1

WORK, POWER, ENERGY & CONSERVATION LAWS


1.

Ø Ø

WORK Whenever a force acting on a body displaces it, work is said to be done by the force. Work done by a force is equal to scalar product of force applied and displacement of the body. Constant

Force 1.1

Variable force

Work done by a constant force : If the direction and magnitude of a force applying on a body is constant, the force is said to be constant. Work done by a constant force, W = Force × component of displacement along force = displacement × component of force along displacement. → If a → F force is acting on a body at an angle θ to the horizontal and the displacement r is along the horizontal, the work done will, be W = (F cos θ) r F

= F (r cos θ)

θ

Fcosθ r

In vector from, W =

→ →

F. r

→

→

If F = î Fx + ˆj Fy + kˆFz and r = î x + ˆj y + kˆz , the work done will be, W = Fx . x + Fy + Fz

→

during displacement of body through it, the force is supposed to be constant. It d r be small displacement of →

→

P2

os

θ

body and → F be the force applying on the body, the work done by force is dW = F .d Fr ...... (i) The total work done in displacing body from P1 to P2 is given by,

or

∫

P2 →

∫P

P2 →

→

dr

1

W=

∫P

F .d r

1

→

→

F .d r

dW =

Fc

1.2

Note : The force of gravity is the example of constant force, hence work done by it is the example of work done by a constant force. Work done by a variable force If the force applying on a body is changing in its direction or magnitude or both, the force is said to be variable suppose a constant force causes displacement in a body from position P1 to position P2. To calculate the work done by the force the path from P1 to P2 can be divided into infinitesimal element, each element is so small that

P1

→

If r and r be the position vectors of the points P1 and P2 respectively, the total work done will be 1 2

W=

r2 →

∫r

→

F .d r

1

Note : When we consider a block attached to a spring, the force on the block is k times the elongation of the spring, where k is spring constant. As the elongation changes with the motion of the block, therefore the force is variable. This is an example of work done by variable force.


1.3

Calculation of work done from force displacement graph : →

Suppose a body, whose initial position is r1, is acted upon by a variable force F and consequently the body acquires its final position r2. From position r to r + dr or for small displacement dr, the work done will be Fdr whose value will the area of the shaded strip of width dr. The work done on the body in displacing it from position r1 to r2 will be equal to the sum of areas of all the such strips P2 F

r2

Thus, total work done, W = ∑ dW

P1

r1

O

M r1

r2

=

dr

N r2

∑ F.dr r1

= Area of P1P2NM The area between the graph between force and displacement axis is equal to the work done. Note : To calculate the work done by graphical method, for the sake of simplicity, here we have assumed the direction of force and displacement as same, but if they are not in same direction, the graph must be plotted between F cos θ and r. (i) Work is a scalar quantity (ii) The dimensions of work : ML2T–2 (iii) Unit of work : there are two types of unit of work (a) Absolute unit : Joule (in M.K.S), Erg (in C.G.S.) (Note : 107 erg = 1 joule) (b) Gravitational unit : Kilogram - metre (in M.K.S), Gram-cm (in C.G.S) (Note : 1 kilogram metre = 9.8 joule = 105 gram cm) 1.4

Nature of work done Although work done is a scalar quantity, yet its value may be positive, negative or even zero

(a)

→ →

Positive work : As W = F . r = F r cos θ ∴ When θ is acute (<90º) cos θ is positive. Hence work done is positive. Example (i) When a body falls freely under the action of gravity θ = 0º, cos θ = + 1, therefore work done by gravity on a body, falling freely is positive.

(b)

Negative work : When θ is obtuse (>90º), cos θ is negative. Hence work done is negative Example →

(i) When a body is through up, its motion is opposed by gravity. The angle θ between the gravitational force F →

and displacement r is 180º. As cos θ = – 1,

therefore, work done by gravity is negative.

(ii) When a body is moved over a rough horizontal surface, the motion is opposed by the force of friction. Hence work done by frictional force in negative. Note that work done by the applied force is not negative (iii) When a positive charge is moved closer to another positive charge, work done by electrostatic force of repulsion between the charges is negative. (c)

→

→

Zero work : When force F or the displacement r or both are zero, work done W, will be zero. Again when →

→

angle θ between F and r is 90º, the work done will be zero.


Example →

(i) When we fail to move a heavy stone, however hard we may try, work done by us is zero, r = 0 (ii) When a collie carrying some load on his head moves on horizontal platform, θ = 90º. Therefore, workdone by the collie is zero. This is because θ = 90º (iii) Tension in the string of simple pendulum is always perpendicular to displacement of the bob. Therefore, work done by tension is always zero. Note : Another way of expressing negative or positive work is that when energy is transferred to the object work done is positive and when energy is transferred from object the work done is negative and hence the work which is a transfer of energy has same dimensions as energy. Example based on W ork Work

→

Ex.1 A position dependent force F = 7 − 2x + 3 x 2 acts on a small body of mass 2kg and displaces it from x = 0 to x = 5 m. The work done in joule will be X2

Sol.

W=

∫

5

Fdx =

X1

∫ 0

5

2 3 (7 − 2x + 3 x 2 )dx = 7x − 2x + 3x  = 135J   2 3 0 

Ex.2 For the force displacement diagram shown in adjoining diagram the work done by the force in displacing the body from x = 1 cm to x = 5 cm is 20 10 F 0 (In dyne) -10 -20

1 2 3 4

5 6 7 8

x-(cm)

Sol. Work = Area under the curve and displacement axis = 10 + 20 – 20 + 10 = 20 erg. Ex.3 A uniform chain of mass M and length L is lying on a frictionless table in such a way that its 1/3 part is hanging vertically down. The work done in pulling the chain up the table is  L Sol. If length x of the chain is pulled up on the table, then the length of hanging part of the chain would be  − x  and 3  

its weight would be

=

ML   − x  g dx L 3 

L/3

∴

w=

ML   − x  g. If it is pulled up further by a distance dx, the work done in pulling up. L 3 

∫ 0

ML MgL   − x  g dx = L 3 18 

Ex.4 The work done in pulling a body of mass 5 kg along an inclined plane (angle 60º) with coefficient of friction 0.2 through 2m, will be Sol. The minimum force with a body is to be pulled up along the inclined plane is mg (sinθ + µcosθ ) r r Work done, W = F . d = Fd cos θº = mg (sinθ + µ cosθ ) × d = 5 × 9.8 (sin 60º + 0.2 cos 60º) × 2 = 98.08 J


r Ex.5 A force F = (7 − 2x + 3 x 2 )N is applied on a 2 kg mass which displaces it from x = 0 to x = 5 m. Work doen in joule is

Sol.

5

5

∫

∫

0

0

W = F dx = (7 − 2 x + 3 x 2 ) dx

5

[7 x]50

2.

