1. LAWS OF MOTION
SYLLABUS
Force and inertia : Newton's Laws of Motion, Conservation of linear momentum and its
applications rocket propulsion, Laws of friction.
FORCE
A force is a push or pull acting on a body. It is a vector quantity i.e. it has both magnitude
and direction.
Unit of Force : Its unit is Newton in SI system and dyne in CGS system.
Dimensions : MLT 2
(a) Contact Forces: Tension, Normal Reaction, Friction etc. Forces that act between
bodies in contact.
(b) Field forces (non-contact forces): Weight, electrostatic forces, etc. Forces that act
between bodies separated by a distance without any actual contact.
CONTACT FORCES
(i) Tension (T)
When a string, thread, wire or a spring
is held taut, the ends of the string or
thread (or wire) pull on whatever
bodies are attached to them in the
direction of the string. This force is
known as Tension.
A
B
O
T
T
T
T
m
R
T
(ii)Normal Reaction
When two surfaces are in contact, then the surfaces exert
forces on each other. The forces are opposite to each other in
direction and are equal to each other. The direction is always
at right angles to the surfaces in contact.
N1
N1 N2
N2
(iii) Friction
Friction is a force that acts between bodies in contact with each
other,
A
B
NEWTON’S FIRST LAW
Every body continues in its state of rest, or of uniform motion in a straight line, unless it is
compelled to change that state by an external force.
INERTIA
Inertia is the property of a body by virtue of which a body retains its state: either of rest or
uniform motion along a straight line.

2. MOMENTUM
It is equal to the product of the mass and the velocity of the body. Its direction is same as
the velocity of the body.
v
m
p


 where p

= momentum of the body.
m = mass of the body.
v

= velocity of the body.
Unit : Its unit is kg-m/s in SI system and gm-cm/s in CGS system.
Dimensions : Its dimensions is MLT-1
NEWTON'S SECOND LAW OF MOTION
Newton’s second law states that the rate of change of momentum of a body is directly
proportional to the applied force.
If a body of mass m is moving with a velocity v

, then its momentum is,
v
m
p


 . . . (i)
According to Newton's 2nd law, if a net force F

acts on the body,
dt
p
d
F


 . . . (ii)
dt
p
d
k
F


 . . .. (iii)
Where k is the constant of proportionality.
Unit force is now defined as that force which produces a unit rate of change of momentum
in a body
i.e. F

= 1 unit if
dt
dp
= 1 unit.
This makes k = 1.
In this system of units,
dt
p
d
F


 . . . (iv)
If the mass of the body m is constant, then
)
v
m
(
dt
d
F


 = m
dt
v
d

= ma

. . .(v)
NEWTON’S THIRD LAW
Every action has an equal and opposite reaction
Example 1: A block of mass m = 10 kg is pulled by a force
F = 100 N at an angle  = 30o with the
horizontal along a smooth horizontal surface.
What is the acceleration of the block? (g = 10
m/s2)

F
m
Solution: The forces that act on the body can be
decomposed along x and y axis.
As there is no acceleration along y-axis, net
force acting along vertical or y axis should be
zero i.e.
F sin 
F cos 
N
mg
y
x
FY = N + Fsin – mg = 0 . . . (1)
The body accelerates along x-axis. Therefore
Fx = Fcos = ma . . . (2)

3. where the acceleration along x-axis is a.
 F Cos  = ma  a =
m
cos
F 
= 2
o
s
/
m
3
5
10
2
3
100
10
30
cos
100



 The acceleration of the block is
2
s
/
m
3
5
Directed towards right. Since F sin < mg & the surface is rigid, the block
remains in equilibrium along y-axis.
Example 2: In the figure, the tension in the horizontal cord is 30N.
Find the weight of the body B.
B
cord 1
cord 2
30
N
P
45
Solution: (i) Isolate the point P
(ii) The forces acting on it are :
Unknown tension T2 in cord 2
Unknown tension T1 in cord 1
Known tension of 30N in the horizontal cord.
(iii)T2 is resolved along x and y axes.
(iv)Condition of equilibrium :
Fx = 0  30 – T2sin45 = 0 (1)
and Fy = 0  T2cos45 - T1 = 0 (2)
For body B
Since the body B is also in equilibrium,
Hence T1 = W (3)
(v) After solving these equations,
we get W = 30N
T2
T1 = W
X
Y
30
N
45
W
T1
Example 3: A block of mass m1 and weight w1 = m1g moves on a
level frictionless surface. It is connected by a light
flexible cord passing over a small frictionless pulley to
a second hanging block of mass m2 and weight w2 =
m2g. What is the acceleration of the system and what
is the tension in the string?
m2
m1
w1=m1g
w2=m2g
T
T
N
Solution: The diagram shows the forces acting on each block. The forces exerted on
the blocks by the cord can be considered an action reaction pair,
fx = T = m1ax
fy = N – mg = m1ay = 0
Applying Newton’s second law to the hanging block, we obtain
fy = m2g – T = m2ay
T = m1a . . . (1)
m2g – T = m2a . . . (2)
Adding two equations eliminates T giving

4. m2g = (m1 + m2)a
a =
2
1
2
m
m
m

g
Substituting the value of a in equation (1) gives the tension as T = g
m
m
m
m
2
1
2
1

