2.classical Mechanics - GATE PDF
2.classical Mechanics - GATE PDF
2.classical Mechanics - GATE PDF
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CLASSICAL MECHANICS SOLUTIONS
GATE- 2010
Q1. For the set of all Lorentz transformations with velocities along the x-axis consider the two
statements given below:
P: If L is a Lorentz transformation then, L-1 is also a Lorentz transformation.
Q: If L1 and L2 are Lorentz transformations then, L1L2 is necessarily a Lorentz
transformation.
Choose the correct option
(A) P is true and Q is false (B) Both P and Q are true
(C) Both P and Q are false (D) P is false and Q is true
Ans: (b)
1 2 3
Q2. A particle is placed in a region with the potential V (x ) = kx x , where k, > 0.
2 3
Then,
k
(A) x = 0 and x = are points of stable equilibrium
k
(B) x = 0 is a point of stable equilibrium and x = is a point of unstable equilibrium
k
(C) x = 0 and x = are points of unstable equilibrium
(D) There are no points of stable or unstable equilibrium
Ans: (b)
1 2 x 3 V k
Solution: V = kx = kx x 2 = 0 x = 0, x = .
2 3 x
2V
= k 2x
x 2
2V k 2V
At x = 0, = + ve (Stable) and At x = , = ve (unstable)
x 2 x 2
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Q3. A meson at rest decays into two photons, which move along the x-axis. They are both
detected simultaneously after a time, t = 10s. In an inertial frame moving with a velocity
V = 0.6c in the direction of one of the photons, the time interval between the two
detections is
(A) 15 s (B) 0 s (C) 10 s (D) 20 s
Ans: (a)
v v
1+ 1
c 1 + 0.6 c
Solution: t1 = t 0 = 10 = 10 2 = 20sec , t 2 = t 0
v 1 0.6 v
1 1+
c c
1 0.6 1
= 10 = 10 = 5sec
1 + 0.6 2
t1 t 2 = 15sec
Statement for Linked Answer Questions 4 and 5:
1 2 2
The Lagrangian for a simple pendulum is given by L = ml mgl (1 cos )
2
Q4. Hamiltons equations are then given by
p p
(A) p = mgl sin ; = (B) p = mgl sin ; =
ml 2 ml 2
p g p
(C) p = m ; = (D) p = ; =
m l ml
Ans: (b)
P2 H H P
Solution: H = + mgl (1 cos ) = P P = mglsin ; = = 2 .Q5. The
2ml 2
P ml
{ }
(A) , = 1 { }
(B) , =
1
ml 2
{ }
(C) , =
1
m
{ }
(D) , =
g
l
Ans: (b)
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{ } P P 1
, = , 2 where = 2 . 2
P 1 1
= 1 2 0 = 2 .
ml ml ml P P ml ml
GATE- 2011
1+ q
Q6. A particle is moving under the action of a generalized potential V (q, q ) = . The
q2
Ans: (c)
d V V 2
Solution: = Fq Fq = 3 .
dt q q q
Q7. Two bodies of mass m and 2m are connected by a spring constant k. The frequency of the
normal mode is
(A) 3k / 2m (B) k/m (C) 2k / 3m (D) k / 2m
Ans: (a)
k k 3k 2mm 2m
Solution: m k
2m = = = where reduce mass = = .
2m 2m 2m + m 3
3
Q8. Let (p, q) and (P, Q) be two pairs of canonical variables. The transformation
Q = q cos(p ) , P = q sin(p )
is canonical for
(A) = 2, = 1/2 (B) = 2, =2 (C) = 1, = 1 (D) = 1/2, = 2
Ans: (d)
Q P Q P
Solution: =1
q p p q
q 2 1 (cos 2 p + sin 2 p ) = 1 q 2 1 = 1 = , = 2 .
1
2
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Q9. Two particles each of rest mass m collide head-on and stick together. Before collision, the
speed of each mass was 0.6 times the speed of light in free space. The mass of the final
entity is
(A) 5m / 4 (B) 2m (C) 5m / 2 (D) 25 m / 8
Ans: (c)
Solution: From conservation of energy
mc 2 mc 2 2mc 2
+ = m1c 2 = m1c 2
2
v2
v2
v
1 1 1
c2 c2 c2
Since v = 0.6c m1 = 5m / 2
GATE- 2012
Q10. In a central force field, the trajectory of a particle of mass m and angular momentum L in
plane polar coordinates is given by,
1 m
= (1 + cos )
r L2
where, is the eccentricity of the particles motion. Which one of the following choice
for gives rise to a parabolic trajectory?
