C 4.

Work done by resistive force = W R = K = =

1 × 20 × 10–3 (1002 – 8002) 2 – 6300 J

So, average resistive force =

RESONANCE

6300 J = 6300 N. 1m

SOLN_Work, Power & Energy - 59

C 6.

Work done by the force =  F ds = F ×

1 2 at 2

1 F × × t2 2 m

=

F×

=

20 2  10 2 F2 t 2 = = 4000 J 25 2m

Now

K = =

1 m(v2 – u2) 2

1 m (2as) 2

=m×

F 1 F × × × t2 m 2 m

F2 t 2 = 4000 J 2m W F = K. =

 C-10.

U = –K

1 2 kx – 2mgx = 0 2  C-11.

x=

4mg . k

(a) Since, gravitational force is conservative, So, work done by it in round trip is zero. 5 1 = 10 2   = 30º W F = mg(sin +  cos) × 

(b) sin =

1 3   = 0.3 × 9.8  2  0.15  2   10 = 18.519 J   (c) W f = –f.s = –mg cos (2) = – 2mg cos = – 2 × 0.15 × 0.3 × 9.8 × 10 ×

3 = –7.638 J. 2

(d) By W.E.T, Kf – Ki = W F + W f + W g  Kf = (18.519 – 7.638)J = 10.880 J. C 12.

Displacement of 4kg block = 2 × 2m = 4m 4kg = 2 × 2m = 4m Final speed of 4kg block = 2 × 0.3 = 0.6 m/s 4kg = 2 × 0.3 = 0.6 m/s W f + W g = K

1 1 × 4 × (0.6)2 + × 2 × (0.3)2 2 2



– × 4 × 10 × 4 + 2 × 10 × 2 =



160  = 40 – (0.72 + 0.09)   =

RESONANCE

40 – 0.81 = 0.2449 160

SOLN_Work, Power & Energy - 60

C-14.

(i) w.r.t. person in the train

Ft m (ii) w.r.t. person on ground, v1 = at =

Ft m (iii) According to person in the train, v = vc + v1 = vc +

1 F2 t 2 mv12 = 2 2m (iv) According to person on ground, K1 =

K =

1 m 2

2

Ft  1  2 v c  m   2 mv c .  

1 2 Ft 2 at = . 2 1 2m (vi) According to person on ground, (v) S1 =

1 F 2 Ft 2 t = + vct. 2 m 2m (vii) According to person in the train work done by F = Fs1 S = vct +

F2 t 2 2m According to person on ground, Work done by F = F.s =

 Ft 2  F =  2m  v c t  .   (viii) Comparing W g = Kg and W c = Kc . (ix) Work–energy theorem holds in moving frame also.

SECTION (D) D 5.

Let

m1 = 2m2



m2 (m1  m 2 ) a = (m  m ) g = 3m g = g/3 1 2 2

So, distance travelled by each block =

Also

2m1m 2 g 4m 2 g T = m m = = 16 3 1 2



12 m2 = g

1 2 at = g/6 2

Hence, loss in gravitation P.E. during first second = (m1 – m2)gh =

(2m2 – m2) g ×

RESONANCE

12 g g = g  g  6 = 2g. 6

SOLN_Work, Power & Energy - 61

D 6.

a=

32 g g= 32 5

Distance covered in fourth second =

(2n  1)a 2

( 2  4  1)  g 25 Hence, work done by gravity = (m2 – m1)gh

=

= (3 – 2)g × = D 7.

=

7g 10

7g 10

7 2 g . 10

W s + W g + W f = K 



1 2 kx + mgx sin37º – mg cos37º × x = 0 2





3 4 1 × 100 × (0.1)2 + 1 × 10 × 0.1 × –  × 1 × 10 × × 0.1 5 5 2



=

1 . 8

SECTION (E) E 4.

Power developed by motor =

E 6.

Power P =



E 7.

P=



E 8.

mgh 400  10  120 = = 1600 W.. t 5  60

mgh t

Pt 2  10 3  60 m = gh = kg = 1200 kg. 10  10 mgh t

t=

200  10  40 mgh = sec. = 8 second. 10  1000 P

20 kg / minute = 20 kg / 60 sec =

1 kg/s 3

 1 10  20 P =   g (20) = watt 3 3    746 W = 1 H.P



P=

100 HP 1119

RESONANCE

SOLN_Work, Power & Energy - 62

SECTION (F) F 1.

  (a) w =  F.ds =  ( x 2 y 2 ˆi  x 2 y 2 ˆj ).(dx ˆi  dyˆj ) =  ( x 2 y 2 dx  x 2 y 2dy ) which is not a perfect integral and hence cannot be integrated without knowing y = f(x) or x = f(y). So, work  done by F depends on path. So, it is non–conservative force. (b) While moving along AB, y = 0 and along BC, x = a. a

a

0

0

2 2 2 2 W ABC =  x y dx   x y dy

a3 a5 = 3 3 While moving along AD, x = 0 and along DC, y = a = 0 + a2 ×

So

a

a

0

0

2 2 2 2 W ADC =  x y dx   x y dy

= 0 + a2 .

a3 a5 = 3 5

Along AC x=y So

a

a

0

0

a

a

0

0

2 2 2 2 W AC =  x y dx   x y dy

2 2 2 2 =  x x dx   y y dy =

F 2.

2a 5 . 5

dU (a) F(y) =  dy =  dU (b) F(y) =  dy = – 3ay2 + 2by dU (c) F(y) =  dy = –U0 cos y..

F 5.

At x = 0, total energy is in form of K.E. since U = 0 and it turns back when its K.E. = 0 So, total energy is in form of P.E.  U = –K   

1 2 kx = 1 2 x2 = 1 × 2 × 2 x = ± 2m

Ans.

PART - II SECTION (A) A 3.

W = (force) (displacement ) = (force) (zero ) = 0

A 6.

W = (2000 sin 15º) × 10 = 5176.8 J

RESONANCE

SOLN_Work, Power & Energy - 63

A 9.

S1 =

1 1 1 g 12 , s2 = g 22 , S3 = g 32 2 2 2

S2 – S1 =

1 1 g 3, S3 –S2 = g5 2 2

W 1 = (mg) S1, W 2 = (mg) (S2 – S1) , W 3 = (mg) (S3 – S2) W1 : W2 : W3 = 1 : 3 : 5 A 10.

T = mg + ma, S =

1 at2 2

WT = T × S =

m(g  a)at 2 2

A 11.* W = K, 0 = K, k remains constant, speed remains constant. A 14.* W = K > 0  K ( = kinetic energy) increases p=

2mk , p  as k.

A 15.* W f + W G + W N = K = 0 As W G = 0, W N = 0 so W f = 0.

SECTION (B) x1

B 1.

W=

 cx

dx = c

x12

o

B 2.

2

F

F2

1

F = K1x1 , x1 = K , W 1 = K x2 = 2 1 1 2 K1 1 similarly

F2 2K2

W2 =

since K1 > K2 , W 1 < W 2

SECTION (C) C 2.

C 5.

a=

h=

1 F 2  t , 2 m

F , m

S=

1 gt2, 2

W = mgh = mg

1 mg2 t 2 = Kf – mu2, 2 2

Kf =

 Ft2   

W F = FS = F  2 m  



gt 2 , W = Kf – Ki 2

1 mg2 t 2 mu2 + 2 2

Hence Ans. is (A) v

C-10.

x

 V2   Kx 2  dV V = – Kx,  2   –  2      dx u 0

V2 – u2 = – Kx2 1 1 1 mu2 – mV2 = mK x2 2 2 2

Loss  x2

RESONANCE

SOLN_Work, Power & Energy - 64

x

C 12.

(mg sin ) x –

 mg cos 

dx = 0

0

x

 x dx

sin  x = o cos 

x tan  = 0

0

x2 , 2

x=

2 tan  0

SECTION (D) D 3.

Ui + 0 = Uf + Ui – Uf =

1 mv2 2

1 mv2 2

U= m= D 4.

1 mv2 2

2U v2

1 mu2 = mgh, u2 = 2gh 2

....(i)

 3h 

mg  4  + K.E. = mgh   K.E. =

mgh 4

mgh / 4 K.E. 1 = 3mgh / 4 = P.E. 3

D 6.

W F + W S = 0, W F – U = 0 , W F = U = E E=

1 1 K x 2 , FxA = K x2 2 A A 2 A A 2E 2F2 , K = A KA E

2F 2F K A = xA , K A =

similarly

KB =

2F2 , EB



...(i)

KA = 2KB



 2F2  2F2   = 2 E  E  B 

 EB = 2E Alter : F = KA xA = KBxB EA =

1 K x2 2 A A

EB =

1 K x2 2 B B

EA  K A   x A   EB  K B   x B

2

  

2

EA 1  1  2   EB 2 2  

RESONANCE

SOLN_Work, Power & Energy - 65

D 7.

D 13.

100 =

1 1 K(2cm)2 , E = K(4cm)2 2 2

so

E =4, 100



E – 100 = 300 J

E = 400 J

1 11  ( 2m)u2   mv 2  2 22 

.... (i)

1 1 (2m) (u + 1)2 = mv2 ....(ii) 2 2

From (i) and (ii) D 14.

1 2 1

W 1 = work done by spring on first mass W 2 = work done by spring on second mass W 1 = W 2 = W (say) W 1 + W 2 = Ui – Uf 2W = 0 – W= –

D 15.

u=

1 Kx2 2

Kx 2 4

W a + W c = K = 0, Wa =







W a – mg  2 – 2 cos 60 º  = 0  

 1 mg  5 = (0.5) (10)  4  = J. 4   4

SECTION (E) E 3.

V = 0 + at, F –  mg = ma , F = mg + ma, P = (mg + ma) at

E 5.

P = TV = 4500 × 2 = 9000 W = 9KW

E 6.

P1 = 80 gh/15 , P2 = 80 gh/20 P1 20 4 P2 = 15 = 3

SECTION (F) F-2.

dU dU   ve,   ve dx x  A dx x B

So, F-5.

F-6.

FA = positive, FB = negative

WC = WC + WC = 5 + 2 = 7 PR PQ QR U = cos (x + y), x

U y = cos (x + y) F = – cos (x + y) ˆi – cos (x + y) ˆj

= – cos (0 +

  ) ˆi – cos (0 + ) ˆj 4 4

|F | = 1

RESONANCE

SOLN_Work, Power & Energy - 66

EXERCISE-2 PART - I F a = (m  m ) 1 2

1.

F f1 = m1a = m1 (m  m ) 1 2

2F – f1 – f2 = m2a – f2 = – 2F + f1 + m2a = m1a + m2a – 2F

F – f2 = (m2 + m1) (m  m ) – 2F = F – 2F = – F 1 2

 f2 = – F

2F – K (m2 + m1)g = (m2 + m1)a

F F 2F – (m2 + m1) (m  m ) = µK (m2 + m1)g  (m  m )g = K 1 2 1 2 W = work done by friction force on smaller block

m1F x (m 2  m1 )

= f 1x = 2.

mg = N + F sin .......(1) N = F cos .......(2)  mg = F cos  + F sin  F=

mg cos    sin 

.......(3)

ì mg cos è (10) 40000 = cos è  ì sin è 5  tan 

WF =

Ans.

1 ì2

F is min. if D = cos  +  sin  is maximum and its maximum value is

Fmin. =

 mg

1 2

1 

WFmin =

mg

mg

=

2

1 2

10

1 ì 2 1 2  = 0.2, mg = 4000 Nt (0.2)( 4000 )10 WFmin =

4.

fK = 



x  4

W=

=–

400  20 ( 1  0.04 )

2

=

8000 = 7692.307 J 1.04

Ans.

m ( – x)g  

W=

=

( 1  ( 0 . 2) 2 ) 2

–ì

m(  – x )g dx 

2  ìmg [(  x ) ] 4 ×  2

ìmg 9 2 × 2 16

=–

RESONANCE

9mg  32 SOLN_Work, Power & Energy - 67

7.