Example 4: Two blocks of masses m1 = 2 kg and m2 = 1 kg
are in contact on a smooth horizontal surface as
shown in the figure. A horizontal force F = 3N is
applied on the block m1. Find the contact force
between blocks.
m1
m2
F
Solution: Let contact force between the blocks be N. Since N is responsible
for the acceleration of m2 hence it will be in the direction of
acceleration of m2 and on m1it will be opposite to the acceleration
Equations of motion
F N = m1a . . . (1)
N = m2a . . . (2)
Where a = acceleration of the blocks.
N1
F
N
m1g
N2
N
m2g
From (1) and (2) , we get
a =
2
1 m
m
F

 N =
2
1
2
m
m
F
m

= 1N.
SPRING FORCE
The extension or compression produced in a spring is directly proportional to the force
applied on it. If x is the deformation in the spring then
F  x
or F = – k x
where k is constant and is called the spring constant or force constant. The negative sign
shows that the force and the extension or compression are opposite in their directions.
FRICTIONAL FORCE
Frictional force comes into play between two surfaces whenever there is relative motion or
a tendency of relative motion between two surfaces in contact. Frictional force has the
tendency to oppose relative motion between the surfaces in contact.
Thus, friction can be classified as
(a) Static Friction: which acts between surfaces in contact
but not in relative motion, it opposes the tendency of
relative motion
(b) Kinetic Friction: which acts between surfaces in
contact which are in relative motion, it opposes the
relative motion between the surfaces.
F
f
45
fk
f
f
(0, 0)
F

5. LAWS OF STATIC FRICTION
Static Friction, acting between the surfaces in contact, (not in relative motion) opposes the
tendency of relative motion between the surfaces.
It is independent of the area of contacting surfaces.
Now, fs(max) N where f = fs(max)
fs(max) = sN
Here s = co-efficient of static friction.
N = normal reaction on the block from the surface.
0  fs  N
When F exceeds f block starts moving and frictional force decreases to a constant value fk.
fk is called kinetic friction and it has unique value which is given by
fk = k N
Here k = co-efficient of kinetic friction.
N = normal reaction.
ANGLE OF FRICTION
The angle made by the resultant reaction force with the vertical
(normal reaction) is known as the angle of the friction.
Now, in the triangle OAB,





tan
AB
OB
cot
OB
AB
or tan =
N
f

N
f
O
A
B
ANGLE OF REPOSE
Consider a body of mass m resting on an inclined plane of
inclination . The forces acting on the body are shown – Ff
being the force of friction. If friction is large enough, the body
will not slide down.
mg

f
N
y
x
Along x: mg sin  – f = 0 …(1)
Along y: N –mg cos = 0 …(2)
i.e. N = mg cos  and f = mg sin 
Thus, f  SN gives,
mg sin   S mg cos 
or, tan   S, the coefficient of static friction between the two surfaces, in order that the
body doesn’t slide down. When  is increased so that
tan  > , then sliding begins, and the angle r = tan-1 , where sliding begins is known as
the angle of repose.
Example 5: A block weighing 2kg rests on a horizontal surface. The coefficient of static
friction between the block and surface is 0.40 and kinetic friction is 0.20.

6. (a) How large is the friction force acting on the block ?
(b) How large will the friction force be if a horizontal force of 5N is
applied on the block?
(c) What is the minimum force that will start the block in motion?
Solution: (a) As the block rests on the horizontal surface and no other force parallel
to the surface is on the block, the friction force is zero.
(b) With the applied force parallel to the surfaces in contact 5N, opposing
friction becomes equal and opposite. Further the limiting friction is sN =
sMg = 8N
 Force of friction is 5N.
(c) The minimum force that can start motion is the limiting one. sN = smg = 8N
Example 6: In the figure suppose that the block weighs 2kg, that the
friction T can be increased to 8N before the block starts to
slide, and that a force of 4N will keep the block moving at
constant speed once it has been set in motion. Find the co-
efficient of static and kinetic friction.
N
2kg
T
Solution: There is no acceleration of the body along y-axis.
Hence fy = N – W = N – 20N = 0
 N =20N
fX = T – fs = 8N – fs = 0
 fs =8N
Hence, we have fs = sN
As the motion takes place at constant velocity and there is no motion
along y-axis.
fy = N – W = 0
 N = mg =20N
There is no acceleration along x-axis
 fX = T – fk = 4N – fk = 0  fs =8N
fk = kN
Hence, 20
.
0
N
20
N
4
N
fk


Example 7: Block A in the figure weighs 0.4 kg and block B
weighs 0.5kg. The coefficient of sliding friction
between all surfaces is 0.25 (a) find the force F
necessary to drag block B to the left at constant
speed. (b) find the tension in the string.
A
B
F
Solution:
For A as there is no motion along horizontal
or vertical. Let m1 be mass of A
g
m
N 1
1  = 0  N1 = m1g
and T – kN1 = 0  T – km1g = 0
 T =km1g= 0.25  0.410 = 1N
T
km1g
N1
m1g

7. For B as there is no motion along vertical.
Let m2 be mass of B
N2 – N1 = m2g  N2 = (m1+m2)g
As there is no acceleration along x-axis
F – f1 – f2 = 0
 F = f1 + f2 = kN1 + kN2 = kq (2m1+m2)
= 0.25 (0.8 + 0.5)10
= 3.25 N
N1
f2
N
m2g
F
N2
f1
FRAME OF REFRENCE
It is a conveniently chosen co-ordinate system, which is used to describe the position and
motion of a body.
Inertial Frame of Reference
Any frame of reference in which Newton’s first laws are valid is an inertial frame (IFR).
Example 8: A block of mass m is placed on an inclined plane. With what acceleration
a, towards right should the system move on a horizontal surface so that m
does not slide on the surface of inclined plane? Assume all surfaces are
smooth.
Solution : the normal reactions and a
pseudo force of magnitude ma
towards left.
Rcos = mg  a = gtan
R sin = ma
R
mg
ma
R cos
R sin
ma
mg
Example 9: A pendulum of mass m is hanging from
the ceiling of a car having an acceleration
ao with respect to the road in the direction
shown. Find the angle made by the string
with the vertical.
a0

