Work Power Energy

WORK, ENERGY,

POWER & MOMENTUM

W O R K

Work is said to be done by a force when the point of application of the force is displaced in the direction of the force. Work is a scalar quantity and is measured by the product of the magnitude of the force and the component of displacement along the direction of the applied force.

Force

Constant Variable

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 force



F is acting on a body at an angle to the horizontal and the displacement r is along the horizontal, the work done will, be W = (F cos ) r

= F (r cos ) _{ Fcos} F r In vector from, W =   r . F If F  iˆF_xjˆF_ykˆF_z  and r iˆxjˆykˆz 

, the work done will be, W = F_x x + F_y y + F_zz

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

The equation W F.S  FS cos is valid when F remains constant but when the force is variable, total work done is obtained by integration method using W 



dW 



F.dS. 

2

(A) WHEN THE FORCE IS SOME FUNCTION OF POSITION

Let a force FF(x) act along the x-axis as shown in the F–x graph in the figure.

Work done during small displacement dx is F dx which is equal to the area of the shaded rectangular strip.

Hence the total work done is B

A

F



F dx Area under the F –x graph.

F(x)

dx

A B x

An example of the variable force is the spring force in which the force F is proportional to the extension x, given by F = – kx, where k = force constant and negative sign indicates opposite direction of spring force F and extension x. Work done by spring =

f 1 x 2 2 f i x 1 F.dx kx dx k(x x ) 2     



 



.

If spring be initially relaxed, then work done by the spring (or, work done against the external agent) for an extension x is obtained by putting xi  and 0 xf  leading tox

2 1

W kx

2

  .

Note that the work done by the spring is negative where as the work done by the external agent producing the extension is positive. The two, however, are equal in quantity.

(B) WHEN THE FORCE IS TIME DEPENDENT: Let FF(t), then dx W F.dx F. dt F.vdt dt     _ _   



 



 



  ,

where F and v are time dependent instantaneous force and velocity vectors expressed as functions of time.

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 be 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

Thus, total work done, W = dW

2 1 r r



dr P1 M O F P₂ N r2 r1 = F.dr 2 1 r r



= Area of P₁P₂NM

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 and r.

(i) Work is a scalar quantity (ii) The dimensions of work : ML2_T–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)}

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 =  

  cos r F r . F

When is acute (<90º) cos is positive. Hence work done is positive. Examples :

When a body falls freely under the action of gravity = 0º, cos  = + 1, therefore work done by gravity on a body, falling freely is positive.

(b) Negative work : When is obtuse (>90º), cos is negative. Hence work done is negative Examples :

(i) When a body is thrown up, its motion is opposed by gravity. The angle  between the gravitational force



F and displacement



r is 180º. As cos = – 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 is 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 between _F and



r is 90º, the work done will be zero. Examples :

(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, = 90º. Therefore,

workdone by the collie is zero. This is because = 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.

4 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 the object the work done is negative and hence the work which is a transfer of energy has same dimensions as energy.

WORK DEPENDS ON THE REFERENCE FRAME

Consider a situation in which a person pushes a block inside a moving train through a distance S. The work done on the block in the train’s frame will be F.S . The work done on the block in the ground frame will be F.(S  S )₀ , where S₀ is the displacement of the train relative to the ground.

In mechanics, we encounter two types of forces : conservative and non-conservative force.

CONSERVATIVE AND NON-CONSERVATIVE FORCES

Conservative Force:

A force is said to be conservative if the work by the force in moving a mass through any closed path is zero. Alternative, if the net work done by the force in moving a mass between two points is path independent i.e., the work done depends only on the location of two points and not on the path followed, the force may be termed conservative.Conservative forces are non-dissipative and can store work (as energy) Examples

Gravitational Force, electrostatic force, ideal spring force are examples of conservative forces.Associated with each conservative force there exists a well defined function known as the potential energy function. In these three cases, we refer to the respective energies as gravitational potential energy, electrostatic potential energy and strain energy.

Non-conservative Force

The forces which do not fulfill the above mentioned conditions are non-conservative. For these the energy isnot stored but dissipated during displacement.

Examples of non-conservative forces are friction, viscous forces etc.

EXAMPLE BASED ON WORK

Example 1. A position dependent force F 72x3x2



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

Solution : 2 1 X 5 2 X 0 W



Fdx



(72x3x )dx 5 2 3 0 2x 3x 7x 135J 2 3   _   _   

Example 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

-1 2 3 4 5 6 7 8 20 10 0 -10 -20 F (In dyne) _x-(cm)

Example 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

Solution : If length x of the chain is pulled up on the table, then the length of hanging part of the chain would be _  x_

3 L

and its weight would be x g. 3 L L M     

  _{If it is pulled up further by a distance} dx, the work done in pulling up.

j x gdx 3 L L M _        



         3 / L 0 18 MgL dx g x 3 L L M w

Example 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

Solution : The minimum force with a body is to be pulled up along the inclined plane is mg (sin + cos) Work done, W F.d

= Fd cos 0º

= mg (sin +  cos ) × d

= 5 × 9.8 (sin 60º + 0.2 cos 60º) × 2 = 98.08 J

Example 5. A particle moves from a point r1(5 m)i (5 m) jˆ ˆ



to another point r₂ (6m)iˆ(4m) jˆ during which a certain force F(5N)i (5N) jˆ ˆ acts on it. Find the work done by the force on the particle during the displacement.

Solution : Displacement vector

S   r₂ r₁ =

 

ˆ ˆij m. Work done = F.S  (5i 5 j).(iˆ ˆ ˆj )

= (5 – 5) = 0.