5

 2x 2   3x 3  −  +  = 135 joule  2  0  3  0

POWER (a) The time rate of doing work is called power →

dw → d x (b) Power = = F. dt dt

In translatory motion :

→ →

P = F.v → →

In rotational motion : P = τ . ω (c) It is scalar quantity (d) Unit : In MKS - J/sec, watt

In CGS - erg/sec, (Note : 1KW = 103 watt, 1 HP = 746 watt) (e) Dimension : [M1L2T–3] Note : Power is the rate at which applied force transfers energy (a) Power P =

w where w work is done in ∆t ∆t

(b) Instantaneous power P =

time

dw , it’s value may change with time. dt

Example based on Power Ex.6 A pump can lift 9000 kg coal per hour from a mine of depth 120 m. Assuming its efficiency to be 75% its power will be (in watts) Sol. Power =

work time

ouput power 3 output power Effeiciency of pump = input power ; = 4 P

∴ Output power =

∴

3P 4

mgh 3 P = t 4

P =

4 3

mgh t

=

4 9000 × 9.8 × 120 × = 3920 W 3 3600


Ex.7 A person of mass 60 kg is capable of taking a 15 kg massive objets to a height of 10 m in 3 miutes. The efficiency of person is work output Sol. % Efficiency η = work input × 100  m   × 100 =  m +M

mgh = (m + M)gh × 100

 15   × 100 = 20% =   15 + 60 

Ex.8 An automobile of mass m accelerates from rest. If the engine supplies constant power p, the velocity at time t is given by Sol. Given that power = Fv = p = constant dv v=p dt

or

m

or

∫ v dv = ∫ m dt

[as F = ma =

mdv dt

p

v2 p = t + C1 2 m Now as initially, the body is at rest i.e. v = 0 at t = 0 so, C1 = 0 ∴

v=

2pt m

Ex.9 In the above problem, the position (s) at time (t) is given by Sol. By definition v =

⇒

∫

ds dt

or 1/ 2

 2pt  ds =    m 

∫

ds  2pt  =  dt  m  1/ 2

dt

 2p  ⇒ s  m

1/ 2

[From (1)]

2 3/2 + C2 t 3 1/ 2

 8p   Now as t = 0, s = 0, so C2 = 0 ⇒ s =   9m 

t3 / 2

Ex.10 A particle moving in a straight line is acted by a force, which works at a constant rate and changes its velocity from u to v in passing over a distance x. The time taken will be Sol. The froce acting on the particle =

mdv dt

 mdv   v = k (constant) Power of the force =   dt 

at

t = 0, v = u ∴

Again

mdv v=k dt

Intergrating,

=

c=

mu 2 2

Now from (1),

⇒

m.v

1 m(v3 – u3) = kx 3

⇒

m

v2 mu 2 = kt + 2 2

m

v2 = kt + c 2 ⇒

....(1)

1 m(v2 – u2) = kt ....(2) 2

dv v=k dx

⇒ mv2 dv = kdx

....(3)

From (2) and (3),

3  v 2 − u2  x t = 2 3 3  v −u 


3.

ENERGY

Ø

The energy of a body is defined as the capacity of doing work.

Ø

There are various form of energy

Ø

(i) mechanical energy

(ii) chemical energy

(iii) electrical energy

(v) nuclear energy

(vi) sound energy

(vii) light energy etc

(iv) magnetic energy

Energy of system always remain constant it can neither be created nor it can be destroyed however it may be converted from one form to another Example Motor

Mechanical energy

→

Generator

Electrical energy

 →

Photocell

Electrical energy

Electrical energy

Heater   →

Heat energy

Electrical energy

Radio  →

Sound energy

Nuclear energy

→

Chemical energy

→

Electrical energy

→

Heat energy

      →

Electric energy

  →

Mechanical energy Light energy

Nuclear Re actor

Electrical energy

Cell

Electrical energy

Secondary Cell

Chemical energy

Incendence nt lamp

Light

Ø

Energy is a scalar quantity

Ø

Unit : Its unit is same as that of work or torque. In MKS : Joule, watt sec In CGS : Erg Note : 1 eV = 1.6 × 10–19 joule 1 KWh = 36 × 105 joule 107 erg = 1 joule

Ø

Dimension [M1L2T–2]

Ø

According to Einstein’s mass energy equivalence principle mass and energy are inter convertible i.e. they can be changed into each other Energy equivalent of mass m is, E = mc2 Where, m : mass of the particle c : velocity of light E : equivalent energy corresponding to mass m.

Ø

In mechanis we are concerned with mechanical energy only which is of two type (a) kinetic energy (ii) potential energy

3.1

Kinetic energy

Ø

The energy possessed by a body by virtue of its motion is called kinetic energy

Ø

If a body of mass m is moving with velocity v, its kinetic energy KE =

1 mv2, for translatory motion 2

KE =

1 Iω2, for rotational motion 2


Ø

Kinetic energy is always positive

Ø

If linear momentum of body is p, KE =

Ø

If angular momentum of body is J, KE =

Ø

p or J α E

p2 1 = mv 2 - for translatory motion 2m 2 J2 1 2 = Iω - for rotational motion 2I 2

p : momentum

E

E

E

1/P

P

E : kinetic energy

P

2

Ø

The kinetic energy of a moving body is measured by the amount of work which has been done in bringing the body from the rest position to its present moving position or

Ø

The kinetic energy of a moving body is measured by the amount of work which the body can do against the external forces before it comes to rest.

Ø

If a body performs translatory and roational motion simultaneously, its total kinetic energy =

1 1 mv 2 + Iω2 2 2

3.1.1 Work energy Theorem :

Ø

For translatory motion : Work done by all the external forces acting on a body is equal to change in its kinetic energy of translation. Work done by all the external forces = change in K.E of translation =

Ø

1 1 m1v12 − m 2 v 22 2 2

For rotational motion : Work done by all the external torque acting on a rigid body is equal to change in its rotational kinetic energy. Work done by all the external torque =

1 2 1 2 Iω1 − Iω2 2 2

Note : In simple words ∆K = Kf – Ki = W in the work energy theorem if only energy changed is kinetic energy. 3.2

Potential energy

Ø

The energy which a body has by virtue of its position or configuration in a conservation force field

Ø

Potential energy is a relative quantity.

Ø

Potential energy is defined only for conservative force field.

Ø

Potential energy of a body at any position in a conservation force field is defined as the workdone by an external agent against the action of conservation force in order to shift it from reference point. (PE = 0) to the present position or.

Ø

Potential energy of a body in a conservation force field is equal to the work done by the body in moving from its present position to reference position.

Ø

At reference position, the potential energy of the body is zero or the body has lost the capacity of doing work.

Ø

Relationship between conservative force field and potential energy (U) F −∇U = −grad (U) = –

→

Ex.

→

(i) U = 3x2 ⇒ F = −6 x î →

(ii) U = 2x2y + 3y2x + xz2

⇒ F = −( 4 xy + 3 y 2 + z 2 ) + (2x 2 + 6 xy) + (2xz)

∂U ˆ ∂U ˆ ∂U ˆ i− j− k ∂x ∂y ∂z

WORK, POWER, ENERGY & CONSERVATION LAWS x2

Ø

∫

dU U = − Fdx If force varies only with one dimension thenF = – or dx x 1

Ø

Potential energy may be positive or negative i) Potential energy is positive, if force field is repulsive in nature ii) Potential energy is negative, if force field is attractive in nature

Repulsion forces U+ve U-ve

r Attraction forces

Ø

If r ↑ (separation between body and force centre), U ↑, force field is attractive or vice-versa.