(a)

Taking F = 40 N, m = 4 kg ,  = 53º ax = (F cos  – mg sin )/m = (40 cos  – 40 × ay =

4 )/4 = 10 cos  – 8 5

F sin á 40 sin á = = 10 sin  m 4

x=0×2+

1 (10 cos  – 8) (2)2 = 20 cos  – 16 2

1 (10 sin ) (2)2 = 20 sin  2 WF = (F cos ) x + (F sin )y WF = (40 cos ) (20 cos  – 16) + (40 sin ) 20 sin  = 800 cos2  – 640 cos  + 800 sin2  WF = 800 – 640 cos  WF  800 – 640 WF  160 J If W F = 160 J then 160 = 800 – 640 cos   cos  = 1 y = 0 and x = 20 – 16 = 4 y=

(b)

W G = (–mg sin ) (4) = (–4 × 10 × (c)

4 )4 5

= –128 J F acts along the x-axis. W G + W F = K –128 + 160 = K Kf = 32 J.

9.

Work energy theorem (Between A & C) W f + W G + W sp = K  mg cos  (5 + 3) + mg 2 sin  = 0 



2 3 tan 37o = 8 16 work energy theorem W sp + W G + W f = K

=

(bet. A & B)

mg 5 sin 37o –  mg 5 cos  –

1 K (0.4)2 = 0 2

16 1 4  3 3 (4 × 10) 5   (5)  = × K 100 2 5 16 5  

K = 9000/ 8 N/m so 10.

x=9

Work energy Theorem on “m” W G + N + W T + W f =K  2

– mg R + O + W T –

 (mg sin ) R d = 0 0



W T = mgR ( + 1)

RESONANCE

SOLN_Work, Power & Energy - 68

11.

13.

W F + W Sp + W fric = K

1 Kx2 –  m1g x = 0 2



Fx –



F–



F = m1g +

& Kx = m2g

1 m2g – m1g = 0 2 m 2 g 2

mg = kx K=

100 mg = = 500 N/m 0 .2 x

1 1 K (0.2)2 + mv2 = m × 10 × 0.2 2 2 1 1 × 500 × 4 × 10–2 + × 10 v2 = 10 × 10 × 0.2 2 2 10 + 5v2 = 20 v2 = 2 v=

2 m/s  Since u is 4 m/s ( ) so block will compress the spring. Let x be the compression of spring.

1 1 1 1 mu 2 + K (0.2)2 + 0 = m(0)2 + Kx2 + mg (x + 0.2) 2 2 2 2  4  1 1 1  = × 10 (4)2 + × 500 ×  × 500 (x)2 + 10 × 10 (x + 0.2)  100  2 2 2

80 + 10 = 250x2 + 100 x + 20 25 x2 + 10 x – 7 = 0 solving this x = 0.36 m So from initial position distance is ( 0.2 + 0.36) m = 56 cm 15.

(i)

mg = T cos  mg =

 x 2mg  2 a  x 2  a  a ×     2 2   a  a x



x=

a 2

(ii)

1 K 2

 2a 

2

+ mga =

   

1 1 K (2a – a)2 + mv2 2 2

1  2mg  1   (2a2 – a2) + mga = mv2 a  2  2 

4ga = v

RESONANCE

SOLN_Work, Power & Energy - 69

1 K 2

(iii)

2

 2a

+ mg a =

2 1  2 2  K  a  y  – mg y  2 

1 1 K2a2 + mg a = K(a2 + y2) – mgy 2 2 1 2mg 1 2mg 2 1 2mg 2 2a2 + mg a = a + y – mg y a a a 2 2 2 3 mg a – mg a =

mg y 2 – mg y a

mg y 2 – mg y a 2a2 = y2 – ay y2 – ay – 2a2 = 0 y2 + ay – 2ay – 2a2 y (a + y) = 2a (y + a)  2 mg a =

  16.

y = 2a

  P = Fext . V  Where V is the vel. of point of application Fext + m, g = T & m2g =T  Fext = m2g – m,g = (m2–m1) g (a)



P = (m2–m1) g v Ans.

(b)

Fext + m,g – T = m,a T–m2g = m2a _____________________________ Fext = (m1+ m2) a (m2–m1)g = m2(g+a) – m1(g – a)  P = (Fext) (0 + at) = {m2(g+a) – m1 (g – a)} at

Ans.

19.

– g [ma + m´a + M

a 1 ] = – g [ m2a + m´0 + Ma] + (M + m + m´)v2 2 2

2g[2ma  ma  Ma 

Ma  m´a] 2 =v

M  m  m' v=

ag

M  2(m – m´) M  m  m´

RESONANCE

Ans.

SOLN_Work, Power & Energy - 70

21.

U (x) = 20 + (x – 2)2 du = 2(x – 2) dx – F = 2(x – 2) F = – 2(x – 2) m (x – 2) = – 2 (x – 2) Let x = x – 2 mx = – 2 x 1x =–2x x=–2x Simple Harmonic Motion Mean position is x = x – 2 = 0  x = 2 W2 = 2 ,

Kinetic energy = =

1 mv2 2

1 (1) (2) (A2 – x2) = x – 2, x = 5 – 2 = 3 2 1 (1) (2) {A2 – 32} 2

20 =

20 = A2 – 9

A2 = 29



A = 29

Aliter : for mean position dU = – 2(x – 2) = 0  dx At x = 5 K.E. = 20 J U(x = 5) = 20 + (5 – 2)2 = 29 J Total energy, T.E. = 20 + 29 = 49 J At amplitude U(x)max = 49 J = 20 + (x – 2)2 29 J = (x– 2)2

F=–

x=2±

29

x=2+

29 , 2 –

xmin = 2 –

x=2

29

29 = –3.38

xmax = 2 +

29 = 7.38 K.E.max when U(x) is minimum at x = 2 U(x)min = 20 J KEmax = 29 J 22.

Using work energy thoerem,

3R 1  4mg   R      W f + mg 2 2  R  2 Wf = 



2

=

1 m 2

2

 3gR 

1 mgR 2

 f = N (as kinetic friction) Wf =

 Fs cos  d x Fs = k 2 R (sec  1) Wf =

f dx =









k 2 R (sec

RESONANCE

( x = 2 R tan

 ; dx = 2 R sec2  d)

 1) cos  2 R sec2  d SOLN_Work, Power & Energy - 71

0

Wf = 4 R

2

k



(sec 2  sec ) d 

0 = tan 1 (3/4)

0

= 4 R2

0

k  tan    n (sec   tan )  = – 4 R2 k [ tan 0  ln (sec 0 + tan 0)] =  0

 4 R2

3

= 23.

5

3 

= R2

k    n     4 4  4 mg 2 R k (3  4  n 2)

=

k [3  4 ln 2] =

mgR 2

mg R 2

1 Ans. 8 (3  4  n 2)

Velocity will be maximum when a = 0 For a = 0, F = 0 This situation occurs for ve following arrangement of springs. Natural length is c = 150 mm Now , Ui + Ki = Uf + Kf

1 1 K{ 5 c – c}2 + K{ 2 c – c}2 2 2 Ki = 0 Ui =

Uf = 2. 

150

150

1 K{ 2 c – c}2 2

1 1 K{ 5 c – c}2 + K{ 2 c – c}2 2 2

1 1 mv2 + 2. K{ 2 c – c}2 2 2 Solving the equation & putting the values we have =

15  ( 5  1)2  ( 2  1)2  v = 2  





1/ 2

m/s = 3.189 ms–1 .

PART - II 3.

W agent + W G = K = 0 W agent = – W G, But W G is independent of the path joining initial and final position. W G is independent of time taken.

5.

W f + W G = K –mgd – mgh = 0 – gd + gh =

1 m v02 2

1 (v02) 2

(0.6) (10) d + 10(1.1) = 18 7.

d=

7 = 1.1666  1.17 6

W S + W f = K – U + W f = – Ki – Uf – mgx = – Ki 1 1 K x 2 + mgx = mu2 2 2

100 x 2 + 2(0.1) (50) (10) x = 50 × 4 x2 + x – 2 = 0 x=1m

RESONANCE

SOLN_Work, Power & Energy - 72

8.

v= s

ds



s

0

s



ds  s , dt



2 s = t

t

  dt

 0

....(1) s = t/2 W = workdone by all the forces = K = 10.

 2 t 2  1 1 1  mv2 = m  2s = m 2   2 2 2 4  

K.E. + P.E. = constant

fu;r = C (say)

K – mg (tu sin  –

1 gt2) = C 2

K = mg [tu sin  –

1 gt2] + C 2

[= parabolic]

C  0 so answer is (B) 12.

dU = positive constant dx

For x < a, F = negative constant and for x > a, F = 0 so, ans. (C) 14.

E=

 1 p2 , ( E) P  =   2m

1 2m

= constant

Rectangular hyperbola (C) 17.

System is block & string. Applying work energy theorem on system (200)10 – 10g(R – R cos60º) =

1 (10)v2 2

2(200 – 10 × 5) = v2 v= 19.

300 = 10 3 .    dW = F . ds where ds = dx ˆi + dyˆj  and F = – K ( y ˆi + xˆj )  dW = – K ( ydx + xdy = – K d (xy) ( a, a )

W=

 20.



( 0, 0 )

( a, a )

dW = – K



( 0, 0 )

(a, a) d ( xy ) = – K [xy] (0, 0)

W = – Ka 2 From given graphs : 3  3 3 t and ay =   t  1  vx = t2 + C 8 4  4 At t = 0 : vx = – 3 C=–3

ax =

 vx =

3 2   dx =  t  3  dt 8 

3 2 t –3 8

 3 2  dy =   t  t  4  dt  8 

Similarly;

.... (1)

.... (2)

As dw = F. ds = F.( dx ˆi  dy ˆj ) W



 0

4

dw 

3

3

   3

  4 t ˆi   4 t  1 ˆj .  8 t 0

RESONANCE

2

  3    3  ˆi    t 2  t  4  ˆj  dt   8   SOLN_Work, Power & Energy - 73

 W = 10 J Alternate Solution : Area of the graph ;

a

x

and

dt = 6

a

y

= V( x ) f  ( 3)  V(x)f = 3.

dt = –10 = V( y ) f  ( 4) V(y)f = – 6. Thus, u = 5 m/s and v =

45 m/s.

Now work done = KE = 10 J 22.*

W G = K, so

23.*

mgh =

1 1 1 1 mv2 – mu2, mu2 + mgh = mv2 2 2 2 2

v > u and v depends upon u.

    dW F = F . ds , if F perpendicular to ds then 

ds  dW F = 0, ds is displacement of point of application of force, v = . dt (A), (C), (D) are true.

EXERCISE-3 1.

The displacement of A shall be less than displacement L of block B. Hence work done by friction on block A is positive and its magnitude is less than mgL. And the work done by friction on block B is negative and its magnitude is equal to mgL. Therefore workdone by friction on block A plus on block B is negative its magnitude is less than mgL. Work done by F is positive. Since F>mg, magnitude of work done by F shall be more than mgL.

2.

The FBD of block is Angle between velocity of block and normal reaction on block is obtuse  work by normal reaction on block is negative. As the block fall by vertical distance h, from work energy Theorem Work done by mg + work done by N = KE of block 1  |work done by N| = mgh – mv2 2 1  mv2 < mgh 2  |work done by N| < mgh (B) Work done by normal reaction on wedge is positive Since loss in PE of block = K.E. of wedge + K.E. of block Work done by normal reaction on wedge = KE of wedge.  Work done by N < mgh. (C) Net work done by normal reaction on block and wedge is zero. (D) Net work done by all forces on block is positive, because its kinetic energy has increased. Also KE of block < mgh  Net work done on block = final KE of block < mgh.

3.

If the particle is released at the origin, it will try to go in the direction of force. Here

du is positive dx

and hence force is negative, as a result it will move towards – ve x-axis. 4.

When the particle is released at x = 2 +  it will reach the point of least possible potential energy (–15 J) where it will have maximum kinetic energy. 

