Date Planned : __ / __ / __ Daily Tutorial Sheet -1 Expected Duration : 90 Min

Actual Date of Attempt : __ / __ / __ Level -1 Exact Duration :_________

+ −

− +

+
+ + +

DTS - 1 1 Level -1 | Dynamics of a Particle

  
 
 
 

=  =  =  = 

DTS - 1 2 Level -1 | Dynamics of a Particle

 

DTS - 1 3 Level -1 | Dynamics of a Particle

 Date Planned : __ / __ / __ Daily Tutorial Sheet -2 Expected Duration : 90 Min
Actual Date of Attempt : __ / __ / __ Level -1 Exact Duration :_________

−
=

=

−
=


−
  

 

 = = 

=  = 

DTS - 2 1 Level -1 | Dynamics of a Particle

 − − −


 

 −  − 

 +   − 

 
( − ) 

− 

 

DTS - 2 2 Level -1 | Dynamics of a Particle

 Date Planned : __ / __ / __ Daily Tutorial Sheet -3 Expected Duration : 90 Min
Actual Date of Attempt : __ / __ / __ Level -1 Exact Duration :_________

  


−

−
=

−
  −
  −
  −
 
       
       
       

−
=

−
=

− −

−
=

−
=

DTS - 3 1 Level -1 | Dynamics of a Particle

 

 l

m




 M
 


− +

+ −

−
=

   

( )
− −
 
 
 
 

DTS - 3 2 Level -1 | Dynamics of a Particle

 Date Planned : __ / __ / __ Daily Tutorial Sheet -4 Expected Duration : 90 Min
Actual Date of Attempt : __ / __ / __ Level -1 Exact Duration :_________

DTS - 4 1 Level -1 | Dynamics of a Particle

 =− =−

= =−

DTS - 4 2 Level -1 | Dynamics of a Particle

 + + = + + =

+ + = + + =

− − − −

DTS - 4 3 Level -1 | Dynamics of a Particle

 Date Planned : __ / __ / __ Daily Tutorial Sheet -5 Expected Duration : 90 Min
Actual Date of Attempt : __ / __ / __ Level -1 Exact Duration :_________

+  −   

−  + 

+  

+ +  

+ 

− 

− ( − ) − −

DTS - 5 1 Level -1 | Dynamics of a Particle

 + − + −

− − + +

( + )
+ −
+ ( + )
 −

=
= = =

L
(  )
 m

 ( − )  ( − )  ( − ) 

  



DTS - 5 2 Level -1 | Dynamics of a Particle

 − −

DTS - 5 3 Level -1 | Dynamics of a Particle

 Date Planned : __ / __ / __ Daily Tutorial Sheet -6 Expected Duration : 90 Min
Actual Date of Attempt : __ / __ / __ Level -2 Exact Duration :_________

= + = +

= + = +

=


     
  +    +    
      
   

=  = 

=  =  =  = 

   

DTS - 6 1 Level -2 | Dynamics of a Particle

 +

T
m x

m N

DTS - 6 2 Level -2 | Dynamics of a Particle

 Date Planned : __ / __ / __ Daily Tutorial Sheet -7 Expected Duration : 90 Min
Actual Date of Attempt : __ / __ / __ Level -2 Exact Duration :_________

 
 −  + +
  +
+  + 

= 

   

A
B
C

  = B A
   = − T
P
  = F2 F1

 

B
  A
 

  ( + ) 


  
 +

DTS - 7 1 Level -2 | Dynamics of a Particle

 = =

=

m
M
37

=

DTS - 7 2 Level -2 | Dynamics of a Particle

 Date Planned : __ / __ / __ Daily Tutorial Sheet -8 Expected Duration : 90 Min
Actual Date of Attempt : __ / __ / __ Level -2 Exact Duration :_________

   
   

 =

 
  −

 
−  +

DTS - 8 1 Level -2 | Dynamics of a Particle

 

 

 

 

= 





=

=
−

−
=

DTS - 8 2 Level -2 | Dynamics of a Particle

 Date Planned : __ / __ / __ Daily Tutorial Sheet -9 Expected Duration : 90 Min
Actual Date of Attempt : __ / __ / __ Level -2 Exact Duration :_________

= =




 + 

 = 
= 

ration of wedge is:

 
=  =
+ 
 
= = 
+

DTS - 9 1 Level -2 | Dynamics of a Particle

 =

+ − + −

= =


−
 =

DTS - 9 2 Level -2 | Dynamics of a Particle

 Date Planned : __ / __ / __ Daily Tutorial Sheet -10 Expected Duration : 90 Min
Actual Date of Attempt : __ / __ / __ Level -2 Exact Duration :_________

−
=

DTS - 10 1 Level -2 | Dynamics of a Particle

   




DTS - 10 2 Level -2 | Dynamics of a Particle

  

= 

=

DTS - 10 3 Level -2 | Dynamics of a Particle

 Date Planned : __ / __ / __ Daily Tutorial Sheet -11 Expected Duration : 90 Min
Actual Date of Attempt : __ / __ / __ Numerical Value Questions Exact Duration :_________
for JEE Main

2 rad s−1.
=

DTS - 11 | Numerical Value Questions for JEE Main 1 Dynamics of a Particle

 −
= +

=− +

DTS - 11 | Numerical Value Questions for JEE Main 2 Dynamics of a Particle

 JEE Main (Archive) Level -1

− − −

= = =
−

+ + −

   

JEE Main (Archive) 124 Dynamics of a Particle

 = =

−
=

 

=− =

 = −  = −  = −  = −

 = =

=  =  =  = 

JEE Main (Archive) 125 Dynamics of a Particle

 −
=

+
+ +

= ( )= −

()


=

JEE Main (Archive) 126 Dynamics of a Particle

 = =

= + =

= − =+
   
 −  − + − − + 
   

JEE Main (Archive) 127 Dynamics of a Particle

 JEE Advanced (Archive) Level -2

 

− − − −

 

 

 
( ) ( )

  +   +
+  
 

JEE Advanced (Archive) 128 Dynamics of a Particle

 


   


(− ) (+ )

(− ) ( )
−
   
− − −

JEE Advanced (Archive) 129 Dynamics of a Particle

 = =

 

=

     

=  

=  + 

=   

=   + 

− + (
   )
=

− =

= 

= 

−
=
= 

 
 
 

JEE Advanced (Archive) 130 Dynamics of a Particle

 =



 

= =

JEE Advanced (Archive) 131 Dynamics of a Particle

 =

=

= =

JEE Advanced (Archive) 132 Dynamics of a Particle

 −

 




( )=− = −
= =

JEE Advanced (Archive) 133 Dynamics of a Particle

 Level -3 JEE Advanced++ Pattern/Solutions

2g
141. sin
3a
mg mvdv
F cos   ma  cos(as ) 
3 ds
v s
g v2 g 2g


0
vdv 
3 
0
cos as ds 
2

3a
sin as  v
3a
sin 

142.(i) 4.2 kg (ii) 10N

The block M1 is moving with constant speed

M1g  T  0  T  M1g …… (i)

