# Chapter 7 Law of Motion Part 1 – Physics free study material by TEACHING CARE online tuition and coaching classes

Chapter 7 Law of Motion Part 1 – Physics free study material by TEACHING CARE online tuition and coaching classes

# Point Mass.

• An object can be considered as a point object if during motion in a given time, it covers distance much greater than its own
• Object with zero dimension considered as a point
• Point mass is a mathematical concept to simplify the

# Inertia.

• Inherent property of all the bodies by virtue of which they cannot change their state of rest or uniform motion along a straight line by their own is called
• Inertia is not a physical quantity, it is only a property of the body which depends on mass of the
• Inertia has no units and no dimensions
• Two bodies of equal mass, one in motion and another is at rest, possess same inertia because it is a factor of mass only and does not depend upon the

# Linear Momentum.

• Linear momentum of a body is the quantity of motion contained in the
• It is measured in terms of the force required to stop the body in unit
• It is measured as the product of the mass of the body and its velocity e., Momentum = mass × velocity.

If a body of mass m is moving with velocity v then its linear momentum p is given by p = mv

• It is a vector quantity and it’s direction is the same as the direction of velocity of the
• Units : kgm/sec [S.I.], g-cm/sec [C.G.S.]

• Dimension : [MLT -1]
• If two objects of different masses have same momentum, the lighter body possesses greater

p = m1v1 = m2v2 = constant

\             v1

v2

m2

m1

i.e.

v µ 1

m

[As p is constant]

• For a given body

p µ v

• For different bodies at same velocities

p µ m

# Newton’s First Law.

A body continue to be in its state of rest or of uniform motion along a straight line, unless it is acted upon by some external force to change the state.

• If no net force acts on a body, then the velocity of the body cannot change e. the body cannot accelerate.
• Newton’s first law defines inertia and is rightly called the law of Inertia are of three types :

Inertia of rest, Inertia of motion, Inertia of direction

• Inertia of rest : It is the inability of a body to change by itself, its state of This means a body at rest remains at rest and cannot start moving by its own.

Example : (i) A person who is standing freely in bus, thrown backward, when bus starts suddenly.

When a bus suddenly starts, the force responsible for bringing bus in motion is also transmitted to lower part of body, so this part of the body comes in motion along with the bus. While the upper half of body (say above the waist) receives no force to overcome inertia of rest and so it stays in its original position. Thus there is a relative displacement between the two parts of the body and it appears as if the upper part of the body has been thrown backward.

Note : @If the motion of the bus is slow, the inertia of motion will be transmitted to the body of the person uniformly and so the entire body of the person will come in motion with the bus and the person will not experience any jerk.

• When a horse starts suddenly, the rider tends to fall backward on account of inertia of rest of upper part of the body as explained
• A bullet fired on a window pane makes a clean hole through it while a stone breaks the whole window because the bullet has a speed much greater than the stone. So its time of contact with glass is So in case of bullet the motion is

transmitted only to a small portion of the glass in that small time. Hence a clear hole is created in the glass window, while in case of ball, the time and the area of contact is large. During this time the motion is transmitted to the entire window, thus creating the cracks in the entire window.

• In the arrangement shown in the figure :

• If the string B is pulled with a sudden jerk then it will experience tension while due to inertia of rest of mass

M this force will not be transmitted to the string A and so the string B will break.

• If the string B is pulled steadily the force applied to it will be transmitted from string B to A through the mass M and as tension in A will be greater than in B by Mg (weight of mass M) the string A will
• If we place a coin on smooth piece of card board covering a glass and strike the card board piece suddenly with a The cardboard slips away and the coin falls into the glass due to inertia of rest.
• The dust particles in a durree falls off when it is beaten with a This is because the beating sets the durree in motion whereas the dust particles tend to remain at rest and hence separate.
• Inertia of motion : It is the inability of a body to change itself its state of uniform motion e., a body in uniform motion can neither accelerate nor retard by its own.

Example : (i) When a bus or train stops suddenly, a passenger sitting inside tends to fall forward. This is because the lower part of his body comes to rest with the bus or train but the upper part tends to continue its motion due to inertia of motion.

• A person jumping out of a moving train may fall
• An athlete runs a certain distance before taking a long This is because velocity acquired by running is added to velocity of the athlete at the time of jump. Hence he can jump over a longer distance.
• Inertia of direction : It is the inability of a body to change by itself direction of

Example : (i) When a stone tied to one end of a string is whirled and the string breaks suddenly, the stone flies off along the tangent to the circle. This is because the pull in the string was forcing the stone to move in a circle. As soon as the string breaks, the pull vanishes. The stone in a bid to move along the straight line flies off tangentially.