1 2 m v max = 25 2

RESONANCE



vmax = 5 m/s

SOLN_Work, Power & Energy - 74

6.

(A) W CL + W f = KE  W CL = KE – W f (a) During accelerated motion negative work is done against friction and there is also change is kinetic energy. Hence net work needed is +ve. (b) During uniform motion work is done against friction only and that is +ve. (c) During retarded motion, the load has to be stopped in exactly 50 metres. If only friction is considered then the load stops in 12.5 metres which is less than where it has to stop. Hence the camel has to apply some force so that the load stops in 50m (>12.5 m). Therefore the work done in this case is also +ve.

7.

W CL|accelerated motion = KE – W friction where W CL is work done by camel on load. =

1  2  2 mv  0     k mg.50   

=

 125  1  1000  5 2  0.1 10  1000  50 = 1000    2  2

similarly,

W CL|retardation = KE – W friction

1   75  2 0  2 mv  – [k mg.50] = 1000  2     



WCL |accelerate d motion 125 5 = = WCL |retarded motion 75 3



5:3

8.

Maximum power = Fmax × V Maximum force applied by camel is during the accelerated motion. We have V2 – U2 = 2as 25 = 02 + 2 . a . 50 a = 0.25 m/s 2 ; for accelerated motion  FC – f = ma  FC = mg + ma = 0.1 × 1000 × 10 + 1000 × 2.5 = 1000 + 250 = 1250 N This is the critical point just before the point where it attains maximum velocity of almost 5 m/s. Hence maximum power at this point is = 1250 × 5 = 6250 J/s.

14.

Potential energy depends upon positions of particles

15.

(i)

The net force on the body may have acute angle with its velocity, but one of the constituent force may have obtuse angle with the velocity. Such a force shall perform negative work on the body even though the kinetic energy of the body is increasing. (ii) A net force that is always perpendicular to velocity of the particle does no work but changes the direction of its velocity. (iii) A force which is always constant is also conservative. (iv) From Work - Energy theorem W all forces = KEfinal – KEinitial

EXERCISE-4 PART - I 1.

  Power P = F . V = FV

 dm   = V  d(  volume  F=V   dt  dt  

 = density

 d( volume  = V   = V (AV) dt   2 = AV  Power P = AV3 or P  V3

RESONANCE

SOLN_Work, Power & Energy - 75

Alternate Solution Power output is proportional to number of molecular striking the blades per unit time [which depends on the velocity V of wind] and also proportional to energy to striking molecules or proportional to square of velocity V2 Therefore, power output P  V3 2.

F=–

dU dx x



dU = – F . dx

3 U(x) = – ( kx  ax ) dx

or ;k

 0

2

U(x) = U(x) = 0

kx 2

–

ax 4

and

4

x = 0 and

x=

2k a

2k a From the given function we can see that F = 0 at x = 0 i.e. slope of U-x graph is zero at x = 0. Therefore, the most appropriate option is (d).

U(x) = negative for

3.

4.

x>

Let x be the maximum extension of the spring. From conservation of mechanical energy : decrease in gravitational potential energy = increase in elastic potential energy 

Mgx =

or

x=

From F = 

1 2 kx 2

2Mg k

dU dx

U( x )

x

x

 dU   Fdx    (kx ) dx 0



U(x) = 

0

0

kx 2 2

as U(0) = 0 Therefore, the correct option is (A). 5.

In horizontal plane Kinetic Energy of the block is completely converted into heat due to Friction but in the case of inclined plane some part of this Kinetic Energy is also convert into gravitational Potential Energy. So decrease in the mechanical energy in second situation is smaller than that in the first situation. So statement-1 is correct. Cofficient of Friction does not depends on normal reaction, In  case normal reaction changes with inclination but not cofficient of friction so this statement is wrong.

6.

RESONANCE

SOLN_Work, Power & Energy - 76

As springs and supports (m 1 and m 2) are having negligible mass. Whenever springs pull the massless supports, springs will be in natural length. At maximum compression, velocity of B will be zero.

And by energy conservation

1 1 (4K) y2 = Kx 2 2 2 7.

2m1m2

T = m m g = 1 2

y 1  Ans. (C) x 2

2  0.72  0.36 × 10 0.72  0.36

T = 4.8 N m1  m2

g

a = m m g = 3 1 2 Work done by T = (T) (S) = (4.8) × 8.

10 =8J 6

s=

1 g 10 1 2   (1)2 = at = 2 3 6 2

Ans.

 Fdt  p  K.E. =

1 1 ×4×3– × 1.5 × 2 = pf – 0 2 2 81 p2 = ;K.E. = 5.06 J 4 22 2m

pf = 6 – 1.5 =



9 2

Ans.

PART - II 1.

Let initial velocity is u and retardation is a So, (vr%)

u2 = u2 – 2a × (0.03) 4

u2 – 2a × S 4 here S is required distance from equation (i) & (ii) S = 0.01 m = 1 cm

0=

2.

...(i)

..(ii)

W C = – U = – (Ufinal – Uinitial) 1 1 2 2 = –   k  15 –  k  5  ] 2 2  W C = 8 Joule

3.

K = 5 × 103 N/m x = 5 cm 1 1 W 1 = k × x12 = 5 × 103 × (5 × 10–2)2 = 6.25 J 2 2  k(x1 + x2)2 W2 = 2

 × 5 × 103 (5 + 10–2 + 5 × 10–2)2 = 25J 2 Net work done = W 2 – W 1 = 25 –6.25 = 18.75 J = 18.75 N-m =

RESONANCE

SOLN_Work, Power & Energy - 77

4.

M 4 = = 2 kg/m L 2 The mass of 0.6 m of chain = 0.6 × 2 = 1.2 kg Mass per unit length =

0.6  0 = 0.3 m 2 Hence, work done in pulling the chain on the table W = mgh = 1.2 × 10 × 0.3 = 1.2 × 10 × 0.3 = 3. 6 J The centre of mass of hanging part =

7.

m T Instantaneous power F = ma =

0   a   T   = F = ma

=

= 8.

m m  . at = . .t T T T m2 T2

.t

Maximum height attained by the particle

H

u2 52 5   m 2g 2  10 4

W g = -MgH = -0.1 × 10 × (5/4) = -1.25 J 9.

Velocity of ball just after throwing v=

2gh =

2  10  2 =

40 m/s

Let a be the acceleration of ball during throwing, then

 10. 11.

40 v2 = = 100 m/s2 2  0 .2 2s

v2 = u2 + 2as = 02 + 2as



F - mg = ma (2) is correct

F = m(g + a) = 0.2(10 + 100) = 22 N

1 mv 2  k  2



a=

1 1  v2  11  K m( v cos 60)2  m    mv 2   2 2  4  42  4

Assuming mass of athlete is between 40 kg to 100 kg here we will consider mass of athlete m = 50 kg 100 = 10 m/sec 10 So, K = 1/2 mv2 1/2 × (50 ×102) = 2500 J So Answer is (C)

V = S/t =

12.

K.E. = ct

1 mv2 = ct 2 P2 = ct 2m

P=

2ctm

RESONANCE

SOLN_Work, Power & Energy - 78

TOPIC : CIRCULAR MOTION EXERCISE-1 PART - I SECTION (A) A 1.

 Given v = 2i – 2j

(a) when moves in clockwise Ans. : First quadrant (b) When moves in counter clockwise

`

Ans. : Third quadrant A 3.

Given 0 = 0 ,  = const

1 2 t 2 for first two seconds  = 0t +

1 ×(2)2 = 2 2 for next two seconds 1 = 0 +

1 1  (4)2 –  (2)2 = 6 2 2 2 / 1 = 3 : 1 Ans. 2 = 4 – 2 =

A 5.

Given  = R = 1 cm ,    V  V2 – V1 V =

t = 15 Second

2V

V = R V=

2  1 = cm/sec. 60 30

a=

V  2 = cm/sec2. Ans. t 30  15

V =

2 cm/sec. 30

SECTION (B) B 1.

R = 0.25 m ,  = 2 rev./sec. = 4 rad/sec. ac = 2R

(at = 0)

= (4)2 × 0.25 = 42 m/s2. Ans.

RESONANCE

SOLN_Circular Motion - 79

B 3.

R = 1.0 cm , V = 2.0 t at t = 1 sec  V = 2.0 cm/sec. ac =

v2 = 4 cm/sec2. R

at =

dv = 2.0 cm/sec2. dt

a=

a c2  a 2t =

2 4 2  2 2 = 2 5 cm/sec . Ans.

SECTION (C) C 1.

m = 200 g = 0.2 kg , g = 2 m/s2 Time period = 2

6  cos  1. 2 = 2 = 2 2 5 g 

mg 13 0.2   2 N Tension = cos  = = 6 12 / 13 C 3.

N=

mv 2 r

given r = 5 m

Ans.

Ans.

, v = 5 5 m/s

for no slipping f  mg µmin N = mg µmin = µmin =

C 5.

 = 2n =

rg mg = 2 N v

5  10 (5 5 ) 2

=

2 Ans. 5

2  1500 rad/sec 60

d = 60 cm = 0.6 m 2 m = 1 g = 10–3 kg r=

2

 2  1500   × 0.6 F = m r = 10 ×  60   2

–3

15  2 = 14.8 Ans. 10 This force is exerted by blade of fan and equal force is exerted by particle on blade in same magnitude but opposite in direction. =

SECTION (D) D 1.

R=

v  2 a

=

u2 sin2  g

RESONANCE

Ans.

SOLN_Circular Motion - 80

SECTION (E) E 1.

Tension is maximum in circular motion in vertical plane at lowest position. At lowest position Tmax – mg = m2R  30 – 0.5 ×10 = 0.5 2 × 2 2 =

E 3.

25 0. 5  2



 = 5 rad/sec. Ans.

When string become slack apply equation for centripetal force.

mv 2 = mg cos 60º a apply energy conservation

v=



ga 2

1 1 mu2 = mv2 + mga(1 + cos) 2 2 from equation (i) & (ii) u=

....(i)

....(ii)

7ga 2

apply equation for centripetal force at lowest position. T – mg =

mu2 a

put the value of u and we get T = 9mg/2 E 5.

Using energy conservation :

1 mv B2 = mgh 2 Also to complite vertical circle vB = 

R=

vB =

2mgh m

vB =

.....(1)

2hg

.....(2)

5gR

2 h = 2 cm 5

Section (F) F 1.

For safe driving vmax =

rg

10 =

rg

for wet road v´ = F 4.

10  rg = = 5 2 m/s Ans. 2 2

v = 48 km/hr = 40/3 m/s. For safe turn without friction v2 h tan  = = rg x

F 7.

T = 2

 cos  geff . = 2

given x   = 1m

v2 2 ( 40 / 3)2 h= = = m rg 45 400  10

Ans.

h geff .

geff. = g + a ; T = 2 put  geff = 20  g + a = 20  a = 10 m/s2. Ans. Retardation = 10 m/s2 2 Ans. 10 m/s

RESONANCE

SOLN_Circular Motion - 81

PART - II SECTION (A) A 1.

Speed v1 =

2ðr t

v 1 2 1 = r  t 1

2ðr t

v2 =

v 2 2 2 = 2r  t 2



...(ii)

From eq. (i) and (ii)

A 3.

r=

...(i)

1 t 2  2 t1

t2 1= t 1

20 m, at = constant 

n = 2nd revolution v = 80 m/s 80 v = = 4 rad/sec 20 /  r  = 2 × 2 = 4 from 3rd equation 2 = 02 + 2  (4)2 = 02 + 2 ×  × (4)

0 = 0, f =

at = r = 2 ×

20 = 40 m/s2 

 = 2 rad/s2

Ans.

A 5.*

In curved path, may be circular or parabolic. In circular path speed and magnitude of acceleration are constant. In parabolic path acceleration is constant.

A 7.

second =

2 2 = rad/sec. 60 T

2 × 0.06 m/s = 2 mm/s 60    mm/s v  v f  v i = 2 v = 2 2 

v = .r =

Ans. Ans.

SECTION (B) B 1.