E N E R G Y

The energy of a body is defined as the capacity of doing work. There are various form of energy

(i) mechanical energy (ii) chemical energy

(iii) electrical energy (iv) magnetic energy

(v) nuclear energy (vi) sound energy

(vii) light energy etc

Energy of system always remain constant it can neither be created nor it can be destroyed, however, it may

6

Electrical energy _Motor_{ Mechanical energy}

Mechanical energy _Generator_{ Electrical energy}

Light energy _Photocell_{ Electrical energy}

Electrical energy



Heater



_{Heat energy}

Electrical energy



Radio Sound energy

Nuclear energy _Nuclear_Re_actor_{ Electrical energy}

Chemical energy _{ }Cell_{ Electrical energy}

Electrical energy _Secondary_Cell_{ Chemical energy}

Heat energy _Incendence_nt_lamp_{ Light energy}

Energy is a scalar quantity

Unit : Its unit is same as that of work . 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} Dimension [M1_L2_T–2_]

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.

In mechanics we are concerned with mechanical energy only which is of two type (a) kinetic energy; (b) potential energy

KINETIC ENERGY

The energy possessed by a body by virtue of its motion is called kinetic energy If a body of mass m is moving with velocity v, its kinetic energy

KE = 2 1

mv2_{, for translatory motion}

KE = 21 I

2_{, for rotational motion} Kinetic energy is always positive

If linear momentum of body is p, KE = 2

2 mv 2 1 m 2 p _

- for translatory motion

If angular momentum of body is J, KE = 2

2

2 1 2

J _{ I}

I - for rotational motion

p or J  E p : momentum

P E

1/P E

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

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.

If a body performs translatory and rotational motion simultaneously, its total kinetic energy = 2 2 2 1 mv 2

1 _{ I}

WORK ENERGY THEOREM

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 = 22 12

1 1

mv mv

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 = 22 21

1 1

2I 2I Note :

In simple words K = K_f – K_i = W in the work energy theorem ?

POTENTIAL ENERGY

The energy which a body has by virtue of its position or configuration in a conservative force field Potential energy is a relative quantity.

Potential energy is defined only for conservative force field.

Potential energy of a body at any position in a conservative force field is defined as the work done by an external agent against the action of conservative force in order to shift it from reference point.

8

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

Relationship between conservative force field and potential energy (U)

F





  U grad (U) = – kˆ z U jˆ y U iˆ x U         Example 6. (i) U = 3x2  iˆ x 6 F   (ii) U = 2x2_{y + 3y}2_{x + xz}2 F (4xy3y2z2) (2x26xy) (2xz) 

If force varies only with one dimension, then F = – dx dU or 2 1 u u

d



U Fdx 2 1 x x



  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

Attraction forces Repulsion forces

r U-ve

U+ve

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

If r , U , force field is repulsive in nature.

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.E_surface = 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. = –

r GMm

, for 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. = –

r Q KQ1 2

, [value of Q₁ and Q₂ are substituted with sign.] Intermolecular potential energy :

Reference point = i.e. P.E_ = 0 U (r) = _{12 6} r b r a  _{, F = – }      _  1 r b / a 2 r b 6 dr dU 6 7 Elastic potential energy :

Which is associated with the state of compression of extension of an elastic object U (x) = 2 1

kx2 _where k = spring constant, x = change in dimensions

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

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 dU

dx = 0 Points P, Q, R and S are the states of equilibrium positions.

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 stable equilibriums position.

Necessary conditions - dU dx = 0, _dx ve U d 2 2  

 Unstable equilibrium : When a particle is displaced slightly from 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 : dx dU

= 0 potential energy is max i.e. = ve dx U d 2 2  

 Neutral equilibrium : In the neutral equilibrium 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

Example 7. A meter scale of mass m initially vertical is dispalced at 45º keeping the upper end fixed, the change in PE will

be-Solution : Work = change in PE = Force × displacement dU = (1 cos ) 2 mg   _{ }/245º G' G = mg × (1 cos45 ) 2 1 _ º (  = 1m) =        2 1 1 2 mg

10

Example 8. If the potential energy function for a particle is U = a – ₂

x c x

b  _{; The force constant for}

oscillation will be.

Solution : U = a – ₂ x c x b  _...(1)  2 3 dU b 2c dx  x x ...(2) and 2 2 3 d U 1 6c 2b x dx x    _  _   ...(3) ` for equilibrium 0 dx dU  x = b c 2

 Substituting this value in (3)

3 2 2 c 2 b dx U d        3 4 c 8 b b / c 2 c 6 b 2 _  _{ } as K dx U d 2 2  K = b4_/8c3

Example 9 On passing through a wooden sheet a bullet looses 1/20 of initial velocity. The minimum number of sheets required to completely stop the bullet will

be-Solution : Use v2 _{= u}2 _{+ 2as}

for a sheet of thickness s v = (19/20)u 2 19 u u 2as 20  _{  }     2as = (361/400)u2 _{– u}2 _{a = –} s 2 u 400 39 2      

suppose for n sheet v = 0 02 _{= u}2 _{+ 2a (ns)}

 n = – as 2 u2 = 11 s s 2 u 400 39 2 u 2 2        

Example 10. The work done in taking out 2 litre of water using a bucket of mass 0.5 kg from a well of depth 6m will

be-Solution : W = mgh

= (m_bucket + m_water)gh [2 Lit water = 2 kg water]

= (0.5 + 2.00) × 9.8 × 6 = 15 × 9.8 = 147 J

Example 11. A body has velocity 200 m/s and its kinetic energy is 200 J. The mass of the body would be

Solution : mv E 2 1 2 _ or 2 ₍₂₀₀₎2 200 2 v E 2 m         2 4 4 10 1 100 4 10      m = 0.01 kg

Example 12. A body of mass 8 kg moves under the influence of a force. The position of the body and time are related as x = t2_{/2 where x is in meter and t in sec. The work done by the force in first two}

seconds.