• The rotating wheel of any vehicle throw out mud, if any, tangentially, due to directional inertia.
• When a car goes round a curve suddenly, the person sitting inside is thrown

Problem 1.   When a bus suddenly takes a turn, the passengers are thrown outwards because of

[AFMC 1999; CPMT 2000, 2001]

• Inertia of motion (b) Acceleration of motion

(c) Speed of motion                                                  (d) Both (b) and (c)

Solution : (a)

Problem 2.       A person sitting in an open car moving at constant velocity throws a ball vertically up into air. The ball fall

[EAMCET (Med.) 1995]

(a) Outside the car                                                    (b) In the car ahead of the person

(c) In the car to the side of the person                            (d) Exactly in the hand which threw it up

Solution : (d)     Because the horizontal component of velocity are same for both car and ball so they cover equal horizontal distances in given time interval.

# Newton’s Second Law.

• The rate of change of linear momentum of a body is directly proportional to the external force applied on the body and this change takes place always in the direction of the applied

• If a body of mass m, moves with velocity v then its linear momentum can be given by p = mv

®

and if force

F is applied on a body, then

r     dp F µ dt

Þ F = K dp

dt

or           r = dp

(K = 1 in C.G.S. and S.I. units)

F       dt

or           r = d

r  = m dv =    r

(As a = dv = acceleration produced in the body)

F

\             F =

dt (mv)

r

ma

dt      ma                             dt

Force = mass ´ acceleration

Problem 3.       A train is moving with velocity 20 m/sec. on this, dust is falling at the rate of 50 kg/min. The extra force required to move this train with constant velocity will be                                                                                   [RPET 1999]

(a) 16.66 N                                   (b) 1000 N                          (c) 166.6 N                             (d) 1200 N

Solution : (a)     Force

F = v dm

dt

= 20 ´ 50 = 16.66 N

60

Problem 4.       A force of 10 Newton acts on a body of mass 20 kg for 10 seconds. Change in its momentum is [MP PET 2002]

(a) 5 kg m/s                                  (b) 100 kg m/s                    (c) 200 kg m/s                       (d) 1000 kg m/s Solution : (b)     Change in momentum = force ´ time = 10 ´ 10 = 100 kg m / sec

Problem 5.       A vehicle of 100 kg is moving with a velocity of 5 m/sec. To stop it in

1 sec, the required force in opposite

10

direction is                                                                                                                         [MP PET 1995]

(a) 5000 N                                    (b) 500 N                             (c) 50 N                                   (d) 1000 N

Solution : (a)

m = 100 kg

u = 5 m / s,

v = 0 t = 0.1 sec

Force mdvm(v u)100(0 5)

# Force.

dt               t

F = -5000N

0.1

• Force is an external effect in the form of a push or pulls which
• Produces or tries to produce motion in a body at
• Stops or tries to stop a moving
• Changes or tries to change the direction of motion of the

 F   u = 0 v = 0 Body remains at rest. Here force is trying to change the state of rest. F F       v     F = mg Body starts moving. Here force changes the state of rest. u = 0 v > 0 In a small interval of time, force increases the magnitude of speed and F u ¹ 0 v > u direction of motion remains same. F u     v < u In a small interval of time, force decreases the magnitude of speed and direction of motion remains same. v In uniform circular motion only direction of velocity changes, speed remains F constant. Force is always perpendicular to velocity. v In non-uniform circular motion, elliptical, parabolic or hyperbolic motion force acts at an angle to the direction of motion. In all these motions. Both magnitude and direction of velocity changes.

• Dimension : Force = mass ´ acceleration

[F] = [M][LT -2 ] = [MLT -2 ]

• Units : Absolute units : (i) Newton (S.I.) (ii) Dyne (C.G.S)

Gravitational units : (i) Kilogram-force (M.K.S.) (ii) Gram-force (C.G.S)

Newton :          One Newton is that force which produces an acceleration of

1m / s 2

in a body of mass

1 Kilogram. \1 Newton = 1kg m / s 2

Dyne :             One dyne is that force which produces an acceleration of 1 gram. \ 1 Dyne = 1gmcm / sec 2

1cm / s 2

in a body of mass

Relation between absolute units of force 1 Newton = 105 Dyne

Kilogram-force : It is that force which produces an acceleration of

\ 1 kg-f = 9.81 Newton

Gram-force :     It is that force which produces an acceleration of

\ 1 gm-f = 980 Dyne

9.8m / s 2

980cm / s 2

in a body of mass 1 kg. in a body of mass 1gm.

Relation between gravitational units of force : 1 kg-f = 107 gm-f

r

F = ma

formula is valid only if force is changing the state of rest or motion and the mass of the body is

constant and finite.

