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    UNIT V

    IMPULSE AND MOMENTUM

    5.1 Work-Energy Method

    In this method, the work done by the forces acting on a particle is related to the

    change in kinetic energy of the particle. The advantage of this method is that the

    determination of the acceleration of the particle is unnecessary. This principle is used in

    the solution of engineering problems.

    5.1.1 Work

    The work done by a force on a moving body is defined as the product of the force

    and the distance moved in the direction of the force. The unit of work done is N-m or

    Joule (J).

    5.1.2 Energy

    It is defined as the capacity to do work. There are many forms of energy like heat

    energy, mechanical energy, electrical energy and chemical energy. In engineering

    mechanics, we are interested in mechanical energy. This energy may be classified into

    potential energy and kinetic energy.

    Potential Energy:

    Potential energy is the capacity to do work due to the position of the body. A body

    of weight w held at a height h possesses an energy wh is called as potential energy.

    Kinetic energy

    Kinetic energy is the capacity to do work due to motion of the body.Mathematically,

    Kinetic energy (K.E)

    2 21 w 1v mv2 g 2

    = =

    where v = velocity of the body

    m = mass of the body.

    5.1.3 Power

    It is defined as time rate of doing work. Unit of power is watt (w) and is defined asone joule of work done in one second. In practice kilowatt (kw) is the commonly used

    unit. Horse power is the unit used in MKS and FPS systems.

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    5.2 Works-Energy Equation or Principle of Work and Energy

    Consider a particle of mass moving along either a rectilinear or curved path under

    the influence of a force. Consider a particle of mass m moving along a rectilinear under

    the influence of a force. Let us consider its movement from point 1 to point 2 as shown

    in figure 5.1.

    O

    1 2

    S 1

    S 2

    Figure 5.1

    Let F = Force applied on the particle

    s1

    = Distance travelled at point 1

    s2

    = Distance travelled at point 2

    v = Velocity of particledsdt

    =

    a = Acceleration of particle

    dv

    dt=

    Using the relation,

    F = ma

    dv dv dsF m m

    dt ds dt= =

    dvF m v

    ds=

    or F ds= m v dv

    Integrating both sides from point 1 to point 2.2 2

    1 1

    s v

    s v

    F ds m v dv=

    ( ) ( )2 22 1 2 1m

    F s s v v2

    =

    Left hand side gives the change in kinetic energy of the particle while moving from

    point 1 to point 2.

    Right hand side gives the change in kinetic energy of the particle while moving

    form point 1 to point 2.

    Work done = change in kinetic energy.

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    5.3 Conservation of Energy:

    Law of Conservation of Energy

    It states that energy can be neither be created nor destroyed, but it may change its

    form and may get converted into another form of energy. In other words the sum ofkinetic energy and potential energy of a particle moving under the influence of

    conservative forces is always a constant.

    i.e.T otal E nergy P.E K = +

    Proof

    Consider a particle of mass m located at a height h above the ground level.

    A ( V =

    B ( V =1

    C ( V =2

    h

    h 1

    Figure 5.2

    Total mechanical energy of the particle at A.

    = P.E + K.E

    21mgh m v2

    = +

    ( )

    1mgh m 0

    2= +

    = mgh ....(i)

    Similarly, total mechanical energy of the particle at B.

    = P.E + K.E

    ( )

    2

    1 1

    1mg h h m v

    2= +

    But, v1

    2 = u2 + 2gh1

    = 0 + 2gh1= 2gh

    1

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    ( ) ( )1 1

    1T.E mg h h m 2gh

    2 = +

    = mgh mgh1

    + mgh1

    = mgh ....(ii)

    Total mechanical energy of the particle C

    = P.E + K.E

    ( )

    2

    2

    1mg 0 m v

    2= +

    But, v2

    2 = 0 + 2gh = 2gh

    ( )

    1T.E 0 m 2gh mgh

    2 = + =

    (iii)

    From equations (i), (ii) and (iii), it can be cleared that total energy possessed by a

    body remains same. Hence the proof.