Solution : Work done = change in kinetic energy or 2 2 2 2 t 2 m 2 1 dt dx m 2 1 mv 2 1               Joules 16 2 2 2 8 2 1 2 _        

Example 13. 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

Solution : mgh mgh' 4 1 _  20 5cm 4 1 4 h ' h   

Example 14. Potential energy function describing the interaction between two atoms of a diatomic molecule

is _{12 6} x b x a ) x (

U   . In stable equilibrium, the distance between them would be Solution : In stable equilibrium potential energy is minimum. For minimum value of U(x)

0 )] x ( U [ dx d _ or 0 x b x a dx d 6 12 _     _ or 0 x b 6 x a 12 7 13    or ( 2a bx ) 0 x 6 6 13    or bx6 – 2a = 0  6 / 1 b a 2 x       

Example 15. Two electrons are at a distance of 1 × 10–12_{m from each other. Potential energy (in eV) of this} system would be

Solution : Potential energy of the system r q Kq U_ 1 2 12 19 19 9 10 1 10 6 . 1 10 6 . 1 10 9           _{= 23.04 × 10}—17 _Joule eV 10 44 . 1 eV 10 6 . 1 10 04 . 23 3 19 17      _

Example 16. The stopping distance for a vehicle of mass M moving with speed v along level road, will be ( is the coefficient of friction between tyres and the road)

Solution : When the vehicle 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  = m MgS cos 180º = – m MgS So by Work-Energy theorem, W = K = K_f – k_i  – MgS = 0 – 1/2 Mv2

12  S = g 2 v2 

Example 17. 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}

Solution : As the particle is moving in a circle, so

2 2 r k r mv _ Now K.E = 2 1 mv2 ₌ r 2 k Now as dr dU F  P.E,



   r dr F U dr r k r 2



         r k   So total energy = U + K.E.

r 2 k r k    r 2 k   Negative energy means that particle is in bound state.

Example 18. 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

Solution : 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

Example 19. 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

-Solution : As the rod is kept in vertical position the shift in the centre of gravity is equal to the half the length = /2 Work done w = mgh = mg 2  = 20 x 9.8 x 2 4 = 392 J

Example 20. 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.

Solution : In first case, W₁ =

2 1 m(v₁)2 _{+ mgh} 2 1  m(12)2 _{+ m × 10 × 12} = 72 m + 120 m

and in second case, W₂ = mgh = 120 m

The percentage of energy saved 100 38%

m 192 m 120 m 192  _{ } 

P O WE R

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

(b) Power =      dw d x F. F.v dt dt In translatory motion : _P_F._v In rotational motion : _{P .}_ _ (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 : [M1_L2_T–3_]

Note : Power is the rate at which applied force transfers energy

(a) Power P w

t  

 where



w

work is done in t time (b) Instantaneous power P =

dt dw

.

EXAMPLE BASED ON POWER

Example 21. A pump can lift 9000 kg coal per hour from a mine of depth 120 m. Assuming its efficiency is 75%, its power will be (in watts)

-Solution : Power = time work

Effeiciency of pump = _inputpower power ouput ; 4 3 = P power output Output power = 4 P 3  P 4 3 = t mgh P = 3 4 t mgh = 3600 120 8 . 9 9000 3 4_   = 3920 W

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Example 22. A person of mass 60 kg is capable of taking a 15 kg massive objets to a height of 10 m in 3 min. The efficiency of person is

-Solution : % Efficiency 100 input work output work    ₌ 100 gh ) M m ( mgh   = 100 M m m _        = 15 60 100 15 _        = 20%

Example 23. An automobile of mass m accelerates from rest. If the engine supplies constant power p, the velocity at time t is given by

Solution : Given that power = Fv = p = constant

or v p dt dv m  [as F = ma = dt mdv ] or dt m p dv v





 1 2 C t m p 2 v _{ }

Now as initially, the body is at rest i.e. v = 0 at t = 0 so, C₁ = 0 

m pt 2 v 

Example 24. In the above problem, the position (s) at time (t) is given by Solution : By definition dt ds v or 2 / 1 m pt 2 dt ds        _{[From (1)]}  dt m pt 2 ds 2 / 1





       _ 1/ 2 3 / 2 2 2p 2 s t C m 3   _{ }    Now as t = 0, s = 0, so C₂ = 0  3/2 2 / 1 t m 9 p 8 s       

LINEAR MOMENTUM

The product of mass and velocity of the body is called the momentum _  v m P

Momentum is a vector quantity and its direction is always along the direction of velocity. Unit : In MKS : kg-m/sec or N-sec

In C.G.S gm-cm/sec or Dyne-sec Dimensions : [M1_L1_T–1_]

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

= dt P d F_ext    for v << c

where c = speed of light but when speed v is very near to c then momentum is given by P = 2 c v 1 mv       

The momentum measure the motion of body

EXAMPLE BASED ON LINEAR MOMENTUM

Example 25. On increasing the momentum of a body by 100% the increase in its KE will be-Solution : E = p2_/2m 2 2 2 2 1 1 E p E p = 2 100 200       = 4 3 1 1 4 E E E 1 1 2 _  _{  100} E E E 1 1 2 _{ = 300%}

Example 26. A jet of water whose cross-section is a, a wall making an angle  with normal and elastically rebounds. The velocity of water of density d is v. Force exerted on the wall is

 v

v Solution : If m kg water strikes the wall in one sec. and rebounds

elastically, then the change of its momentum

= 2mv cos  (perpendicular to wall) per sec.