Angular velocity of every particle of disc is same aP = 2rp , aQ = 2rQ  rP > rQ  aP > aQ Ans.

B 3.

ac =

v2 , radius is constant in case (a) and increase in case (b). So that magnitude of acceleration is r

constant in case (a) and decrease in case (b).

SECTION (C) C 1.

r = 144 m, m = 16 kg, Tmax = 16 N T=

v=

mv 2 r

Tr = M

16  144 = 12 m/s 16

RESONANCE

Ans.

SOLN_Circular Motion - 82

C 3.

Uniformly rotating turn table means angular velocity is constant. New radius is half of the original value. r´ = r/2 and  = constant v´ = r= r/2 = v/2 = 5 cm/s Ans. a´ = 2 r = r/2 = a/2 = 5 cm/s2 Ans.

C 5.

T1 – T2 =

M 2 L  2 2

T 1 > T2

Ans.

SECTION (D) D 1.

At t = 0

a = g cos ,

R=

v2 u2 = a g cos 

SECTION (E) E 1.

Let the car looses the contact at angle  with vertical

mv 2 mv 2  N = mg cos  – R R During descending on overbridge  is incerese. So cos  is decrease therefore normal reaction is decrease. mg cos  – N =

E 3.

mv 2 ....(1) r from energy conservation.

T – mg cos  =

(from centripetal force)

1 1 mu2 = mv2 + mgr (1 – cos ) (here u is speed at lowest point) 2 2 from (1) and (2) T= E 5.*

mu2 + 3mg cos  – 2mg r

for  = 30º & 60º 

T 1 > T2

For normal reaction at points A and B.

mv 2 mv 2  N = mg – r r NA > NB and normal reaction at C is NC = mg, so NC > NA > NB

mg – N =  E-7._

either | T | = 0 a= So

V2 0 r

or

mg 2 +

|a|=0

or

 = 90º

 for whole motion there is velocity.

T = 0,

T + mg =

mV 2 

T=0 T=

for

mV 2 – mg 

1 1 mV2 = mu2 2 2

V2 = u2 – 4 gl E 9_

Ans.

  T . a = | T | | a | cos = 0

T=

mu 2 – 5 mg 

T=0

or

T F2 Ans.

N2 = mg –

m( v  R)2 R

g R

mg = m2 R ,  =

EXERCISE-2 PART - I 1.

Change in velocity when particle complete the half revolution : v = vf – vi = 2v Time taken to complete the half revolution t =

average acceleration =

3.

ac = a cos 30º = 25 v2 R

v 2v 10 2v 2 2  52 = = = = m/s2 t R / v  R 5

3 m/s2 2

Ans.

Ans.



v2 = aCR = 25

 3  v = 125 4  

m/s

Ans.

at = a sin 30º =

25 m/s2 2

ac =

R v

3 × 2.5 2

1/ 2

5.

Ans.

mv 2 R

(i)

The normal reaction by wall on the block is N =

(ii)

The friction force on the block by the wall is f = µN =

(iii)

The tangential acceleration of the block =

(iv)

dv µv 2 =– dt R

µmv 2 R

f µv 2 = m R v

integrating we get

RESONANCE

or

dv µv 2 v =– ds R



v n v = – µ 2 0



v0

or

dv v =–

2R

 0

µ ds R

v = v0 e–2µ

SOLN_Circular Motion - 84

7.

Centripetal acceleration m2 r = T1 cos  + T2 cos  .... (1) apply Newton law in vertical direction T1 sin  = mg + T2 sin  .....(2) given m = 4 kg, T1 = 20 kgf = 200 N, r = 3m 3 4 , sin  = 5 5

cos  =

Put in equation (2) T2 = 150 N Put in equation (1) we get 210 35 = 43 2

2 =

n= 9.

 1 = 2 2

Ans.

35 rad/s 2

=

35 rev/sec. 2

n=

30 

35 rev/min. Ans. 2

Time take by ring to fall on ground. T=

2h g

from centripetal force m2x = ma = mv

2x = v

dv dx

x =  . T = 

dv dx 



0

v

2 x d x  vdv

2



0

2h g

vy = L

L2 v 2  2 2

y = T = 

distance of one ring from center is =

vx = L

2h g

y 2  ( x  )2

distance between the point on the ground where the rings will fall after leaving the rods. = 2 y 2  ( x  )2 12.

(i)  

where

x = y = 

2h g

CP = CO = Radius of circle (R)  COP =  CPO = 60º  OCP is also 60º

Therefore, OCP is an equilateral triangle. Hence, OP = R Natural length of spring is 3R/4.  Extension in the spring x=R–



3R R = 4 4

 mg   R  mg    = Spring force, F = kx =  R 4     4

The free body diagram of the ring will be a shown.

mg 4 and N = Normal reaction Here,

C

F = kx =

RESONANCE

O

P F mg

SOLN_Circular Motion - 85

(ii) Tangential acceleration ar = The ring will move towards the x-axis just after the release. So, net force along x-axis :  mg   Fx = F sin 60º + mg sin 60º =   4 

Fx =

3 + mg 2

 3    2   

5 3 mg 8

Therefore, tangential acceleration of the ring. aT = ax = aT =

Fx 5 3 = g m 8

5 3 g 8

Normal Reaction N : Net force along y-axis on the ring just after the release will be zero. Fy = 0  N + F cos 60º = mg cos 60º N = mg cos 60º – F cos 60º =



N= 14.

mg mg mg  1  mg   = – – 8 2 4 2 2

3mg 8

(a) at equator T + m2 R = mg. %

 4  2  6400  1000  2R T   100 = 0.65 % Ans. = =  2  g T  ( 24  60  60 )  9 .8 

mg ....(1) 2 T + m2R = mg ....(2) from (1) and (2) (b) T =

2R = g/2

T= 16.



=

g 2R

2R 2 = 2 g = 2hr 

Ans.

Block B rotate in vertical plane. Tension is maximum in string at lowest position. When block B at lowest position and block A does not slide that means block A not slide at any position of B. At lowest position T – mg =

mv 2  

T = mg +

mv 2 

....(1)

From energy conservation mg(1 – cos ) =

1 mv2 2

...(2)

from equation (1) and (2) T = mg + 2mg (1 – cos ) = 3mg – 2mg cos  for no slipping. T = mg = 3mg – 2mg cos  min = 3 – 2 cos  Ans.

RESONANCE

SOLN_Circular Motion - 86

18.

Constant speed = 18 km/hr = 5m/sec. m = 100 kg, r = 100 m (a)

at B mg – NB =

100  5 2 mv 2 = = 25 100 r

NB = 975 N

mv 2  ND = 1025 N r at B & D friction force act is zero. at D ND – mg =

(b)

Ans.

Ans.

1 at C  f = mg sin 45 = 100 × 10 (c)

2

( v = constant) = 707 N

Ans.

for BC part mg cos 45 – NBC =

mv 2 R



NBC = 682 N

mv 2 R



NCD = 732 N

for CD part NCD – mg cos 45 = (d)

f N position where its maximum and N is minimum which is in part BC at C position. f N







mg sin 45 º mv mg cos 45 º  r



2

707 = 1.037 682

Ans.

PART - II 1.

QP = 2 – 5 = – 3 rad/s RP = 3 – 5 = – 2 rad/s Time when Q particle reaches at P = t1 =

3.

/2 1 = sec. 3 6

t2 =

5 / 2 5 = sec. 3 6

t3 =

9 / 2 3 = sec. 3 2

Time where R particle reaches at P.

t1 =

 1 = sec. 2 2

t2 =

3 3 = sec. 2 2

Common time to reaches at P is

3 sec. Ans. 2

at loose contact N = 0 mv 2 R from energy conservation

mg cos  =

mgR(1 – cos ) =

....(1)

1 mv2 2

....(2)

from (1) & (2) cos  =

2 3



tangential acceleration

RESONANCE

sin  =

5 3

= g sin  =

5g 3

Ans.

SOLN_Circular Motion - 87

6.

For M to be stationary T = Mg Also for mass m, T cos  = mg

.... (1) .... (2)

mv 2 .... (3)  sin 

T sin  =



 Tcos

m

dividing (3) by (2)

v2 tan  = v= g  sin 

T

g . sin  cos 

M

2 R = v

g . sin  cos 

From (1) and (2) cos  =

9.

12.

m M

then time period = 2 

F = kx, T1 = ka = m2 2a  = Time period =

mg

Mg

2  sin 

Time period =

Tsin

m gM

k 2m

2 2m = 2 =T  k

T2 = 2ka = m23a



Time period = 2

3m = T´ 2k

=

2k 3m

 3   T´ =  2  T  

Ans.

T – mg cos  =

mv 2 

(i) at  angle at = g sin  from centripetal acceleration

...(1)

From energy conservation : 0 + mg cos  = from (1) & (2) a= (ii)

1 mv2  v = 2g cos  2 T = 3mg cos  aC = 2g cos 

a 2t  a c2 = g

....(2)

1  3 cos 2 

Vertical component of sphere velocity is maximum when acceleration in vertical is zero that means net force in vertical direction is zero. Net force in vertical at  angle mg cos  and tension also from equation T = 3mg cos  ....(4) from (3) & (4)

T cos  = mg

3 mg cos  =

T=

mg cos 

...(3)

1 

cos  =

3

T = mg (iii)

Ans. 3 Total acceleration is directed along horizontal that means avertical = 0

1 cos  =

RESONANCE

3

Ans. SOLN_Circular Motion - 88

14.

For vertical circular motion, in lower half circle tension never be zero anywhere. Tension is maximum at lowest point of oscillation. Tension decrease both side in same amount. Therefore correct option is (D).

16.

Maximum retardation a = g For apply brakes sharply minimum distance require to stop. 2

0 = v – 2gs



v2 s= 2g

For taking turn minimum radius is g =

v2 , r



r=

v2 , g

here r is twice of s

so apply brakes sharply is safe for driver. 19.

 = 2 

d 2d  = = 2 × 0.4 = 0.8 rad/s dt dt

vAC = r = 0.8 ×

1 = 0.4 m/s 2

1 = 0.32 m/s2 2 a = aC = 0.32 m/s2 (at = 0) aC = 2r = (0.8)2

21.

Energy conservation from initial and final position

 1   = 1 mv2 + 1 mv2 mgr + mg r 1  2 2 2 

v=

2gr 

1

gr

Ans.

2

Normal reaction at bottom position A 2

N – mg = 23.

mv 2 r

N=

mv 2 r

 gr  m  2gr   2  + mg =  r

mg + mg = 3 mg –

2

= 2.29 mg

The acceleration vector shall change the component of velocity u|| along the acceleration vector. r=

v2 an

Radius of curvature rmin means v is minimum and an is maximum. This is at point P when component of velocity parallel to acceleration vector becomes zero, that is u|| = 0. u|| = 0 

R=

u 2 42 = = 8 meter.. a 2 T cos 60o

2

mv T 3 = (  3 / 2) 2

25.

T = mg 2 Hence T = 2 mg , So (B) holds From (1) & (2) V2 = 3 g/2 3  9.8  1.6 2



V=



V = 2.8

........(1) 60o 

.......(2)

60o

/ 2

T  3/2

V o

T sin 60

mg

3 m/s . So (C) hold

RESONANCE

SOLN_Circular Motion - 89

ac = V2/r =  t=

27.

( 3 g  / 2) (  3 / 2)

3 × g = 9.8 3 m/s2

=

(D) holds 2  3 /2 2 r = (3 g  / 2) v

Speed of cage =

t = 4/7

 (A) holds.

gr = const.

Normal reaction at (weight reading) NA – mg =

mv 2 r

NA = 2mg = 2w Ans. Weight reading at G & C = mg = w Ans. weight reading at E mg – NE =

mv 2 r

NE = 0 29.

Ans.

Tangential acceleration = at = gsin Normal acceleration = an = g cos at = an g sin = g cos   = 45°  vy = vx uy – gt = ux 20 – (10)t = 10 t = 1 sec. During downward motion at = an vy = – vx 20 – 10 t = – 10  t = 3 sec.