Solution: Since bob of the pendulum is stationary relative to
car
Hence
T sin = mao (pseudo force) (i)
T cos = mg (ii)
Dividing (i) by (ii), we get
tan =
g
ao
  = tan-1
g
ao
mao
mg
Tsin
Tcos
T

CENTRIPETAL FORCE & CENTRIFUGAL FORCE
If a body is moving with a constant speed in a circle, as seen from an inertial frame it is
continuously accelerated towards the centre of rotation. The magnitude of the centripetal
acceleration for a body moving with a tangential velocity v is given by v2/r. If the angular
velocity of the body is  then the centripetal acceleration is mr2.
According to Newton’s law force causes acceleration and so, the net centripetal force

8. F = ma =
r
mv2
= m2 r
a

= a

T + a

R
 ma

= ma

T + ma

R
Where T
a

is tangential and, R
a

is centripetal are radial acceleration and a

is total
acceleration
Example 10: Find the ratio of radius of curvature at the highest point of projectile to that
just after its projection if the angle of projection is 300.
Solution : If 0
v

is the initial velocity vp = v0 cos 
Normal acceleration at O = g cos 
Normal acceleration at P = g
Hence if r0 and rp be radii of curvatures at O & P
respectively.
r0 =

cos
g
v2
0
,
rp =
g
cos
v 2
2
0 
Hence the required ratio =
0
p
r
r
cos3  =
8
3
3
.
O

P
v0

vP

(an)p
x
y
Example 11: A 1200 kg automobile rounds a level curve of radius 200m, on an
unbanked road with a velocity of 72 km/hr. What is the minimum
coefficient of friction between the tyres and road in order that the
automobile may not skid (g= 10m/s2)?
Solution: In an unbanked road the centripetal force is provided by the frictional
force.
 ffriction =
r
mv2
But flimiting friction  ffriction
or mg  ffriction
or mg 
r
mv2
 min = 2
.
0
200
10
20
20
gr
v2




Rocket
1. Acceleration of rocket
0
M
u
t
a
M
M t
t
d
×
d
=
d
æ ö
- ×
ç ÷
è ø
d
2. Thrust on rocket
M
F u
t
d
=
d
3. Impulse of force av
F t p
I = ´ = D

9. 4. Recoil velocity of the gun b b
g
g
m v
v
m
=
OBJECTIVE
1. A reference frame attached to the earth is:
(A) an inertial frame sometimes (B) an inertial frame always
(C) a non-inertial frame (D) may be inertial or non-inertial.
Ans. (C)
Solution: As the earth itself is rotating on its axis the frame attached to it should be a
non-inertial one.
2. A vehicle is moving on a rough road in a straight line with uniform velocity.
(A) No force is acting on vehicle
(B) A force must act on the vehicle
(C) An acceleration is being produced in the vehicle
(D) An centriprital acceleration is being produced in the vehicle
Ans. (A)
Solution: As the vehicle is moving with uniform velocity, net force acting on it should be
zero.
3: Two identical particles A and B, each of mass m, are interconnected by a spring of
stiffness k. If the particle B experiences a force F & the elongation of the relative
spring is x, the relative acceleration between the particles is equal to:
(A) F/2m (B)
F - kx
m
(C)
F - 2kx
m
(D)
kx
m
Ans. (C)
Solution : Equation of motion for A :
T = ma 
a =
m
kx
m
T

for B : F – T = ma
 a =
m
kx
F 
A B
F
m
T
T
m
 The relative acceleration ar = | a  a | =
m
kx
2
F 
.
4. A 60 kg man stands on a spring balance in a lift. At same instant he finds that the
reading on the scale has changed from 60kg to 50 kg for a while and then comes
back to the original mark. What is his conclusion?
(A) The lift was in constant motion upwards
(B) The lift was in constant motion downwards
(C) The lift while in motion downward suddenly stopped.
(D) The lift while in motion upward suddenly stopped.
Solution: When lift accelerates up W = W 








g
a
1

10. When lift decelerates up W = W 








g
a
1
Here weight decreases i.e. lift decelerates up. Initially lift was moving with
constant velocity up and then it suddenly stopped i.e. decelerated,
5. A body is accelerated by applying a force of 30N. The change in the momentum of
the body after 2sec is:
(A) 7.5 kg-m/s (B) 30 kg-m/s
(C) 120 kg-m/s (D) 60 kg-m/s
Solution: P = Ft = 302 = 60 kg m/s
Hence D
6. Two blocks of masses 2kg and 5kg are connected by a light string passing over a
frictionless pulley. The tension in the cord connecting the masses will be
(A) 20N (B) 30N
(C) 28 N (D) 50 N
Solution: T = g
m
m
m
m
2
2
1
2
1

8
.
9
2
5
2
5
2




 = 28N
Hence C
7. A vessel containing water moves with constant acceleration towards the right, along
a horizontal straight line. Which of the following figures: a, b and c represents the
surface of the liquid.
None of the above
(A) (B) (C) (D)
Solution: When the vessel accelerates forward, the pseudo force acts on water
backward and the resultant of the weight of water and the pseudo force, acts
on the left downward. The surface of water is perpendicular to this resultant.
Hence C
8: A bob hanging from the ceiling of the car acts as an accelerometer. Then the relation
expressing horizontal acceleration a of the car and the angle  made by bob with the
vertical is
(A) a = gtan (B) a = gsin
(C) a = gcot (D) None
Hence A
Solution: tan =
Mg
Ma
 a = gtan  
Ma
a
Mg

