2. classical mechanics net-jrf june 2011 - june 2014

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CLASSICAL MECHANICS SOLUTIONS

NET/JRF (JUNE-2011)

Q1. A particle of unit mass moves in a potential ( ) 22 xbaxxV += , where a and b are positive

constants. The angular frequency of small oscillations about the minimum of the potential

is

(a) b8 (b) a8 (c) ba /8 (d) ab /8

Ans: (b)

( )2

2

xbaxxV += V 0

x = 3

2b2ax 0x

= 4 0ax b = 14

0 .bxa

= Since

mk=

, m 1= and

0

2

2x x

Vkx =

= where ox is stable equilibrium point.

Hence aa

bba

xba

xVk 86262 4

02

2

=+=+== at 4

1

0

==abxx .

Thus 8a = .

Q2. The acceleration due to gravity (g) on the surface of Earth is approximately 2.6 times that

on the surface of Mars. Given that the radius of Mars is about one half the radius of

Earth, the ratio of the escape velocity on Earth to that on Mars is approximately

(a) 1.1 (b) 1.3 (c) 2.3 (d) 5.2

Ans: (c)

Escape velocity = gR2

e e

m m

g REscape velocity of Earth 2.3Escape velocity of Mass g R

= = where e

m

R 2R

= and em

g 2.6.g

=





Q3. The Hamiltonian of a system with n degrees of freedom is given

by ( )tppqqH nn ;,.......;,..... 11 , with an explicit dependence on the time t. Which of the following is correct?

(a) Different phase trajectories cannot intersect each other.

(b) H always represents the total energy of the system and is a constant of the motion.

(c) The equations iiii qHppHq == /,/ are not valid since H has explicit time dependence.

(d) Any initial volume element in phase space remains unchanged in magnitude under

time evolution.

Ans: (a)

Q4. The Lagrangian of a particle of charge e and mass m in applied electric and magnetic

fields is given by evAevmL += GGG221 , where A

G and are the vector and scalar

potentials corresponding to the magnetic and electric fields, respectively. Which of the

following statements is correct?

(a) The canonically conjugate momentum of the particle is given by vmp GG =

(b) The Hamiltonian of the particle is given by epAme

mpH ++= GGG

2

2

(c) L remains unchanged under a gauge transformation of the potentials

(d) Under a gauge transformation of the potentials, L changes by the total time derivative

Ans: (d)

20

LVt

+ = ,





Q5. Consider the decay process + in the rest frame of the -. The masses of the and , are M, M and zero respectively.

A. The energy of is

(a) ( )

McMM

2

222 (b) ( )

McMM

2

222 +

(c) ( ) 2cMM (d) 2cMM Ans: (b)

+ From conservation of energy 2M c E E . = +

2 2 2 2 4E p c M c = + and 2 2 2E p c = since momentum of and is same. 2M c E E = + , 2 4 2 2M c E E =

2 4

2

M cE EM c

=

2M cE EM

= and 2E E M c + =

( )2 2 2M M cE .

2M

+ =

B. The velocity of is

(a) ( )

22

22

MMcMM

+

(b) ( )

22

22

MMcMM

+

(c)

McM

(d)

McM

Ans: (a)

Velocity of ( )2 2 2

2

2

M M M cE2M v1

c

+= =

( )2 22

22 2 2

4M Mv1c M M

= + ( )

2 22

22 2 2

4M Mv 1c M M

=

+

( )4 4 2 2 2 22

22 2 2

M M 2M M 4M Mvc M M

+ + =+

2 2

2 2

M Mv cM M

= + .





Q6. The Hamiltonian of a particle of unit mass moving in the xy-plane is given to be:

22

21

21 yxypxpH yx += in suitable units. The initial values are given to be

( ) ( )( ) ( )1,10,0 =yx and ( ) ( )( ) = 21,210,0 yx pp . During the motion, the curves traced out by the particles in the xy-plane and the yx pp plane are

(a) both straight lines

(b) a straight line and a hyperbola respectively

(c) a hyperbola and an ellipse, respectively

(d) both hyperbolas

Ans: (d)

22

21

21 yxypxpH yx +=

.