(a) = 0 (b) = 1 (c) 0 < < 1 (d) > 1
Ans: (b)
l m
Solution: = (1 + cos ) for parabolic trajectory = 1 .
r l2
Q11. A particle of unit mass moves along the x-axis under the influence of a potential,
V ( x ) = x( x 2) . The particle is found to be in stable equilibrium at the point x = 2. The
2
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V 2
V ( x ) = x( x 2 ) = ( x 2 ) + 2 x( x 2 ) = 0 x = 2, x =
2 2
x 3
2V 2V
= 2( x 2 )2
+ 2( x 2 ) + 2 x = 2 2 = 4
x 2 x 2
2V 2
= = =2 T = .
x 2 x=2
T
Q12. A rod of proper length l0 oriented parallel to the x-axis moves with speed 2c/3 along the
x-axis in the S-frame, where c is the speed of the light in free space. The observer is also
moving along the x-axis with speed c/2 with respect to the S-frame. The length of the rod
as measured by the observer is
(a) 0.35l0 (b) 0.48l0 (c) 0.87l0 (d) 0.97l0
Ans: (d)
u2x
Solution: l = l0 1 = 0.97 l0
c2
Q13. A particle of mass m is attached to fixed point O by a weightless inextensible string of
length a. It is rotating under the gravity as shown in the figure. The
z
Lagrangian of the particle is
1
( )
L( , ) = ma 2 2 + sin 2 2 mga cos where and are the
2 a
1 2 p2 g
(a) H = p + mga cos (b)
2ma 2 sin 2
1 2 p2
H= p + + mga cos
2ma 2 sin 2
(c) H =
1
2ma 2
( p2 + p2 ) mga cos (d) H =
1
2ma 2
( p2 + p2 ) + mga cos
Ans: (b)
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Solution: H = P + P L = P + P ma 2 2 + sin 2 2 + mga cos
2
( )
L P L P
= P ma 2 = P = 2 and P = = ma 2 sin 2 =
ma ma 2 sin 2
2
2
2
P P 1 2 P
P
H = P + P ma + sin + mga cos
ma 2 ma 2 sin 2 2 ma 2 ma 2 sin 2
P2 P2 P2 P2
H= + + mga cos
ma 2 2ma 2 ma 2 sin 2 2ma 2 sin 2
1 2 P2
H= P + + mga cos
2ma 2 sin 2
x
The Lagrangian for this particle is given by
m(1 + 4a 2 x 2 )x 2 mgax 2
1 2 1
(a) L = mx mgax 2 (b) L =
2 2
(c) L =
1 2
2
mx + mgax 2 (d) L =
1
2
( )
m 1 + 4a 2 x 2 x 2 + mgax 2
Ans: (d)
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Solution: Equation of constrain is given by y = ax 2 , K.E T = m ( x 2 + y 2 )
2
1 1
y = 2axx T = m ( x 2 + 4ax 2 x 2 ) = mx 2 (1 + 4ax 2 )
2 2
negative.
L = T V L =
1
2
( )
m 1 + 4a 2 x 2 x 2 + mgax 2
Q15. The Lagranges equation of motion of the particle for above question is given by
(a) x = 2 gax (b) m (1 + 4a 2 x 2 ) x = 2mgax 4ma 2 xx 2
( )
(c) m 1 + 4a 2 x 2 x = 2mgax + 4ma 2 xx 2 (d) x = 2 gax
Ans: (c)
d dL dL
Solution: = m(1 + 4a 2 x 2 ) x = 4ma 2 xx 2 + 2mgax
dt dx dx
GATE- 2013
Q16. In the most general case, which one of the following quantities is NOT a second order
tensor?
(a) Stress (b) Strain
(c) Moment of inertia (d) Pressure
Ans: (b)
Solution: Strain is not a tensor.