But m = vad

 Force = rate of change of momentum = 2av2d cos 

Example 27. 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)

Solution : The change in momentum of the sand of mass m kg to gain a velocity v. = mv per sec.

 required force = mv

Example 28. In the above example,. what must be the velocity to have same kinetic energy (in m/s) Solution : The kinetic energy of first truck = 1/2 mv2 _{and the kinetic energy of second Jruck = 1/2 m’v’}2

Now 2 m'v'2 2 1 mv 2 1 _  5000 ) 20 ( 000 , 10 ' m mv ' v 2 2 2_{ }  _{v’ = 20 2 m/s}

LAW OF CONSERVATION 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.

16  dt P d F_ext    If  ext F = 0  dt P d  = 0 dP = 0 Change in momentum = 0 Momentum = constant If p ,p ,p ,...1 2 3   

, be the linear momentum of elements of a system, then p₁p₂p₃...

  

= constant Hence if the external force acting on a system is zero, the resultant momentum remains conserved.

The above equation is equivalent to three scalar equations. p_1x + p_2x + p_3x + ...+ p_nx = constant

p_1y + p_2y + p_3y + ...+ p_ny = constant p_1z + p_2z + p_3z + ...+ p_nz = 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 an axis, then the component of the linear momentum of the system along that axis cannot change.

EXAMPLE OF CONSERVATION OF LINEAR MOMENTUM

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 VM

 0 = mv VM _{V = –}  v M m  V m v M   _

The direction V is opposite to that of



v

(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 = M m mv 

Initial kinetic energy =

2 1

mv2 _{= E} i

Final kinetic energy = 2 1 2 2 f

m v

E

m M





M V v m i f i f _{1 E E} M m m E E _{  } 

 _{as some part of energy gets dissipated.}

(ii) When the bullet of mass m comes out after penetrating the block of mass M : In this case, mv = MV + mv₁ ...(1) Loss of energy = 2 1 mv2 _{– (} 2 1 MV2 ₊ 2 1 mv₁2₎ v m v1 m V M

= (initial energy) – (Final energy) l 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  2 2 2 1 m m v v  

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.

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

According to conservation of linear momentum mv = (m + M) V h V m V = ) M m ( mv 

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 = mg 2 V ) M m (  2

EXAMPLE BASED ON LINEAR MOMENTUM CONSERVATION

Example 29. A bullet of mass m moving horizontally with a velocity V strikes a block of mass M suspended by a string of length L and gets embedded in it. After the collision the maximum angular deviation of block from the vertical is

-Solution : From the conervation of momentum (m + M) v₁ = mv  v =

M m

mv 

(velocity of combined system after impact)



1

18

2 1

(M + m)u2 = (M + m)gL (1 – cos) or [1 – cos] = gL 2 1 2 M mmv       or                    2 1 M m mv gL 2 1 1 cos

Example 30. 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

Solution : Momentum of a bullet

= 0.25 × 400 = 100 kg –m/s

From momentum conservation law (0.25 + 4.75)v = 100 or 20m/s 5

100

v 

Example 31. -particle is emitted with a speed of 2.34 × 108 _{m/s from a stationary uranium nucleus, (U}238_). The velocity of residual nucleus will be

Solution : From momentum conservation law

234 × v = 4 × 2.34 × 108 _{4 10 m/s} 234 10 34 . 2 4 v 6 8     

Example 32. 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

Solution : From momentum conservation law m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

2 × 40 + m₂ × 0 = 2 × 25 + m₂ × 25 m₂ = 1.2 kg

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 direction 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 law of conservation of mechanical energy may be conserved.

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



On the basis of direction One-dimensional collision or Head on collision or Direction collision

On the basis of conservation of kinetic energy

TYPES OF COLLISIONS

Two dimensional collision or Oblique collision Elastic collision In-elastic-collision Perfectly inelastic collsion T h e c o l l i s i o n , i n which the particles move along the same straight line beforeand after the collision, is defined as one di mensi o nal co ll i si on

The collision, in which the particles move in the same plane at different angles before and after colli sion, is defined as oblique collision. A co ll i s i o n i s s a i d t o b e el as t i c, if th e t o t a l k i n e t i c en er gy b ef o re and after collision remains the same

A c o l l i s i o n i s s a i d t o b e e l as t i c, i f t h e t o t a l k i n e t i c energy does not remains constant

T h e c o l l i s i o n , i n which particles gets s t i k e d t o g e t h e r after the collision, is c a l l e d p e r f e c t l y i n el as t i c c o l l i si on . In this type of inelastic collision, lose of energy i s m a x i m u m .

Note :

Linear momentum is essentially conserved in any collision.