Solving Hamiltonion equation of motion

xH px

= x xp x p = and yH py

= y yp y p = .

x

H xp = x x = and y

H yp = y y = .

After solving these four differential equation and eliminating time t and using boundary

condition one will get 1xy

= and x

y

1 1p2 p

=





NET/JRF (DEC-2011) Q7. A double pendulum consists of two point masses m attached by strings of length l as

shown in the figure: The kinetic energy of the pendulum is

(a) [ ]2221221 +ml (b) ( )[ ]212122212 cos2221 ++ ml (c) ( )[ ]212122212 cos2221 ++ ml (d) ( )[ ]212122212 cos2221 +++ ml

Ans: (c)

Let co-ordinate ( )11 , yx and ( )22 , yx . ( ) ( )22222121 2121.. yxmyxmEK +++= 11 sinlx = , 11 cosly = 111 cos lx = , 111 sin ly =

212 sinsin llx += , 212 coscos lly += 22112 coscos llx += , ( ) ( ) 22112 sinsin += lly

Put the value of 2211 ,,, yxyx in K.E equation, one will get

( )[ ]21212221 cos221 ++= mT .

Q8. A constant force F is applied to a relativistic particle of rest mass m. If the particle starts

from rest at t = 0, its speed after a time t is

(a) mFt / (b)

mcFtc tanh (c) ( )mcFtec /1 (d)

2222 cmtFFct+

Ans: (d) dp Fdt

= p Ft c = + .

At t=0, p=0 so c=0 p Ft =2

2

mu Ftu1c

=

2

F tmu

Ft1mc

= +

.

1

2 l

l

m

m





Q9. The potential of a diatomic molecule as a function of the distance r between the atoms is

given by ( ) 126 rb

rarV += . The value of the potential at equilibrium separation between

the atoms is:

(a) ba /4 2 (b) ba /2 2 (c) ba 2/2 (d) ba 4/2 Ans: (d)

( )126 rb

rarV += , for equilibrium 0=

rV ( ) 7 13a 12b6 0r r = 0

1261 67 =

r

bar

01266=

rba

ab

rb

61212

6= 6

1

612

=

abr

61

2

=abr

16

22b a bV r

2ba 2ba a

= = +

2 2 2a a a2b 4b 4b

= + = .

Q10. Two particles of identical mass move in circular orbits under a central

potential ( ) 221 krrV = . Let l1 and l2 be the angular momenta and r1, r2 be the radii of the

orbits respectively. If l1/l2 = 2, the value of r1 / r2 is:

(a) 2 (b) 2/1 (c) 2 (d) 1 / 2

Ans: (a)

22

2

21

2kr

mrJVeff += where J is angular momentum.

Condition for circular orbit 0=

rVeff 03

2

=+ krmrJ 2 4J r 2J r .

Thus 2

1 1 1 1 1

2 2 2 2 2

J r r J r 2J r r J r

= = = since 1

2

J 2J

= .





Q11. A particle of mass m moves inside a bowl. If the surface of the bowl is given by the

equation ( )2221 yxaz += , where a is a constant, the Lagrangian of the particle is

(a) ( )222221 garrrm +

(b) ( )[ ]222221

21 rrram ++

(c) ( )2222222 sin21 garrrrm ++

(d) ( )[ ]2222221

21 garrrram ++

Ans: (d)

( ) mgzzyxmL ++= 22221 where ( )22

21 yxaz += .

It has cylindrical symmetry. Thus x r cos= , y r sin= , ( )21z a r2

= .

x r cos r sin= , y r sin r cos= + and ( )z a rr= . So ( )2 2 2 2 2 21L m 1 a r r r gar

2 = + + .

Q12. A planet of mass m moves in the inverse square central force field of the Sun of mass M.

If the semi-major and semi-minor axes of the orbit are a and b, respectively, the total

energy of the planet is:

(a) ba

GMm+ (b)

+

baGMm 11

(c)

aba

GMm 11 (d) ( )

+ 2ba

baGMm

Ans:

a is semi-major axis and b is semi-minor axis so 21 eab = . If planet moves under inverse square law total energy of planet is given by

akE

2=

where GMmk = . So we have to check all the options and will find whether e is smaller than 1 or not.