• If m is not constant

r = d       r = m dv + r dm

F      dt (mv)

dt      v dt

• If force and acceleration have three component along x, y and z axis, then

r        ˆ     ˆ

ˆ and r

ˆ     ˆ     ˆ

F = Fxi + Fy j + Fzk             a = axi + ay j + azk

From above it is clear that

Fx   = max , Fy   = may , Fz

= maz

• No force is required to move a body uniformly along a straight

F = ma

\ F = 0

(As a = 0 )

• When force is written without direction then positive force means repulsive while negative force means

Example : Positive force – Force between two similar charges

Negative force – Force between two opposite charges

• Out of so many natural forces, for distance 10 -15 metre, nuclear force is strongest while gravitational force

weakest.

Fnuclear > Felectromagnetic

• Fgravitational

• Ratio of electric force and gravitational force between two electron

Fe / Fg

= 1043

\ Fe

>> Fg

• Constant force : If the direction and magnitude of a force is It is said to be a constant force.
• Variable or dependent force :
• Time dependent force : In case of impulse or motion of a charged particle in an alternating electric field force is time

• Position dependent force : Gravitational force between two bodies

or                                   Force between two charged particles =

• Velocity dependent force : Viscous force (6phrv)

Force on charged particle in a magnetic field (qvB sinq )

Gm1m2 r 2

q1q2     .

4pe 0r 2

• Central force : If a position dependent force is always directed towards or away from a fixed point it is said to be central otherwise non-central.

Example : Motion of earth around the sun. Motion of electron in an atom. Scattering of a-particles from a nucleus.

• Conservative or non conservative force : If under the action of a force the work done in a round trip is zero or the work is path independent, the force is said to be conservative otherwise non

Example : Conservative force : Gravitational force, electric force, elastic force. Non conservative force : Frictional force, viscous force.

• Common forces in mechanics :
• Weight : Weight of an object is the force with which earth attracts It is also called the force of gravity or the gravitational force.
• Reaction or Normal force : When a body is placed on a rigid surface, the body experiences a force which is perpendicular to the surfaces in Then force is called ‘Normal force’ or ‘Reaction’.
• Tension : The force exerted by the end of taut string, rope or chain against pulling (applied) force is called the The direction of tension is so as to pull the body.
• Spring force : Every spring resists any attempt to change its This resistive force increases with

change in length. Spring force is given by (unit N/m).

F = –Kx ; where x is the change in length and K is the spring constant

# Equilibrium of Concurrent Force.

• If all the forces working on a body are acting on the same point, then they are said to be
• A body, under the action of concurrent forces, is said to be in equilibrium, when there is no change in the state of rest or of uniform motion along a straight
• The necessary condition for the equilibrium of a body under the action of concurrent forces is that the vector sum of all the forces acting on the body must be

• Mathematically for equilibrium åFnet = 0

or   å Fx

= 0 ; å Fy = 0 ; , å Fz = 0

• Three concurrent forces will be in equilibrium, if they can be represented completely by three sides of a triangle taken in

• Lami’s Theorem : For concurrent forces

F1

sina

=   F2

sin b

=   F3

sin g

Problem 6. Three forces starts acting simultaneously on a particle moving with velocity v. These forces are represented in magnitude and direction by the three sides of a triangle ABC (as shown). The particle will now move with velocity                                                                                                                               [AIEEE 2003]

• v remaining unchanged
• Less than v
• Greater than v
• v in the direction of the largest force BC

Solution : (a)     Given three forces are in equilibrium i.e. net force will be zero. It means the particle will move with same velocity.

Problem 7.       Two forces are such that the sum of their magnitudes is 18 N and their resultant is perpendicular to the smaller force and magnitude of resultant is 12. Then the magnitudes of the forces are                                                     [AIEEE 2002]

(a) 12 N, 6 N                            (b) 13 N, 5N                       (c) 10 N, 8 N                         (d) 16 N, 2 N

Solution : (b)     Let two forces are F1 and F2 (F1 < F2 ) .