11. 9. A weight W is attached to two weightless strings AB
and AC as shown in figure. The tension in each string
will be:
(A)
W
4
(B)
W
2
(C)
3 W
2
(D) W
T
T
C
B
120
W
A
Solution: For vertical equilibrium
2T cos60 = W
T =

60
cos
2
W
= W
Hence D
10. A body of mass 2kg is acted upon by two forces each of magnitude 1N, making an
angle 60 with each other. The net acceleration of the body in m/s2 is
(A) 0.1 (B) 1.0
(C) 3 2 (D) 2 3
Solution: Net force F = 

 60
cos
F
F
2
F
F 2
1
2
2
2
1 = N
3
2
1
1
2
1
1 2
2





a = 2
s
/
m
2
3
2
3
M
F


Hence C
11. A scooter of mass 120kg is moving with a uniform velocity of 108km/hr. The force
required to stop the scooter in 10s is
(A) 180N (B) 208N
(C) 360 N (D) 720 N
Solution: F =
t
mv
t
mv
0
t
P







=
 
10
18
5
108
kg
120 






= 360N Ans. (C)
12. Two blocks A and B of masse 4kg and 12 kg are placed on
a smooth horizontal plane a force F of 16N is applied on A,
as shown, then the contact force between the blocks is
(A) 4N (B) 8N
(C) 12N (D) 16N
F = 16N
A
B
Solution: Acceleration

12. a =
12
4
16
M
M
F
2
1 


= 1m/s2
P = 12a (for block B)
 F = 16 – M1a = 16 – 4  1 = 12N
Hence C
13. N bullets, each of mass m, are fired with a velocity v at the rate of n bullets/sec upon
a wall. The bullets are stopped by the wall. The reaction offered by the wall to the
bullets is
(A)
Nmv
n
(B) nNmv
(C)
Nv
n
m
(D) nmv
Solution: F =
t
Δ
P
Δ
Moment transferred to the wall,
P = mv, t =
n
1
 F = nmv
A
1
mv

Hence D
14. A knife edge of mass M is dropped from a height ‘h’ on a wooden floor with its tip
pointing of downward. If the blade penetrates a distance ‘s’ into the wood, the
average resistance offered by the wood to the blade is
(A) Mg (B) Mg
h
1+
s
 
 
 
(C) Mg
h
1-
s
 
 
 
(D) Mg
2
h
1+
s
 
 
 
Solution: Velocity of knife edge when it hits the floor
v = gh
2
Retardation, a =
s
gh
s
2
gh
2
s
2
v2


 Equation R – Mg = Ma
R = Mg + Ma
= Mg + M 







s
h
1
Mg
s
gh
Hence B
R
Mg
a
15. A rocket, set for vertical launching, has a mass of 50 kg and contains 450 kg of fuel.
It can have a maximum exhaust speed of 2 km s–1. If g = 10 ms–2, what should be
the minimum rate of fuel consumption to just lift it off the launching pad?
(A) 2.5 kg s–1 (B) 5 kg s–1
(C) 7.5 kg s–1 (D) 10 kg s–1
Sol: (A) When the rocket is at the launching pad, M = M0 = 50 + 450 = 500 kg. Also u =
2kms–1 = 2000 ms–1.
M
F u
t
d
=
d

13. But F = Mg. Therefore
M
u Mg
t
d
=
d
or
2
1
1
M 500 kg 10 ms
2.5 kg s
t 2000 ms
-
-
-
d ´
= =
d
16. In above Q. what should be the minimum rate of fuel consumption to give an initial
acceleration of 20 ms–2 to the rocket ?
(A) 2.5 kg s–1 (B) 5 kg s–1
(C) 7.5 kg s–1 (D) 10 kg s–1
Sol: (C)
M
F u Mg
t
d
= -
d
But F = Ma, where a is the initial acceleration. Thus
M
Ma u Mg
t
d
= -
d
or
2
1
M M(a g) 500 kg (20 10) ms
t u 2000 ms
-
-
d + ´ +
= =
d
= 7.5 kg s–1
17. A block can slide on a smooth inclined plane of inclination  kept on the floor of a lift.
When the lift is descending with a retardation a. the acceleration of the block relative
to the incline is:
(A) (g + a) sin  (B) (g – a)
(C) g sin  (D) (g – a) sin 
Solution:
N
mg
ma

m(g a)cos
  m(g a)sin
 
net accel. =
m(g a)sin
(g a)sin
m
 
   Ans. (A)
18. A force of 5 N acts on a body of weight of 10 N. What is the acceleration in m/s2
(A) 50 (B) 5
(C) 0.5 (D) 2
Solution:
5N
mass =
10
1
10

A = 2
5
5m/ s
(10/10)

Ans. (B)

14. 19. A man of mass m stands on a crate of mass M. He pulls on a light rope passing over
a smooth light pulley. The other end of the rope is attached to the crate.
For the system to be is equilibrium, the force exerted by the man on the
rope will be:
(A) mg (B) Mg
(C)
1
(M m)g
2
 (D) (m + M)g
Solution:
T
m
M
T
2T = (M + m)g
T =
(M m)g
2

Ans. (C)
20. A uniform rope of mass M and length L is pulled along a frictionless inclined plane of
angle  by applying a force F (> Mg) parallel to incline. Tension in the chain at a
distance x from the end at which force is applied is
(A)
x x
F Mgsin
L L
  (B)
x
F 1-
L
 