(a) ba

GMm+

21 eaa

GMm+

= =a

GMm2

, when e = 0. Which is condition for circlular orbit.

(b)

+=ba

GMma

GMm 112

+=21

1121

eaaa 2111

21

e+= =

21

11e

e is imaginary.

(c)

aba

GMm 11 is dimensionally incorrect for energy. It must be

ab

GMm 11 .

=ab

GMma

GMm 112 aeaa

111

21

2

= 1

11

21

2

=

e 211

23

e=

22 1 ee =9412 = e 1

352 12e 1>e .

So equation is given by hyperbola.

Q13. An annulus of mass M made of a material of uniform density has inner and outer radii a

and b respectively. Its principle moment of inertia along the axis of symmetry

perpendicular to the plane of the annulus is:

(a) ( )( )2244

21

ababM

+ (b) ( )2221 abM

(c) ( )2221 abM (d) ( )22

21 abM +

Ans: (d)





Q14. The trajectory on the zpz-plane (phase-space trajectory) of a ball bouncing perfectly

elastically off a hard surface at z = 0 is given by approximately by (neglect friction):

(a) (b)

(c) d)

Ans: (a)

mgzm

PH z +=2

2

and mgzm

PE z +=2

2

.

ZP

z

ZP

z

ZP

z

ZP

z





NET/JRF (JUNE-2012)

Q15. The bob of a simple pendulum, which undergoes small oscillations, is immersed in water.

Which ofthe following figures best represents the phase space diagram for the pendulum?

(a) (b)

(c) (d)

.

Ans: (d)

When simple pendulum oscillates in water it is damped oscillation so amplitude

continuously decrease and finally it stops.

Q16. Two events separated by a (spatial) distance 9 109m, are simultaneous in one inertial

frame.The time interval between these two events in a frame moving with a constant

speed 0.8 c (where the speed of light c = 3 108 m/s) is

(a) 60 s (b) 40 s (c) 20 s (d) 0 s

Ans: (b)

x

p

p

x

p

x

p

x





mxx 9'1'2 109= and 0'1'2 = tt . Then

+

+=

2

2

'12

11

2

2

'22

'2

12

11cv

xcvt

cv

xcvt

tt( ) ( )

2

2

'1

'2

2

2

2

'1

'2

2

2

2

'1

'2

12

111cv

xxcv

cv

xxcv

cv

tttt=

+

= .

Put cv 8.0= sec4012 tt

Q17. If the Lagrangian of a particle moving in one dimensions is given by ( )xVx

xL =2

2 the

Hamiltonian is

(a) ( )xVxp +221 (b) ( )xV

xx +2

2 (c) ( )xVx +2

21 (d) ( )xV

xp +2

2

Ans: (a)

Since LxpH x = and xpxL = x

pxx = xpx x= .

( )2x xH p x V x2x= + ( ) ( )

2x

x x

p xH p p x V x

2x = + ( )xVxpH x +=

2

2

.

Q18. A horizontal circular platform mutes with a constant angular velocity directed

vertically upwards. A person seated at the centre shoots a bullet of mass m horizontally

with speed v. The acceleration of the bullet, in the reference frame of the shooter, is

(a) 2v to his right (b) 2v to his left

(c) v to his right (d) v to his left

Ans: (a)

Velocity of bullet = jv , Angular velocity= k . There will be coriolis force ( )= vmF 2 . ivaivmF 22 == .





Q19. The Poisson bracket { }pr , has the value (a) pr (b) pr (c) 3 (d) 1

Ans: (b) kzjyixr ++= , ( ) 2/1222 zyxr ++= , kpjpipp zyx ++= , ( ) 2/1222 zyx pppp ++=

{ }pr , =

xp

pr

pp

xr

xx

+yp

pr

pp

yr

yy z z

r p r pz p p y

+

yx zpp px y z

r p r p r p+ + ( )r p r p

r p =

G G

Q20. Consider the motion of a classical particle in a one dimensional double-well

potential ( ) ( )22 241 = xxV . If the particle is displaced infinitesimally from the minimum

on the x-axis (and friction is neglected), then

(a) the particle will execute simple harmonic motion in the right well with an angular

frequency 2= (b) the particle will execute simple harmonic motion in the right well with an angular

frequency =2

(c) the particle will switch between the right and left wells

(d) the particle will approach the bottom of the right well and settle there

Ans : (b) ( ) ( )22 241 = xxV ( ) 022

42 2 ==

xxxV 0= x , 2=x .