According to problem:

F1 + F2 = 18

…..(i)

Angle between F1 and resultant (R) is 90°

F sinq

\     tan 90 =        2             = ¥

F1 + F2 cosq

Þ F1 + F2 cosq = 0

Þ cosq = – F1

F2

and R 2 = F 2 + F 2 + 2F  F

cosq

…..(ii)

1         2           1 2

144 = F 2 + F 2 + 2F F

cosq

…..(iii)

1         2           1 2

by solving (i), (ii) and (iii) we get F1 = 5 N and F2 = 13N

Problem 8.       The resultant of two forces, one double the other in magnitude, is perpendicular to the smaller of the two forces. The angle between the two forces is                                                                 [KCET (Engg./Med.) 2002]

(a)

60 o

(b)

120 o

(c)

150 o

(d)

90 o

Solution: (b)      Let forces are F and 2F and angle between them is q and resultant makes an angle a with the force F.

tana =     2F sinq     = tan 90 = ¥

F + 2 F cosq

Þ F + 2F cosq = 0             \ cosq = -1 / 2 or q = 120°

Problem 9. A weightless ladder, 20 ft long rests against a frictionless wall at an angle of 60o with the horizontal. A 150 pound man is 4 ft from the top of the ladder. A horizontal force is needed to prevent it from slipping. Choose the correct magnitude from the following                                                                                               [CBSE PMT 1998]

(a) 175 lb                                 (b) 100 lb                            (c) 70 lb                                  (d) 150 lb

Solution: (c)      Since the system is in equilibrium therefore å Fx

= 0 and å Fy   = 0 \

F = R2 and W = R1

Now by taking the moment of forces about point B.

F.(BC) + W.(EC) = R1(AC)

[from the figure EC= 4 cos 60]

F.(20 sin 60) + W(4 cos 60) = R1(20 cos 60)

10 3F + 2W = 10R1

[As R1 = W]

\ F =

8W  =

10

8 ´ 150

10

= 70lb

Problem 10. A mass M is suspended by a rope from a rigid support at P as shown in the figure. Another rope is tied at the end Q, and it is pulled horizontally with a force F. If the rope PQ makes angle q with the vertical then the tension in the string PQ is

• F sinq
• F / sinq
• F cosq
• F / cosq

Solution: (b)      From the figure

For horizontal equilibrium

T sinq = F

T =   F

sinq

Problem 11. A spring balance A shows a reading of 2 kg, when an aluminium block is suspended from it. Another balance B shows a reading of 5 kg, when a beaker full of liquid is placed in its pan. The two balances are arranged such that the Al – block is completely immersed inside the liquid as shown in the figure. Then [IIT-JEE 1985]

• The reading of the balance A will be more than 2 kg
• The reading of the balance B will be less than 5 kg
• The reading of the balance A will be less than 2 kg. and that of B will be more than 5 kg
• The reading of balance A will be 2 kg. and that of B will be 5 kg.

Solution: (c)      Due to buoyant force on the aluminium block the reading of spring balance A

will be less than 2 kg but it increase the reading of balance B.

Problem 12.     In the following diagram, pulley

value of angle q will be

• 60o
• 30o
• 45o
• 15o

P1 is movable and pulley P2

is fixed. The

Solution: (b)      Free body diagram of pulley P1 is shown in the figure

For horizontal equilibrium T1 cosq = T2 cosq

and T1 = T2 = W

For vertical equilibrium

\ T1 = T2

T1 sinq + T2 sinq = W

Þ W sinq + W sinq = W

\ sinq = 1

2

or q = 30°

Problem 13.     In the following figure, the pulley is massless and frictionless. The relation between T1 , T2 and T3

will be

(a) T1 = T2 ¹ T3

(b)

T1 ¹ T2

= T3

• T1 ¹ T2 ¹ T3
• T1 = T2 = T3

Solution : (d)     Since through a single string whole system is attached so W2 = T3 = T2 = T1

Problem 14.     In the above problem (13), the relation between W1 and W2

will be

(a) W   =     W1

(b)

2W cosq

(c)

W   = W

(d) W

2 cosq

2     2 cosq                            1

2         1                                        2         W1

Solution : (a)     For vertical equilibrium

T1 cosq + T2 cosq = W1

2W2 cosq = W1

[As T1 = T2 = W2 ]

\ W2 =

W1    .

2 cosq

Problem 15.     In the following figure the masses of the blocks A and B are same and each equal to m. The tensions in the

strings OA and AB are mg on B. The T1 is

• mg

T2 and

T1 respectively. The system is in equilibrium with a constant horizontal force

• 2 mg

• 3 mg

• 5 mg

Solution : (b)     From the free body diagram of block B

T1 cosq1 = mg ……(i)

T1 sinq1 = –mg …..(ii)

 1

by squaring and adding T 2 (sin 2 q1

• cos 2 q1

) = 2(mg)2

\ T1 =

2mg

Problem 16.     In the above problem (15), the angle q 1 is

(a) 30o                             (b) 45o                          (c) 60o                            (d)

tan1 æ 1 ö

 ç   ÷

2

è ø

Solution : (b)     From the solution (15) by dividing equation(ii) by equation (i)

T1 sinq1mg

T1 cosq1      mg

\ tanq1 = 1 or q1 = 45°

Problem 17.     In the above problem (15) the tension T2 will be

(a) mg                                       (b)