NEWTON’S LAW OF COLLISION

collision before velocity lative Re collision after velocity lative Re

= –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

1 2 1 2 u u v v   = – e e = 1, for elastic collision e < 1, for inelastic collision

e = 0, for perfectly inelastic collision Now we discuss the collision as follows

DIRECT ELASTIC COLLISION

m1 A u1 m1 B u2 F A B F m1 A v1 m2 B v2

Before collision During collision After collision

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

1 1 2 2 1 1 2 2

m u m u m v m v

   

   [where, p stands for linear momentum b.c for before collision a.c for after collision ]

20  2 2 2 2 1 1 2 2 2 2 1 1 m v 2 1 v m 2 1 u m 2 1 u m 2 1 _{  }

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

Important formula and features for direct elastic collision : (i) The velocity of first body after collision

2 2 1 2 1 2 1 2 1 1 u m m m 2 u m m m m v _                

(ii) The velocity of second body after collision

2 2 1 1 2 1 2 1 1 2 u m m m m u m m m 2 v _                

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

1 2 1 2 1 1 u m m m m v    _and 1 2 1 1 2 2m v u m m  

When a particle of mass m₁ moving with velocity u₁ collides with another particle with m₂ at rest and if

-  

(a) m₁ = m₂ (b) m₁ >> m₂ (c) m₁ << m₂

In this case, In this case, In this case,

v₁ = 0 and v₂ = u₁ v₁ = u₁ and v₂ = 2u₁ v₁ = – u₁ and v₂ = 2_mm u1 0 2

1 _

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

DIRECT INELASTIC COLLISION

In this case, (p)_a.c = (p)_b.c  m u1 1 m u2 2 m v1 1 m v2 2

   

   ,

(K.E)_b.c (K.E.)_a.c  (K.E)_b.c + Q = (K.E.)_a.c (where Q = heat energy, sound energy etc.)

 2 2 2 2

1 1 2 2 1 1 2 2

1 1 1 1

m u m u m v m v

2  2  2 2

According to Newton’s law, for inelastic collision we have, v₁ – v₂ = –e (u₁ – u₂) In inelastic collision, velocity of first body after collision :

2 2 1 2 1 2 1 2 1 1 u m m ) e 1 ( m u m m em m v _                 

and velocity of second body, v₂ = 2

2 1 1 2 1 2 1 1 _u m m em m u m m ) e 1 ( m     

Loss of energy in inelastic collision : 1 2 2 2

k 1 2 m m 1 E   (u u ) (1 e)   _{ }   

DIRECT, PERFECTLY INELASTIC COLLISION

m1 A u₁ m2 B u₂ Before collision A B After collision V

In this type of collision,

(p)_b.c = (p)_a.c.  m₁u₁m₂u₂ (m₁m₂)V

 

(K.E)_b.c. (K.E)_a.c.  (K.E)_b.c. + Q = (K.E)_a.c

 2 2 2 2 1 1 m u 2 1 u m 2 1  = 1 2 2 1 (m m )V Q 2  

According to Newton’s law for this collision : v₁ = v₂ (e = 0) Velocity after collision of the combined body : V =

2 1 2 2 1 1 m m u m u m   Loss of energy : 1 2 2 2 1 2 1 k (u u ) m m m m 2 1 E    

If u₂ = 0, ratio of final energy to initial energy : 1

m m m u m 2 1 v ) m m ( 2 1 E E 2 1 1 2 1 1 2 2 1 i f _     E_f < E_i

i.e there is loss of kinetic energy (f) If u₂ = 0,

) m m ( m E E 2 1 2 i k   

LINE OF IMPACT

It is important to know the line of impact during the collision. The line of impact is the line along which the impulsive force acts on the bodies. To find it draw the tangent at the point of contact of the two bodies. Draw a normal to the tangent at the point. This normal line is known as line of impact.

Line of impa ct v2 v2 v₁ v1 n B A t

OBLIQUE COLLISION

Let us now consider the case when the velocities of the two colliding spheres are not directed along the line of impact as shown in the figure. As already discussed, the impact is oblique. Since velocitiesv ' and1 v ' of2

22

four independent equations.

We choose as coordinate axes the n -axis along the line of impact, i.e. along the common normal to the surfaces in contact, and the t-axis along their common tangent. Assuming that the spheres are perfectly smooth and frictionless, we observe that the only impulses exerted on the spheres during the impact are due to contact forces directed along the line of impact i.e. along the line n. It follows that

m2v2 n B A t m1v1

+

B A t n _n B A t m1v1 m₂v₂

=

(1) The component along the t-axis of the momentum of each particle, considered separately, is conserved; hence the t component of the velocity of each particle remains unchanged. We can write

(v₁)_t= (v ' )1 t; (v2)t = (v ' )2 t.

(2) The component along the n axis of the total momentum of the two particles is conserved. We write. m₁ (v

1)n + m2(v2)n = m1(v )1 n + m2(v )2 n.

(3) The component along the n-axis of the relative velocity of the two particles after impact is obtained by multiplying the n-component of their relative velocity before impact by the coefficient of restitution.

(v )2 n - (v )1 n = e [(v1)n - (v2)n].

We have thus obtained four independent equations, which can be solved for the components of the velocities of A and B after impact.

Note :

Definition of the coefficient of restitution can be applied along common normal direction in the case of oblique collisions.

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.

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.

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 :

According to law of conservation of momentum :

p₃ p₃cos p₃sin p₁ p₂ 

Along horizontal direction : p₃cos = p₁ Along vertical direction : p₃sin = p₂

Magnitude of p3  p12p22 and its, direction from horizontal, tan  =

1 2 p p = tan–1       1 2 p p Direction of p₃, from the direction of motion of first part = [+tan–1

1 2

p p

] and that from the direction motion second part = [        2 + tan–1 1 2 p p ]

EXAMPLES BASED ON COLLISION

Example 33. A block of mass 12kg moving at 20 cm/s collides with an identical stationary block. If the coefficient of restitution is 3/5 the loss in K.E during collision is

-Solution : The loss in K.E. 2 _{1 2} 2

2 1 2 1 _{(1 e )(u u )} m m m m 2 1 E      Here m₁ = 12kg, m₂ = 12kg, e = 3/5 u₁ = 20 cm/sec = 0.2 m/s, u₂ = 0

On substiuting the values 

                25 9 1 12 12 12 12 2 1 E _(0.2)2 = 7.7 10 J 10 2 10 2 25 16 6 2 1_{     } 2

Example 34. 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

Solution : The height h₁ up to which the ball rises after the first rebound is given by h₁ = e2_h

After second rebound, h₂ = e4_h

After rebounding n times, h_n = (e2₎n_{h Total distance described} s = h + 2h₁ + 2h₂ + ... + 2h_n + ... = h + 2e2_{h + 2e}4_{h + ... + 2e}2h_{h + ...} = h + 2e2_{h (1 + e}2 _{+ e}4 + .... ) = h e 1 e 1 e 1 h e 2 h 2 2 2 2             

Example 35. 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 w.r.t ground velocity of 800 m/s. What velocity does rifle man acquire at the end of 10 shots.