Ø

If r ↑, U ↓, force field is repulsive in nature.

3.2.1 Different types of forces and corresponding P.E (a)

Gravitational potential energy : For small distances above or below the earth surface : Reference point = Earth surface i.e. P.Esurface = 0 P.E above the earth surface is positive and below the earth surface is negative and in magnitude it is equal to mgh For greater distance : Reference point = ∞ i.e. P.E∞ = 0 P.E. = –

GMm , for r > R r

Where R = radius of earth, r = distance of body from the centre of earth, m = mass of body, M = mass of earth Electrostatic potential energy : Reference point = ∞ i.e. P.E∞ = 0 P.E. = –

KQ1Q 2 , [value of Q1 and Q2 are substituted with sign.] r

Intermolecular potential energy : Reference point = ∞ i.e. P.E∞ = 0  dU 6b  2a / b a b  = − 1 F=– U (r) = 12 − 6 , 7 6 dr r  r  r r Elastic potential energy : Which is associated with the state of compression of extension of an elastic object U (x) = spring constant, x = change in dimensions 3.2.2 Potential energy curve A graph plotted between the PE of a particle and its displacement from the centre of force field is called PE curve Using graph, we can predict the rate of motion of a particle at various positions. Force on the particle is F(x) = –

dU dx

1 kx2 where k = 2


Case : I On increasing x, if U increase, force is in (–)ve x-direction i.e. attraction force. Case : II On increasing x, if U decreases, force is in (+)ve x-direction i.e. repulsion force. Different positions of a particle Position of equilibrium If net force acting on a body is zero, it is said to be in equilibrium for equilibrium dU = 0 Points P, Q, R and S are the states of equilibrium positions. dx

3.2.3 Types of equilibrium : Stable equilibrium - When a particle is displaced slightly from a position and a force acting on it brings it back to the initial position, it is said to be in state equilibriums position.

Necessary conditions -

Ø

d2U dx 2

= + ve

Unstable equilibrium : When a particle is displaced slightly from a position a position and force acting on it tries to displace the practice further away from the equilibrium position, it is said to be in unstable equilibrium.

Condition :

Ø

dU = 0, dx

dU d2U = − ve = 0 potential energy is max i.e. = dx dx 2

Neutral equilibrium : In the neutral equilibrium potential energy is constant when a particle is displaced from its position it does not experiences any force to acting on it and continues to be in equilibrium in the displaced position, it is said to be in neutral equilibrium. Example based on Energy

Ex.11 A meter scale of mass m initially vertical is dispalced at 45º keeping the upper and fixed, the charge in PE will beSol. Work = change in PE = Force × displacement

l

l dU = mg (1 − cos θ) 2

= mg ×

1 (1 − cos 45 º ) 2

l/2 45º

G'

G

(Q l = 1m)

mg  1   = 2 1 − 2 

Ex.12 If the speed of a car increases 4 times, the stopping distance for this will increase by Sol. Work = Change in KE ∴ FS =

S' v' 2 = S v2

1 1 2 mv2 – 0 = mv 2 2

⇒

S' = 16 S

⇒ S’ = 16 S

Ex.13 If the potential energy function for a particle is U = a –

Sol. U = a –

∴

b c + x x2

dU b 2c =− 2 − 3 dx x x

.........(1)

.........(2)

b c + x x2

the force constant for oscillation will be.


and `

d2U 1  6c  = 3  −2b +  2 x  dx x 

.........(3)

for equilibrium dU =0 dx

2c b

∴ x=

∴ Substituting this value in (3)

as

d2U dx 2

=K

d2U

 b  =  2  2c  dx

3

6c  b 4  − + 2 b =  2c / b  8c 3 

∴ K = b4/8c3

Ex.14 On passing through a woodn sheet a bullet looses 1/20 of initial velocity. The minimum number of sheets required to completely stop the bullet will beSol. Use v2 = u2 + 2as for a sheet of thickness s v = (19/20)u

 19  2  20 u  = u + 2as  

2

 39  u 2  a=–   400  2s

2

2as = (361/400)u – u

suppose for n sheet v = 0 ∴ 02 = u2 + 2a (ns) n = –

u2 = 2as

u2  39  u 2 2 ×s   400  2s

≈ 11

Ex.15 The work done in taking out 2 lit of water using a bucket of mass 0.5 kg from a well of dpeth 6m will beSol. W = mgh = (mbucket + mwater)gh

[2 Lit water = 2 kg water]

= (0.5 + 2.00) × 9.8 × 6 = 15 × 9.8 = 147 J Ex.16 A body has velocity 200 m/s and its kinetic energy is 200 J. The mass of the body would be Sol.

1 mv 2 = E 2

or

 2E  2 × 200 m= 2=  v  (200 )2

=

4 × 102 1 = 4 100 4 × 10

∴

m = 0.01 kg

Ex.17 A body of mass 8 kg moves under the influence of a force. The position of the body and time are related as x = 1/ 2t2 where x is in meter and t in sec. The work done by the force in first two seconds. Sol. Work done = change in kinetic energy 2

or

1 1  dx  1  2t  mv 2 = m  = m  2 2  dt  2 2 2

=

2

1 2× 2 ×8×   = 16Joules 2  2 


Ex.18 A body falls on the surface of the earth from a height of 20 cm. If after colliding with the earth, its mechanical energy is lost by 75%, then body would reach upto a height of Sol.

1 mgh = mgh' 4 h' =

∴

h 1 = × 20 = 5cm 4 4

Ex.19 Potential energy function describing the interaction between two atoms of a diatomic molecule is U( x ) = a − b x 12 x 6 In stable equilibrium, the distance between them would be Sol. In stable equilibrium potential energy is minimum. For minimum value of U(x) d [U( x )] = 0 dx

or

or

d  a b   12 − 6  = 0 dx  x x 

6 x 13

or

( −2a + bx 6 ) = 0

or

−12a x13

+

6b x7

=0

bx6 – 2a = 0

1/ 6

∴

 2a  x=   b 

Ex.20 Two electrons are at a distance of 1 × 10–12m from each other. Potential energy (in eV) of this system would be Sol. Potential energy of the system U=

=

=

Kq1q2 r

9 × 10 9 × 1.6 × 10 −19 × 1.6 × 10 −19 1× 10 −12

23.04 × 10 −17 1.6 × 10 −19

= 23.04 × 10—17 Joule

eV = 1.44 × 10 3 eV

Ex.21 Potential energy function U(r) corresponding to the central force F =

K r2

would be

Sol. Central force is conservative. Therefore r dU rˆ F(r ) = − VU = − dr

∴

∫

∫

U = dU = − F(r ) dr = −

r r dU = −F(r ). dr = −F(r ) dr

or K

∫ 2 dr

= −K

1

∫r

2

dr = Kr −1 + C

If at r = ¥, U = 0, then C = 0 U = Kr–1 =

K r

Ex.22 The stopping distance for a vehicle of mass M moving with speed v along level road, will be (µ is the coefficient of friction between tyres and the road)