The Block M 2 is also moving with constant speed, we get

T  T   M 2g sin   f  T   M 2 g sin   N 2  f  N  M g cos 

 2 2 2 
 T  T   M 2g sin   M 2 g cos  …… (ii)

For M 3 we get

T   f 3  M3  0  T   f 3  M 3 g …… (iii)

Putting the value of T and T  from (i) and (iii) in (ii) we get
M1g  M 3 g  M 2g sin   M 2 g cos 

After substituting the values, we get, M1  4.2 kg

The tension in the horizontal string will be
T   M 3 g  0.25  4  10  T   10 N

2 2 mg
143. (i) zero (ii) mg (iii)
3 3 2
(i) For finding direction of friction first assume there is no friction anywhere. In the absence of
friction the block B will move down the plane and the block A will move up the plane. Frictional
force opposes this motion.
F.B.D. of the block A
 T  mg sin 45  f 1  ma … (1)

From F.B.D of B
and 2 mg sin 45  f 2  T  2 ma … (2)

Adding (1) and (2), we get

Dynamics of a Particle 1 DTS -12 | Solutions
 
mg sin 45  f 1  f 2  3 ma 
For a to be non-zero mg sin 45° must be greater
than the maximum value of  f1  f 2 
4
  f1  f 2 max  1N1  2N 2   1m1  22m 2  g cos 45  3 mg cos 45
 mg sin 45  f1  f 2  max ; Hence block will remain stationary

(ii) F.B.D. of the block B

1 2
f 2(max)  2 N 2  2mg cos 45  mg
3 3 2
Component of weight of B parallel to inclined plane
2mg
W B  2 mg sin 45  2
2 mg sin 45  f 2

 max  , therefore, block B has tendency to slide down the plane.
For block B to be at rest

mg  2  4mg 2 2
T  f2  2mg sin 45  T  2     T  mg
 max  2 3  3 2 3

(iii) Component of weight of A parallel to inclined plane

mg
W  A  mg sin 45  2

As T  W  A
Hence block A has tendency to move up the plane, therefore frictional force on the block A will be
down the plane.
For A to be at rest
F.B.D. of A mg sin 45  f  T

2 2mg mg mg
 f  T  mg sin 45    f 
3 2 3 2

 3
144. (i) 1m / s (ii)   s
3 4 
 

 
The particle of mass m  10 2 kg is moving along positive x-axis

K  dv  K
(i) F x    or mv  
2x 2  dx  2x 2
 
v x K
Now integrating both side, m
0
vdv  

1 2x 2
dx

Dynamics of a Particle 2 DTS -12 | Solutions

 x
m 2 K  K 1 
or      1
x 
2  2x  1 2  

K 1  K 1 
or v2    1 or v    1 (1)

m x  
m x 
 

K 1  K 10 2
When x  0.5 m , v    1    1
m  0.5 
 m 10 2
As the force is acting along negative x-direction, therefore, the velocity will be in  x direction.

Hence v   1m / s and v   1i m / s .
K dx 1 x
(ii) As  1m / s , hence from (1) v   [we have chosen–sign because velocity is in –x
m dt x
direction]
x 0.25 x t

1 x
dx   dt 
1 1 x
dx 
  dt
0
2
Put x  sin , dx  2 sin  cos  d 
/6
So
/2
2 sin 2  d    t cos 2  1  2 sin 2  ; 2 sin 2   1  cos 2

/6
/6  sin 2 
/2 1  cos 2 d    t ,  


2  /2
 t

 1   1   3
  sin     sin     t  t    sec
6 2 3   2 2  3 4 
   
10 
145.
3g

Acceleration of ball and rod are as shown (from constraint relation).

For ball : 2T  mg  ma . . . . (i)
For rod : mg  T  m (2a ) . . . . .(ii)
(i) + (ii) × 2
2mg – mg = ma + 4ma
g
 a 
5
  
a BR  a B  a R  a  ( 2a )  3a (taking upward direction as +ve)

S BR   

1 2 2
  (3a )t 2  t  
2 3a 3( g / 5)

1
146. cot 1( k )
2
ma  mg sin   k mg cos   a  g (sin   k cos  )
1  2
at 2   t2 
2 cos  
g cos  sin   k cos  
For least time, denominator should be maximum

Dynamics of a Particle 3 DTS -12 | Solutions

 d 
 sin  cos   kcos 2   0
d   
 sin (  sin  )  cos (cos  )  k (2 cos  ) (  sin  )  0

  sin 2   cos2   k 2 sin  cos   0

 cos 2  k sin 2  0
 cot 2   k
1
  cot 1( k )
2

Dynamics of a Particle 4 DTS -12 | Solutions

 Level -3 JEE Advanced++ Pattern/Solutions

4 16
147. (i) m/s2 (ii) N
3 3
Equation of motion for mass M1
T  M1g sin 37  f 1  M1a
T  M1g sin 37  1M1g cos 37  M1a …… (i)
Equation of motion of mass M 2
M 2 g sin 37  T  f 2  M 2a
M 2 g sin 37  T  2 M 2 g cos 37  M 2a …… (ii)
Adding (i) and (ii)
    
M1  M 2 g sin 37  1M1g  2 M 2 g cos 37  M1  M 2 a 
 4  210  0.6  0.75  4  10  0.25  2  10   0.8   4  2 a
8 4
36  30  5  0.8  6a  ms 2
a  
6 3
Substituting this value in equation (i), we get
16 16
T  24  24   T  N
3 3

3
148. (i) f1 = 30 N, f2 = 15 N (ii) F = 60 N, T = 18 N, a  m/s2
5
Free body diagram of M is shown in the figure below.

(i) Supposing all the blocks are in motion

f1  1N1  1m 1g  0.3  20  10  60 N
max

and f 2  2 N 2  2m 2 g  0.3  5  10  15 N

max

Friction between m 2 and ground will be maximum which is 15 N. Given that f1  2f 2 ,

so f 1  2  15  30 N

The block m 1 can not move on M.

Let all the blocks are at rest, then

For M : F  f 1  0, for m 1 : T  f 1  0 and for m 2 : T  f 2  0 which gives f 1  f 2 which

does not satisfy the given condition
Since m 1 cannot move over the block M so m 2 can’t move relative to M, therefore all

the blocks move together and T  f 1  30 N and f 2  15 N

Dynamics of a Particle 1 DTS -13 | Solutions

 for m 1 : 30  T  20 a (i)
for M : F  30  50 a (ii)
and for m 2 : T  15  5 a (iii)

3
After solving these equations, we get ; a  m / s 2; T  18 N and F  60 N
5

mg ( sin  cos )
149. (i) tan 1 (ii)
1  2
At the moment of just sliding,
F cos   mg sin   N
mg(sin    cos  )
 F cos   mg sin   (mg cos   F sin )  F  . . . (i)
(cos    sin )
d
For Fmin : [cos    sin ]  0  tan       tan 1 
d
mg(sin    cos  )
Fmin  [from (i)]
1  2

3mg 9mg
150. (i) (A) (B)
a a
Friction between B and surface appears first.
( f BS )max  (3mg )
Slipping between B and surface starts when F  ( f BS )max
3mg
 at  3mg  t 
a
Maximum acceleration with which A and B move together is
2mg
a max   2g ( ( f AB )max  2mg )
m
Slipping between A and B starts after this
 Fmax  3mg  3ma max
mg
 Fmax  3mg  6mg  9mg  t 
a

2mg
(ii) (a) (B) Never
a
( f AB )max  2mg
( f BS )max  3mg
 ( f AB )max  ( f BS )max
 B will never slip
A slips on B when
2mg
F  ( f AB )max  at  2mg  t 
a

Dynamics of a Particle 2 DTS -13 | Solutions

 4ML
151.