Solution : Let m₁ and m₂ be the masses of bullet and the rifleman and v₁ and v₂ their respective velocities after the first shot. Initially the rifleman and bullet are at rest, therefore initial momentum of system = 0.

i.e. initial momentum = final momentum = m₁v₁ + m₂v₂

or kg 100 ) s / m 800 )( kg 10 10 ( m v m v 3 2 1 1 2      = - 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.

Example 36. 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)

24

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



m M V



2

2 , which is lost due to work done on it by the force of friction F.

Since force of friction F =  (M+m)g and the work done is given by Fd, we have 1/2 (M + m) V2 =  (M + m) gd or  = gd V 2 1 2 3 1 ) 15 ( ) 10 ( 10 2 1_ 2 _  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.

Example 37. 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 two 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.

Solution : 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 2 2 2 1 2 _(4)v 2 1 v ) 4 ( 2 1 15000 ) 50 ( ) 8 ( 2 1 _{  }  30º m =4kg2 v1 v2 m =4kg1 m = 8kg u = 50m/s or 25000 = 2(v₁2 _{+ v} 2 2₎ ....(1)

Applying the law of conservation of momentum along x-axis and y-axis respectively, we get

8(5) = 4 v₁ cos + v₂ cos 30º ....(2)

and 0 = 4v₁ sin = 4 v2 sin 30º = 2v2 ....(3)

or 1 2 v 2 v sin ....(4) From eq. (2) 100 = v₁ cos + v₂ cos 30_

or _                  2 3 v v 4 v 1 v 100 ₂ 2 1 2 2 1 or 2 3 v 2 v v 4 v 100 2 2 2 2 1 1    ....(5)

Solving equs. (1) and (5) for v₁ and v₂ we get v₁ = 51.56 m/s and v₂ = 99.2 m/s Now 56 . 51 2 2 . 99 sin    Solving we get  = 74º8.

SYSTEM OF VARIABLE MASS

In our discussion of the conservation of linear momentum, we have so far dealt with systems whose masses are constant. In this section, we will study systems which gain or lose mass during their motion e.g. The motion of a rocket depends upon the constant ejection of hot gases from its rear.

For solving such problems, you need to include the thrust (F )t



acting on the rocket in addition to all the other forces. This thrust is given by :

t rel dm F v dt    _{ }    _ Here,vrel 

is the velocity of the mass (ejected or gained if it is a different problem) relative to the socket, and

dm

dt is the rate at which mass is changing.

y x O v m at time t y x O at time t+ t v d v m-dm dm r v system

STEPS TO SOLVE THE PROBLEM

1. List all the forces acting on the body and show them on a diagram. 2. Show the additional thrust force F_t on the body..

t r dm F v dt    _ _  

26 net dv F m dt   

(m = mass at a particular instant) 4. Integrate it with proper limits to find velocity at any time.

If a external force also acts on the system then equation will be ext thrust

dv

M = F + F

dt

Example 38. A feed hopper releases grains at a ratedm

dt onto a conveyor belt that moves at a constant speed v. What is the power of the motor driving the belt ?

v

Solution : We consider a fixed length of the belt plus the grain lying on it as the system.

The mass M of the system increases as it receives grain falling from the hopper.

So dM dm

dt  dt

Since the grains fall vertically, u in the horizontal direction.0

 vrel u – vv.

Since the speed is constant,

dv 0 dt   ext dm F v dt  

whereF is the force needed to maintain constant speed because the mass is increasing._ext The power P = F.v 2dm P v dt  _.

MOTION OF ROCKET

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

Hot-Gas Jet v u D ir e c ti o n o f J e t D ir e c ti o n o f ro c k e t

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.

Time Velocity Mass

t v m

t + dt v + dv m – dm

The relative velocity of mass dm relative to rocket will be v_r = 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 – u) or dv = u m dm  dv f i v v



= f i m m dm u m



Where i and f represents initial and final states Therefore v_f – u_i = u log

i f

m m

Therefore if initial velocity be zero, the final velocity will be

f i i f f m m log u m m log u v   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

Example 39. The rate of burning of fuel in a rocket is 50gm /sec. and com es out w ith velocity 4 x 103_{m/s. The} force exerted by gas on rocket will be

Solution : The rate of change of momentum is equal to force dt dp F dt dm v  (Here v is constt.) Here v = 4 × 103 _{m/s &} dt dm = 50 × 10–3 _kg/s  F = 4 × 103 _{× 50 × 10}–3 _{= 200 N}

CENTRE OF MASS

In a system of many particles, the centre of mass is that point in the system, which alone 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.

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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

   3 2 1,r ,r

r ... with respect to the centre of mass, m₁r₁m₂r₂m₃r₃...0

   or 0 r m_{i i} 

  , where r_i = position vector of ith _{particle with respect to the centre of mass, m}

i = mass of the i th particle.