Sol. When the vehical of mass m is moving with velocity v, the kinetic energy of the where K = 1/2 mv2 and if S is the stopping distance, work done by the firction W = FS cos θ = m MgS cos 180º = – m MgS So by Work-Energy theorem, W = DK = Kf – ki ⇒ – µ MgS = 0 – 1/2 Mv2 ⇒

S=

v2 2µg

Ex.23 A particle of mass m is moving in a horizontal circle of radius r, under a centripetal force equal to (–k/r2), where k is constant. The total energy of the particle is Sol. As the particle is moving in a circle, so mv 2 k = 2 r r

Now

Now as F = − dU dr

P.E, U = − F dr

K.E =

1 k mv2 = 2 2r

r

∫

∞

r

k  = +  2  dr r  ∞

∫

=−

k r

=−

k 2r

So total energy = U + K.E. =−

k k + r 2r

Negative energy means that particle is in bound state. Ex.24 The work done by a person in carrying a box of mass 10 kg. through a vertical height of 10 m is 4900J. The mass of the person is Sol. Let the mass of the person is m . Work done, W = P.E at height h above the earth surface. = (M + m) gh or 4900 = (M + 10) 9.8 x 10 or M = 40 kg Ex.25 A uniform rod of length 4m and mass 20kg is lying horizontal on the ground. The work done in keeping it vertical with one of its ends touching the ground, will be - . Sol. As the rod is kept in vertical position the shift in the centre of gravity is equal to the half the length = l/2 Work done w = mgh = mg

= 20 x 9.8 x

4 2

l 2

= 392 J

Ex.26 A man throws the bricks to the height of 12 m where they reach with a speed of 12 m/sec. If he throws the bricks such that they just reach this height, what percentage of energy will he save Sol. In first case,

W1 = =

1 m(v1)2 + mgh 2

1 m(12)2 + m × 10 × 12 2

= 72 m + 120 m and in second case, W2 = mgh = 120 m The percentage of energy saved =

192m − 120m × 100 = 38% 192m


4.

LINEAR

MOMENTUM

Ø

→ The product of mass and velocity of the body is called the momentum → P =mv

Ø

Momentum is a vector quantity and its direction is always along the direction of velocity.

Ø

Unit : In MKS : kg-m/sec or N-sec In C.G.S gm-cm/sec or Dyne-sec

Ø

Dimensions : [M1L1T–1]

Ø

The rate of change of momentum of a body is equal to the magnitude of applied external force →

→

= Fext =

dP for v << c dt

where c = speed of light but when speed v is very near to c then momentum is given by

mv P=

Ø

v 1−   c

2

The momentum measure the motion of body Example based on Linear Momentum

Ex.27 On increasing the momentum of a body by 100% the increase in its KE will beSol. E = p2/2m 2 E2 p22  200  = 2 =   =4 E1 p1  100 

E 2 − E1 4 − 1 E 2 − E1 = =3 ⇒ × 100 = 300% E1 1 E1 Ex.28 A jet of water whose cross-section is a, strickes a wall making an angle θ with normal and elastically rebounds. The velocity of water of density d is v. Force exerted on the wall is v

θ

θ

v

Sol. If m kg water strikes the wall in one sec. and rebounds elastically, then the change of its momentum = 2mv cos θ (perpendicular to wall) per sec. But

m = vad

∴

Force = rate of change of momentum = 2av2d cos θ

Ex.29 Sand is falling on a conveyor belt at the rate of m kg per second. The force needed to maintain its velocity v m/s is (in newton) Sol. The change in momentum of the sand of mass m kg to gain a velocity v. = mv per sec. ∴ required force = mv Ex.30 If the increase in kinetic energy of a body is 300% then the increase in its momentum would be Sol. Initial kinetic energy E1 = 1/2 mv12 Final kinetic energy


E2 =

∴

or

1  1 1  1 mv 12 + 3 mv 12  = 4  mv 12  = mv 22 2  2 2  2

1 2 1 mv 22 4 mv 1  E2 2   =4 2 = = 1 1 E1 mv 12 mv 12 2 2 v 22 v12

=4

or

v2 =2 v1

or

mv2 = 2mv1

or

mv 2 =2 mv 1

∴

Increase in momentum = 2mv1 – mv1 = mv1

∴

% increase in momentum =

mv 1 × 100 = 100 % mv 1

Ex.31 The momentum of a 10,000 kg truck whose velocity is 20m/s will be Sol. The momentum of truck, p = mv = 10, 000 x 20 = 2 x 105 kg m/s Ex.32 In the above question, the velocity which a 5000kg truck attain in order to have the same momentum, will be Sol. If m’ is the mass of second truck and v’ its velocity, then v' =

p 2 × 10 5 = = 40 m / s m' 5000

Ex.33 In the above example,. what must be the velocity to have same kinetic energy (in m/s) Sol. The kinetic energy of first truck = 1/2 mv2 and the kinetic energy of second Jruck = 1/2 m’v’2 Now

1 1 mv 2 = m' v' 2 2 2

⇒

v'2 =

⇒

v’ = 20 2 m/s

mv 2 10,000( 20 ) 2 = m' 5000

Ex.34 The rate of burning of fuel in a rocket is 50gm/sec. and comes out with velocity 4 x 103m/s. The force exerted by gas on rocket will be Sol. The rate of change of momentum is equal to force F=

Here

dp dt

=v

v = 4 × 103 m/s &

dm (Here v is constt.) dt

dm = 50 × 10–3 kg/s dt

∴ F = 4 × 103 × 50 × 10–3 = 200 N


5.

Ø

LAW

OF

OF

LINEAR

MOMENTUM

According to this principle, when the value of external force acting on a particle or system is zero, its linear momentum remains conserved. On the other hand in the absence of external force, the linear momentum of a particle or system remains unchanged. This is known as law of conservation of linear momentum. →

→

∴ Fext

Ø

CONSERVATION

dP = dt

→

→

dP → = 0 ⇒ dP = 0 dt Change in momentum = 0

If Fext = 0 ⇒

Momentum = constant →

→

→

→

→

→

If p1,p2 ,p3 ,.................. , be the linear momentum of elements of system, then p1+ p 2 + p 3 + ........ = constant Hence if the external force acting on a system is zero, the resultant momentum remains conserved.

Ø

The above equation is equivalent to three scalar equations. p1x + p2x + p3x + ...........+ pnx = constant p1y + p2y + p3y + ...........+ pny = constant p1z + p2z + p3z + ...........+ pnz = constant On the other hand in the absence of external force, the components, of momentum in different direction remains conserved or the momentum along X axis and Z axis remains conserved. Note : Depending upon forces acting on a system, linear momentum might be the conserved in one direction (i.e. Px = 0) or two direction (i.e. Px = 0, Py = 0) or in all direction (i.e. Px = 0, Py = 0, Pz = 0). If a component of the net external force on a closed system is zero along on axis, then the component of the linear momentum of the system along that axis cannot change.