 M m g 
V = initial vol of both the blocks.
Due to friction from the ground, M tends to retard and hence its
velocity becomes less than that of m. So m exerts forward friction on
M.

mg  ma1 (a1 and a2 are towards left)
2


 m M g  2
mg  Ma 2

 a1 
g
, a 2  g
 M  m / 2
2 M
g g
a rel  a12  a1  a 2 
2
    M  M  m / 2  
g 
 M  m g
Let us take left-positive : a rel 
2M
 M  2 M  m   2M
1
Srel  a rel t 2 (left-positive)
2

L 

1  M  m  gt 2  t 
4 ML
2 2M 
 M m g 
152. g (sin   2  cos  )

2N cos  mg cos 
4

mg sin   2N  ma

 a  g sin   2g cos 

Dynamics of a Particle 3 DTS -13 | Solutions

 Level -3 JEE Advanced++ Pattern/Solutions

g g
153. (i) h  R (ii)
2 R

(i) Let m be the mass of the particle and N the normal reaction acting on it. Resolving N in horizontal
and vertical direction
rw 2 r r 2 g
N cos   mg, N sin   mr 2  tan       R h
g R h g 2
g
 h R
2
g g
(ii) When h  0, R  0  
2 R

When  is greater than this value, then h has a
positive value.

g
Therefore   , in order to have a non-zero
R
g
positive value of h.  min 
R

154. (i) 36 N (ii) 11.67 rad/s (iii) 0.1 m

(i) The tension T of the string acting on m 2 is responsible for rotation of the mass m 2 .

 T  m 2r 22  5  0.176  10  10  88N

Since m 1 is also rotating in a circular path, the required centripetal force on m 1

 Fc  m 1r 12  10  0.124  10  10  124 N

Out of this 124 N required, 88 N will be provided by the tension T in the string and rest will be
provided by frictional force between block and surface. Therefore frictional force acting on m 1

f  124  88  36N
(ii) Let  be the maximum angular speed for which no slipping of masses occurs (or we may say
that  is the minimum angular speed for which slipping occurs)
f max  m 1g  0.5  10  9.8  49 N

The equation for m 1 to move in circular motion is T  f  m 1r 12

When f  f max then T  T  (say ) and T   m 2r 22 (for mass m 2 ) and   

f max
 m 2r 22  f max  m 1r 12   
m 1r 1  m 2r 2

49
    11.67 rad / s
10  0.124  5  0.176
(iii) For no friction force acting on mass m 1
The tension should be sufficient to provide centripetal force for both the masses. Then

Dynamics of a Particle 1 DTS -14 | Solutions

 r1 m2 5 1
T  m 1r 12 and T  m 2r 22     … (i)
r2 m1 10 2

But r 1  r 2  length of thread  r 1  r 2  0.3 m … (ii)

Solving equation (i) and (ii) we get

r 1  0.1m and r 2  0.2 m

b  0 gb
155. (i) (ii)
2 2
When the speed is maximum for a particular radius, then
v2  r 
g   v 2  gr  0 gr 1  
r  b 

d (v 2 )
To find maximum value of v, let us put 0
dr
b 0 gb
This gives r  and v2  .
2 4
b 0 gb
So, for maximum velocity, r  and v max  .
2 2
1/4
 2 2
 g  1
156.  2     
 
 r    
 
Before slipping occurs, the block is undergoing a circular motion with constant 

a t  r , a r  2r  (t )2r  a net  a r2  a t2  ( 2rt 2 )2  (r )2

This acceleration in horizontal plane is provided by friction. So

f  ma  m ( 2rt 2 )2  ( r )2
Slipping starts when.

f  f max  N  mg  m ( 2rt 2 )2  ( r )2  mg

1/4 1/4
 2g 2  2r 2   2 2
 g  1
 t         
  4r 2   2r  
      

rg  l r  
157. (i) 1  cos  (ii) sin 1  1  cos l 
l  r l  r  

(i) Initial tangential acceleration of all elements will be same
For the element of mass dm :
m  m
 T  dT   dm  g sin   T  dm  a t  dT   rd   g sin   rd at
 l  l
Integrating between 1 and 2
T2 l /r l /r
m m


T1
dT 
l
rg

0
sin  d  
l
rat
 d
0

m  l m l
  T2  T1   l
rg 1  cos  
r l
rat
r


Dynamics of a Particle 2 DTS -14 | Solutions

 But at free ends tension is zero
rg  l
 T1  T2  0  at  1  cos 
l  r
(ii) Integrating between 1 and  :

m mr  rg  l 
T  0  rg 1  cos     1  cos  
l l  l  r  
Differentiating w.r.t.  :

dT m mr 2  l
0 rg sin   g 1  cos 
d l l2  r

dT r 
For Tmax , 0  sin   1  cos l 
d l r

Mgcot
158.
2

d
2T sin  Td  towards centre.
2

dN  sin   dm  g  

 
T d
 cot 
dN cos   Td    
dm g

Td 
  cot 
M
2R
 
R d g

2T Mg cot 
  cot   T 
Mg 2

Dynamics of a Particle 3 DTS -14 | Solutions

 Level -3 JEE Advanced++ Pattern/Solutions

2mM
159.
( M  m )t 2

T  Mg  Ma1

T  mg  ma 2

1 1
2
a 21t 2  S21 
2
a 2  a1  t 2  
1  T  T  2mM
   g    g  t 2    T 

2  m    ( M  m )t 2
 M 

160.(3)

From constaints between A and B,

(1) a sin   a 0 cos 

From Newton’s Laws

9m
(2) N sin   a
64
(3) 2mg  T  N cos   2ma 0

ma 0
(4) 2T  mg 
2

Solve to get a  8m/s2 and a 0  6m/s2

 Acceleration of block C  3m/s2.