Suppose a system consists of n particles with masses m₁, m₂, m₃ ...m_n and their position vector with respect to chosen reference frame will be





                dm r dm m r m ... m m m ... r m r m r m r i i i 3 2 1 3 3 2 2 1 1 cm m₁ m₂ m3 C r₁ r₂ r₃

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

-dm dmx m x m m m . ... x m x m X i i i 2 1 2 2 1 1 cm _          1 1 2 2 i i cm 1 2 i m y m y ... m y dmy Y m m m dm           m (x ,y ,z )_{1 1 1 1} m (x ,y ,z )_{2 2 2 2} m (x ,y ,z )_{3 3 3 3} r2 r3 r1 O _X Z Y dm dmz m z m m m . ... z m z m Z i i i 2 1 2 2 1 1 cm _         

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 Note :

The centre of mass of a system is an imaginary point where the whole masses of the system may suppose to be acting.

The centre of mass of a body may lie inside or outside the body.

The centre of mass always lie on the axis of symmetry of the body if it exists. For a body in which there are two or more axes of symmetry, then the centre of mass lies at their point of intersection.

In a system of n particles, the centre of mass of the system may or may not coincide with any of the particles.

CENTRE OF MASS OF SOME COMMMON SYSTEMS

 A system of two point masses. The centre of

mass lies closer to the heavier mass

The centre of mass lies closer to the heavier mass. L m1 m₂ cm m L2 m1+m2 m L1 m1+m2  A Rectangular Plate C b x 2  C L y 2  b L cm yc xc O x y  A Triangular plate C h y 3  cm yc h y x O  A Solid Cone C 3h y 4  h yc cm  A semi-circular Ring C 2R y   C x 0 R cm O x ycm y  A semi-circular disc c 4R y 3   c x 0 R cm O x ycm y  A Hemispherical shell c R y 2  c x 0 R cm O x ycm y

30

If particles of body of masses m₁, m₂... are moving with velocities v , v , v1 2 3   

... respectively then velocity of centre of mass is given by

1 1 2 2 3 3 1 i i cm 1 2 i i m V m V m V ... m V p V m m ... m m                   = mass Total system of momentum Total

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

-1 2 i i 1 x 2 x x x x 1 2 i i m v m v ... mv P Vcm m m ... m m            i y i y 2 1 y 2 y 1 y m P m mv ... m m ... V m v m Vcm 1 2 i i            i z i z 2 1 z 2 z 1 z m P m mv ... m m ... V m v m Vcm 1 2 i i           

If Fx_net 0,Px_i  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 no external force acts on the body, state of motion of its centre mass remains the same i.e. it it is moving with same 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

   3 2 1,a ,a

a ..., acceleration of centre of mass is given by,,

1 1 2 2 i i cm 1 2 i m a m a ... m a a m m ... m    _{  }      

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

i x i 2 1 x 2 x 1 cm m a m ... m m ... a m a m a 1 2 i x         i y i 2 1 y 2 y 1 cm m a m ... m m ... a m a m a 1 2 i y _        i z i 2 1 z 2 z 1 cm m a m ... m m ... a m a m a 1 2 i z         If a₁a₂ a₃a_cm a

EXAMPLES BASED ON CENTRE OF MASS

Example 40. 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 ₂0

L m

(L + x). Find the position of centre of mass of this system. Discuss the case when x = 0.

Solution : Mass of element of length dx is dm = ₂0 L m (L + x) . dx 9 L 5 dx ) x L ( L m x . dx ) x L ( L m dm x . dm X L 0 2 0 L 0 2 0 L 0 L 0 cm      









Example 1. Find the centre of mass of the four point-masses as shown in the figure Solution : The coordinates of C.M. can be calculated as follows :

1 1 2 2 3 3 4 4 cm 1 2 3 4 x m x m x m x m X m m m m        = 2 3 4 3 5 ( 4) 1( 3) 12         = 5 m 0.42m 12    1 1 2 2 3 3 4 4 cm 1 2 3 4 m y m y m y m y Y m m m m        =2( 1) 4 3 5 4 1 ( 2) 12         = 28 2.3 cm 12 -4 -2 2 4 -4 -2 4 2 m = 2 kg1 m = 4 kg2 m = 5 kg3 m = 1 kg4

The position vector of C.M. rx i_cmˆy_cmˆj.



ˆ ˆ



r  0.42i2.3 j m 

.

Example 2. A thin rod of length 3L is bent a right angles at a distance L from one end. Locate the C.M. with respect to the corner. Take L = 1.2 m.

Solution : The centre of mass of each arm is at its mid-point. The C.M. of the two arms can be found by treating each arm as a point particle at its own C.M. From the figure, we see that

1 1

L

x , y 0

2

  and x₂ 0, y₂L If we take m ₁ = m then m ₂ = 2m, we get

1 1 2 2 cm L m 2m(0) m x m x 2 L 1.2 X 0.2m 3m 3m 6 6       _{ }      1 1 2 2 cm m y m y m(0) 2m(L) 2L 2 1.2 Y 0.8 m 3m 3m 3 3        

The position vector of C.M. is r_cm 0.2iˆ0.8 j mˆ

2L

L x

y

(x y )2, 2

(x y )1, 1

Example 3. A body of mass m₁ is projected vertically upwards with velocity v₁. Another body of same mass is projected at an angle of 450_{. Both reach the same maximum height. What is the ratio of their} kinetic energies at the point of projection?