EXERCISE-3 1.

From graph (a)  = k angular acceleration = 

where k is positive constant

d = k × k = k2 d

 angular acceleration is non uniform and directly proportional to .  (A) q, s From graph (b)  2

d =k d

or

From graph (c)  angular acceleration = From graph (d)  angular acceleration =

2 = k . 

Differentiating both sides with respect to .

d k = d 2

Hence angular acceleration is uniform.  (B) p

 = kt d =k dt  = kt2

d = 2kt dt

Hence angular acceleration is uniform  (C) p

Hence angular acceleration is non uniform and directly proportional to t.

 (D) q,r

RESONANCE

SOLN_Circular Motion - 90

2.

v = 2t2 Tangential acceleration at = 4t Centripetal acceleration ac =

Angular speed  =

v2 R



4t 4 R

v 4t = , R R

tan  =

4tR R at  3 = 4t 4 t ac

Sol. 3 to 5. The angular velocity and linear velocity are mutually perpendicular    v   = 3x + 24 = 0 or x=–8 5 v 1 = = meter  10 2 The acceleration of particle undergoing uniform circular motion is    a    v = ( 8 ˆi  6 ˆj )  (3 ˆi  4 ˆj ) =  50 kˆ

The radius of circle r =

6.

mu02 r Now, along vertical mg =

r=

1 2 gt 2



u0 =



t=

gr

2r g

Along horizontal ; OP = 2u0t = 2 2 r 7.

As at B it leaves the hemisphere,  N=0

A

mV 2 mg cos = r

N

u0/3 B

r

c mg



h mV 2 mg = r r mv2 = mgh .............(1) By energy conservation between A and B

 os



v

h

O

2

1  u0  1 mgr + m   = mgh + mv2 2  3  2 Put u0 and mv2 8.

9.



h=

19r 27

v2 = g cos r  at = g sin   anet = g Alternate Solution : when block leave only the force left is mg.  anet = g. As ac =

   geff  g  a

–a

 Tension would be minimum when it (tension) is along geff mg

tan  = 3

4

mg

=

4 3

RESONANCE

g 

 = 53º .

geff

SOLN_Circular Motion - 91

10.

Vmin =

geff =



5 g 4

=

= 6 mgeff

5g . 2

(geff =

5 15 g) = mg 4 2

11.

Tmax

12.

For conical pendulum of length , mass m moving along horizontal circle as shown T cos = mg .... (1) T sin = m2 sin .... (2) From equation 1 and equation 2,

 cos =

g

2  cos is the vertical distance of sphere below O point of suspension. Hence if  of both pendulums are same, they shall move in same horizontal plane. Hence statement-2 is correct explanation of statement-1. 13.

The normal reaction is not least at topmost point, hence statement 1 is false.

14.

Let the minimum and maximum tensions be Tmin and Tmax and the minimum and maximum speed be u and v.  Tmax = Tmin =

mu2 + mg R mv 2 – mg R

 u2 v 2     T = m  R  R  + 2 mg.   From conservation of energy

u2 v 2  = 4g  is indepenent of u. R R and T = 6 mg.  Statement-2 is correct explanation of statement-1. 15.

Statement-2 is wrong. R =

v2 , where a is acceleration component perpendicular to velocity.. a

and as particle goes up, v2 decreases and a  increases so radius of curvature R decreases hence statement -1 is true 16.

(i) False. (ii) True. (iii) True.

It has tangential as well as radial acceleration. The angle is less than 180°. The angle between velocity and radial acceleration is 90°. It has no acceleration in verticall direction

(iv) False.

=

(v) False.

aT =

RESONANCE

initial  final is valid only for constant angular acceleration. 2  dv  a c2     dt 

2

> ac

SOLN_Circular Motion - 92

17.

(i)

Given that tangential acceleration = at = 3 m/s2 Centripetal acceleration = ac = a=

a c2  a 2t =

2 4 2  3 2 = 5 m/s

`

Now

(ii)

= average acceleration =

v 2 R 2R2 = = t R / R 

= 2R

Instantaneous acceleration

(iii)

20 2 v2 = = 4 m/s2 100 r

a 2 = a  Tension before cutting T sin  = mg

Ans.

mg sin  Tension after cutting. T2 = mg sin 

T1 =

T2 2 T1 = sin  (iv)

Ans.

h

tan  = (v2/rg) =

]

2

b  h2 1/ 2

Ans : (v)

 ghr   2 2  b h

   

Acceleration at lowest position v2 R From energy conservation

aL =

mgR (1 – cos ) =

mv 2 2

v2 = 2g(1 – cos) R aL = 2g (1 – cos) acceleration at highest position. aH = g sin  according to problem aL = aH 2g(1 – cos ) = g sin  2 (1 – cos ) = sin   2(1 – 1 + 2 sin2 /2) = 2 sin /2 cos /2

tan

1  = 2 2

tan =

2 tan 2 1  tan 2

RESONANCE

2  2



21 

1

1 4

=

4 3



 = 53º Ans.

SOLN_Circular Motion - 93

EXERCISE-4 PART - I 1.

2.

 Net acceleration a of the bob in position B has two components.  (i) an = radial acceleration (towards BA)  (ii) ar = tangential acceleration (perpendicular to BA)  Therefore, direction of a is correctly shown in option (C).

A //////////////////////////

an a

d  h =  R   (1 – cos) 2   velocity of ball at angle  is

B

(a)

at

d  v2 = 2gh = 2  R   (1 – cos)g 2  

.......(1)

Let N be the total normal reaction (away from centre) at angle . Then mg cos – N =

mv 2 d  R   2  

h 

 v mg

Substituting value of v2 from equation (1) we get mg cos – N = 2mg (1 – cos)  N = mg (3 cos – 2) Ans. (b) The ball will lose contact with the inner sphere when N=0

or

3cos – 2 = 0

or

2  = cos–1   3

After this it makes contact with outer sphere and normal reaction starts acting towards the centre. Thus for 2   cos–1   : 3 NB = 0

and

NA = mg (3 cos – 2)

and

for

2   cos–1   3

NA = 0 and NB = mg (2 – 3cos) The corresponding graphs are as follows NB

NA

5mg

mg 2mg

-1

3.

2/3

+1

cos -1

2/3

+1

cos

By energy conservation, 1 1 mu2 = mv2 + mg(1 – cos) 2 2

V2 = U2 – 2g (L – L cos)

5gL = 5gL – 2gL (1 – cos) 4 5 = 20 – 8 + 8 cos cos = –

7 8

3 m2 vb = –(20 + 10) = –30 m/sec.

2d t = v (time for succeesive collision) 0 N × t = dP = mv0 – (–mv0)

2d N × v = 2mv0 0 N=

mv 02 d

RESONANCE

SOLN_Centre of Mass - 102

F-14.

If mass = m first ball will stop  v = 0 so K.E. = 0 (min) (K.E. can't be negative )

SECTION (G) G 1.

F=

dm dt

210 = 300 ×

dm dt



dm = 0.7 kg/s. dt

EXERCISE-2 PART - I 1.

2H g =

2  80 = 16 = 4 s. 10

2h g

=

2 (80  60 ) = 10



t=

t = T – t = 4 – 2 = 2 s



V=

2 d = = 1 m/s t 2

Mv = mv



v=

60  1 = 30 m/s Ans. 2

T=

40 = 2 s. 10

R = vt = 30 × 2 = 60 m Ans.

4. By momentum conservation Mu = mV (i) V=

M u m

By energy conservation mgh =

1 1 Mu2 + mV2 2 2

=

1 1 M Mu2 + m   2 2 m

mgh =

1 1 M2 2 Mu2 + u 2 2 m

=

2  1 2  Mm  M  u   m 2  

2

u2

2m 2gh = u2 (Mm + M2) 2 m2gh



Mm  M2

u=m

= u 2.

2 gh

...(ii)

Mm  M2

By momentum conservation mV = (M + m) V1 V1 =

mV Mm

...(iii)

By energy conservation 1 1 mV2 = (m + M) V12 + mgh1 2 2 1 1  mV   mV2 = (m + M)  2 2 Mm

RESONANCE

2

+ mgh1



1 1 m2 V 2 mV2 – = mgh1 2 2 (M  m)

SOLN_Centre of Mass - 103

2   1 2 m – m  V  M  m  = mgh1 2  

 2gh

M

Put V = –  m  × u and u = m  

Mm  M2

put value of V from eqn (iv) to (iii)

1 MmV 2 = mgh1 2 (M  m)



V=M



h' =

2gh Mm  M2

...(iii)

....(iv)

M2h (M  m)2

6.

By mechancnical energy conservation 1 1 ( 2m)Vb2  mVr2 = 2 mgl 2 2

2 Vb2 + Vr2 = 4gl using momentum conservation mVr = 2 mVb Vr = 2 Vb 2Vb2 + 4Vb2 = 4 gl

....(1) ....(2)

6Vb2 = 4gl

Vb =

2 gl 3

2 gl 3

(a) Vr = 2 Vb = 2

when string be comes vertical velocity of block wrt to string. Vbr = Vb – (– Vr) = 3Vb = 3 (b) T – 2 mg = 8.

2m( Vbr )2 

2 gl 3

T = 2 mg +

9  2gl  ( 2)m = 14 mg 3

m = 20 × 10–3 kg ; M = 5 kg u = 400 d = 0.2 m V = 200 =? PBullet = PBlock m (u – v) = 20×10–3 (400 – 200) = 4 kg. m/s. 2

KEBlock =

4 P2 = = 1.6 J =  Mgd 2M 25

1 .6

1 .6

 = Mgd = = 0.16 Ans. 5  10  0 . 2 10.

string will taut when A waves a distance of (.7 – .25) m at that Pt VA =

2gh and Now B starts on using with same velocity as A.

let us suppose it is u.

 T dt

= mB u + 0

....(1)

–  T dt = mA (u –VA)

....(2) m A VA

6

from (1) and (2) u = m  m = 5 A B

RESONANCE

6



 Tdt  3  5

= 3.6 m/s

SOLN_Centre of Mass - 104

12.

R1 = V cos  T 1 R2 = V cos  T 2 R3 = V cos  T 3 R = R1 + R2 + R3 = u cos  [T 1 +T 2 + T 3]

 2 u sin  2 eu sin  2 e 2u sin      = V cos   g g  g  (1  e  e 2 ) V 2 sin 2 R= g 14.

( V )2 2a (a) e = 1 so after collision VA = 0 and VB = 5 m/sec V2 = U2 + 2as

0 = (V)2 – 2as

s=

(5 ) 2  s = 6.25 m 2  0.2  10 (b) when e = 0 applying momentum conservation m × 5 + 0 = (m + m) × V V = 2.5m/sec So mg = ma s =

so V2 = u2 + 2as 16.

s=



Px = 5×2 = 10

; P = Px ˆi + Py ˆj

Py = 10 3

= 10 ˆi + 10 3 ˆj = (5 + 10) V

(2.5)2  2 2  0.2  10



a=

g 2



s = 3.12 m.

10  ˆ   i  3ˆj    15

=

2 ˆ i  3ˆj  V = 3  

V =

4 m/s. 3

Ans.

1 1 1 2 2 H = E = Ei – Ef =  2 m1v1  2 m2u2  – (m1+m2) V2 2  

= 18.

1 1 × 5 × 22 + ×10 × 2 2

(a) V= In x dir

2

 3

–

(b)

20.

= 3 mV

v=

V 2 – 2gl(1 – Cos )

=

2

V =

Ans.

2gl

v 2m ×

1 40 35 × (10+5) (4/3)2 = 25 – = 2 3 3

3 2

V

=

2

 2gl

v = 3 gl

Ans.