23 222

= x

xV . At 0=x , 02

2





Q21. What is proper time interval between the occurrence of two events if in one inertial frame

events are separated by 7.5 108m and occur 6.5 s a part? (a) 6.50 s (b) 6.00 s (c) 5.75 s (d) 5.00 s

Ans: mxx 812 105.7 = and stt 5.612 = .

( ) ( )2

2

122

2

2

12

2

2

121

2

2

222'1

'2

1111cv

xxcv

cv

tt

cv

xcvt

cv

xcvt

tt

=

+

+= .

In this problem v is not given so its not possible to calculate proper time interval.

NET/JRF (DEC-2012)

Q22. A solid cylinder of height H, radius R and density , floats vertically on the surface of a

liquid of density 0 . The cylinder will be set into oscillatory motion when a small instantaneous downward force is applied. The frequency of oscillation is

(a) Hg

0 (b)

Hg

0

(c) Hg

0

(d) Hg

0

Ans: (d)

From Newtons law of motion 0ma mg Agh= where A is area of cross section, m AH= .

0AHa AHg Agh = 01 gha H =

0gH

=





Q23. Three particles of equal mass (m) are connected by two identical massless springs of

stiffness constant (K) as shown in the figure

If x1, x2 and x3 denote the horizontal displacement of the masses from their respective

equilibrium positions the potential energy of the system is

(a) [ ]23222121 xxxK ++ (b) ( )[ ]31223222121 xxxxxxK +++ (c) ( )[ ]312232221 2221 xxxxxxK +++ (d) ( )[ ]3122221 2221 xxxxxK ++

Ans: (c) ( ) ( )2 22 1 3 21 12 2V K x x K x x= + , ( ) ( )2 2 2 22 1 2 1 3 2 3 21 12 22 2V K x x x x K x x x x= + + + ( )2 2 21 2 3 2 1 31 2 22V K x x x x x x = + + +

Q24. A planet of mass m moves in the gravitational field of the Sun (mass M). If the semi-

major and semi-minor axes of the orbit are a and b respectively, the angular momentum

of the planet is

(a) ( )baGMm +22 (b) ( )baGMm 22 (c) ba

abGMm

22 (d) ba

abGMm+

22

Ans:

All the options given in this question are incorrect.

We know that Eccentricity, 2221

mkEJe += where (J) is the angular momentum of planet.

akE

2= 2

2

22

1mk

Jak

e

+=

mkaJe

2

1=mkaJe

22 1=

21 eab = mkaJ

ab 2

2

2

11 = 22

2 1abe = =

mkaJ

ab 2

2

2

2

22

amkabJ =

amkbJ

2

= where GMmk = then a

bGMmJ22

= .

m

K

m

K

m





Q25. The Hamiltonian of a simple pendulum consisting of a mass m attached to a massless

string of length l is ( ) cos12 2

2

+= mglmlpH . If L denotes the Lagrangian, the value of

dtdL is:

(a) sin2 plg (b) 2sinpl

g

(c) cosplg (d) cos2lp

Ans: (a)

[ ]tLHL

dtdL

+= , where ( ) cos1

2 22

+= mglmlpH .

HpHqpL ii

i == , 2mlppH == ,

( ) cos12

22

= mglmlL . Hence we have to calculate [ ]HL, which is only defined into phase space i.e. p and . Then ( ) cos12 2

2

= mglmlpL

[ ] sin2, plgH

pL

pHLHL =

= and 0=

tL sin2 pl

gdtdL =

Q26. Two bodies of equal mass m are connected by a massless rigid rod of length l lying in the

xy-plane with the centre of the rod at the origin. If this system is rotating about the z-axis

with a frequency , its angular momentum is

(a) 4/2ml (b) 2/2ml (c) 2ml (d) 22ml Ans: (b) Since rod is massless i.e. M = 0.