 
 
(C)
x x
F 1- Mgsin
L L
 
 
 
 
(D) None of these
Solution:
Mgsin
Mgcos

N F
F – Mg sin = MA (1)
F –
M M
x gsin T x A
L L
   
  
   
   
F–
m M F Mgsin
xgsin T x
L L M
 
   
  
   
   
T = F
x
1 N
L
 

 
 
Ans. (D)
21. In the system shown, the acceleration of the wedge of
mass 5M is (there is no friction anywhere)
(A) zero (B) g/2
(C) g/3 (D) g/4
Solution: 2 1
2 1
m g (m m )g
A
m m m
  

 
M
m
5M
M
2M
600

15. (6 0.4(m 4))g
A 0
m 6 4
 
 
 
60
m 4
0.4
 

44
m 11kg
4
 
Ans. (C)
22. What is the maximum value of the force F such that the
block shown in the arrangement, does not move?
(A) 20 N (B) 10 N
(C) 12 N (D) 15 N
Solution:
60
F 1
r
F
mg
Fsin60
Fcos60
N
N = F sin60 + Mg (1)
F cos60 = 1
r
F = N = (F sin60 + Mg)
F =
Mg
cos60 sin60

 
=
1
3 1
5 4
2 3 20N
1
1 1 3
2 2
2 3
 

 
 
=
5 4
20N
1


Ans. (A)
23. A force vector F = 6 i

– 8 j

+ 10 k

newton applied to a body accelerates it by 1 ms–2.
What is the mass of the body?
(A) 10 2 kg (B) 2 10 kg
(C) 10 kg (D) 20 kg
Solution: F = m a
|F| = 10 2
10 2 = m × A = m × 1  m = 10 2 kg. Ans. (A)
24. A flexible chain of weight W hangs between two fixed
points A and B at the same level. The inclination of the
chain with the horizontal at the two points of support is
. What is the tension of the chain at the end point.
(A)
W
cosec
2
 (B)
W
sec
2

(C) W cos (D)
W
sin
3

Solution:
600
F
m 3kg

1
2 3
 
A
W
B



16. T
F 

T
2T sin = w  T =
w w
cosec
2sin 2
 

Ans. (A)
25. Two blocks A and B of masses m1 and m2 are placed in contact with each other on a
horizontal platform. The coefficient of friction between the platform and the block A is
1 and that between block B is 2. The platform moves with a constant acceleration a
as shown in fig. The force of interaction between the
blocks exists if
(A) 1 = 2 (B) 1  2
(C) 1 < 2 (D) none of these
Solution:
Form of indication will exhibit
By 1 2
r r
1 2
F mg F mg
m m
 

1 2
r r
1 2
F F
m m
  1 1 1 2
1 2
m g m g
m m
 

 1  2
Ans. (D)
26. For the system shown in the figure, the pulleys are light and frictionless. The tension
in the string will be
(A)
2
mgsin
3
 (B)
3
mgsin
2

(C)
1
mgsin
2
 (D) 2mgsin
Solution: mg sin – T = mA (1)
T = mA (2)
(1) + (2)  mg sin = 2mA
 A =
gsin
2

T =
mgsin
2

Ans. (C)
27. Two masses M and m are connected by the arrangement shown in figure. What is
the downward acceleration of mass M.
a
m1
m2
2
r
F
1
r
F
a
1
m 2
m
m
 m
m
M

17. (A)
2M m
g
3M m
 

 

 
(B)
2M m
2g
4M m
 

 

 
(C)
M m
g
M m
 

 

 
(D)
M
g
2M m
 
 

 
Solution: Mg – T = MA (1)
2T – mg =
mA
2
(2)
2T
m
T
A
M
2 × (1) + (2)
(2M – m) g = A(2M +
m
2
)
A =
2(2M m)g
4M m


Ans. (B)
28. In a rocket, the mass of the fuel is 90% of the total mass. The rocket is blasted from
the launching pad. If the exhaust gases are ejected at a speed of 2 km s–1, what is
the maximum speed attained by the rocket ? (Neglect the effects of gravity and air
resistance).
(A) 2 ln (10) kms–1 (B) 2 ln (30) kms–1
(C) 2 ln (60) kms–1 (D) 2 ln (90) kms–1
Sol: (A) The mass of the container = 10% of the total mass of the rocket, 90% being the
mass of the fuel. Hence M0/Mc = 10. The rocket will attain the maximum speed vb
when all its fuel is burnt. This speed is given by
1 1
0
b
c
M
v u ln 2 kms ln(10) 2ln(10)kms
M
- -
æ ö
= = =
ç ÷
è ø
.
29. Blocks A and C start from rest and move to the
right with acceleration aA = 12t m/s2 and aC = 3
m/s2. Here t is in seconds. The time when
block B again comes to rest is
(A) 2 s
(B) 1 s
(C) 3/2 s
(D) ½ s
Solution: B
A C

18. 12t
3
A
12t
3
A =
12t 3
2

And A = 0, t =
3
12
=
1
4
Ans. (E)
30. At some instant, the 10 kg mass has acceleration of 12 m/s2, what is the acceleration
of 20 kg mass
(A) 2.5 m/s2
(B) 4.0 m/s2
(C) 3.6 m/s2
(D) 12 m/s2
Solution:
200N
(1) (2)
10 20
1
r max
F = 0.1 × 10 ×10
= 10 N
2
r max
F = 20 × 0.1 × 10 = 20 N
For 10 kg block Kx – 10 = 10A = 10 ×12
Kx = 120 + 10 = 130 N.
For 20 kg block = 200 – 130 = –20 = 20A
 A =
50
2.5m/ s
20