161. (i) g/2 down the plane, g , g 

(ii) 0, 0, 2g 

70
162.
13
Let x1 to x 5 be displacement of each body shown from
the ceiling, measured downward positive.
Also, let us assume acceleration of all bodies in the
downward direction (a1 to a 5 )
Newton’s second law
(1) 10 g  T  10a 3 (2) 5g  2T  5a 4 (3) 10 g  2T  10a 5

Constraint equations
x1  x 2  constant  a1  a 2  0

Dynamics of a Particle 1 DTS -15 | Solutions

 ( x 3  x1 )  ( x 4  x1 )  ( x 4  x 2 )  ( x 5  x 2 )  x 5  constant
 2a1  2a 2  a 3  2a 4  2a 5  0
Combining the two equations:  a 3  2a 4  2a 5  0.
7 g
Solve to get : a4 
13
7g
 Acceleration of 5 kg block is m/s2 in upwards direction.
13

2 M  m g
163.
 3M  2m 

FBD of A and C are w.r.t. ground and FBD of B is w.r.t. A

For C : mg  N1  ma1 …(1) [in vertical]

N2
For A :  Ma 2 …(2) [in horizontal]
2
Ma 2 Mg N1
For B : N 2    …(3) [perpendicular to incline]
2 2 2
Mg N1 Ma 2
   Mar …(4) [parallel to incline]
2 2 2
Acceleration of B w.r.t ground is
a2
 a r sin 45  a1 (from constraint between B and C) 45°

 a r  2a1 …(5) ar

Putting N 2 from (2) in (3):

Ma 2 Mg N1
2 Ma 2   
2 2 2
 Mg  N1  3Ma 2 …(6)

Putting a r from (5) in (4) :

Mg N1 Ma 2
   M 2 a1
2 2 2
 Mg  N1  2Ma1  Ma 2 …(7)

(1) × 4 + (6) + (7) × 3:

2 M  m  g
4 mg  4 Mg  4 ma1  6 Ma1 ;  a1 
 3M  2m 

Dynamics of a Particle 2 DTS -15 | Solutions

  2 2 2 2 
 v 2 v1  v2 v1 v1  v 2 
164.  , 
 2g 2g 
 
Relative retardation a r  g

vr
 Time when slipping will stop is t 
ar

v12  v 22
or t  … (i)
g

vr2 v12  v 22
sr  
2a r 2g

 v2  v2   2 2
x r  sr cos     1 2 
v2   v 2 v1  v2
 2g 
 2 2  2g
   v1  v2 


 v2  v2   2 2
yr  sr sin    1 2 
v1   v1 v1  v2
 2g 
 2 2  2g
   v1  v2 


v 2 v12  v 22
In time t, belt will move a distance s  v 2t or in x-direction
g

v2 v12  v22 v1 v12  v22

Hence, coordinate of particle, x  x r  s  and y  yr 
2g 2g

Dynamics of a Particle 3 DTS -15 | Solutions

 Miscellaneous Question Bank Level -1, 2, 3

1. A body of mass 10 kg is being acted on by a force of 3t2 N and an opposing constant force of 32 N. The
initial speed is 10 m/s. The velocity of the body after 5 seconds is:
(A) 14.5 m/s (B) 6.5 m/s (C) 4.5 m/s (D) 3.5 m/s

2. A 5 kg sphere is accelerated upwards by a string whose breaking strength is 20 kg wt. The maximum

acceleration with which the sphere can move without the string breaking is: (Take g  10 ms 2 )
(A) 10 m/s2 (B) 15 m/s2 (C) 30 m/s2 (D) 50 m/s2

3. A cricketer hits a ball of mass 1 kg coming towards him with a velocity of 20 m/s and the ball bounces
back with the same speed. If the time of impact is 1/50 second, the average force exerted is:
(A) 500 N (B) 1000 N (C) 2000 N (D) 4000 N

4. The mass string system shown is in equilibrium. If the coefficient of

friction between A and the table is 0.3, the frictional force on A is:
(A) 9.8 N (B) 2.94 N
(C) 1.96 N (D) 0.59 N

*5. A block is placed over a plank. The coefficient of friction between the block and plank is s  0.3 and

k  0.2 . Initially both are at rest, suddenly the plank starts moving towards east with acceleration a0 =

4m/s2. The displacement of the block in 1 s is: (Take g  10 ms 2 )

(A) 1 m towards east relative to ground (B) 1 m towards west relative to plank
(C) zero relative to plank (D) 2 m relative to ground

*6. A monkey of mass m kg slides down a light rope attached to a fixed spring balance, with an acceleration
a. The reading of the spring balance is W kg. (g = acceleration due to gravity)
(A) The force of friction exerted by the rope on the monkey is m (g – a) N
Wg
(B) m 
g a

 a
(C) m  W 1  
 g 

(D) The tension in the rope is WgN

PARAGRAPH FOR QUESTIONS 7 - 8

A rod of length  (< 2R) is kept inside a smooth spherical shell as shown in
figure. Mass of the rod is m.

7. Keeping mass to be constant, if length of the rod is increased (but

always < 2R) the normal reactions at two ends of the rod:
(A) will remain constant (B) will increase
(C) will decrease (D) may increase or decrease

8. The normal reaction when l = R is :

mg mg mg mg
(A) (B) (C) (D)
2 4 2 3 3

MEQB 161 Level -1, 2, 3 | Dynamics of a Particle

9. Two blocks of same mass m = 10 kg are on rough horizontal surface as
10kg 10kg
shown in figure. Initially tension in the massless string is zero and string
B A F
 
is horizontal. A horizontal force F  40 sin  / 6 t is applied as shown string
2  0.3 1  0.2
on the block A for time interval t = 0 to t = 6 sec. here F is in Newton and
t in second. Friction coefficient between block A and ground is 0.20 and
between block B and ground is 0.30. (Take g = 10 m/sec2). Match the
statements in column I with the time interval (in seconds) in column II.
COLUMN-I COLUMN-II
Friction force between block B and ground is zero in the time
(A) (P) 0<t<1
interval
(B) Tension in the string is non-zero in the time interval (Q) 1<t<3
(C) Acceleration of block A is zero in the time interval (R) 3<t<5
Magnitude of friction force between A and ground is decreasing in
(D) (S) 5<t<6
the time interval

10. A 4 kg block A is placed on the top of 8 kg block B which rests on a smooth table. A just slips on B when
a force of 12 N is applied on A. Then the maximum horizontal force F applied on B to make both A and B
move together, is x N. Find value of x.

11. A block of mass 5 kg is kept on a horizontal floor having coefficient of friction 0.09. Two mutually
perpendicular horizontal forces of 3 N and 4 N act on this block. The acceleration of the block is x m/s2.
Find value of x. (Take g  10 ms 2 )

1
12. A chain is lying on a rough table with a fraction   of its length hanging down from the edge of the
n 
 
table. It is just on the point of sliding down from the table, then the coefficient of friction between the
table and the chain is:
1 1 1 n 1
(A) (B) (C) (D)
n n 1 n 1 n 1

13. A block of mass M lying on a rough horizontal plane, is acted on by a

horizontal force P and another force Q inclined at an angle  to the vertical.
The block will remain in equilibrium, if the coefficient of friction between the
block and surface is more than:
P  Q sin  P cos   Q
(A) (B)
Mg  Q cos  Mg  Q sin 
P  Q sin  P sin   Q
(C) (D)
Mg  Q sin  Mg  Q cos 

14. A block placed on a rough horizontal table is acted upon by an external force P. The graph of frictional
force f against external force P is:

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

MEQB 162 Level -1, 2, 3 | Dynamics of a Particle

15. A car starts from rest on cover a distance‘s’. The coefficient of kinetic friction between the road and tyres
is  . The minimum time in which the car can cover the distance‘s’ is proportional to:
1 1
(A)  (B)  (C) (D)
 

16. A string of negligible mass going over a clamped pulley of mass m supports a
block of mass M as shown in fig. The force on the pulley by the clamp is given
by:
(A) 2 Mg (B) 2 mg

 2   2 
(C) 
 M m

  m2 g

(D) 
 M m

  M2 g


*17. Two men of unequal masses hold on to the two sections of a light rope passing
over a smooth light pulley. Which of the following are possible?
(A) The lighter man is stationary while the heavier man slides down with
some acceleration
(B) The heavier man is stationary while the lighter man climbs up with
some acceleration
(C) The two men slide down with the same acceleration
(D) The two men move with accelerations of the same magnitude in opposite directions

PARAGRAPH FOR QUESTIONS 18 - 20

Study figure and answer the following questions accordingly. Neglect
all friction and masses of the pulleys.