    5.4 Impulse-Momentum

    5.4.1 Impulse and Impulsive force

    When a constant force F is applied to a particle for a time t, then the product Ft

    is called the impulse.

    When the force is large and the time interval is very short (i.e., instantaneous), then

    the force is called an impulsive or impact force. The resulting motion is callled an

    impulsive motion. Practical situations where impulsive forces arise are collisions,

    explosions, the driving of a pile, the hitting of a cricket ball, etc.

    5.4.2 Momentum (or) Linear momentum

    It is the product of mass of the particle and linear velocity of the particle with

    which the mass is in motion.

    Mathematically,

    Momentum m v=

    where, m = mass of the particle

    v = velocity of the particle.

    5.4.3 Impulse-Momentum Principle

    This principle can be used to solve problems involving force, mass, velocity and

    time. This principle is particularly useful for problems involving impulsive motion orimpact.

    Impulse-Momentum Principle states that the component of the resultant linear

    impulse along any direction is equal to change in the component of momentum in that

    direction.

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    Proof

    Consider a particle of mass m acted upon by a force F.

    Newtons second law of motion states that the rate of change of momentum isdirectly proportional to the external force acting on a body.

    F = ma

    But,

    dva

    dt=

    dvF m

    dt =

    or F dt = m dv

    Integrating on both sides,

    We get,

    2

    1

    t v

    t u

    F dt m dv=

    I = m (v u)

    = mv mu

    where, I = Impulse

    u = Velocity of the body during t1

    v = Velocity of the body during t2.

    Impulse = Change in momentum

    = Final momentum - Initial momentum

    Hence the proof.

    5.4.4 Principle of Conservation of Momentum

    It states that when the sum of the impulses of all external forces acting on a particle

    is zero, the total linear momentum of the system is remains constant. This law is mainly

    used to study collision of elastic bodies and also to find the recoil velocity of gun.

    Consider a system comprising of a man and a boat, the action of the man is equal

    and opposite to the reaction of the boat. Hence the resultant force is zero in the system.

    According to law of conservation momentum,

    Momentum before jumping = Momentum after jumping(m

    mu

    m+ m

    bu

    b) = m

    mv

    m+ m

    bv

    b

    But, um

    = ub

    (i.e., before jumping) = u

    ( )m b m m b bm m u m v m v + = +

    where mm

    = Mass of man

    mb

    = Mass of boat

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    u = Initial velocity of man and boat

    vm

    = Final velocity of man

    vb

    = Final velocity of boat.

    SOLVED PROBLEMS ON WORK-ENERGY METHOD

    1. Find the work done in drawing a body i) Weighing 500 N through a distance

    10 m along a horizontal surface by a horizontal force of 200 N ii) Weighing

    500 N through a distance 10 m along a horizontal surface by a force of 200

    N whose line of action makes an angle of 30 with the horizontal.

    Solution:

    F = 2 0 0 N

    W = 5 0 0 N

    F = 2 0 0 N

    W = 5 0

    3 0

    Case (i)

    Work done = Force Distance= 200 10

    = 2000 N-m

    Case (ii)

    Work done = Force Distance

    = 200 cos 30 10

    = 1732.05 Nm.

    2. Calculate the energy possessed by an aeroplane of mass 8 tonnes whileflying at the rate of 10 m/sec at altitude of 300 m.

    Solution:

    Given, m = 8 tonnes = 8000 kg

    v = 10 m/sec, h = 300 m

    Energy possessed by an aeroplane (i.e., Total energy)

    = Potential energy + Kinetic energy

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    But, Potential energy = mgh = 8000 9.81 300

    = 2.354 107 N-m

    Kinetic energy( )

    221 1mv 8000 102 2

    = = =4.0 105 N-m.

    Total energy, T.E= 2.354 107 + 4.0 105 = 2.394 107 N-m.

    3. Two weights 50 N and 30 N are connected by a thread and move along a

    rough horizontal plane under the action of a force 40 N applied to the first

    weight of 50 N as shown in figure. The co-efficient of friction between

    the sliding surfaces of the weights and plane is 0.4. Determine the acceleration

    of the weights and the tension in the thread using work-energy equation.