2gl – 2gl  2glCos

For  = 60°

V =

Vx =

(at heighest point)

gl Cos 60°

3

gl

Vx =

gl 2

Applying momentum conservation in horizontal direction mV0 = Mu M = 2m u=

mV0 V0  M 2

Eqn of e along normal V cos   u sin  e= V0 sin  V

1

e = V cot  + 2 0

RESONANCE

V V cos   0 sin  2 = V0 sin 

...(i) SOLN_Centre of Mass - 105

Along incline surface of wedge friction is negligible so change in momentum mV0 cos  = mV sin  V V0 = cot 

...(ii)

Put value of (ii) in (i) e = cot2 +

1 2

given tan  = 2 =

1 1 3   4 2 4

Ans.

(b) h = (ut) tan  By (2)nd eq. of motion or

– u tan = V – 2

– h = Vt – 1 gt 2

– (ut) tan = Vt –

2

V0



2V0 tan  2e V0 tan  t= (e) = g g

1

t = g V0 tan (cot2 + ) 2 2



1 gt2 2

1 gt = V + u tan 2



t = g (V0 cot + tan ) 2

22.

1 gt2 2

substituting values :

3  10  2 4 = 3sec 10

(a) At highest point V = 50 cos  ...(i) After striking bullet get embedded with bob so by momentum conservation. MV = 4Mu from (i)

u=

u=

V 4

....(ii)

50 cos  4

By energy conservation after collission 1 (4m) u2 = 4mg (1 + cos 60º) 2

50  50 cos2  = 100 16

cos 2 =



(b) Max height R

y=

1  50 cos     2  4 

 16 25

1  u2 sin 2  502  2 sin 37 º cos 37 º  g 2  10 

(c) x = 2  2  

=

=

cos =



( 50 sin 37 º )2 2g

2

10  10  1 1   3 2   4 5



 = 37º

1 3 3  50  50   = 5 × 9 = 45 20 5 5

= 120 m Ans

(a)  = 37º (b) x = 120 m and y = 45 m

PART - II 1.

COM can lie anywhere, within or at the radius r.

3.

Since no external force is acting on the system hence VCM remain constant.

5.

when cylinder reaches pt B. then block get shifted by x  but since than there is no ext force therefore com remain at its position [(R–r) – x]m = Mx x=

8.

m (R – r ) Mm

Pi = 0 Pf = MV – mV1

...(i) ....(ii)

MV – mV1 = 0



using

u=

M V.. m

V12 = u2 + 2ax. a = g.

RESONANCE

 MV     m 

2

= 0 + 2g x.



x=

M2 V 2 2m2g

SOLN_Centre of Mass - 106

10.

Taking the origin at the centre of the plank. m 1x 1 + m 2 x 2 + m 3x 3 = 0 (  x CM = 0) (Assuming the centres of the two men are exactly at the axis shown.) 60(0) + 40(60) + 40 (–x) = 0 , x is the displacement of the block.  x = 60 cm i.e. A & B meet at the right end of the plank.

12.

yCM = 0 yCM = 

m 3m y1 + y2 4 4 y2 = –5 cm

y1 = + 15

14.

I. Since velocity of both R and S is positive they will move in same direction. II. At mid point velocities of R and S are same. III. Change in velocity of R is small as compare to change in velocity of S. But change in momentum is same for both in magnitude. Hence mass of R should be greater than S. Hence all three are correct.

16.

If we treat the train as a half ring of mass 'M' then its COM will be at a distance the circle. Velocity of centre of mass is : VCM = RCM .

18.

2R  V  .   R

=

2R . 



VCM =



The linear momentum of the train =

=

(  =

2R from the centre of 

V ) R

2MV 2V  MVCM =   As the linear momentum of any system = MVCM

I = f × t

and

F=

2MV 

Ans.

m( 2gh 2  2gh1 ) t

100  10 –3 ( 2  9.8  0.625  2  9.8  2.5 ) 0.01 F = 105 N

F=

20.

Using momentum conservation     p1  p 2  p 3  p 4  0     p1  – p 2 – p 3 – p 4



 p1  p 22  p 32  p 24

p 2  p 32  p 24 p12  2 = E0 + E0 + E0 2m 2m Total energy = 3E0 + E0 + E0 + E0 = 6E0

K. E1 =

v sin  22.

e=

2gh cos 

apply conservation of momentum m 2gh sin = m vcos e

2gh cos × m = mv cos

tan  = cot. e  e = tan2

......(i) ......(ii)

on solving

RESONANCE

SOLN_Centre of Mass - 107

25.

m2vcos = 3vy

vy v cos 

vy

Also e =

27.

v cos 

=

2 . 3

R/2 ;  = 30º R Both have equal mass it means along LOI particle transfer it velocity to disc which is vcos. sin =

so 29.

2 3

=

VD = Vcos  = Vcos 30º =

3V 2

v r  v mc

vr = v m – v c = v – u = 0.  since vr = 0 so Ft = Fnet = m 30.

dv dt

vrdm = 0. dt

F + 0 = (m0 – t)

dv dt

 F = (m0–t)

dv . dt

Neglecting gravity,

m 

0 v = un  m  ;  t 

u = ejection velocity w.r.t. balloon.

m0 = initial mass

mt = mass at any time t.

 m0   = 2n2. = 2n    m0 / 2  34.

mv = nvm

 v =

v n

time for first collisen is t1 = 2nd collisions

t2 =

L (2nd block) V

2 = 2t1 V

(3rd block)

so t = t1 + 2t1 + 3t1 + at1 ...........(n–1) t1. t = t1 [1 + 2 + 3] .......................(n–1)] =

36.

(n – 1) (n – 1  1) n(n – 1) = 2 2

f m 2 v1 = 0 + 2ad a=

vb12 =

so t =

L n (n – 1). 2V

for elastic collission e = 1

2F .d m

vb1 =

2Fd m

after collisin vb2 = 0.

RESONANCE

SOLN_Centre of Mass - 108

39. Pi = mv (i) Pf = (m + m) v at maximum conservation Pi = Pf v' = v/2 By energy compression

1 1 1 mv2 + 0 = (2m) (v)2 + kx2 2 2 2 at maximum compression k =



kx2 =

mv 2 2

x=



m v. 2k

1 v2 × 2m ×  k = mv'2 = mv2/4. 2 4

EXERCISE-3 1.

(A) If velocity of block A is zero, from conservation of momentum, speed of block B is 2u. Then K.E. of 1 2

block B = m(2u)2 = 2mu2 is greater than net mechanical energy of system. Since this is not possible, velocity of A can never be zero. (B) Since initial velocity of B is zero, it shall be zero for many other instants of time. (C) Since momentum of system is non-zero, K.E. of system cannot be zero. Also KE of system is minimum at maximum extension of spring. (D) The potential energy of spring shall be zero whenever it comes to natural length. Also P.E. of spring is maximum at maximum extension of spring. 2.

(A) (B)

(C) (D) 3.

Initial velocity of centre of mass of given system is zero and net external force is in vertical direction. Since there is shift of mass downward, the centre of mass has only downward shift. Obviously there is shift of centre of mass of given system downwards. Also the pulley exerts a force on string which has a horizontal component towards right. Hence centre of mass of system has a rightward shift. Both block and monkey moves up, hence centre of mass of given system shifts vertically upwards. Net external force on given system is zero. Hence centre of mass of given system remains at rest.

(a) The acceleration of the centre of mass is

F 2m The displacement of the centre of mass at time t will be a COM =

x =

1 Ft 2 a COM t 2 = 2 4m

Ans.

4&5 Suppose the displacement of the first block is x 1 and that of the second is x 2. Then, x =

mx1  mx 2 2m

or,

Ft 2 x1  x 2  4m 2

or,

x1 + x2 =

Ft 2 2m

...(i)

Further, the extension of the spring is x 1 – x 2. Therefore, x1 – x2 = x0 From Eqs. (i) and (ii),

RESONANCE

1 x1 = 2

 Ft 2    x0   2m   

and

...(ii)

1 x2 = 2

 Ft 2    x0   2m   

SOLN_Centre of Mass - 109

6.

During collision, forces act along line of impact. As collision is elastic and both the balls have same mass, velocities are exchanged along the line of impact. Therefore ball B moves with velocity VB||, that is equal to u cos 30°. Ball A moves perpendicular to the line of impact with velocity VA  = u cos60°. Along the line of impact, ball A does not have any velocity after the collision. Therefore velocity of ball A in vector form after the collision y

VA 30°

VA||

u

x

30°

R 60°

VB||

= VA cos60°i + VA cos 30°j = (u cos 60°) cos60°i + (u cos 60°) cos 30°j

1 1 1 3 .j = 4. . . i + 4. . 2 2 2 2 7.

= (i  3 j) m/s

x

Using impulse-momentum equation for ball B     N dt  p f  p i and as pi  0

 N dt VB||



 N dt  p f

B

= (mu cos 30°) cos 30 i – (mu cos30°) cos 60° j = m. 4 . 8.

3 3 3 1 .i – m. 4 . . . . j = (3 m i – 2 2 2 2

m 3 m j) kg s

Suppose V2 is velocity of ball B along the line of impact and V1 is velocity of ball A along the line of impact, after the collision, as shown. Then

1 (Velocity of approach) = Velocity of separation 2

1 3  . u  2  2  = V2 – V1

.... (1)

A

V1

Conserving momentum along the line of impact m.

3 u = m. V2 + mV1 2

V2

B .... (2)

Solving and using u = 4 m/s V2 =

3 3 m/s  2

9 3 3  3 3 3 3  V2  cos 30 i  cos 60 j =  4 i  4 2 2 

 j  m/s  

EXERCISE-4 PART - I 1.

vCOM =

=

m1v 1  m 2 v 2 m1  m 2

10  14  4  0 10  4

RESONANCE

= 10 m/s.

SOLN_Centre of Mass - 110

2.

Angular speed of particle about centre of the circle,

v2 v2 ,  = t = t R R

=

 v p = (– v2 sin  ˆi + v2 cos  ˆj )

  v v v p =   v 2 sin 2 t ˆi  v 2 cos 2 t ˆj 

or



and

R

R



 v m = v1 ˆj

 linear momentum of particle w.r.t. man as a function of time is    L pm = ( Vp – Vm )    v2  v2 = m  v 2 sin t ˆi   v 2 cos t  v 1 ˆj  R R  

3.

(i)

X1 = V0 t – A (1– cost) Xcm =

(ii)

a1 =

m1x1  m 2 x 2 = V0 t m1  m 2 d2 x1 dt 2



m1 X2 = 0 t + m A (1–cos t) 2

Ans.

= – 2 A cos t

The separation X2 – X1 between the two blocks will be equal to 0 when a1 = 0 or cos t = 0 x2 – x1 =

m1 A (1–cos t) + A (1– cos t) m2

Thus the relation between 0 and A is, 8.

0 =

 m1     m  1 A  2  m

(cos t = 0)



1 0 =  m  1 A  2 

According to Newton’s Law   v 2  v1 e = u  u 1 2 For elastic collision cofficient of restitution e = 1 so     Statement - 1 is correct v 2  v 1 = u1  u 2

9.

Linear momentum is conserved in both elastic & non elastic collision but it’s not the explanation of statement -1 so it is not the correct explanation of the statement A.   P2  p ˆi P1  p ˆi as there is no external force so momentum will remain conserved     P1  P' 2  P1  P2   P1  P2  0 Now from option   (A) P1  P2 = (a1  a 2 ) ˆi  (b1  b 2 )ˆj  c 1kˆ   (B) P1  P2 = (c 1  c 2 ) kˆ   (C) P1  P2 = (a1  a 2 ) ˆi  (b1  b 2 )ˆj   (D) P1  P2 = (a1  a 2 ) ˆi  2 b1 ˆj and it is given that a1 b1 c 1 , a2, b2, c 2,  0

  in case of A and D it is not possible to get P1  P2 = 0 Hence Ans. (A) and (D)

RESONANCE

SOLN_Centre of Mass - 111

10.

11.

At point B there is perfectly inelastic collision so component of velocity  to incline plane becomes zero and component parallel to second surface is retained velocity immediately after it strikes second incline V=

2gh cos 30

V=

45 m/s

=

2  10  3 ×

3 = 2

2  10  9 4

At point ‘C’

VC2  VB2  2gh VC2 = 45 + 2 × 10 × 3 VC = 12.