18. What is the tension in the string?

700 450 500 900
(A) N (B) N (C) N (D) N
11 11 11 11

19. What is the acceleration of block B?

(A) 30 / 77 ms 2 (B) 60 / 77 ms 2 (C) 80 / 77 ms 2 (D) 120 / 77 ms 2

20. What is the acceleration of block A?

(A) 60 / 77 ms 2 (B) 80 / 77 ms 2 (C) 180 / 77 ms 2 (D) 60 / 77 ms 2

21. The rear side of a truck is open and a box of mass 20 kg is placed on the truck 4 m away from the open
end.   0.15 and g  10 m / s 2 . The truck starts from rest with an acceleration of 2 m/s2 on a straight
road. The box will fall off the truck when it is at a distance from the starting point equal to x meter. Find
value of x.

22. A block of mass 1 kg is stationary with respect to a conveyor belt that

is accelerating with 1 m/s2 upwards at an angle of 30° as shown in
figure. Which of the following statement(s) is (are) correct ?
(Take g  10 ms 2 )
(A) Force of friction on the block is 1 N upwards
(B) Force of friction on the block is 1.5 N upwards
(C) Contact force between the block and the belt is 10.5 N
(D) Contact force between the block and the belt is 5 3 N

MEQB 163 Level -1, 2, 3 | Dynamics of a Particle

23. A roller coaster is designed such that riders experience 'weightlessness' as they go round the top of a hill
whose radius of curvature is 20 m. The speed of the car at the top of the hill is between:
(Take g  10 ms 2 )

(A) 14ms 1 and 15ms 1 (B) 15 ms 1 and 16ms 1

(C) 16 ms 1 and 17ms 1 (D) 13ms 1 and 14ms 1

24. A particle moves in the X-Y plane under the influence of a force such that its linear momentum is

p(t )  A[i cos(kt )  j sin(kt )] , where A and k are constant. The angle between the force and the momentum
is:
(A) 0° (B) 30° (C) 45° (D) 90°

25. A man of mass m stands on a frame of mass M. He pulls on a light rope,

which passes over a pulley. The other end of the rope is attached to the
frame. For the system to be in equilibrium, what force must the man exert on
the rope?
1
(A)
2

M m g  (B) (M + m)g

(C) (M - m)g (D) (M + 2m)g

26. Assertion (A): Action and reaction acts on same body.

Reason (R): Newton’s third law states that action and reaction are equal.
(A) Both A and R are individually true and R is the correct explanation of A
(B) Both A and R are individually true and R is not the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true

27. A person is sitting in a lift accelerating upwards. Measured weight of person will be :
(A) Less than actual weight (B) Equal to actual weight
(C) More than actual weight (D) None of the above

*28. A block of mass m is placed on a rough horizontal surface. The coefficient of friction between them is .
An external horizontal force is applied to the block and its magnitude is gradually increased. The force
exerted by the block on the surface is R.
(A) The magnitude of R will gradually increase

(B) R  mg 2  1
(C) The angle made by R with the vertical will gradually increase
(D) The angle made by R with the vertical  tan 1 

*29. A simple pendulum with a bob of mass m is suspended from the roof of a car moving with a horizontal
acceleration a.
(A)  
The string makes an angle of tan 1 a / g with the vertical
 a
(B) The string makes an angle of tan 1 1   with the vertical
 g 


(C) The tension in the string is m a 2  g 2

(D) The tension in the string is m g 2  a 2

MEQB 164 Level -1, 2, 3 | Dynamics of a Particle

30. Figure represents a painter in a crate which hangs alongside a building. When the
painter of mass 100 kg pulls the rope, the force exerted by him on the floor of the

crate is 450 N. If the crate weighs 25 kg, find the acceleration (in ms 2 ) of the
painter.

31. Block A is given an acceleration 12 ms 2 towards left

as shown in figure. Assuming block B always remains

horizontal, find the acceleration (in ms 2 ) of B.

32. A block of mass m is placed on the inclined smooth surface of a wedge

of mass M as shown. An external force is applied on the wedge such the
block falls freely along vertical direction. This minimum value of F is:
(A) Mg cos  (B) Mg sin 

(C) Mg cot  (D) Mg tan 

33. A sphere is resting on a smooth V-shaped groove and subjected to a spring force
as shown in the figure. The spring is compressed to a length of 50 mm from its
free length of 100mm. If spring constant is k = 2N/mm. Then normal reaction R1
and R2 exerted at point 1 and 2 on sphere are :
(A) R1 = 131.24 N and R2 = 60 N (B) R1 = 131.24 N R2 = 80 N
(C) R1 = 121.24 N and R2 = 70 N (D) R1 = 121.24 N R2 = 80 N

34. Friction co-efficient between block and inclined surface is s . The

maximum value of M for which will remain equilibrium is:

(A) m (sin   s cos )

(B) 2m sin 
(C) 2m (sin   s cos )

(D) m sin 

35. A block of mass 2 kg is placed on large plank of mass 4 kg

placed on rough horizontal surface. A variable force F = 5t(N)
is acting on the plank. Then the friction force exerted by plank
on the block is best respected by :

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

MEQB 165 Level -1, 2, 3 | Dynamics of a Particle

PARAGRAPH FOR QUESTIONS 36 - 40

Consider the situation shown in the figure. (Take g  10 ms 2 )

36. Find acceleration of 1 kg.

50 70 35 120
(A) (B) (C) (D) m/s2
11 11 11 11

37. Find acceleration of 2 kg.

35 70 50
(A) (B) (C) 3 (D) m/s2
11 11 11

38. Find acceleration of 3 kg.

35 70 70
(A) (B) (C) (D) 3m/s2
11 24 11

39. Find tension T1 .

60 120 240 10
(A) N (B) N (C) N (D) N
11 11 11 11

40. Find tension T2 .

120 60 240 10
(A) N (B) N (C) N (D) N
11 11 11 11

41. Block B, of mass mB = 0.5 kg, rests on block A, with mass mA = 1.5 kg, which
in turn is on a horizontal tabletop (as shown in figure). The coefficient of
kinetic friction between block A and the tabletop is k  0.4 and the

coefficient of static friction between block A and block B is s  0.6 . A light

string attached to block A passes over a frictionless, massless pulley and

block C is suspended from the other end of the string. What is the largest
mass mC (in kg) that block C can have so that blocks A and B still slide
together when the system is released from rest?