5.1

Example of conservation of linear momentum

Ø

When a bullet fired from the gun : M – Mass of gun with man m – mass of bullet →

V – velocity of gun (with respect to ground)

→

v – velocity of bullet (with respect to ground)

The initial momentum of system = 0 →

→

The final momentum = – m v + M V

→ → m → m→ v v ⇒V= ∴ 0 = m v +MV ⇒ V = – M M →

→

The direction V is opposite to that of v

Ø

(i) When a bullet of mass m with velocity v pierces into a wooden block of mass M and gets embedded in it : The block is free to move on frictionless surface. Now both, the bullet & block have same velocity V, then Momentum before collision = Momentum after collision mv = (m + M) V V=

mv m+M


Initial kinetic energy =

1 mv2 = Ei 2

m2 v 2 = Ef m+M

1 Final kinetic energy = 2

M m v V

Ef m = < 1 ⇒ E f < Ei some part of energy gets dissipated. Ei m + M

Ø

(ii) When the bullet of mass m comes out after penetrating the block of mass M : In this case, mv = MV + mv1 ............(1)

M m

1 1 1 mv2 – ( MV2 + mv12) Loss of energy = 2 2 2

m V1

v V

= (initial energy) –

Ø

(Final energy)

When a bomb burst suddenly Let a bomb initially at rest, suddenly bursts into two pieces of masses m1 and m2 with velocity v1 and v2 respectively. As there is no external force acting on it, therefore the linear momentum remains conserved. ∴ 0 = m1v1 + m2v2

⇒

v1 m =− 2 v2 m2

Thus the velocities of pieces are inversely proportional to their initial masses. Note : If the bomb initially is in motion, the initial momentum will not be zero.

Ø

When a block of mass M is tied to a string of length l and a bullet of mass m strikes to it with velocity v and gets embedded in it : According to conservation of linear momentum

mv = (m + M) V

m

h

V

V=

mv (m + M)

As bullet gets embedded in the block, so the collision is not elastic. There is loss of energy. Now if the block rises to a height h, then mgh = (1/2) (m + M)V2 h=

(m + M)V 2 2mg

Example based on Linear Momentum Conservation Ex.35 A bullet of mass m moving horizontally with a velocity V strikes a block of mass M being suspended by a string of length L and gets embedded in it. After the collision the maximum angular deviation of block form the vertical is Sol. From the conervation of momentum (m + M) u = mv ∴u=

mv m+M


(velocity of combined system after impact)

θ

from the conservation of energy 1  mv  1  (M + m)u2 = (M + m)gL (1 – cosθ) or [1 – cosθ] = 2gL  2 m +M

2

2  1  mv   −1   θ = − cos 1   or  2gL  m + M  

Ex.36 A bullet moving with a speed of 400 m/s is stopped after penetrating into a bag of sand. Mass of bullet and bag are 0.25 kg and 4.75 kg respectively. If the bag is free to move then its speed would be Sol. Momentum of a bullet = 0.25 × 400 = 100 kg –m/s 100 = 20 m / s 5 Ex.37 α-particle is emitted with a speed of 2.34 × 108 m/s from a stationary uranium nucleus, (U238). The velocity of residual nucleus will be

(0.25 + 4.75)v = 100 or v =

From momentum conservation law

Sol. From momentum conservation law 4 × 2.34 × 10 8 = 4 × 10 6 m / s 234 Ex.38 A body of mass 2 kg moving with a speed of 40 m/s collides with another stationary body. After collision both bodies move simultaneously with a speed of 25 m/s. The mass of the other body is v=

234 × v = 4 × 2.34 × 108

Sol. From momentum conservation law m1u1 + m2u2 = m1v1 + m2v2 2 × 40 + m2 × 0 = 2 × 25 + m2 × 25

m2 = 1.2 kg

Ex.39 A particle of mass m1 moving with a speed v, collides inelastically with a stationary particle of mass m2(m2 > m1). The fraction of initial kinetic energy converted into heat would be Sol. From momentum conservation law v=

∴

m1v1 + 0 = (m1 + m2) v

Total kinetic energy before collision = Total kinetic energy after collision  m1v 1  1  = (m1 + m 2 )  2  m1 + m 2 

∴

=

2

m1v 1 (m1 + m 2 )

1 1 m1v12 + 0 = m1v12 2 2 1 (m1 + m 2 ) v 2 2

=

1 m12 v 12 2 (m1 + m2 )

=

1 m1v 12 (m1 + m2 − m1 ) 2 (m1 + m2 )

Loss of kinetic energy

=

1 1 m12 v 12 m1v 12 − 2 2 (m1 + m2 )

Fraction of initial kinetic energy converted into heat.

=

1 m1m 2 v12 1 / m1v 12 2 (m1 + m 2 ) 2

=

m2 m1 + m 2

=

1 m1m2 v 12 2 (m1 + m2 )


6.

COLLISION

OF

BODIES

The event or the process, in which two bodies either coming in contact with each other or due to mutual interaction at distance apart, affect each others motion (velocity, momentum, energy or direct of motion) is defined as a collision. In collision (a) The particles come closer, before collision and after collision, they either stick together or move away from each other. (b) The particles need not come in contact with each other for a collision. (c) The law of conservation of linear momentum is necessarily conserved in a collision, where as they the low of conservation of mechanical energy is not. →

Note - If F is the average of the time varyfing force during collision and ∆t is the duration of collision then impulse →

J = F ∆t . TYPES OF COLLISIONS

On the basis of conservation of kinetic energy

On the basis of direction One-dimensional collision or Head on collision or Direction collision The collision, in which the particles move along the same straight line beforeand after the collision, is defined as one dimens ional colli sion

Two dimensional collision or Oblique collision

The collision, in which the particles move in the same plane at different angles before and after collision, is defined as oblique collision.

Elastic collision

A c o ll i s io n is said to be elas ti c, if the total kinetic ener gy bef ore and after collision remains the same

In-elasticcollision

Perfectly inelastic collsion

A collision is said to be e l as t i c , i f t h e total kinetic energy does not remains constant

Th e co lli si on, in which particles gets stiked together after the collision, is called perfectly i nel as t i c c ol li s i on . In this type of inelastic collision, lose of energy is maximum .