Moment of inertia of the system 2222

11 rmrm += , mmm == 21 and 221lrr ==

44

22 mlml +=2

2ml= . Angular momentum, IJ = and 2

2mlJ = .





Q27. Which of the following set of phase-space trajectories is not possible for a particle

obeying Hamiltons equations of motion?

(a) (b)

(c) (d)

Ans: (b)

Phase curve does not cut each other

Q28. The muon has mass 105 MeV/c2 and mean life time 2.2 s in its rest frame. The mean

distance traversed by a muon of energy 315 MeV before decaying is approximately,

(a) 3 105 km (b) 2.2 cm (c) 6.6 m (d) 1.98 km

Ans: (d)

Since MeVE 315= and 20 105 cMeVm = .

2mcE =2

2

20

1cv

cmE

=2

2

20

1315

cv

cm

=

2

2

1

105315

cv

= cv 94.0= .

P

x

P

x

P

x

P

x





Now, st

cv

tt 2.2,

10

2

2

0 =

= =

981

102.2 6t t = 6.6 s

Now the distance traversed by muon is 6106.694.0 = cvt = km86.1= .

NET/JRF (JUNE-2013)

Q29. The area of a disc in its rest frame S is equal to 1 (in some units). The disc will appear

distorted to an observerO moving with a speedu with respect to S along the plane of the

disc. The area of the disc measured in the rest frame of the observer O is ( c is the speed

of light in vacuum)

(a) 2/1

2

2

1

cu (b)

2/1

2

2

1

cu (c)

2

2

1cu (d)

1

2

2

1

cu

Ans: (a) Area of disc from S frame is 1 i.e. 12 =a or 1= aa

Area of disc from S frame is ba 22

1cuaa = 2

2

11cu= 2

2

1cu=

where 22

1cuab = .

Q30. A planet of mass m and an angular momentum L moves in a circular orbit in a potential,

( ) ,/ rkrV = where k is a constant. If it is slightly perturbed radially, the angular frequency of radial oscillations is

(a) 32 2/ Lmk (b) 32 / Lmk (c) 32 /2 Lmk (d) 32 /3 Lmk

Ans: (b)

rk

mrLVeff = 2

2

2. For circular orbit

2

3 2 0effV L kr mr r

= + =

232

rk

mrL = . Thus

mkLrr

2

0 == mk= ,

effV

r





0

2

2

rr

eff

drVd

k=

=0

34

2 23

rrrk

mrL

=+= 3242

2 23

=mkL

k

mkLm

L = 643

6

433L

kmL

km 643

Lkm=

m

drVd

rr 0

2

2

== 32

Lmk= .

Q31. The number of degrees of freedom of a rigid body in d space-dimensions is

(a) d2 (b) 6 (c) ( ) 2/1+dd (d) !d Ans: (c)

Q32. A system is governed by the Hamiltonian

( ) ( )2221

21 bxpaypH xx +=

where a and b are constants and yx pp , are momenta conjugate to x and y respectively.

For what values of a and b will the quantities ( )ypx 3 and ( )xp y 2+ be conserved? (a) 2,3 == ba (b) 2,3 == ba (c) 3,2 == ba (d) 3,2 == ba

Ans: (d) Poisson bracket [ ]3 , 0xp y H = and 2 , 0yp y H + = 2( 3) (3 ) 0yp b x b b + = and 2( 2) (2 ) 0xp a y a a+ + =

3,2 == ba Q32. The Lagrangian of a particle of mass m moving in one dimension is given by

bxxmL = 221

whereb is a positive constant. The coordinate of the particle ( )tx at time t is given by: (in following 1c and 2c are constants)

(a) 212

2ctct

mb ++ (b) 21 ctc +





(c)

+

mbtc

mbtc sincos 21 (d)

+

mbtc

mbtc sinhcosh 21

Ans: (a)

Equation of motion 0=

xL

xL

dtd

( ) 0=+ bxmdtd 0=+ bxm bxm =

mb

dtxd =2

2

1Ctmb

dtdx += 21

2

2CtCt

mbx ++=

NET/JRF (DEC-2013)