Ans. (A)
31. A block slides down an inclined plane of slope  with constant velocity. It is then
projected up the plane with an initial speed v0. How far up the incline will it move
before coming to rest
(A)
2
0
v
4gsin
(B)
2
0
v
gsin
(C)
2
0
v
2gsin
(D)
2
0
v
2g
Solution:
mgsin

r
F
For constant velocity the net force acting on the block will be O.
10 kg 20 kg
F=200 N
0.1


19.  Fr = mg sin
o
V
r
F mgsin
 
mA = mg sin + mg sin
= 2mg sin
 A = 2g sin
O = V0
2 – 2As
s =
1 2
0 o
V V
2A 2gsin


32. A force of 100N is applied on a block of mass 3kg as shown in the
figure. The coefficient of friction between the wall and block is 0.6.
The magnitude of the force exerted by the wall on the block is
(A) 15N downwards (B) 25N upwards
(C) 20N downwards (D) 30N upwards
Solution:
F cos30 = N.  N = 6 3 10.39

F sin 30 = 6N < sin
So. Fr will act upward
And Fr max = N
= 0.6 × 10.39= 6.23 N.
So force exerted by wall = 2 2
(6.23) (10.4)

= 15N down ward
Ans. (A)
33. A force acts for 8 seconds on a body of mass 10 kg initially at rest. The force then stops
acting and the body moves 80 m in next 5s. Calculate the force.
(A) 10 N (B) 20 N
(C) 30 N (D) 40 N
Solution : After the force stops acting, the body moves with a uniform velocity given by
v =
taken
time
travelled
distance
=
5
80
= 16 m/s
Initial velocity of the body, u = 0
From v = u + at
16 = 0 + at
or a = 2 m/s2
As F = ma
 F = 10  2 = 20N
12N
100N
30

20. Ans. (B)
34. A block is gently placed on a conveyor belt moving horizontally with constant speed.
After t = 4s, the velocity of the block becomes equal to the velocity of the belt. If the
coefficient of friction between the block and the belt is  = 0.2, then the velocity of
the conveyor belt is
(A) 2ms–1 (B) 4ms–1
(C) 64ms–1 (D) 8ms–1
Solution:
0
V
Fr = mg = 0.2 mg.
A = 0.2g = 2m/s
Let. velocity of belt = V0
So, vel. of block with respect belt will be V0
U = V0, A = 2, t = 4, v = 0
0 = V0 – 2 × 4
V0 = 8 m/s
Ans. (D)
35. A block of mass M is connected to a massless pulley and massless
spring of stiffness k. The pulley is frictionless. The spring
connecting the block and spring is massless. Initially the spring is
untstretched when the block is released. When the spring is
maximum stretched, then tension in the rope is
(A) zero (B) Mg
(C) 2Mg (D) Mg/2
Solution:
m
k
by consternation of energy
mgx =
1
kx
2
x =
2mg
k
T = kx = 2mg Ans. (c)
36. A balloon has a mass of 10 gram in air. The air escapes from the balloon at a
uniform rate with a velocity of 5 cm/s and the balloon shrinks completely in
2.5s. Calculate the average force acting on the balloon.
(A) 10 Dyne (B) 20 Dyne
(C) 30 Dyne (D) 40 Dyne
M
K

21. Solution : Here, m = 10 gm, v = 5 cm/s
5
.
2
10
dt
dm
 = 4 gm/s.F = ?
F =  
mv
dt
d
dt
dp
 = v
dt
dm






F = 4  5 = 20 dyne.
Ans. (B)
37. The weights W1 and W2 are suspended from the ends of a light string passing over a
smooth fixed pulley. If pulley is pulled up at an acceleration g, the tension in the
string will be
(A) 1 2
1 2
4W W
W W

(B) 1 2
1 2
2W W
W W

(C) 1 2
1 2
W W
W W

(D)
 
1 2
1 2
W W
2 W W

Solution:
1
w
2
w
g
consider w2 and w1 with respect to bully
2w2 – T = 2
w
A
g
 
 
 
(1)
T – 2w1 = 1
w
A
g
(2)
(1) × w1 – (2) × w2  T(– w1 – w2) = – 2gw1w2
T = 1 2
1 2
4w w
w w

Ans. (A)
38. A block of mass m is placed on a smooth wedge of inclination . The whole system
is accelerated horizontally so that the block does not slip on the wedge. The force
exerted by the wedge on the block has a magnitude
(A) mgtan (B) mgsin
(C) mgcos (D) mgsec
Solution:

m
A
consider F. B. D of m with respect to m

22. mAcos
N
mA sin
mgsin
mgcos
mg
mA
mg sin = mA cos
 A = g tan
and N = mgcos + mAsin
= m(gcos +
2
gsin
cos


)
= mg sin
Ans. (D)
39. A cube is resting on an inclined plane. What must be the value of coefficient of
friction between cube and plane so that cube topples before sliding
(A)
1
2
 (B)  < 1
(C)  > 1 (D)
1
2

Solution:
The given condition on only take place when line of force of mg passes
troughs
 = 45° …(i)
and mg sin  fr max =  mg cos
   tan
   1
(C)
40. A block is placed on a rough horizontal
plane. A time dependent horizontal force
F = K t acts on the block. Here K is a
positive constant. Acceleration-time
graph of the block is
(A)
a
t
(B)
a
t
10 kg 20 kg
F=200 N
0.1