Note : Linear momentum is essentially conserved in any collision. Newton’s law of collision : Re lative velocity after collision Re lative velocity before collision = –e, where e = coefficient of restitution

Let u1 and u2 be the velocities of two bodies before collision and v1 and v2 that after the collision, then v 2 − v1 u2 − u1 = – e

e = 1, for elastic collision e < 1, for inelastic collision e = 0, for perfectly inelastic collision Now we discuss the collision as follows 6.1

Direct elastic collision A m1

u1

B m1

Before collision

u2

A F

B F

During collision

A m1

v1

B

After collision

m2

v2


Ø

In this collision, both momentum and kinetic energy remains conserved. Thus for such collision (p)b.c. = (p)a.c ⇒

→

→

→

→

m1 u1 + m2 u2 = m1 v1 + m2 v 2 [where, p stands for linear momentum b.c for -

before collision a.c for - after collision ] and (K.E)b.c = (K.E)a.c ⇒

Ø

[K.E for - kinetic energy]

1 1 1 1 m1u12 + m 2u 22 = m1v12 + m 2 v 22 2 2 2 2

Newtons’ law for elastic direct collision : v2 – v1 = – (u2 – u1)

Ø

Important formula and features for direct elastic collision : (i) The velocity of first body after collision  m − m2   2m 2   u1 +  u 2 v1 =  1 +  m1 m 2   m1 + m 2 

(ii) The velocity of second body after collision  2m1   m − m1   u1 +  2 u 2 v 2 =   m1 + m 2   m1 + m 2 

(iii) If the body with mass m2 is initially at rest i.e u2 = 0 v1 =

Ø

m1 − m 2 u1 m1 + m 2

and

v2 =

2m 2 u1 m1 + m 2

When a particles of mass m1 moving with velocity u1 collides with another particle with m2 at rest and if ↓ ↓ ↓ m1 = m 2

m1 >> m2

m1 <<<< m2

In this case,

In this case,

In this case,

v1 = 0 and v2 = u1

v1 = u1 and v2 = 2u1

2m1 v1 = – u1 and v2 = m u1 ≈ 0 2

Ø

When m1 = m2 = m but u2 ≠ 0, then v1 = u2 and v2 = u1 i.e. the particles mutually exchange their velocities.

Ø

Exchange of energy is maximum, when m1 = m2. This fact is utilized in atomic reactor in slowing down the neutrons. To slow down the neutrons, these are made to collide with nuclei of almost similar mass. For this hydrogen nuclei are most appropriate . Note : To solve problems based on direct elastic collisions, the momentum conservation law and Newton’s law of collision are to be applied. In special circumstances law of conservation of kinetic energy should be applied.

6.2

Direct inelastic collision

Ø

In this case, (p)a.c = (p)a.c

→

→

→

→

⇒ m1 u1 + m2 u2 = m1 v1 + m2 v 2 (K.E) b.c ≠ (K.E.) a.c ⇒ (K.E) b.c + (K.E.) a.c + Q,

(where Q = heat energy, sound energy etc.) ⇒

1 1 1 1 m1u12 + m 2 u 22 ≠ m1v 12 + m 2 v 22 2 2 2 2

Ø

According to Newton’s law, for inelastic collision we have, v1 – v2 = –e (u1 – u2)

Ø

In inelastic collision, velocity of first body after collision :


 m − em 2   m (1 + e)   u1 +  2  u 2 v1 =  1 + m m 2   1  m1 + m 2  m1(1 + e) m 2 − em1 and velocity of second body, v2 = m + m u1 + m + m u 2 1 2 1 2

Ø

1  m1m2  2 Loss of energy in inelastic collision : ∆Ek = 2  m + m  (u1 − u2 )  1 2 

6.3

Direct, perfectly inelastic collision A m1

u1

B m2

A

u2

(1− e)2

B V

Before collision

After collision

In this type of collision,

Ø

(p)b.c = (p)a.c.

⇒

→

→

m1 u1+ m 2 u 2 = (m1 + m 2 )V

(K.E)b.c. ≠ (K.E)a.c. ⇒ (K.E)b.c. = (K.E)a.c + Q ⇒

1 1 1 m1u12 + m 2u 2 2 = (m1 + m 2 )V 2 + Q 2 2 2

Ø

According to Newton’s law for this collision : v1 = v2 (Q e = 0)

Ø

Velocity after collision of the combined body : V =

Ø

Loss of energy :

Ø

∆Ek =

m1u1 + m 2u 2 m1 + m 2

1 m1m 2 (u1 − u 2 )2 2 m1 + m 2

1 (m1 + m 2 )v 2 m1 Ef 2 = = <1 If u2 = 0, ratio of final energy to initial energy : Ei 1 m1 + m 2 m1u12 2

⇒ Ef < E i

m2 ∆E k i.e there is loss of kinetic energy (f) If u2 = 0, E = (m + m ) i 1 2

Note : In some cases the kinetic energy of combined body gets increased, but this is compensated by the lose in potential energy of the body. 6.4

Oblique elastic collision :

In this case, both momentum and K.E remains conserved, Momentum is a vector quantity and it can be resolved into any tow perpendicular directions (say, x and y)


Ø

Applying law of conservation of linear momentum : along x-axis : (px)b.c = (px)a.c ⇒ m1u1 cosα1 + m2u2 cosα2 = m1v1 cosβ1 + m2v2 cos β2 and along y-axis : (py)b.c = (py)a.c. ⇒ m1u1 sinα1 + m2u2 sinα2 = m1v1sinβ1 + m2u2 sinβ2

Ø

Applying law of conservation of K.E : (K.E)b.c = (K.E)a.c ⇒

1 1 1 1 m1u12 + m 2u 22 = m1v12 + m 2 v 22 2 2 2 2

Ø

If m1 = m2 and (α1 + α2) = 90º, (β1 + β2) = 90º. Which means, if two particles of same mass moving at right angle to each other collide elastically, after collision also they move at right angle to each other

Ø

If a body A collides elastically with another body B of same mass at rest at a glancing angle, than after collision the two bodies move at right angle to each other i.e. α + β = 90º

Ø

If a stationary body breaks due to some interaction in three parts, out of which the first two parts, move at right angles to each other with momenta p1 and p2 respectively, then the momenta of third part is determined as follows : p2

p3cosθ

p1

θ

According to law of conservation of momentum : p3

p3sinθ

Along horizontal direction : p3cosθ = p1 Along vertical direction : p3sinθ = p2 p2 p  Magnitude of p3 = p12 + p22 and its, direction from horizontal, tan θ = p ⇒ θ = tan–1  2  1  p1  p2 –1 Direction of p3, from the direction of motion of first part = [π +tan p ] and that from the direction motion second 1 p2 π –1 part = [   + tan p ] 1 2

Exampled based on Collision Ex.40 A block of mass 12kg moving at 20 cm/s collides with an identical stationary block. If the coefficient of restitution in 3/5 the loss in K.E during collision is 1 m1m 2 (1 − e 2 ) (u1 − u 2 )2 2 m1 + m 2 Here m1 = 2kg, m2 = 2kg, e = 3/5

Sol. The loss in K.E

∆E =

u1 = 20 cm/sec = 0.2 m/s, u2 = 0 On substiuting the values

∆E =

1  12 × 12   9  16 2 2 1  (0.2)2 = × 6 ×  1 −  × × ≅ 7.7 × 10 − 2 J 2  12 + 12   25  25 10 10 2

Ex.41 A ball is dropped from a height h on a stationary floor and rebounds several times until it stops. If the coefficient of restitution is e, then the total distance covered by the ball before it stops, would be Sol. The height h1 up to which the ball rises after the first rebound is given by h1 = e2h After second rebound,

h2 = e4h

After rebounding n times, hn = (e2)nh

Total distance described

s = h + 2h1 + 2h2 + ....... + 2hn + ..... = h + 2e2h + 2e4h + ....... + 2e2hh + ...... = h+