Q17. An electron is moving with a velocity of 0.85c in the same direction as that of a moving
photon. The relative velocity of the electron with respect to photon is
(a) c (b) c
(c) 0.15c (d) 0.15c
Ans: (b)
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Q18. The Lagrangian of a system with one degree of freedom q is given by L = q 2 + q 2 ,
where and are non-zero constants. If p q denotes the canonical momentum
Ans: (d)
L L
Solution: = pq but 0
q q
Q19. The relativistic form of Newtons second law of motion is
mc dv m c 2 v 2 dv
(a) F = (b) F =
c 2 v 2 dt c dt
mc 2 dv c 2 v 2 dv
(c) F = (d) F = m
c 2 v 2 dt c2 dt
Ans:
mv dP dv 1 1 1 2v dv
Solution: P = F= =m + mv 3/ 2
2
v2 dt dt v2 2 v2 c dt
1 1 2 1 2
c2 c c
v 2
2 2
dv 1 1 dv 1 v
F =m 1+ c =m 2c 2
v 2 2 1 v dt v 2 3 2
2
dt
1 2
2
c c 1 c 2
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Q20. Consider two small blocks, each of mass M, attached to two identical springs. One of the
springs is attached to the wall, as shown in the figure. The spring constant of each spring
is k . The masses slide along the surface and the friction is negligible. The frequency of
one of the normal modes of the system is,
3+ 2 k
(a)
2 M
3+ 3 k
(b)
2 M
k k
3+ 5 k M M
(c)
2 M
3+ 6 k
(d)
2 M
Ans: (c)
1 2 1 2
Solution: T = mx1 + mx 2 ,
2 2
V =
1 2 1
2
1 1 1
(
kx1 + k ( x 2 x1 ) = kx12 + k x 22 + x12 2 x 2 x1 = k 2 x 2 + x 2 2 x 2 x1
2
2
2 2 2
) ( )
m 0 2k k
T = ; V =
0 m k k
2k 2 m k 3+ 5 k
= 0 ( 2 k 2 m )( k 2 m ) k 2 = 0 =
k k m2
2 m
GATE- 2014
Q21. If the half-life of an elementary particle moving with speed 0.9c in the laboratory frame is
5 10 8 s, then the proper half-life is _______________ 10 8 s. c = 3 10 8 m / s ( )
Ans: 2.18
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t0 v2
Solution: t = , t0 = t 1 2 = t0 = 5 10 8 .19 2.18 108 s
v2 c
1
c2
Q22. Two masses m and 3m are attached to the two ends of a massless spring with force
constant K . If m = 100 g and K = 0.3 N / m , then the natural angular frequency of
oscillation is ________ Hz .
Ans: 0.318
1 k m1.m2 3m.m 3m 4k
Solution: f = = = = = = 2 = 0.318 Hz
2 m1 + m2 4m 4 3m
Q23. The Hamiltons canonical equation of motion in terms of Poisson Brackets are
(a) q = {q, H }; p = {p, H } (b) q = {H , q}; p = {H , p}
2V 2a 4ar02 2a 4ar02 2a
= 3 + 5 = 3 + 5 = 3
r 2 r r r0
r0 r0 r0
2V
r 2 mr03
= T = 2
r0
m 2a
Q26. A planet of mass m moves in a circular orbit of radius r0 in the gravitational potential
k
V (r ) = , where k is a positive constant. The orbit angular momentum of the planet is
r
(a) 2r0 km (b) 2r0 km (c) r0 km (d) r0 km
Ans: (d)
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J 2
k dVeffect J 2
k
Solution: Veffctive = 2
= 3 + 2 =0 at r = r0
2mr r dr mr r
so J = r0 km
Q27. Given that the linear transformation of a generalized coordinate q and the corresponding
momentum p ,
Q = q + 4ap
P = q + 2p
is canonical, the value of the constant a is _________________
Ans: 0.5
Q P Q P
Solution: . . = 0 1.2 4a.1 = 0 a = 0.5
q p p q
p2 q2
Q28. The Hamiltonian of particle of mass m is given by H = .which one of the
2m 2
following figure describes the motion of the particle in phase space?
(a) (b)
p p
q q
(c) (d)
p p
q q
Ans: (d)
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GATE- 2015
Q29. A satellite is moving in a circular orbit around the Earth. If T ,V and E are its average
kinetic, average potential and total energies, respectively, then which one of the
following options is correct?
(a) V = 2T ; E = T (b) V = T ; E = 0
T T 3T T
(c) V = ;E = (d) V = ;E =
2 2 2 2
Ans.: (a)
n +1
Solution: From Virial theorem T = V where V r n +1
2
k 1
V = V n = 2 V = 2 T
r r
Q30. In an inertial frame S , two events A and B take place at (ct A = 0, rA = 0) and
(ct B = 0, rB = 2 y ) , respectively. The times at which these events take place in a frame
S moving with a velocity 0.6cy with respect to S are given by
3
(a) ct A = 0; ct B = (b) ct A = 0; ct B = 0
2
3 1
(c) ct A = 0; ct B = (d) ct A = 0; ct B =
2 2
Ans.: (a)
Solution: Velocity of S ' with respect to S is v = .6c
v
tA y
t A' = c2 for event A t A = 0, y = 0 so ct A' = 0
v2
1 2
c
v
tB y
c2 3
t B' = for event B t B = 0, y = 2 so ct B' =
v2 2
1 2
c
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Q31. The Lagrangian for a particle of mass m at a position r moving with a velocity v is given
m
by L = v 2 + Cr .v V (r ) , where V (r ) is a potential and C is a constant. If pc is the
2
canonical momentum, then its Hamiltonian is given by
1 1
(a) ( pc + Cr )2 + V (r ) (b) ( pc Cr )2 + V (r )
2m 2m
pc2 1 2
(c) + V (r ) (d) pc + C 2 r 2 + V (r )
2m 2m
Ans.: (b)
m 2
Solution: L = v + Cr.v V ( r ) where v = r
2
m 2
H = r pc L = rpc L where L = r + Cr.r V ( r )
2
L p Cr
= pc = ( mr + Cr ) r = c
r m
2
p Cr m pc Cr pc Cr
H = c pc cr +V (r )
m 2 m m
2
p Cr m pc Cr
H = c ( pc Cr ) +V (r )
m 2 m
( p Cr ) ( p Cr )
2 2
1
+V (r ) ( pc Cr ) + V ( r )
2
H = c c H =
m 2m 2m
Q32. The Hamiltonian for a system of two particles of masses m1 and m2 at r1 and r2 having
1 1 C
velocities v1 and v2 is given by H = m1v12 + m2v22 + z ( r1 r2 ) , wrong where
( r1 r2 )
2
2 2
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Solution: Lagrangian is not function of time so energy is conserve and component of ( r1 r2 ) are
Q33. A particle of mass 0.01 kg falls freely in the earths gravitational field with an initial
velocity (0) = 10ms 1 . If the air exerts a frictional force of the form, f = kv , then for
k = 0.05 Nm 1 s , the velocity (in ms 1 ) at time t = 0.2 s is _________ (upto two decimal
m 0.05 .05
ln 10 u ln 10 10 = 0.2
k 0.01 .01
m
k
{ln (10 5u ) ln ( 40 )} = 0.2
ln ( 40 ) can not be defined. So given data are not correct.