2 mg

(c)

3 mg

(d)

5 mg

Solution : (d)     From the free body diagram of block A

For vertical equilibrium T2 cosq 2 = mg + T1 cosq1

T2 cosq 2 = mg +

T2 cosq 2 = 2mg

2mg cos 45°

…..(i)

For horizontal equilibrium T2

sinq 2

= T1 sinq1 =

2mg sin 45°

T2 sinq 2

= mg

…..(ii)

by squaring and adding (i) and (ii) equilibrium

 2

T 2 = 5(mg)2 or

T2 =

5mg

Problem 18,     In the above problem (15) the angle q 2 will be

(a) 30o                             (b) 45o                          (c) 60o                            (d)

tan1 æ 1 ö

 ç   ÷

2

è ø

Solution : (d)     From the solution (17) by dividing equation(ii) by equation (i)

sinq 2 = mg  Þ

tanq   = 1

\ q 2 = tan1é 1 ù

cosq 2

2mg                   2     2

êë 2 úû

Problem 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 in equilibrium, the force exerted by the men on the rope will be

• (M + m)g

1 (M + m) g

2

• Mg
• mg

Solution : (b)     From the free body diagram of man and crate system: For vertical equilibrium

2T = (M + m)g

T (M + m)g

2

Problem 20.     Two forces, with equal magnitude F, act on a body and the magnitude of the resultant force is between the two forces is

F . The angle

3

(a)

cos -1 æ- 17 ö

(b)

cos 1 æ- 1 ö

(c)

cos 1 æ 2 ö

(d)

cos 1 æ 8 ö

ç 18 ÷

ç   3 ÷

ç 3 ÷

ç 9 ÷

è          ø                                       è        ø                                     è   ø                                             è ø

Solution : (a)     Resultant of two vectors A and B, which are working at an angle q , can be given by

R =                                                                                             [As

A = B = F and R = F ]

3

æ F ö 2

ç 3 ÷

= F 2

• F 2

+ 2F 2

cosq

è     ø

F 2 = 2F 2 + 2F 2 cosq         Þ   17 F 2 = 2F 2 cosq Þ cosq = æ  17 ö

or q = cos1æ  17 ö

ç          ÷

9                                           9                                          è 18 ø

ç 18 ÷

 è
 ø

Problem 21. A cricket ball of mass 150 gm is moving with a velocity of 12 m/s and is hit by a bat so that the ball is turned back with a velocity of 20 m/s. The force of blow acts for 0.01s on the ball. The average force exerted by the bat on the ball is

(a) 480 N                                 (b) 600 N                             (c) 500 N                                (d) 400 N

Solution : (a)

v1 = -12m / s

and

v2 = +20m / s

[because direction is reversed]

m = 150gm = 0.15kg , t = 0.01sec

Force exerted by the bat on the ball F = m[v2 – v1 ] = 0.15[20  (12)]

= 480 Newton

# Newton’s Third Law.

t                          0.01

To every action, there is always an equal (in magnitude) and opposite (in direction) reaction.

• When a body exerts a force on any other body, the second body also exerts an equal and opposite force on the
• Forces in nature always occurs in A single isolated force is not possible.
• Any agent, applying a force also experiences a force of equal magnitude but in opposite The force applied by the agent is called ‘Action’ and the counter force experienced by it is called ‘Reaction’.
• Action and reaction never act on the same If it were so the total force on a body would have always been zero i.e. the body will always remain in equilibrium.

• If (Reaction)

F AB = force exerted on body A by body B (Action) and

F BA = force exerted on body B by body A

Then according to Newton’s third law of motion

F AB = –F BA

• Example : (i) A book lying on a table exerts a force on the table which is equal to the weight of the This is the force of action.

The table supports the book, by exerting an equal force on the book. This is the force of reaction.

As the system is at rest, net force on it is zero. Therefore force of action and reaction must be equal and opposite.

• Swimming is possible due to third law of
• When a gun is fired, the bullet moves forward (action). The gun recoils backward (reaction)
• Rebounding of rubber ball takes place due to third law of
• While walking a person presses the ground in the backward direction (action) by his feet. The ground pushes the person in forward direction with an equal force (reaction). The component of reaction in horizontal direction makes the person move
• It is difficult to walk on sand or
• Driving a nail into a wooden block without holding the block is

Problem 22.     You are on a frictionless horizontal plane. How can you get off if no horizontal force is exerted by pushing against the surface

• By jumping (b) By splitting or sneezing

(c) By rolling your body on the surface                         (d) By running on the plane

Solution : (b)     By doing so we can get push in backward direction in accordance with Newton’s third law of motion.