105 m/s

The block coming down from incline AB makes an angle 30° with incline BC. If the block collides with incline BC elastically, the angle of block after collision with the incline shall be 30°. Hence just after collision with incline BC the velocity of block shall be horizontal. So immediately after the block strikes second inclined, its vertical component of velocity will be zero.

13.



ycm =

m1y1  m2 y 2  m3 y 3  m4 y 4  m5 y 5 m1  m2  m3  m4  m5

ycm =

6m(0)  m(a )  m(0 )  m( a)  m(– a ) m  m  m  m  6m

=

a . 10

14.

Since masses of particles are equal and collisons are elastic, so particles will exchange velocities after each collision. The first collision will be at a point P and second at point Q again and before third collision the particles will reach at A.

15.

from momentum conservation : 9m = (2m) V1 – (m)V2  9 = 2V1 – V2 ..... (1) e=

V1  V2  1 ......(2) 9

from eqn(1) and eqn(2) V1 = 6 m/sec. for second collision between second block and third block : (2m) 6 + m(0) = (2m + m) VC  VC = 4 m/sec.

16*.

Since collision is elastic, so e = 1 Velocity of approach = velocity of separation So, u=v+2 .............(i) By momentum conservation : 1 × u = 5v – 1 × 2 u = 5v – 2 v + 2 = 5v – 2 So, v = 1 m/s and u = 3 m/s

RESONANCE

SOLN_Centre of Mass - 112

Momentum of system = 1 × 3 = 3 kgm/s Momentum of 5kg after collision = 5 × 1 = 5 kgm/s

1 So, kinetic energy of centre of mass = (m1 + m2) 2

Total kinetic energy =

17.

2h g

R= u



2

2

 m1u   1 3    = 1 (1 + 5)   = 0.75 J m  m 2  6  2   1

1 × 1 × 32 = 4.5 J. 2

20 = V1

25 25 and 100 = V2 10 10

 V1 = 20 m/s , V2 = 100 m/sec. Applying momentum conservation just before and just after the collision (0.01) (V) = (0.2)(20) + (0.01)(100) V = 500 m/s

18.

 = 0.1



1 1 mu 2 = mg × 0.06 + kx2 2 2

1 × 0.18 u2 = 0.1 × 0.18 × 10 × 0.06 2 0.4 =

N 10

N = 4 Ans.

PART - II 14.

If initial momentum of particles is zero, then they loss all their energy in inelastic collision but here initial momentum is not zero. Principle of conservation of momentum holds good for all collision.

RESONANCE

SOLN_Centre of Mass - 113

TOPIC : RIGID BODY DYNAMICS EXERCISE-1 PART - I SECTION (A) A 1.

i = 0  = it +

t = 5 sec

 = 50 (2) rad.

1  t2 2

(50) (2) = 0 +

1  (5)2 2

(50) (2) = 0 +

25  2

(50 )(2)(2) = 4 (2) = 4 rev/ se2 25 f = i + t f = 0 + 4(5) = 20 rev/ sec

=

SECTION (B)

B 1.

For first solid spheres IAB = Iam + Md2 7 2 2 MR2 + MR2 =  MR  5 5  Similar way for second sphere

IAB =

IAB =

7 MR2 5

 14 2 I = 2 IAB =  MR   5 

B 4.

 mR 2    I0 =  2    0 = cm + md2  4R  mR 2  = cm + m   3  2  cm

2

 4R  mR 2  = –m   3  2

2

2  MR 2  4R    M   ICM =   2  3  

RESONANCE

SOLN_RIGID BODY DYNAMICS - 114

SECTION (C) C 2.

F1 = 2 ˆi – 5 ˆj – 6 kˆ



at point



at point

F2 = – ˆi + 2 ˆj – kˆ r0(–1,0,5) 

r = (1 ˆi + ˆj + 0 kˆ ) – (– ˆi + 0 ˆj + kˆ ) 1 

`

  = 2ˆi  ˆj – kˆ  × 2ˆi – 5 ˆj – 6kˆ 

r = 2ˆi  ˆj – kˆ 1 





 = r F 1 1 1 

 = ( – 10 kˆ + 12 ˆj – 2 kˆ – 6 ˆi – 2 ˆj – 5 ˆi ) 1 

 = (–11 1 ˆi + 10 ˆj – 12 kˆ ) 1 



2



Total 



T





C 4.

T





 

= r  F = ˆi  ˆj  kˆ × – ˆi  2ˆj – kˆ 2 2 



T





1

2

=  + 

   = – 14 ˆi  10 ˆj – 9kˆ 

 



 

= 2ˆi  ˆj – kˆ × 2ˆi – 5 ˆj – 6kˆ + ˆi  ˆj  kˆ × – ˆi  2ˆj – kˆ

(a)  0 = mg R/2 = mg

 = mg 0



 v 2 sin 2      2g  

 mv 2 sin 2  v 2 sin 2   =   2 2g  

 = (mv2 sincos) 0

(b)

 = mgR 0 = (2mv2 sincos)

RESONANCE

SOLN_RIGID BODY DYNAMICS - 115

SECTION (D) D 2.

The F.B.D. of rod is as shown For rod to be in translational equilibrium N1 = P ....(1) N2 = W = mg ....(2) For rod to be in rotational equilibrium, net torque on rod about any axis is zero. Net torque on rod about B is zero

 i.e.,

 cos  – N2  cos  + P  sin  = 0 2 from equation (2) and (3) solving we get mg

P= D 3.

.......(3)

mg cot  2

For translational equilibrium Fx = 0 Fy = 0 N1 = f N2 = 75g + 24g = 99g = 990 N Rotational equilibrium  = 0 (about any point) B = 0 N1 × 6 = 24g (5cos 37º) + 75g (8cos 37º)

4 4 ) + 75g (8 × ) 5 5 N1 × 6 = (96g + 480 g) N1 = 96g = 960 N f = N1 N2 = N1

N1 × 6 = 24g (5 ×

N1 96g 32  = N = 99g = 33 2 Ans.

990 N, 960 N ,

32 33

SECTION (E) E 1.

(a) Torque about hinge  (m1g – m2g)   = .  2

=

(m1  m 2 )g( / 2) 2

2

  m1   m 2   2   2

2(m1  m 2 )g  = (m  m )  1 2 =

2(6  3)10 10 = rad/sec2. 2(6  3) 3

RESONANCE

SOLN_RIGID BODY DYNAMICS - 116

(b)

If mass of rod is 3 Kg Torque about hinge (m1g – m2g)

' =

' =

 = '' 2

 (m1  m 2 )g   2 2    2 m 2   m1   m 2    3  12  2   2  

2(m1  m 2 )g m    m1  m 2  3  3  

=

2(6  3)10 2 3  = 3 rad/s  2 6  3   3 

For m1 block m1g – T1 = m1a m1    T1 =  m1g  2  

623 = 42 N 2 For m2 block T2 – m2g = m2a T1 = 60 –

T2 = m2g + m2 E 3.

 323 = 30 + 2 2

T2 = 39 N

Let  be the angular acceleration of the pulley system. For 6 kg block 6 g – T1 = 6 (2) .........(i) for 3 kg block T2 – 3g = 3 .........(ii) for pulley system  2T1 – T2 =  = 3 .........(iii) From equation (i) and (ii) putting the values of T1 and T2.  2[6g – 12] – [3g + 3] = 3  12 g – 24 – 3g – 3 = 3  30  – 9g 

=

90 = 3 rad/s2 Ans. 30

SECTION (F)

F 2.

initial positial

final position

Using Energy conservation Ki + Ui = Kf + Uf

1  = 2 2 2  = (1 + 2) 0 + 3mg

[ =

m 2 + m2] 3

RESONANCE

SOLN_RIGID BODY DYNAMICS - 117

1  3mg = 2 2 3g = F-4.

 m 2    m 2  2 + 0  3   

4 2  3

=

 2  2  3g=  3    2  

9g 4

Initial and final positions are shown below

5R   5mgR  = Decrease in potential energy of mass ‘m’ = mg 2  4   2  R mgR Decrease in potential energy of disc = mg 2   = 4  2 Therefore, total decrease in potential energy of system

=

1 2 2  = moment of inertia of system ( disc + mass ) about axis PQ. = moment of inertia of disc + moment of inertia of mass

Gain in kinetic energy of system W here

5mgR mgR + = 3 mgR 2 2

=

2 2  mR 2 R   5R  15mR 2   m       = + m  =  4  4   8  4  From conservation of mechanical energy Decrease in potential energy = Gain in kinetic energy

2 16 g 1  15mR  2   3 mgR =  =  8  5R 2   Therefore, linear speed of particle at its lowest point  5R  5R 16 g  = v =  or v = 5 gR 5R  4  4

SECTION (G) G 2.

– 3x 5     y  4 4 

3x + 4y = 5 P = mv = 2 × 8 = 16 (kg – m/s) L = (5/4) × mv cos 37º L = 5/4 × 2 × 8 ×

4 = 16 kg m2/s 5

RESONANCE

SOLN_RIGID BODY DYNAMICS - 118

G 3. initial position

Final position

No external torque so



L = cont.

Li = Lf (i0 = f0)  mr 2 mr 2      2 2  4 4  0 = ( + mr + mr )  2        mr    2         2mr 2  0          

G 5.

Angular momentum conservation about O  = mvR MR 2  = mvR 2 MR = 2mv

v M

o

m

R 

 MR   v =   2m  With respect to bord man's rotation v + R velocity so in one rotation when velocity v + R angle taken by man (2).  2R  t   V  R   2  Angular velocity bord is  so at the same time angle covered by disc =  . t   R  .  R  V 

2R  4m    MR M  2m   R  2m

SECTION (H) H1

VA

= (Vcm –  /2) = 50 – 5 × 5 = 25 m/s

VB

=  Vcm   2 









= 50 + 25 = 75 m/s H4

(a) vA sin  = v0 cos  vA =

v0 4v 0 = tan  3

 4v  3v 0  4 0  v 0 sin   v A cos  9v 0  16 v 0 5v 0   3  = (b)  = =  15 3 5

(c) vx = vy =

  v Ax  vBx  v   = 0  2  2 

1 v Ay  v By = 2





2v 0 3

RESONANCE

SOLN_RIGID BODY DYNAMICS - 119

SECTION (I) I-3.

For linear motion : mg – T = ma For angular motion :

............(i)

 mR 2  T.R. =  2 

   

mR 2 For no sliping : a = R From equation (i), (ii) & (iii) T=

a= I-4.

............(ii) ............(iii)

2 g. 3

Let R & r be the radii of hemispherical bowl & disc respectively From energy conservation,

1 1 mv2 + 2 2 2 For pure rolling, v = r mg(R – r) =

1 1 2  v  1 mg(R – r) = mv2 +  mr    22 2 r  3 mv2 4 From FBD of bottom : mg(R – r) =

N – mg =

2

...........(i)

mv 2 (R  r )

...........(ii)

From equ. (i) & (ii), N= I-5.

7 mg. 3

Let v1 & v2 be minimum speed of ring of bottom & top of cylindrical part At top of path mv 22 (R – r ) for minimum speed N = 0 v22 = g (R – r) .......... (i) From energy conservation between bottom & top point of cylindrical part

N + mg =

1 1 1 1 mv12 + 12 = 2 mg (R – r) + mv22 + 22 2 2 2 2 For pure rolling 1 =

v1 v2 , 2 = r r

v 12 v 22 1 1 1 1 mv12 + (mr2) 2 = 2 mg (R – r) + mv22 + (mr2) 2 2 2 2 2 r r  mv12 = 2 mg (R – r) + mv22 .......... (ii) from equation (i) & (ii)  mv12 = 2 mg (R – r) + mg (R – r) 



v1 =

3g(R – r )

RESONANCE

SOLN_RIGID BODY DYNAMICS - 120

I-6.