23. (C)
a
t
(D)
a
t
Solution:
a
t
F = kt
Fnet = F – fr = 0 up to time Kt = mg = 1
r max
F
And for t >
mg
kt

Fnet = kt – mg
Ans. (C)
For Q. no. 41 to 44: A 20 kg box is dragged across a rough level floor (k = 0.3) by means
of a rope which is pulled upward at an angle of 30° to the horizontal. The
pulling force has a magnitude 80 N.
41. What is the normal force exerted by the floor on the block?
(A) 160 N (B) 140 N
(C) 180 N (D) 40 N
42. What is the frictional exerted by the floor on the block force?
(A) 48 N (B) 50 N
(C) 52 N (D) 60 N
43. What is the acceleration of the box?
(A) 1.06 m/s2 (B) 2.06 m/s2
(C) 2.08 m/s2 (D) 3.09 m/s2
44. If the force is reduced until the acceleration becomes zero, what is the tension
in the rope?
(A) 54.42 N (B) 55.42 N
(C) 58.42 N (D) 60.42 N
Solution for Q. 41 to 44: Let the normal
contact force be N
(41)  F sin  + N = mg
 N = mg – F sin 
= 20  10 – 80 sin 300 = 160 N.
Ans (A)
(42) fk = k N = (0.3 ) (160) = 48 N. Ans
(A)
F

mg
N
m
fk
(43) a = (Fcos   fk)/m =
)
20
(
48
3
40 
= 1.06 m/s2. Ans (A)

24. (44) F' cos  = fk
 F' = fk sec  = 48  (2 / 3) = (96/3) = 55.42 N. Ans (B)
45. An elevator accelerates upward at a constat rate. A uniform spring of length L and
mass m supports a small block of mass M that hangs from the ceiling of the elevator.
The tension at distance 1 from the ceiling is T. The acceleration of the elevator is
(A)
T
g
M m ml/L

 
(B)
T
g
2M m ml/L

 
(C)
T
g
M ml


(D)
T
g
2M m ml/L

 
Solution:
m
system
mass of system
m
(L ) M g
L
 
 
 
 
m m
T (L ) M g (L ) M A
L L
   
     
   
   
A =
T
g
m
(L ) M
L

 
46. A particle of mass 70 g, moving at 50 cm/s, is acted upon
by a variable force opposite to its direction of motion. The
force F is shown as a function of time t.
(A) Its speed will be –50 cm/s after the force stops
acting.
(B) Its direction of motion will reverse.
(C) Its average acceleration will be 1 m/s2 during the interval in which the force
acts.
(D) Its average acceleration will be 10 m/s2 during the interval in which the force
acts.
Solution: Total area under curve = (10 – 2 + 4  10–3) ×
10
2
= (10–2 + 0.4 × 10–2) × 5
–
2
1.4 10
5
10

 


 

 
= 0.07(V – 0.5)
V = –0.5 m/s
= – 50 cm/s
Ans. (A)
10
F(N)
t(s)
O 4 x 10-3 10-2
8 x 10-3

25. 47. Two masses M and m(M>m) are joined by a light string passing over a smooth light pulley
then what is the tension on block
(A) The acceleration of each block is
M m
g.
M m
 

 

 
(B) The tension in the string is
2Mmg
.
M m

(C) The centre of mass of the ‘M plus m’ system moves down with
an acceleration of
2
M m
g .
M m
 

 

 
(D) The tension in the string by which the pulley is attached to the roof is (M + m)g.
Solution:
(M m)
A g
M m



m
M
And for m: Mg – T = MA
T = Mg – MA = Mg – Mg
(M m)
M m


=
Mg(2m) 2mMg
M m (M m)

 
Ans. (A, B)
48. In the figure, the blocks A, B and C of mass m each have acceleration a1, a2 and a3
respectively. F1 and F2 are external forces of magnitudes 2mg and mg respectively.
(A) a1 = a2 = a3
(B) a1 > a3 > a2
(C) a1 = a2, a2 > a3
(D) a1 > a2, a2 = a3
Solution:
2mg
m
T
mg
2mg
2mg
2mg
T
m
T
m T
mg
mg
I II III
2mg
2mg – T = 2mA1  2mg –2mA1 = T
m
M
m m
2m
m
m
A B
F1
= 2mg
C
F2
= mg

26. Case II and Case III are identical to each other.
So a2 = a3 < a1
Ans. (D)
49. Block A is placed on block B, whose mass is greater than
that of A. There is friction between the blocks, while the
ground is smooth. A horizontal force P, increasing linearly
with time, begins to act on A. The accelerations a1 and a2
of A and B respectively are plotted against time (t). Choose
the correct graph.
(A)
t
a2
a1
(B)
t
a2
a1
a1
a2
(C)
t
a2
a1
a1
a2
(D)
t
a1
a1
a2
a2
Solution: Ans. will be B because upto some moment both will move with each other and
after that B will move with constant accel. due to kinetic friction and aceel. Of
A will increase
50. In the figure, the ball A is released from rest when the spring is at its
natural (unstretched) length. For the block B of mass M to leave contact
with the ground at some stage, the minimum mass of A must be
(A) 2M
(B) M
(C)
M
2
(D) a function of M and the force constant of the spring
Solution:
M
x
m
o
mg x = 2
1
kx
2
 x =
2mg
k
(1)
and kx = Mg
x =
Mg
k
Mg 2mg
k k
  m = (M/2)
Ans. (c)
B
A P
B M
A