# Frame of Reference.

• A frame in which an observer is situated and makes his observations is known as his ‘Frame of reference’.
• The reference frame is associated with a co-ordinate system and a clock to measure the position and time of events happening in space. We can describe all the physical quantities like position, velocity, acceleration etc. of an object in this coordinate

• Frame of reference are of two types : (i) Inertial frame of reference (ii) Non-inertial frame of

## (i)  Inertial frame of reference :

• A frame of reference which is at rest or which is moving with a uniform velocity along a straight line is called an inertial frame of
• In inertial frame of reference Newton’s laws of motion holds
• Inertial frame of reference are also called unaccelerated frame of reference or Newtonian or Galilean frame of
• Ideally no inertial frame exist in For practical purpose a frame of reference may be considered as inertial if it’s acceleration is negligible with respect to the acceleration of the object to be observed.
• To measure the acceleration of a falling apple, earth can be considered as an inertial

• To observe the motion of planets, earth can not be considered as an inertial frame but for this purpose the sun may be assumed to be an inertial

Example : The lift at rest, lift moving (up or down) with constant velocity, car moving with constant velocity on a straight road.

## (ii)  Non inertial frame of reference :

• Accelerated frame of references are called non-inertial frame of
• Newton’s laws of motion are not applicable in non-inertial frame of

Example : Car moving in uniform circular motion, lift which is moving upward or downward with some acceleration, plane which is taking off.

# Impulse.

• When a large force works on a body for very small time interval, it is called impulsive

An impulsive force does not remain constant, but changes first from zero to maximum and then from maximum to zero. In such case we measure the total effect of force.

• Impulse of a force is a measure of total effect of

• I

=   t2 F dt .

 ò

t1

• Impulse is a vector quantity and its direction is same as that of
• Dimension : [ MLT -1 ]
• Units : Newton-second or Kg-m- s-1 (S.I.) and     Dyne-second or gm-cm- s-1 (C.G.S.)
• Force-time graph : Impulse is equal to the area under F-t

If we plot a graph between force and time, the area under the curve and time axis gives the value of impulse.

I = Area between curve and time axis

= 1 ´

2

Base ´ Height

• If

Fav

= 1 F t

2

is the average magnitude of the force then

I =    t2 F dt = F       t2 dt = F  Dt

 1
 1

òt                              av òt                        av

• From Newton’s second law ®= dp

F       dt

 ò

or            tF dt =

t1

p2 dp

 ò

p1

r

Þ I = p 2

• p1

= Dp

i.e. The impulse of a force is equal to the change in momentum. This statement is known as Impulse momentum theorem.

• Examples : Hitting, kicking, catching, jumping, diving, collision etc.

In all these cases an impulse acts.

I = ò

F dt = Fav . Dt = Dp = constant

So if time of contact Dt is increased, average force is decreased (or diluted) and vice-versa.

• In hitting or kicking a ball we decrease the time of contact so that large force acts on the ball producing greater
• In catching a ball a player by drawing his hands backwards increases the time of contact and so, lesser force acts on his hands and his hands are saved from getting
• In jumping on sand (or water) the time of contact is increased due to yielding of sand or water so force is decreased and we are not However if we jump on cemented floor the motion stops in a very short interval of time resulting in a large force due to which we are seriously injured.
• An athlete is advised to come to stop slowly after finishing a fast So that time of stop increases and hence force experienced by him decreases.
• China wares are wrapped in straw or paper before

Problem 23.     A ball of mass 150g moving with an acceleration

20m / s 2

is hit by a force, which acts on it for 0.1 sec. The

impulsive force is                                                                                                              [AFMC 1999]

(a) 0.5 N-s                               (b) 0.1 N-s                           (c) 0.3 N-s                               (d) 1.2 N-s

Solution : (c)     Impulsive force = force ´ time

= ma ´ t

= 0.15 ´ 20 ´ 0.1 = 0.3 N-s

Problem 24.     A force of 50 dynes is acted on a body of mass 5 g which is at rest for an interval of 3 seconds, then impulse is

[AFMC 1998]

(a)

0.15 ´ 10 3 N s

(b)

0.98 ´ 10 3 N s

1.5 ´ 10 3 Ns

2.5 ´ 10 3 N s

Solution : (c)

Impulse = force ´ time

= 50 ´ 10 5 ´ 3

= 1.5 ´ 103 N – s

Problem 25.     The force-time (F t) curve of a particle executing linear motion is as shown in the figure. The momentum acquired by the particle in time interval from zero to 8 second will be                                           [CPMT 1989]

• – 2 N-s

(b) + 4 N-s

• 6 N-s
• Zero

Solution : (d) Momentum acquired by the particle is numerically equal to the area enclosed between the Ft curve and time Axis. For the given diagram area in a upper half is positive and in lower half is negative (and equal to the upper half). So net area is zero. Hence the momentum acquired by the particle will be zero.