42. A block of mass 4 kg is placed on rough surface and is being acted

upon by a time dependent force F = 4t(N) (where t in second). The co-
efficient of friction between the block and horizontal surface is   0.5 .

Then velocity of block at t = 10s is:

(A) 25 m/s (B) 50 m/s (C) 12.5 m/s (D) 10 m/s

MEQB 166 Level -1, 2, 3 | Dynamics of a Particle

43. Two block of mass 10 kg and 8 kg are linked by thread passing over a pulley.
A small block of mass 2kg is placed on 10kg block then normal reaction on 2
kg block is:

(A) 20 N

(B) 16 N

(C) 24 N

(D) 100 N

44. A body of mass 5 kg is suspended by a spring balance on a smooth

inclined plane as shown in figure. The spring balance measures:
(A) 50 N (B) 25 N
(C) 500 N (D) 10 N

45. In the given figure if both block A and B remain in

contact with ground then find the value of v if string
remains taut.
(A) 10 m/s (B) 8 m/s
16
(C) m /s (D) 16 m/s
3

*46. A block of weight W is suspended from a spring balance. The lower surface of the block rests on a
weighing machine. The spring balance reads W1 and the weighing machine reads W2. (W, W1, W2 are in
the same unit)
(A) W = W1 + W2 if the system is at rest
(B) W > W1 + W2 if the system moves down with some acceleration
(C) W1 > W2 if the system moves up with some acceleration
(D) No relation between W1 and W2 can be obtained with the given description of the system

47. Two blocks A and B of masses 1 kg and 2 kg respectively are placed on a smooth F = 2t N
horizontal surface. They are connected by a massless inextensible string going
over a pulley as shown. The pulley is being acted upon by a vertical force of
magnitude varying with time as F = 2t N. Which of the following represents the
velocity time variation of A and B?

v v B
A
A 1 kg 2 kg
(A) B (B) B
A

t t

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

t t
MEQB 167 Level -1, 2, 3 | Dynamics of a Particle
*48. Two blocks A and B of equal mass m are connected through a massless string and arranged as shown in
figure. Friction is absent everywhere. When the system is released from rest:
mg
(A) tension in string is
2
mg
(B) tension in string is
4
g
(C) acceleration of A is
2
3
(D) acceleration of A is g
4

49. Column-II gives certain situations involving two blocks of mass 2 kg and 4 kg. The 4 kg block lies on a
smooth horizontal table. There is sufficient friction between both the blocks and there is no relative
motion between both the blocks in all situations. Horizontal forces act on one or both blocks as shown.
Column-I gives certain statement related to figures given in column-II. Match the statement in Column-I
with the figure in column-II.
COLUMN-I COLUMN-II

(A) Magnitude of frictional force is maximum (P)

(B) Magnitude of friction force is least (Q)

(C) Friction force on 2 kg block is towards right. (R)

(D) Friction force on 2 kg block is towards left. (S)

50. An inclined plane makes an angle of 30° with the horizontal A groove OA
= 5m cut in the plane makes an angle of 30° with OX. A short smooth
cylinder is free to slide down under the influence of gravity. The time
taken by the cylinder to reach from A to O is W sec. Find W.

51. A suitcase is dropped on a conveyor belt moving at 3 m/sec. If the coefficient of friction between the belt
and the suitcase is 0.5, the displacement of the suitcase relative to the conveyor belt before slipping
between the two is stopped is W m. Find W.

52. A window scrubber is used to brush up a vertical window as shown in figure.

The brush weight 12 N and coefficient of friction of 0.15. Calculate minimum
value of F required to scrub the window.
(A) 15 N (B) 10.2 N
(C) 16.9 N (D) 18.1 N

MEQB 168 Level -1, 2, 3 | Dynamics of a Particle

53. A rope is stretched between two boats at rest. A sailor in the first
boat pulls the rope with a constant force of 100 N. First boat with
the sailor has a mass of 250 kg whereas the mass of second boat is
double of this mass (see figure). If the initial distance between the
boats was 30m. the time taken for two boats to meet each other is:
(neglect water resistance between boats and water)
(A) 20 s (B) 10 s (C) 5s (D) 15 s

54. A plumb bob is hung from the ceiling of a train compartment. The train moves on an inclined track of
inclination 30° with horizontal. Acceleration of train up the plane is a = g/2. The angle which the string
supporting the bob makes with normal to the ceiling in equilibrium is:
(A) 30° (B) tan 1(2 / 3 ) (C) tan 1( 3 / 2) (D) tan 1(2)

55. A main is raising himself and the crate on which he stands with an acceleration
of 5m/s2 by a massless rope – and-pulley arrangement. Mass of the man is 100
kg and that of the crate is 50 kg. If g = 10 m/s2, then the tension in the rope is :
(A) 2250 N (B) 1125 N
(C) 750 N (D) 375 N

56. Figure shows an arrangement in which three identical blocks are

joined together with an inextensible string. All the surfaces are
smooth and pulleys are massless. If aA, aB, and aC are the
respective accelerations of the blocks A, B, and C, then the value
of aB in terms of aA and aC is :
a A  aC
(A) a A  aC (B)
2
a A  aC
(C) (D) a A  aC
2

57. In the system in the figure, the friction coefficient between ground and bigger
block is  . There is no friction between both the blocks. The string connecting
both the blocks is light; all three pulleys are light and frictionless. Then the
minimum value of  so that the system remains in equilibrium is:
1 1 2 3
(A) (B) (C) (D)
2 3 3 2

PARAGRAPH FOR QUESTIONS 58 - 61

A block of mass m = 3.0 kg is experiencing two forces acting on it as F2
shown.
1 1 
(Take s  and k  . g = 10 m/s2). If F2 = 20 N and  = 60.
3 3 m F1
58. Determine the minimum value of F1 so that the block remains at rest.
(A) 2.68 N (B) 3.57 N (C) 7.32 N (D) 20 N

59. Determine the maximum value of F1 so that the block remains at rest.
(A) 10 N (B) 12.68 N (C) 17.32 N (D) 20 N

60. Calculate the magnitude and direction of frictional force on the block if F1 = 12 N.
(A) 2 N Right (B) 2 N Left (C) 4.68 N Left (D) 4.68 N Right

MEQB 169 Level -1, 2, 3 | Dynamics of a Particle

61. If F1 = 20 N, then:
(A) The block is at rest
(B) The force of friction on the block is 7.32 N in magnitude
(C) The force of friction on the block is 4.23 N in magnitude
(D) Block accelerates towards left

62. Two blocks A and B of masses 6 kg and 3 kg rest on a smooth horizontal

surface as shown in the figure. If coefficient of friction between A and B
is 0.4, the maximum horizontal force which can move them without
separation is:
(A) 72 N (B) 40 N
(C) 36 N (D) 20 N