32

Solution : From Work-Energy principle, 2 1 1 1 mgH 0 mv K 2      or _{K1 = mgH.}

In the projectile motion

o 2 2 2 2 1 1 mgH m(v cos 45 ) mv 2 2    2 2 2 K K K 2 2      _{K2 = 2mgH}  1 2 K K 1 2  .

Example 4. An object of mass m is tied to a string of length and a variable horizontal force is applied on it, which initially, is zero and gradually increases until the string makes an angle  with the vertical. Work done by the force F is

Solution : Since change in K.E. in zero, WgrWF 0  WFWgr   ( mgh)= mg (1- cos ) θ .

Example 5. A particle of mass m is fixed to one end of a light spring of force constant k and unstretched length . The system is rotated about the other end of the spring with an angular velocity  , in gravity free space(see figure). The increase in length of the spring will be

k m (a) 2 m l k (b) 2 2 m m   l k -(c) 2 2 m m    l k (d) none of these.

Solution : The required centripetal force is provided by the spring force. 2 0 0 m (l x )kx ,  x0 2 2 m l = k - m ω ω . Ans. (b)

Example 6. In the figure, a 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

B M (a) 2 M

(b) M (c) M/2

(d) A function of M and the force constant of the spring. Solution : Applying work energy principle,

2 1 mgx kx 0 2    Minimum value of x = 2 mg/k.

The block will leave contact with ground provided kx = Mg

or k.2mg Mg

k 

 mM

2 . Ans. (c)

Example 7. A particle of mass m moves in a conservative force field where the potential energy U varies with position coordinate x as U = U₀ (1 – cos ax), U₀ and a being positive constants. Which of the following statement is true regarding its motion

(a) the acceleration is constant (b) the kinetic energy is constant

(c) the acceleration is directed along the position vector (d) the acceleration is directed opposite to the position vector.

Solution : 0 dU F aU sin ax dx      acceleration F aU0sinax m m    2 0 a U x m

  for small value of x  acceleration   .x

Ans. (d)

Example 8. One end of a spring of natural length h and spring constant k is fixed at the ground and the other is fitted with a smooth ring of mass m which is allowed to slide on a horizontal rod fixed at height h as shown in the figure. Initially, the spring makes an angle of 37° with the vertical when the system is released from rest. Find the speed of the ring when the spring becomes vertical.

37o

h

Solution : The forces acting on the system are the spring force (conservative) and other constraint forces.

Therefore, mechanical energy is conserved. Initial tension in the spring = 5h h

4  = h

4 Applying energy (mechanical) conservation,

2 2 1 1 h mv k. 2 2 4    _{ }   or v = h k 4 m.

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Example 9. A rocket of initial mass 5000 kg ejects gas at a constant rate of 60 kg/s with a relative speed of 2050 m/s. Acceleration of the rocket 15 second after it is blasted off from the surface of earth will be (g = 10 m/s2) (a) 2 10m/s (b) 2 20 m/s (c) 2 30 m/s (d) 2 40 m/s Solution : Mass of the rocket after 15 s = 5000 – 60154100 kg

Thrust force rel

dm v 60 2050 12300N dt     Hence acceleration = 123004100 4100 = 2 20 m/s .

Example 10. Hailstones are observed to strike the surface of a frozen lake at an angle of 30º with vertical and rebound at 60 with the vertical. Assuming the contact to be smooth, the coefficient of restitution e is

(a) 1

3 (b) 0.25

(c) 0.5 (d) 0.75.

Solution : Let u and v be the velocity before and after impact respectively. The components of u and v along the horizontal direction are u sin 30º and v sin 60º respectively. Similarly, the components along vertical directions are u cos 30º and v cos 60º respectively by Newton’s Law

o v cos 60eu cos 30 ....(1) and o o v sin 60 u sin30 ....(2)  o o o o cos 60 cos 30 e sin 60  sin30 o o cot 60 e cot 30 e = (1/3).

Example 11. A constant force is applied on the wedge horizontally as shown in figure. Find the power required to push the w edge horizontally w ith constant speed V₀ to raise the rod A of mass m.

Solution : tan y x    y = x tan

B

F



A

or 0 dy dx tan V tan dt dt   _{ }      .

Required power = Force Velocity = mg. dy

dt mg V tan0  .

Example 12. In the figure shown the system is initially at rest and the light spring is elongated by a length x. The supports A and B fixed with the large block are light. Find the maximum speed of the block M after the thread is cut. Assume that friction is absent and all the surfaces are horizontal.

A

B

k

m

thread

M

Solution : When the string is burnt, the P.E. stored in the spring will appear as K.E. of the two blocks. The velocity of the blocks will be maximum when the initial tension in the spring is reduced to zero. Since no internal force acts on the given system, the centre of mass will remain at rest.

 CM mv MV V 0 M m        mv MV 0, or v MV m     . Since the total energy is conserved,

2 2 2 1 1 1 kx mv M.V 2 2 2 = 2 2 1 MV 1 m MV 2 m 2  _{ }      2 2 M kx MV 1 m    _  _   





1/ 2 2 mkx V M m M       _      .

Example 13. Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the centre of mass is

(a) 30 m/s (b) 20 m/s (c) 10 m/s (d) 5 m/s. Solution : cm 1 1 2 2 1 2 m v m v 10 14 4 0 v m m 10 4          10 m/s Ans. (c)

Example 14 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

Solution : The force acting on the particle = mdv dt Power of the force = mdv v k

dt   _     (constant) 2 kt c v m 2   ....(1) at t = 0, v = u  2 mu c 2  Now from (1), 2 mu kt 2 v m 2 2    2 1 m(v2 _{– u}2_{) = kt ....(2)} Again mdvv k dt   m.v dx dv v = k