# Law of Conservation of Linear Momentum.

If no external force acts on a system (called isolated) of constant mass, the total momentum of the system remains constant with time.

• According to this law for a system of particles

F dp

dt

In the absence of external force

F = 0

then

p = constant

i.e.,

p = p1 + p 2 + p 3 +….. = constant.

®                     ®                    ®

or          m1 v1 + m2 v2 + m3 v3 +….. = constant

This equation shows that in absence of external force for a closed system the linear momentum of individual particles may change but their sum remains unchanged with time.

• Law of conservation of linear momentum is independent of frame of reference though linear momentum depends on frame of
• Conservation of linear momentum is equivalent to Newton’s third law of

For a system of two particles in absence of external force by law of conservation of linear momentum.

p1 + p 2 = constant.

\             m1v1 + m2v2 =

constant.

Differentiating above with respect to time

m1 dv1 + m2 dv2 = 0

Þ m1a1 + m2a2 = 0

Þ F 1 + F 2 = 0

dt                dt

\             F 2 = –F 1

i.e. for every action there is equal and opposite reaction which is Newton’s third law of motion.

• Practical applications of the law of conservation of linear momentum
• When a man jumps out of a boat on the shore, the boat is pushed slightly away from the

• A person left on a frictionless surface can get away from it by blowing air out of his mouth or by throwing some object in a direction opposite to the direction in which he wants to move.
• Recoiling of a gun : For bullet and gun system, the force exerted by trigger will be internal so the momentum of the system remains

Let mG  =

mass of gun, mB =

mass of bullet,

vG   = velocity of gun, vB   =

velocity of bullet

Initial momentum of system = 0

Final momentum of system = mGvG + mBvB

By the law of conservation linear momentum

mGvG  + mBvB   = 0

So recoil velocity

r   = – mB  r

 v
 vG

mG     B

• Here negative sign indicates that the velocity of recoil vG

is opposite to the velocity of the bullet.

• v

µ   1

G            m

i.e. higher the mass of gun, lesser the velocity of recoil of gun.

G

• While firing the gun must be held tightly to the shoulder, this would save hurting the shoulder because in this condition the body of the shooter and the gun behave as one body. Total mass become large and recoil velocity becomes too
 G

v     µ       1

mG + m

man

• Rocket propulsion : The initial momentum of the rocket on its launching pad is zero. When it is fired from the launching pad, the exhaust gases rush downward at a high speed and to conserve momentum, the rocket moves

Let m0 =

initial mass of rocket,

m = mass of rocket at any instant ‘t’ (instantaneous mass)

mr =

residual mass of empty container of the rocket

u = velocity of exhaust gases,

v = velocity of rocket at any instant ‘t’ (instantaneous velocity)

dm =

dt

rate of change of mass of rocket = rate of fuel consumption

= rate of ejection of the fuel.

• Thrust on the rocket :

F = –u dm mg dt

Here negative sign indicates that direction of thrust is opposite to the direction of escaping gases.

F = –u dm

dt

(if effect of gravity is neglected)

• Acceleration of the rocket : a = u

m

dm – g dt

and if effect of gravity is neglected a = u dm

m dt

• Instantaneous velocity of the rocket : v = u log

æ m0 ö – gt

 e ç          ÷

m

and if effect of gravity is neglected v = u log

è

æ m0 ö

ø

= 2.303u log

æ m0 ö

e ç m ÷                      10 ç m ÷

è      ø                         è      ø

æ m0 ö

• Burnt out speed of the rocket : vb= vmax

= u log e ç        ÷

 m

è    r ø

The speed attained by the rocket when the complete fuel gets burnt is called burnt out speed of the rocket. It is the maximum speed acquired by the rocket.

Problem 26.     A wagon weighing 1000 kg is moving with a velocity 50 km/h on smooth horizontal rails. A mass of 250 kg is dropped into it. The velocity with which it moves now is                                                                 [MP PMT 1994]

• 5 km/hour (b) 20 km/hour                  (c) 40 km/hour                      (d) 50 km/hour Solution : (c)      Initially the wagon of mass 1000 kg is moving with velocity of 50 km/h

So its momentum = 1000 ´ 50 kg ´ km

h

When a mass 250kg is dropped into it. New mass of the system = 1000 + 250 = 1250kg

Let v is the velocity of the system.