Q33. Let BA, and C be functions of phase space variables (coordinates and momenta of a

mechanical system). If { }, represents the Poisson bracket, the value of { }{ } { }{ }CBACBA ,,,, is given by

(1) 0 (2) { }{ }ACB ,, (3) { }{ }BCA ,, (4) { }{ }BAC ,, Ans: (4)

We know that Jacobi identity equation

{ }{ } { }{ } { }{ } 0,,,,,, =++ BACACBCBA Now { }{ } { }{ } { }{ } { }{ }BACACBCBACBA ,,,,,,,, ==

Q34. A particle moves in a potential2

222 zyxV ++= . Which component(s) of the angular

momentum is/are constant(s) of motion?

(1) None (2) yx LL , and zL (3) only xL and yL (4) only zL

Ans: (4) A particle moves in a potential 2

222 zyxV ++=

( ) 22222222 cos2

sinsincossin,, rrrrV ++=

( ) 2222 cos2

sin,, rrrV +=

Now is cyclic-co-ordinate ( )p i.e zL is constant of motion.





Q35. The Hamiltonian of a relativistic particle of rest mass m and momentum p is given

by ( )xVmpH ++= 22 , in units in which the speed of light 1=c . The corresponding Lagrangian is

(1) ( )xVxmL += 21 (2) ( )xVxmL = 21 (3) ( )xVxmL += 21 (4) ( )xVxmL = 2

21

Ans: (2)

( )xVmpH ++= 22 ( )12 2 21 22

H pxp p m

= = + ( ) pmpx =+ 2/122

21xmp

x =

Now = HpxL Hpx = ( )2 2xp p m V x= + Put value

21xmp

x=

( )xVxmL = 21 Q36. A pendulum consists of a ring of mass M and radius R suspended by a massless rigid

rod of length l attached to its rim. When the pendulum oscillates in the plane of the ring,

the time period of oscillation is

(1) g

Rl +2 (2) ( ) 4/1222 Rlg

+

(3) ( )lRglRlR

+++ 22 222 (4) ( ) 4/122 222 lRlR

g++

Ans: (3)

The moment of inertia about pivotal point is given by 2 2 2

. ( )c mI I Md MR M l R= + = + + If ring is displaced by angle then potential energy is ( ) cosMg l R + The Lagrangian is given by





21 ( )2

L I V = = 2 2 21 ( ( ) ) ( ) cos2

MR M l R Mg l R + + + +

0d L Ldt

= 2 2( ( ) ) ( )sin 0MR M l R Mg l R + + + + =

For small oscillation sin = 2 2( ( ) ) ( ) 0MR M l R Mg l R + + + + =

Time period is given by ( )lRglRlR

+++ 22 222 .

Q37. Consider a particle of mass m attached to two identical springs each of length l and

spring constant k (see the figure below). The equilibrium configuration is the one where

the springs are unstretched. There are no other external forces on the system. If the

particle is given a small displacement along the x -axis, which of the following describes

the equation of motion for small oscillations?

(1) 023

=+lkxxm (2) 0=+ kxxm (3) 02 =+ kxxm (4) 0

2

=+l

kxxm

Ans: (1)

The lagrangian of system is given by

21 ( )2

L mx V x= The potential energy is given by

( ) ( )2 21 12 2 2 22 2( ) 2 2k kV x x l l x l l = + + +

y

o

o x

y

o

o x

x

l

l





( ) 212 2 2( )V x k x l l = + For small oscillation one can approximate potential by Taylor expansion

212 2

22( ) 1 1xV x kll

= +

22 42

2 4

1 1( ) 1 12 8

x xV x kll l

= +

222

2( )xV x kll

=

4

2( ) 4xV x kl

= .

So Lagrangian of system is given by 4

22

12 4

xL mx kl

=

The Lagranges equation of motion 0d L Ldt x x

= 023

=+lkxxm .