63. A block of metal weighing 2 kg is resting on a frictionless plane. It

is struck by a jet releasing water at a rate of 1kg/s and at a speed
of 5m/s. The initial acceleration of the block is:
5 25 25 5
(A) m / s2 (B) m / s2 (C) m / s2 (D) m / s2
3 4 6 2

64. Find the least horizontal force P to start motion of any part of the system of
the three block resting upon one another as shown in the figure. The
weights of blocks are A = 300 N, B = 100 N and C = 200 N. Between A and
B, coefficient of friction is 0.3, between B and C is 0.2 and between C and
the ground is 0.1.
(A) 60 N (B) 90 N (C) 80 N (D) 70 N

65. Determine the time in which the smaller block reaches

other end of bigger block in the figure.
(A) 4s (B) 8s
(C) 2s (D) 1s

66. Two blocks A and B each of mass m are placed on a smooth

horizontal surface. Two horizontal force F and 2F are applied on
both the blocks A and B, respectively, as shown in the figure.
The block A does not slide on block B. then the normal reaction
acting between the blocks is:
F
(A) F (B) F/2 (C) (D) 3F
3

*67. A 10-kg block is placed on a horizontal surface. The coefficient of friction between them is 0.2. A
horizontal force P  15 N first acts on it in the eastward direction. Later, in addition to P a second
horizontal force Q = 20 N acts on it in the northward direction.
(A) The block will not move when only P acts, but will move when both P and Q act
(B) If the block moves, its acceleration will be 0.5 m/s2
(C) 
When the block moves, its direction of the force of friction acting on the block will be tan 1 4 / 3 
east of north
(D) When both P and Q act, the direction of the force of friction acting on the block will be

 
tan 1 3 / 4 west of south

MEQB 170 Level -1, 2, 3 | Dynamics of a Particle

*68. A block of mass 1 kg is at rest relative to a smooth wedge moving leftwards
with constant acceleration a = 5 m/s2. Let N be the normal reaction between

the block and the wedge. Then, (Take g  10 ms 2 )

(A) N 5 5N (B) N = 15 N

(C) tan   1/ 2 (D) tan   2

PARAGRAPH FOR QUESTIONS 69 - 70

The coin is rotating with a plate without sliding. The coefficient of friction
between the coin and plate is   0.75 .

g
69. If the angular frequency of rotation of the plate is   , the friction force acting on coin is:
2R
3 mg mg mg
(A) mg  (B)  (C)  (D) 
4 4 2 2

70. If the plate is rotating alone and the coin is gently placed on the rotating plate, the frictional force on the
coin is:
3 mg 3
(A) mg (B) mg (C) (D) mg
2 2 4

71. A car is going at a speed of 6 m/sec when it encounters a slope of angle

37°. The length of the sloping slide is 7 m. The friction coefficient between
the road and the type is 0.5. The driver applied brakes. The minimum
speed of the car with which it can reach the bottom is W m/sec. Find W.

72. In which of the following cases, force on the particle is zero?

(A) particle is going in a circle
(B) particle is accelerating along a straight line
(C) the momentum of the particle is constant
(D) acceleration of the particle is non zero constant

73. A coin of mass 10 g is placed over a book of length 50 cm. The coin is on the verge of sliding when one
end of the book is lifted 10 cm up. The coefficient of static friction between the book and the coin is:
(A) 1.0 (B) 0.4 (C) 0.3 (D) 0.2

74. To give the system of bodies (as shown in figure) a velocity of 3 m/s after moving 4.5 m from rest the
value of the constant force P should be : ( k = 0.2 for all surfaces in contact)
45N
(A) 163.8 N P A
180N
(B) 255.5 N B 90N
3  C
(C) 195.3 N 4

(D) None of theses

MEQB 171 Level -1, 2, 3 | Dynamics of a Particle

*75. A solid block of mass 2 kg is resting inside a cube as shown in the

figure. The cube is moving with a velocity v  5t iˆ  2 ˆj m / s . Here, t is
time in second. The block is at rest with respect to the cube and
coefficient of friction between the surface of cube and block is 0.6.
Then, (Take g  10 ms 2 )
(A) force of friction on the block is 10 N
(B) force of friction acting on the block is 4 N
(C) the total force exerted by the block on the cube is 14 N
(D) the total force exerted by the block on the cube is 10 5 N

*76. A trolley of mass 8 kg is standing on a frictionless surface inside

which an object of mass 2 kg is suspended. A constant force F starts
acting on the trolley, as a result of which the string stood at an angle
37° from the vertical (bob at rest relative to trolley). Then,
(A) acceleration of the trolley is 2 m/sec2
(B) force applied is 60 N
(C) force applied is 75 N
(D) tension in the string is 25 N

77. The coefficient of friction between the masses 2m and m is 0.5. All other
surfaces are frictionless and pulleys are massless. Column-I gives the
different values of m1 and column-II gives the possible acceleration of
2m and m. Match the columns.
COLUMN-I COLUMN-II
(A) m1 = 2m (P) Acceleration of 2m and m are same.
(B) m1 = 3m (Q) Acceleration of 2m and m are different.
(C) m1 = 4m (R) Acceleration of 2m is greater than m.
(D) m1 = 6m (S) Acceleration of m is less than 0.6 g.

78. A man thinks about 4 arrangements as shown to raise two small bricks each having mass m. Which of
the arrangement would take minimum time?

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

79. In the system shown, the inclined plane is fixed and smooth. Mass of
block A is 50 kg and mass of block B is 75 kg. Tension in the string
connected to block A is: (Take g  10 ms 2 )

(A) 500 N
(B) 400 N
(C) 300 N
(D) 600 N

MEQB 172 Level -1, 2, 3 | Dynamics of a Particle

80. In the arrangement shown in the figure, the coefficient of friction
between the blocks is   1/ 2 . The force of friction acting between
the block is W N. Find W.

81. The masses of blocks A and B are m = 1 kg and M = 2 kg. Between

A and B, there is a constant frictional force F = 4 N. But B can
slide without friction on the horizontal surface (see figure). A is set
in motion with velocity v0  6 m/sec. while B is at rest. The
distance moved by A relative to B before they move with the same
velocity is W m. Find W.

*82. Two masses m1 = 4 kg and m2 = 2 kg are connected with an inextensible,

massless string that passes over a frictionless pulley and through a slit, as
shown. The string is vertical on both sides and the string on the left is acted
upon by a constant friction force 10 N by the slit as it moves.
(Take g  10 ms 2 )
5
(A) Acceleration of mass m1 is m / s 2 , downwards
3
(B) Tension in the string is same throughout
0
(C) Force exerted by the string on mass m2 is N
3
(D) If position of both the masses are interchanged, then 2 kg mass moves up with an acceleration
10
m / s2
3

*83. Figure shows a block of mass m placed on a smooth wedge of mass M.