By the conservation of linear momentum : Initial momentum = Final momentum Þ 1000 ´ 50 = 1250 ´ v

\                                                                                          v = 50,000 = 40km / h.

1250

Problem 27.     The kinetic energy of two masses m1 and m2 are equal their ratio of linear momentum will be                       [RPET 1988]

(a) m1/m2                         (b) m2/m1                      (c)                                   (d)

Solution : (c)     Relation between linear momentum (P), man (m) and kinetic energy (E)

P =                   Þ  P µ            [as E is constant]

\ P1 =

P2

Problem 28.     Which of the following has the maximum momentum

(a) A 100 kg vehicle moving at 0.02 ms–1                  (b) A 4 g weight moving at 10000 cms–1

(c) A 200 g weight moving with kinetic energy 10–6 J (d) A 20 g weight after falling 1 kilometre Solution : (d)                        Momentum of body for given options are :

(a) P = mv = 100 ´ 0.02 = 2kgm / sec

(b) P = mv = 4 ´ 103 ´ 100 = 0.4kgm / sec

• P =

• P= m

=

= 20 ´ 10 3 ´

= 6.3 ´ 10 4 kgm / sec

= 2.82kgm / sec

So for option (d) momentum is maximum.

Problem 29.     A rocket with a lift-off mass 3.5 ´ 104 kg is blasted upwards with an initial acceleration of 10m / s 2 .

Then the

initial thrust of the blast is                                                                                                         [AIEEE 2003]

(a)

1.75 ´ 105 N

(b)

3.5 ´ 105 N

(c)

7.0 ´ 105 N

(d)

14.0 ´ 105 N

Solution : (c)      Initial thrust on the rocket F = m(g + a)

= 3.5 ´ 104 (10 + 10)

= 7.0 ´ 105 N

Problem 30.     In a rocket of mass 1000 kg fuel is consumed at a rate of 40 kg/s. The velocity of the gases ejected from the rocket is 5 ´ 104 m / s . The thrust on the rocket is                                                                           [MP PMT 1994]

(a)

2 ´ 103 N

(b)

5 ´ 104 N

(c)

2 ´ 106 N

(d)

2 ´ 109 N

Solution : (c)     Thrust on the rocket F = udm

dt

= 5 ´ 104 (40)

= 2 ´ 106 N

Problem 31.     If the force on a rocket moving with a velocity of 300 m/s is 210 N, then the rate of combustion of the fuel is

[CBSE PMT 1999]

(a) 0.7 kg/s                                 (b) 1.4 kg/s                            (c) 0.07 kg/s                              (d) 10.7 kg/s

Solution : (a)     Force on the rocket = udm \ Rate of combustion of fuel æ dm ö = F = 210 = 0.7kg / s.

dt                                                             ç dt ÷      u       300

è        ø

Problem 32.     A rocket has a mass of 100 kg. 90% of this is fuel. It ejects fuel vapours at the rate of 1 kg/sec with a velocity of

500 m/sec relative to the rocket. It is supposed that the rocket is outside the gravitational field. The initial upthrust on the rocket when it just starts moving upwards is                                                              [NCERT    1978]

(a) Zero                           (b) 500 N                             (c) 1000 N                              (d) 2000 N

 è
 ø

Solution : (b)     Up thrust force F = u æ dm ö = 500 ´ 1 = 500 N

ç dt ÷

# Free Body Diagram.

In this diagram the object of interest is isolated from its surroundings and the interactions between the object and the surroundings are represented in terms of forces.

Example :

# Apparent Weight of a Body in a Lift.

When a body of mass m is placed on a weighing machine which is placed in a lift, then actual weight of the body is mg.

This acts on a weighing machine which offers a reaction R given by the reading of weighing machine. This reaction exerted by the surface of contact on the body is the apparent weight of the body.

 Condition Figure Velocity Acceleration Reaction Conclusion Lift is at rest LIFT   R     Spring Balance   mg   LIFT   R     Spring Balance   mg   LIFT   R     Spring Balance   mg   LIFT   R     Spring Balance   mg     LIFT   R     Spring Balance   mg a               g                   a v = 0 a = 0 R – mg = 0 \ R = mg Apparent weight = Actual weight Lift moving upward or downward with constant velocity v = constant a = 0 R – mg = 0 \ R = mg Apparent weight = Actual weight Lift accelerating upward at the rate of ‘a’ v = variable a < g R – mg = ma \R = m(g + a) Apparent weight > Actual weight Lift accelerating upward at the rate of ‘g’ v = variable a = g R – mg = mg R = 2mg Apparent weight = 2 Actual weight Lift accelerating downward at the rate of ‘a’ v = variable a < g mg – R = ma \ R = m(g – a) Apparent weight < Actual weight