27. 51. Neglecting friction and mass of pulley, what is the acceleration of
mass B
(A) g/3 (B) 5g/2
(C) 2g/3 (D) 2g/5
Solution: mg – T = mA (1)
2T – mg =
mA
2
(2)
A
2 A
B
2T
T
A
2
A
(1) × 2 + (2)
mg =
5mA
2
A =
2g
5
Ans. (D)
52. In the arrangement shown in figure, pulley is smooth and
massless and all the strings are light. Let F1 be the force
exerted on the pulley in case (i) and F2 the force in case (ii).
Then
(A) F1>F2 (B) F1<F2
(C) F1 = F2 (D) F1 = 2F2
Solution: (c) F1 = F2 because cases are identical.
53. A block of mass 2kg is pushed against a rough vertical wall with a force of 40N, the coefficient
of static friction being 0.5. Another horizontal force of 15N, is applied on the block in a
direction parallel to the wall. Will the block move? If yes, in which direction and with what
minimum acceleration ? If no, find the frictional force exerted by the wall on the block. ?
(A) 2 m/s2 (B) 2.5 m/s2
(C) 3 m/s2 (D) 4 m/s2
4m
2m
(i)
4m
m
(ii)
m
A
B
m
m

28. Solution : The force which may cause the tendency of
motion or motion in the body is its own
weight and the applied horizontal force of
15N. The resultant of the forces
F = N
25
15
20 2
2


In a direction tan-1 





20
15
= 370 with the
vertical.
The friction will, by its very virtue of
opposing the tendency will act in a direction
opposite to the resultant.
40N
15N
mg
The acceleration is minimum when the resultant force is minimum = F –N
(as, N is the maximum frictional force)
= 25 – 0.5 x 40 = 5 N
 Minimum acceleration is 2
s
/
m
5
.
2
2
5
 . Ans (B)
54. An inclined plane makes an angle 30 with the
horizontal. A groove OA = 5m cut in the plane makes
an angle 30 with OX. A short smooth cylinder is free
to slide down the influence of gravity. The time taken
by the cylinder to reach from A to O is (g = 10 m/s2)
(A) 4 s (B) 2 s
(C) 2 2s (D) 1 s
Solution: Accel. of
2
2
mgsin 30
gsin 30
m

s = 5 = 2
1
4 t
2
 

10 4
t 2sec
10

 
55. Two blocks of masses m1 = 4kg and m2 = 6 kg are
connected by a string of negligible mass passing over a
frictionless pulley as shown in Fig. The coefficient of
friction between block m1 and the horizontal surface is
0.4. When the system is released, the masses m1 and
m2 start accelerating. What additional mass m should
be placed over mass m1 so that the masses (m1 + m)
slide with a uniform speed?
(A) 9 kg (B) 10 kg
(C) 11 kg (D) 12 kg
Solution: For m2 T = m2g.
For (m1 + m)
T = (m1 + m)g = 0.4(m1 + m) g = m2g
300
3
0
0
O X
A
cylinder
m
m2
m1

29.  2
1
m
m m
0.4
 

6 10
m 4 11kg
0.4

  
Ans. (C)
56. A block of mass 4 kg is suspended through two light spring balances A and
B as shown in the figure. Then balance A and B will respectively read
(A) 4 kg and zero kg
(B) zero kg and 4 kg
(C) 4 kg and 4 kg
(D) 2 kg and 2 kg
Solution: (c) If we consider 4 kg block as our system then it will be the reading
of B and if we consider (spring B 4kg block) as our system then
spring A will give 4kg reading.
57. A block of mass m is placed on another block of
mass M lying on smooth horizontal surface, as
shown in the figure. The co-efficient of friction
between the blocks is . What is the maximum
horizontal force F that can be applied to the block
M so that the blocks move together ?
(A) (M + m) g
(B) (M – m) g
(C) (M – mg) 
(D) None of these
F
m
M
Solution: If there is no relative motion between the blocks then acceleration of the
blocks is a =
m
M
F

F.B.D. of m relative to M
For vertical equilibrium N = mg (1)
For horizontal equilibrium fl  ma
 mg  m
 
m
M
F

 F  (M + m)g
 Fmax = (M + m) g. Ans (A)
ma
mg
f
N
58. When a force F acts on a body of mass m, the acceleration
produced in the body is a. If three equal forces F1 = F2 = F3 = F
act on the same body as shown in Figure the acceleration
produced is
(A)  
2 1 a
 (B)  
2 1 a

(C) 2 a (D) None of these
Solution: Fnet = 2 2 2 2 2
F F (1 sin45) F cos 45
  
= F 1 1 1 2sin45
  
= F 3 2

A
B
4 kg
F2
F1
F3
m
1350
900

30. a =
F 3 2
m

Ans. (D)
59. A body is moving down a long inclined plane of angle of inclination . The coefficient
of friction between the body and the plane varies as  = 0.5 x, where x is the
distance moved down the plane. The body will have the maximum velocity when it
has travelled a distance x given by
(A) x = 2 tan  (B)
2
x=
tanθ
(C) x= 2cot (D)
2
x=
cot 
Solution:
mgcos
mgsin
N
r
F

N = mg cos
Velocity will be maximum when Fr max  N  mg sin
0.5x – mg cos = mg sin
x = 2tan
Ans. (A)
60. A man of mass 60 kg is standing on a horizontal
conveyer belt. When the belt is given an
acceleration of 1 ms–2, the man remains stationary
with respect of the moving belt. If g = 10 ms–2, the
net force acting on the man is
(A) 0.6 N (B) 6 N
(C) 60 N (D) 600 N
Solution: 1
r
F 60A 60 1
  
Fnet = 1
r
F = 60N
Ans. (C)
a = 1 ms-2