NET/JRF (JUNE-2014)

Q38. The time period of a simple pendulum under the influence of the acceleration due to

gravity g is T . The bob is subjected to an additional acceleration of magnitude g3 in

the horizontal direction. Assuming small oscillations, the mean position and time period

of oscillation, respectively, of the bob will be

(a) o0 to the vertical and T3 (b) o30 to the vertical and 2/T

(c) o60 to the vertical and 2/T (d) o0 to the vertical and 3/T

Ans: (c)

glT 2=

( ) 223 ggg += 24g= g2=

glT

22= 12

2lTg

= 2

TT =

T

g g

g3





mgT =cos , sin 3T mg = tan 3 = o60 = Q39. A particle of mass m and coordinate q has the Lagrangian 22

221 qqqmL =

where is a constant. The Hamiltonian for the system is given by (a) 2

22

22 mqp

mp + (b) ( )qm

p2

2

(c) ( )222

22 qmqp

mp

+ (d) 2

qp

Ans: (c)

H qp L= where 22 2121 qqqmL = L p mq qqq

= = ( )p q m q = pq

m q =

H qp L = ( )( )

( ) ( )22 2

2 212 2

pp pm qm q m q m q

= +

H qp L = ( ) ( ) ( )2 2

22p p m q

m q m q =

H qp L = ( ) ( )2 2

2p p

m q m q = ( )2

2pH

m q =

Q40. The coordinates and momenta ( )3,2,1, =ipx ii of a particle satisfy the canonical Poisson bracket relations { } ijji px =, . If 23321 pxpxC += and 12212 pxpxC += are constants of motion, and if { } 1331213 , pxpxCCC +== , then (a) { } { } 213132 , and, CCCCCC == (b) { } { } 213132 , and, CCCCCC == (c) { } { } 213132 , and, CCCCCC == (d) { } { } 213132 , and, CCCCCC ==





Ans: (d)

1 2 3 3 2 ,C x p x p= + 2 1 2 2 1,C x p x p= 3 1 3 3 1C x p x p= +

{ } 3 3 3 3 3 32 2 2 2 2 22 31 1 1 1 2 2 2 2 3 3 3 3

,C C C C C CC C C C C CC C

x p p x x p p x x p p x = + +

{ }2 3,C C ( )( ) ( ) ( )2 3 2 3 1 1 10 0 0 0p x x p x x x= + + ( )2 3 2 3 1p x x p C= + = { } 3 3 3 3 3 31 1 1 1 1 13 1

1 1 1 1 2 2 2 2 3 3 3 3

,C C C C C CC C C C C CC Cx p p x x p p x x p p x

= + +

{ }3 1,C C ( ) ( ) ( ) ( )3 3 3 3 1 2 1 2 1 2 2 1 20 0 0 0p x x p p x x p x p x p C= + + = = Q41. The recently-discovered Higgs boson at the LHC experiment has a decay mode into a

photon and a Z boson. If the rest masses of the Higgs and Z boson are 2GeV/c125 and 2GeV/c90 respectively, and the decaying Higgs particle is at rest, the energy of the

photon will approximately be

(a) GeV335 (b) GeV35 (c) GeV30 (d) GeV15

Ans: (c)

B H BH P Z + From conservation of momentum 1 2 1 20 P P P P= + =

G G G G1 2P P =

Now B H BH P Z

E E E= + 2H B BP Z H

E E M c + = 0221

2 += cPEHP

and 422222 cMcPE

BB ZZ+=

( )( ) 2 4B H B H BZ P Z P ZE E E E M c + = 21 PP = 2 4 2 2

2B B

B H

B B

Z ZZ P

H H

M c M cE E

M c M = =

2B H BZ P H

E E M c+ =

2 222 B

H B

B

zP H

H

M cE M c

M = ( )2 2 2B B

H

B

H zP

H

M M cE

M =

4

4

125 125 90 90 30.12 125HP

cE GeVc

= =





Q42. A canonical transformation relates the old coordinates ( )pq, to the new ones ( )PQ, by the relations 2qQ = and qpP 2/= . The corresponding time independent generating function is

(a) 2/ qP (b) Pq2 (c) Pq /2 (d) 2qP

Ans: (b) 2 ; / 2Q q P p q= =

2 2 2F Fp P qq q

= = 2

2 ( )F q P f P = +

22F Q qP

= =2

2 ( )F q P f q = + comparing both side ( ) ( ) 0f q f P= = 22F q P =

2. classical mechanics net-jrf june 2011 - june 2014

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