Calculate the minimum value of M' and tension in the string, so that the
block of mass m will move vertically downward with acceleration 10ms 2 .
M cot 
(A) The value of M' is
1  cot 
M tan 
(B) The value of M' is
1  tan 
Mg
(C) The value of tension in the string is
tan 
Mg
(D) The value of tension is
cot 

*84. The velocity-time graph of the figure shows the motion of a wooden block
of mass 1 kg which is given an initial push at t = 0 along a horizontal
table. (g = 10 m/s2)
(A) The coefficient of friction between the block and the table is 0.1
(B) The coefficient of friction between the block and the table is 0.2
(C) If the table was half of its present roughness, the time taken by the
block to complete the journey is 4 s
(D) If the table was half to its present roughness, the time taken by the
block to complete the journey is 8 s

MEQB 173 Level -1, 2, 3 | Dynamics of a Particle

85. In the arrangement shown, neglect the mass of the ropes and pulley.
What must be the value of m to keep the system in equilibrium? There is
no friction anywhere.
(A) M M
(B) 2M 30

(C) M/2
(D) M/4

m
86. Figure shows a system in which all surfaces are smooth. The
F
acceleration of mass m is . Find the value of  .
m

87. A body is launched up an inclined plane with inclination  to horizontal. It is observed that the time of
ascent is half of the time of descent along length of incline. The coefficient of friction between the body
and the incline is x tan  . Find the value of x.

88. A ball is held at rest in position A and another identical ball

is released from rest in position B. The ratio of tensions
T1/T2 at this moment is given by:
1
(A) 1 (B)
2
4 3
(C) (D)
3 4

89. A boy X (mass = 20 kg) is climbing up on a strong string 1. Attached to a

hook on bag of boy another string 2 is hanging which is held by another
boy Y (mass = 10 kg). The breaking strength of string 2 is 200 N. Find the
maximum force (in N) which boy X can exert on string 1 to carry boy Y
with him.

*90. Two blocks A and B of mass 4 kg and 2 kg respectively connected by a

spring of force constant k = 100 N/m are placed on an inclined plane of
inclination 30° as shown in figure. If the system is released from rest,
which one of the following statement(s) is (are) correct?
(Take g  10 ms 2 )
(A) There will be no compression/elongation in the spring if all surfaces are smooth
(B) Maximum compression of the spring is 10 cm if all surfaces are smooth
(C) Maximum elongation in the spring is 60 cm if all surfaces are smooth
(D) There will be elongation in the spring if A is smooth and B is rough

91.  
A body under the action of a force F  6iˆ  8 ˆj  10kˆ N, acquires an acceleration of 1 m/s2. The mass of

the body must be:

(A) 10 kg (B) 20 kg (C) 10 2 kg (D) 2 10 kg

MEQB 174 Level -1, 2, 3 | Dynamics of a Particle

*92. A 3-kg block of wood is on a level surface where s  0.25 and k  0.2 . A force of 7 N is being applied
horizontally to the block. Mark the correct statement(s) regarding this situation. (take g = 10 m/s2)
(A) If the block is initially at rest, it will remain at rest and friction force will be about 7 N
(B) If the block is initially moving, then it will continue its motion forever if the force applied is in the
direction of the motion of the block
(C) If the block is initially moving and the direction of applied force is same as that of motion of
block, then block moves with an acceleration of 1/ 3 ms 2 along its initial direction of motion
(D) If the block is initially moving and direction of applied force is opposite to that of initial motion of
the block, then the block decelerates, comes to a stop, and starts moving in the opposite
direction

93. Match column I with Column II:

COLUMN I COLUMN II
(A)  between the wall and the block is 0.5, (P) Force of friction is
  0.5
F  16 2 N ,   45 and m  1kg . m  1kg static

45
F
(B) A block of mass m = 1 kg is placed on the (Q) Force of friction is zero
inclined surface (inclination   37 ) of a
wedge which is moving horizontally with m
a
acceleration a = 4g/3 m/s2 as shown. 
Coefficient of friction between the block
and the inclined surface is   0.5 .
(C) A block of mass 1.2 kg is placed on a fixed (R) Force of friction is
inclined surface (inclination   30 ). kinetic

Coefficient of friction between block and
30
the surface is 0.8.
(D) A block of mass 1.2 kg is placed on a (S) Net force on the block
fixed inclined surface (inclination is zero

  60 ). Coefficient of friction between
60
block and the surface is 1.
(T) Net force on the block
is 10 N

94. A right circular cone is fixed with its axis vertical and vertex down. A particle
is in contact with its smooth inside surface and describes circular motion in a
horizontal plane at a height of 20 cm above the vertex. Find its velocity in
m/s. (Take g = 10 m/s2)
(A) 2m/s (B) 2m / s (C) 50 m / s (D) 2 2m /s

*95. A wedge of mass M and a block of mass m are kept on an incline of

inclination . The system is released from rest. The surface of the wedge
on which the block rests is horizontal. All the surfaces are smooth. Then:
(A) The acceleration of the block is along the incline
(B) The acceleration of the block is in vertical direction
(C) The acceleration of the wedge is greater than g sin 
(D) Normal force exerted by incline plane on wedge is less than ( M  m )g cos 

MEQB 175 Level -1, 2, 3 | Dynamics of a Particle

PARAGRAPH FOR QUESTIONS 96 – 97
A smooth semicircular wire-track of radius R is fixed in a vertical plane. One
end of a massless spring of natural length 3R/4 is attached to the lowest point
O of the wire-track. A small ring of mass m, which can slide on the track, is
attached to the other end of the spring. The ring is held stationary at point P
such that the spring makes an angle of 60° with the vertical, the spring
constant K  mg / R . Consider the instant when the ring is released.

5 3
96. If the tangential acceleration of the ring is g , find the value of .


97. If the normal reaction between ring and track is mg , find the value of .

98. A hockey player is moving northward and suddenly turns westward with the same speed to avoid an
opponent. The force that acts on the player is:
(A) frictional force along westward (B) muscle force along southward
(C) frictional force along south-west (D) muscles force along south-west

99. 
A body of mass 2 kg travels according to the law x t  pt  qt 2  rt 3 where p  3 ms 1, q  4 ms 2 and

r  5 ms 3 . The force acting on the body at t = 2 seconds is:

(A) 136 N (B) 134 N (C) 158 N (D) 68 N

100. A block is placed on a rough inclined plane. Angle of inclination  of the plane is varied
starting from zero. The coefficient of static friction and kinetic friction between the block and
the plane is s and k respectively (s  k ). Column II shows the graphs which necessarily
contains  taken on x-axis. Column I represents the quantities taken on y-axis of column I.
Match the quantities of column I with graphs of column II.
Column I Column II

(A) Friction force between the block and plane (P)

(B) Normal force between the block and the plane (Q)

Total contact force between the block and the

(C) (R)
plane

(S)

(T)

101. Two blocks A and B of equal masses are placed on rough inclined
plane as shown in the figure. Initially the block A is 2 m behind
the block B. Coefficient of kinetic friction for the blocks A and B are
0.2 and 0.3 respectively ( g  10 m / s 2 ) . By the time the two blocks
come on the same line on the inclined plane if they are released
simultaneously, the distance moved by B is  m . Find the value of
.

MEQB 176 Level -1, 2, 3 | Dynamics of a Particle