 1+ e2   h = 1 − e 2  1 − e 2  2e 2h

= h + 2e2h (1 + e2 + e4 + .... ∞)


Ex.42 A rifle man, who together with his rifle has a mass of 100 kg, stands on a smooth surface fires 10 shots horizontally. Each bullet has a mass 10 gm a muzzle velocity of 800 m/s. What velocity does rifle man acquire at the end of 10 shots. Sol. Let m1 and m2 be the masses of bullet and the rifleman and v1 and v2 their respective velocities after the firstr shot. Initially the rifleman and bullet are at rest, therefore initial momentum of system = 0. i.e. initial momentum = final momentum

or

v2 =

= m1v1 + m2v2

m1v1 (10 × 10 −3 kg)(800 m / s) =− m2 100 kg

= - 0.08 m/s Velocity acquired after 10 shots = 10 v2 = 10 × (–0.08) = – 0.8 m/s i.e., the velocity of rifle man is 0.8 m/s in a direction opposite to that of bullet. Ex.43 A bullet of mass 10 g travelling horizontally with a velocity 300 m/s strikes a block of wood of mass 290 g which rests on a rough horizontal floor. After impact the block and the bullet move together and come to rest when the block has travelled a distance of 15 m. The coefficient of friction between the block and the floor will be (Duration of impact is very short) Sol. Let the mass of block and bullet be M and m respectively. If v is the velocity of bullet and V is the velocity of block with bullet embedded in it, Now according to conservation of momentum, mv = (M + m) V (10 x 10–3) (300/ = (290 x 10–3 + 10 x 10–3) V or

V = 10 mls

The kinetic energy just after impact is 1/2(M + m) V2, which is lost due to work done on it by the force of friction F. Since force of friction F = m(M+m)g and the work done is given by Fd, we have 1/2 (M + m) V2 = µ (M + m) gd or

m=

1 V2 2 gd

=

1 10 2 1 × = 2 (10 ) (15 ) 3

NOTE- Here an external horizontal force due to friction is present however as it has been assumed that impact lasted for such a small interval of time that the block could not move appreciably no work was done by friction during impact. Here during impact the presence of friction cannot be ignored. Ex.44 A block of mass m1 = 150 kg is at rest on a very long frictionless table, one end of which is terminated in a wall. Another block of mass m2 is placed between the first block and the wall, and set in motion towards m1 with constant speed u2 Assuming that all collisions are completely elastic, find the value of m2 for which both blocks move with the same velocity after m2 has collided once with m1 and once with the wall. (The wall has effectively infinite mass.) Sol. Let after the collision, v1 = speed of mass m1 towards left v2 = speed of mass m2 towards right. Hence, momentum before collision = momentum after collision m2u2=m1v1 – m2v2

.....(1)

The mas m2 rebounds elastically from the wall and its speed its reversed after the collision with the wall. According to the problem, the mass m2 has the same speed as that of mass m1 after its collision with the wall ie. v2 = v1 . From eq. m2u2 = (m1 – m2) v1

.....(2)


1 1 1 m 2u 22 = m1v12 + m 2 v12 2 2 2

Since the collision is elastic, then m2u22 = (m1 + m2) v12 From eq. (2),

v1 =

......(3)

m 2u 2 m1 − m 2

Substituting this value of v1 in eq. (3), we get

m 2u22 =

(m1 + m 2 )(m2u2 )2 (m1 − m2 )2

or

(m1 – m2)2 = (m1 + m2) (m2) or m12 + m22 – 2m1 m2 = m1 m2 + m22 or m12 = 3 m1 m2 m1 = 3m2

or

m2 =

m1 150 = = 50.kg 3 3

Ex.45 A ball moving with a speed of 9 m/s strikes with an identical stationary ball such that after the collision the direction of each ball makes an angle of 30° with the original line of motion. Find the speeds of the two balls after the collision. Is the kinetic energy conserved in this collision process? Sol. Initial momentum of the balls =m×9+m×0=9m

....(1)

where m is the mass of each ball. Let after collision their velocities are v1 and v2 respectively. Final momentum of the balls after collision along the same line = mv1 cos 30 + mv2 cos30

=

mv1 3 mv 2 3 + 2 2

.....(2)

According to law of conservation of momentum

9m = 9×2 3

mv1 3 mv 2 3 + 2 2 = v1 + v 2

9m/s

Stationary ball

....(3)

(a) Before collision

The initial momentum of the balls along perpendicular direction = 0. Final momentum of balls along the perpendicular direction = mv1 sin 30 – mv2 sin 30 =

m (v – v2) 2 1

Again by the law of conservation of momentum (m/2)(v1 – v2) = 0 v1

∴ (v1 – v2) = 0

Solving equations (3) and (4), we have v1 = 3 3 m/s and v2 = 3 3 m/s

.....(4)

30º 30º

v2

(b) After collision


According to law of conservation of energy. Energy before collision = Energy after collision 1 1 1 1 mu12 + m u22 = m v12 + m v 22 2 2 2 2

⇒

1 1 1 m(9)2 + 0 = m(3 3 )2 + m(3 3 )2 2 2 2

81m 54 m = L.H.S. ≠ R.H.S. 2 2

i.e., energy is not conserved in this collision or this is a case of inelastic collision. Ex.46 A sphere of mas 8 kg moving at constant speed 50 m/s, contains a compressed light spring with strain energy 15,000 joule. At a given instant, the spring breaks and causes the sphere to explode into tow pieces of equal masses. If one piece flies off at 30° to the original velocity of the sphere, find the direction of motion of the other piece and magnitudes of the velocities of the two pieces. Assume the energy of the compressed spring is completely imparted to the two pieces all kinetic energy. Sol. The situation is shown in fig. Let v1 and v2 be the velocities of two pieces after explosion. Applying the law of conservation of energy, we .have v1

m 1=4kg m = 8kg θ 30º

1 1 1 (8 ) (50 )2 + 15000 = ( 4 )v12 + ( 4)v 22 2 2 2

u = 50m/s m2=4kg

or

25000 = 2(v12 + v22)

v2

....(1)

Applying the law of conservation of momentum along x-axis and y-axis respectively, we get 8(5) = 4 v1 cos q + v2 cos 30º and

0 = 4v1 sin q = 4 v2 sin 30º = 2v2

or

sin θ =

....(2) ....(3)

v2 2v 1

....(4)

From eq. (2) 100 = v1 cos q + v2 cos 30

or

  3 v2   100 = v 1 1 − 2  + v 2  2  2   4 v   1  

or 100 = v1

4v12 − v 22 2

Solving equs. (1) and (5) for v1 and v2 we get v1 = 51.56 m/s and v2 = 99.2 m/s Now

sin θ =

99.2 2 × 51.56

Solving we get q = 74º8

+

v2 3 2

....(5)


7.

ROCKET

MOTION

The motion of rocket is based on Newton’s third law of motion and conservation of momentum.