For linear motion, F = ma For angular motion,

..........(i)

2 2 F.R. =  mR   5  

=

5F 2mR

..........(ii)

1 2 at 2

 = 0t +

 8mR  1  5F  2   t  t2 =    5F  2  2mR  Distance covered by sphere during one full rotation

2 = 0 +

S = ut +

S=

1 2 at 2

=0 +

1  F   8mR      2  m   5F 

4R 5

SECTION (J) J 2.

(a) Pi = m2v Pf = (m1 + m2) Vcm m2v = (m1+m2) Vcm

 m2v  Vcm =  m  m  2   1 (b) v1 = (u – Vcm)  m1u m 2u V1 = v – m  m =  m  m 1 2 2  1  – m1u (c) V1 = – Vcm =  m  m 2  1

  

  

L m1(0)  m 2   m 2L 2 (d) Xcm = = 2(m1  m 2 ) (m1  m 2 )

L1 =

m 2L L – 2(m  m ) 2 1 2



L1 =

1 2

 m1L   m1  m 2

  

momentum of particle  m 2m12u    m1L m 2u   1 1    m ( u  V ) L  m u   Pi =  2  =  cm 2  2 2(m1  m 2 )  m1  m2   2(m1  m 2 )   2

Momentum for rod = m1Vcm  L cm 

m1L m2 u 2 (m1  m 2 ) 2

(e) For particle : 2

1 =

m2L2

m 2m1 L2 = 4(m1  m 2 )2

 = 1 + 2 =

RESONANCE



  m L2 m2 L  2 = 1  m1 12 2 ( m  m ) 1 2  

2

m1(m1  4m 2 )L2 12(m1  m 2 ) SOLN_RIGID BODY DYNAMICS - 121

(f) Velocity of centre of mass

 m2v  =  m  m  2   1 Using angular momentum conservation m2v × Lcm = cm  m1L = m 2u 2(m  m ) = cm . 1 2 m1L m1(m1  4m 2 )L2 = m 2u 2(m  m ) = × 12(m1  m 2 ) 1 2



6m2 v  = (m  4m )L . 1 2

SECTION (K) K 1.

Force balance N= mg cos  f = mg sin  Torque balance (about centre of mass) Nx = f ×

a mg sin  amg sin  a tan  a = and x = = 2 2mg cos  2 2

Torque of normal force Nx = mg sin 

a 2

PART - II SECTION (A) A 1.

0 = 3000 rad/min 3000 rad/sec = (50 rad/sec) 60 t = 10 sec f = 0 f = 0 + t  = 50 –  (10)  = 5 rad/sec2

0 =

 = o t +

1  t2 2

1 (–10) (10)2 2  = 500 – 250 = 250 rad

 = (50) (10) +

A 3.*

Sphere is rotating about a diameter so , a = R but, R is zero for particles on the diameter.

SECTION (B) B 3.

B > A B > A so, If the axes are parallel

B 6.

Moment of inertia of the elliptical disc should be less than that of a circular disc having radius equal to the major axis of the elliptical disc. Hence (D)

RESONANCE

SOLN_RIGID BODY DYNAMICS - 122

B 7.

0 = 1 + 2 2

2

m / 2  

m / 2  

2 3

0 =

2 3

+

m 2 12

=

SECTION (C) C 1.





 F2 = –2i – 3j – 4k F1 = 2i + 3j + 4k    Net force Fnet  F1  F2  0 the body is in translational equilibrium. 



r2 = i

r1 = 3i + 3J + 4k 





= (3 ˆi  3ˆj  4kˆ ) × (2ˆi  3ˆj  4kˆ )

1 = r1 × F1 

1 = 9kˆ – 12ˆj – 6ˆj  12ˆi  8ˆj – 12ˆi 







1 = – 4ˆj  3kˆ  2 = r2 × F2 = ( ˆi ) × ( – 2ˆi – 3ˆj – 4kˆ )     1 2  –4ˆi  3kˆ – 3kˆ  4ˆj  0     

= –3 kˆ + 4 ˆj

body in rotational equilibrium C 3.

 ˆj ˆ ˆ F = 2 i + 3 – k at point (2,–3,1) torque about point (0, 0, 2)



r = 2ˆi – 3ˆj  kˆ – 2 kˆ











 = r × F = (2ˆi – 3ˆj – kˆ )  (2ˆi  3ˆj – kˆ )



 = (6 ˆi  12kˆ ) 

 = (6 5 )

SECTION (D) D 2.

N1 =  N2 , N1 + N2 = mg ,  A = o 

3 N2 – 4 N1 – Hence  =

3 mg = o 2

1 Ans. 3

Aliter Using force balance f1 = –N1 f2 = N2

N1 + f2 = mg (1) N2 = f 1 N2 = N1 (2)

Using aq (1) N1 + N2 = mg N1 + N1 = mg

 mg   N1 +  2   1   torque about point B  B = 0 f1 × 4 + mg (5/2 cos 53º) = 3N1

RESONANCE

For rotational equilibrium

SOLN_RIGID BODY DYNAMICS - 123

4N1 +

3mg = 3N1 2

3mg = (3 – 4) 2

 mg     1 2   



3mg = (3 – 4) N1 2



 3 – 4  3  =  2  1   2  

3 + 32 = 6 – 8 32 + 8 – 3 = 0 32 + 9 –  – 3 = 0 3( + 3) –1 ( + 3)  ( = 1/3)

w1

w D4

x

–x

weight of object = w

 w ( – x) = w1x ...........(i) If weight is kept in another pan then : w2( – x) = wx ...........(ii) By (i) & (ii)

w w1 = w2 w w=



w2 = w1 w2

w 1w 2 .

SECTION (E) E-3.  2  N =  m 2   

E-4.

Initial velocity of each point onthe rod is zero so angular velocity of rod is zero. Torque about O  =  20g (0.8) =



m 2  3

 20g (0.8) =

20 (1.6 )2  3

3g =  = angular acceleration 3 .2



=

15 g 16

SECTION ( F ) F 2.

By energy conservation 2

1  7 m  2  2   mg = .   4 2  48 0 =

7 ml 2 48

RESONANCE



[ (about O)

=

24g 7

m 2   m  ] = 12 4

Ans.

SOLN_RIGID BODY DYNAMICS - 124

SECTION ( G ) G 3.

x = v0 cos 45º × t =  = mgx =

L=

mgv 0 t

mgv 0 2

v0t

=

2

2 dL dt

v0 / g



t dt =

0

mv 30 2 2g



G 5.

external torque ext = 0 11 = 22 when he stretches his arms  so 1 < 2 then (1 > 2) so, (L = constant)

G 7.*

External force will act at hinge so linear momentum of system will not remain const. but torque of external 

force is zero about hinge so L = const., collision is elastic so K.E = const.

SECTION (H) H 3*.

for pure rolling V = R VA = 2V VB = 2 V (VC = 0)

SECTION (I) -3.

mg sin – f = ma  mg sin  – f  a=  .......(i)  m   a is same for each body.

f.R = 



For solid sphere k2 =



f .R mk 2

2 R2 is minimum there fore  is maximum hence, k.E. for solid sphere will be max 5

at bottom. -5.

mg sin  – f = ma

mg sin  – f m a is equal for each body so all the object will reach at same time. a=

-7.

There is no relative motion between sphere and plank so friction force is zero then no any change in motion of sphere and plank.

SECTION (J) J-2.*

at the moment when ring is placed friction will act between them due to relative motion. Friction is internal force between them so angular momentum of system is conserved. I11 = I22  mR 2  2 mR 2  0 =  2  mR  2  

RESONANCE



=

0 3

SOLN_RIGID BODY DYNAMICS - 125

J-3.

Conservation of angular momentum about C.O.M. of m and loop of mass m gives 2 2  R  R  mVR m R 2  m    m    =   2    2   2 

V = 3 R





=

V 3R

J-4.

velocity of COM after collision is V friction will act such that  = o at some intant after some time (V = R)

SECTION (K) K-2.

For no slipping µmg cos mg sin For toppling

.........(1)

h a  mg cos. 2 2 for minimum µ (by dividing) mg sin

µ.

.........(2)

2 2 = a h

µmin =

a . h

[ Ans.: a/h ] Sol.(2) If f > mg sin  mg cos  > mg sin  ( > tan ) block will topple before sliding torque about point A A =0

a mg sin  h 2 = mg cos 2

  tan  = a h   > a h 

If  > tan  (block will slide)

N

a a/2

K-3. b/2

mg The block will not topple if mg acts from within the base area of the block. So,

a b cos   2 2

RESONANCE

 cos  

b a

SOLN_RIGID BODY DYNAMICS - 126

EXERCISE-2 PART - I 3.

linear density  =

m 

m



dm =   dx    

AB = 

m 

4.

 0



dm · x

2

m

2

  dx · ( x cos 45)

=

o

m x2 dx = 2 2

 x3   3 

   

 m2     6    0 



dm = (2xdx) R

=

 dm· x

2

=

 (2xdx).x

2

o

R

 = 2 x 3 dx

 0

R

R

3  = 2 x ·(    x )dx = 2 

 0

 o

R

 x 3 dx   x 4 dx

 o

 R 4  R 5     = 2  4  5    5.

h y  R r

r=

R y h

dm =  (r2dy) dAB =

1 (dm) r2 2 h

 AB =

1

2

 2 r dy r

2

y 0

=

 R 4 2 h4

 h5     5 

     m    3  4 m    = = R h .  1 2  ...........    mR2 10 10 R h 1   2 R h   3   3  

RESONANCE

SOLN_RIGID BODY DYNAMICS - 127

 m   2  R 

7.

 = 

For small ring friction force dfr = K(2rdr)g Torque of the friction  = (–rdfr) = – 2krgr2dr R

 = –2krg

2

 r dr = –

2 krgR3 3

0

For rotation about z-axis ( = ) –

2 ( R 2 )( ) 2 krgR3 = R  3 2



 –4kg    3R 

= 

From equation of motion  = 0 +  t  –4kg 

0 = 0 +  3R  t   8.

 3R 



m

0 t =    4kg 

m

=    (a = R)



m1 = x =    x   m 1g – T = m 1a

............ (i)

2

 MR   

T R =  2   + (m – m1) (R2)    MR  + (m – m1) R  2 Ma m 1g – – (m – m1) a = m1a 2

T=

 Ma

 

............(ii) Ma + (m – m1) a 2 Ma m 1g – – ma + m1a = m1a 2

T=



m1g =  2  ma    2m1a  = (M  2m)

10.



 mx  2 g     = (M  2m)R



2mgx



    (M  2m)R   

Using energy conservation mgh =

1 1 1 kx2 + 2 + mv2 2 2 2

String does not slip So(V = r) mg x =

1 1  v2  1 kx2 +   2  + mv2 2 2 r  2

x = 0.1m m = 11 kg 11 × 10 × 0.1 =

 = 0.1 kg – m2 r = 0.1 m

K = 100 N/m

V2 1 1 1 × 100 × (0.1)2 + × 0.1 × 11 × V2 2 + (0.1) 2 2 2

22 = 1 + 10 V2 + 11 V2 21 V2 = 21 V = 1 m/s

RESONANCE

SOLN_RIGID BODY DYNAMICS - 128

11.

(a)

Energy conservation loss in P.E. = gain in rotational K.E.

(b)

1 m2 2  (1 – cos ) =  2 3 2



mg



2 =



=

 = mg

 sin  = 0 2



mg

 m2 sin  =  2 3



=

3g sin  3g = sin  2  4

3g (1 – cos ) 

3g (1  cos ) 

fy = may 

mg – N2 = mat sin 



N2 = mg – mat sin  = mg – m

3g sin2  4

2 N2 = mg 1  3 sin  



4



fx = max 

N1 = ma1 cos  =

Ans.

Normal reaction = where N1 =

12.

m3 g sin  . cos  4

N12  N22

3mg sin  cos  4





N2 = mg 1  

3 sin2    4 

(a) About the axis of rotation of rod, the angular momentum of the system is conserved velocity of the flying bullet is V 

2



3 

M  mv =  m2   

=



 3mv  mv  (m