Q34. Consider the motion of the Sun with respect to the rotation of the Earth about its axis. If
Fc and FCo denote the centrifugal and the Coriolis forces, respectively, acting on the
Sun, then
(a) Fc is radially outward and FCo = Fc
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Q35. A particle with rest mass M is at rest and decays into two particles of equal rest masses
3
M which move along the z axis. Their velocities are given by
10
(a) v1 = v 2 = (0.8c )z (b) v1 = v 2 = (0.8c )z
(c) v1 = v 2 = (0.6c )z (d) v1 = (0.6c )z; v 2 = ( 0.8c )z
Ans.: (b)
3 3
Solution: M M+ M
10 10
From momentum conservation
0 = P1 + P 2 P1 = P 2 P1 = P2
From energy conservation E = E1 + E2
3 Mc 2 3 Mc 2 3 Mc 2
Mc 2 = + Mc 2 =
10 v 2 10 v2 5 v2
1 2 1 2 1 2
c c c
v2 9 v 2 16
1 2 = 2 = v = 0.8c
v 25 v 25
GATE-2016
Q36. The kinetic energy of a particle of rest mass m0 is equal to its rest mass energy. Its
m0c 2
Solution: = 2m0 c 2 E
2
v
1
c2
E 2 = p 2 c 2 + m02 c 4 4m02c 4 m02c 4 = p 2 c 2 p = 3m0 c = 1.732m0 c
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Q37. In an inertial frame of reference S , an observer finds two events occurring at the same
time at coordinates x1 = 0 and x 2 = d A different inertial frame S moves with velocity
v with respect to S along the positive x -axis. An observer in S also notices these two
events and finds them to occur at times t1 and t 2 and at positions x1 and x2 respectively.
1
If t = t 2 t1 , x = x 2 x1 and = , which of the following statements is true?
v2
1 2
c
d
(a) t = 0, x = d (b) t = 0, x =
vd vd d
(c) t = , x = d (d) t = , x =
c2 c 2
Ans.: (c)
vx vx
t2 22 t1 21
Solution: t2' t1' = c c t ' = t vx it is given t = 0, x = d
v2 v2 c2
1 2 1 2
c c
vx
t ' =
c2
x vt2 x1 vt1
x2 x1 = 2
' ' x ' = ( x vt ) it is given t = 0, x = d
v2 v2
1 2 1 2
c c
x ' = d
Q38. The Lagrangian of a system is given by
L=
1 2 2
2
[ ]
ml + sin 2 2 mgl cos , where m, l and g are constants.
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fiziks
InstituteforNET/JRF,GATE,IITJAM,JEST,TIFRandGREinPHYSICALSCIENCES
Ans.: (a)
L
Solution: is cyclic coordinate so = p ml 2 sin 2 is constant hence m, l and g are
M 450
600
2
Ans.: 1.40
Solution: p 2 c 2 + M 2 c 4 = E1 + E2 = 1.82GeV
p 2 c 2 + m 2c 4 = 3.312
Headoffice Branchoffice
fiziks,H.No.23,G.F,JiaSarai, AnandInstituteofMathematics,
NearIIT,HauzKhas,NewDelhi16 28B/6,JiaSarai,NearIIT
Phone:01126865455/+919871145498 HauzKhas,NewDelhi16
Website:www.physicsbyfiziks.com
Email:fiziks.physics@gmail.com18