Ø

Diagram Oxidizing substance Fuel

v

Combustion Chamber

Hot-Gas Jet u

Ø

Direction of Jet Direction of rocket

Ø

There is a chamber in Rocket in which fuel is filled in the form of solid or liquid. It also has a combustion chamber. When fuel begins to burn with oxidizing substance, the pressure increases in the combustion chamber due to heat create in it. The hot gases ejects with high velocity in the form of jet. The jet causes a reaction force on the rocket, then the rocket accelerates against the jet. The velocity and mass of (rocket + fuel) Time

Velocity

Mass

t

u

v

t + dt

v + dv

m – dm

The relative velocity of mass dm relative to rocket will be vr = u, the velocity relative to earth will be (v + dv – u) There is no external force acting on the system, therefore in the earth’s coordinate system mv = (m – dm) (v + dv) + dm + (v + dv – v) mf

vf

or

dm dv = v m

⇒

∫ dv

vi

=

∫

mi

dm m

Where i and f represents initial and final states mf Therefore vf – ui = u log m i

Therefore it initial velocity be zero, the final velocity will be v f = u log

mf m m = −u log i = v log i f mi mf mf

Note : First rocket equation Ru = ma where R = positive rate of fuel consumption u = speed of exhaust products relative to rocket. m = mass of rocket a = acceleration produced 8.

CENTER

OF

MASS

In a system of many particles, the centre of mass is that point in the system, which along describe the linear motion of the system as a whole. Centre of mass is that point in the body, at which the total mass of the body is supposed to be concentrated for description to linear motion of the body. Centre of mass is a point, about which the sum of moments of masses (i.e. multiplication of mass and its position vector with respect to centre of mass) of all the particles in the body is zero. If m1, m2, m3..... are situated →→ →

→

→

→

→

at position vector r , r , r ...... with respect to the centre of mass, m r + m r + m r ... = 0 or ∑ m r = 0 , 1 2 3 1 1 2 2 3 3 i i


where ri = position vector of ith particle with respect to the centre of mass, mi = mass fo the ith particle. Suppose a system consists of n particles with masses m1, m2, m3 ......mn and their position vector with respect to chosen reference frame will be m1 →

rcm

→

→

→

m r + m 2 r2 + m 3 r3 + ..... ∑ miri = = 1 1 = m1 + m 2 + m 3 + ..... ∑ mi

→

r1

∫ ∫ dm

dm r

C

r2

m2

r3 m3

If masses m1, m2, m3......... are situated at (x1, y1, z1), (x2, y2, z2), (x3, y3, z3) ......, co-ordinate of centre of mass in the same reference are given by X cm =

m1x1 + m 2 x 2 + .......... . ∑ mi x i ∫ dmx = = ∑ mi m1 + m 2 ∫ dm

m1(x1,y1,z1)

Y r1

Ycm =

m1y 1 + m 2 y 2 + ........... ∑ m i y i ∫ dmy = = m1 + m 2 ∑ mi ∫ dm

m2(x2,y2,z2) r2

r3

O

m3(x3,y3,z3) X

Z

Z cm =

m1z1 + m 2 z 2 + .......... . ∑ mi z i ∫ dmz = = ∑ mi m1 + m 2 ∫ dm

For regular shape and configuration, the centre of mass coincide with geometrical centre only if the mass distribution of body with respect geometrical centre of configuration is uniform. If mass distribution is non uniform, centre of mass and geometrical centre are not the same point. In uniform gravitational field, centre of mass and centre of gravity are the same point → →

→

If particles of body of masses m1, m2..... are moving with velocities v 1, v 2 , v 3 ...... respectively then velocity of centre of mass is given by →

Vcm

→

→

→

→

Total momentum of system m V + m 2 V2 + m3 V3 + ...... ∑ m1 Vi ∑ pi = = = 1 1 = Total mass m1 + m2 + ....... ∑ mi ∑ mi

Now components of velocity of centre of mass along x, y and y axes can be written as -

Vcm x = Vcm y =

Vcm z =

m1v x1 + m 2 v x 2 + ..... m1 + m 2 + ....... m1v y1 + m 2 Vy 2 + ..... m1 + m 2 + ....... m1v z1 + m2 Vz2 + ..... m1 + m2 + .......

= =

=

∑ mv xi ∑ mi ∑ mv yi ∑ mi ∑ mv zi ∑ mi

= =

=

∑ Pxi ∑ mi ∑ Pyi ∑ mi ∑ Pzi ∑ mi

If ∑ Fxnet = 0, ∑ Pxi = constant i.e if net external force acting on system along x axis is zero, total momentum of the system along x-axis remain constant, therefore velocity of centre of mass along x-axis remain unaffected during subsequent motion. Same analysis can be done for y and z axis. Motion of centre of mass is not affected by the internal forces. Therefore, if a shell moving under gravity explodes into pieces moving in different direction, even then the centre of mass moves along the same (previous) path. Note : Due to mutual interaction forces, velocity of CM does not change. If not external force acts on the body, state of motion of its centre mass remains the same i.e. it it is moving with some velocity then it keep on moving in the same direction with same velocity and if CM is at rest, it remain at rest. → → →

If particles of system are moving with acceleration a , a , a ....., acceleration of centre of mass is given by, 1 2 3


→

acm

→

→

m a + m2 a 2 + ......... ∑ miai = = 1 1 m1 + m 2 + ........ ∑ mi

Components of centre of mass along x, y and z axes can be written as

acmx = acmz =

m1a x1 + m2a x2 + ....... m1 + m 2 + ....... m1a z1 + m2a z2 + ....... m1 + m2 + .......

= =

∑ mia xi

acmy =

∑ mi ∑ mia zi

→

m1a y1 + m 2a y 2 + ....... m1 + m 2 + ....... →

→

→

=

∑ mia yi ∑ mi

→

If a = a = a = a = a 1 2 3 cm

∑ mi

Total K.E of system of particles about a fixed reference frame = Total K.E of particles with respect to CM + K.E of CM with respect to the fixed reference frame (f.r.f_ ie.

T.E =

1 1 Iω2 + mv2 2 2

K.E wrt CM K.E or Cm wrt f.r.t Angular momentum of a system of particles with respect to a fixed axis = angular momentum of particles with respect to CM + angular momentum of CM with respect to the fixed axis. Exampled based on Center of Mass Ex.47 Center of mass for the following figure are

(i)

(ii)

X cm =

X cm =

0 xm + mxa + mx m+m+m

a 2 =a, 2

0xm + axm + 2mx m + m + 2m

Ycm =

a 2 = a, 2

Ycm =

a 3 2 =a 3 m+m+m 6

0 xm + 0 xm + mx

a 3 2 =a 3 m + m + 2m 4

0xm + 0xm + 2mx

Ex.48 A rod of length L has nonuniformly distributed mass along its length. For its mass perunit length varying with distance x from.one end as

m0 L2

(L + x). Find the position of centre of mass of this system. Discuss the case

when x = 0.

Sol. Mass of element of length dx is dm = L

X cm

L2

(L + x) . dx

∫ dm . x ∫ mL (L + x)dx . x 5L = = = m 9 (L + x )dx dm ∫ L ∫ L

∴

m0

0 2

0

0

L

L

0

0

0 2

1

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