    3 0 N 5 0 N 4 0

    Solution:

    Consider the FBD of the two weights.

    3 0 N

    T

    R = 3 0 N

    ( 0 . 4 3 0 )

    5 0 N

    4 0

    R = 5 0 N

    ( 0 . 4 5 0 )

    FBD of the weights

    Let x = Distance moved by the system

    u and v = Initial and final velocities with the action of 40 N force.

    Fx

    = Net horizontal force acting along the direction of movement

    = 40 (0.4 30) (0.4 50)

    = 8 N

    Now applying work-energy equation

    Work done = change in K.E

    ( ) ( )2 21 2x1 w w

    F x v u2 g

    + =

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    4. Two blocks A and B of weight 150 N and 80 N are hung to the ends of a rope,

    which is passing over an ideal pulley as shown in figure . The velocity of

    the system is increased from 1 m/sec to 1.5 m/sec. How much the distance,

    these block will move? Also calculate the tension in the string. Use energy

    method.

    1 5 0 N

    8 0 N

    A

    B

    Solution:

    When the system is released from the rest, 150 N moves down and 80 N blockmoves up. But both the bodies will have the same velocity and displacement.

    Let x = Distance moved by each block.

    Work done by 150 N block = 150 x

    Similarly, Work done by 80 N block = 80 x

    ( sign indicate that the work done by 80 N is opposite to the 150 N).

    Total work done = (150 80) x = 70 xChange in kinetic energy of the system

    ( )2 21 21 w w

    v u2 g

    +=

    ( )2 21 150 80

    1.5 12 9.81

    + =

    Apply work energy equation,

    Work done = Change in K.E

    ( )2 2

    1 150 8070 x 1.5 1

    2 9.81

    + =

    70 x = 14.65

    x = 0.209 m

    To find T

    Let T = Tension in the string. Consider the work done by any one of blocks.

    Work done by 150 N Block = (150 T) x

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    Change in K.E( )2 2

    1 1501.5 1

    2 9.81

    =

    Apply work energy equation,

    Work done = change in K.E

    ( ) ( )

    2 2

    1 150150 T 0.209 1.5 12 9.81 =

    (150 T) 0.209 = 9.56

    150 T = 9.56/0.209 = 45.72

    T = 150 45.72 = 104.27 N.

    5. A block of mass 60 kg slides down at 30 inclined plane from rest as shown

    in figure . After moving 1.3 m, the block strikes a spring whose modulus

    is 25 N/mm. Determine the maximum deformation of the spring. Take the

    co-efficient of kinetic friction between the block and plane is 0.3. Also

    calculate the velocity with which the block strikes the sping.

    Solution:

    Let x = maximum deformation of the spring,

    m. Consider the FBD of the

    60 kg block.

    Fy

    = 0, R 588.6 cos 30 = 0

    R = 509.73 N.

    Let Fx

    = Net force acting along the plane.

    = 588.6 sin 30 0.3 R

    = 588.6 sin 30 0.3 509.73

    = 141.38 N.

    3 0

    3 0

    6 0 9 . 8 1 N

    = 5 8 8 . 6 N

    588 .

    6 sin

    30

    588 .

    6 c o

    s30

    R

    0 .3R

    3 0

    6 0k g

    1 . 3m

    x

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    Fx

    = 141.38 N.

    But, Fx

    = ma (i.e., Newtons second law)

    141.38 = 60 a

    a = 2.36 m/sec2

    Using the relation,

    v2 = u2 + 2 a s

    where, u = Initial velocity of block= 0

    v = Final velocity of block = ?

    s = Distance moved by the block before striking the spring

    = 1.3 m (givesn)

    v2 = 0 + 2 2.36 1.3 = 6.136 m/sec.

    or v = 2.48 m/sec.

    Velocity with which the block strikes the spring = 2.48 m/sec.

    6. A body weighing 200 N is pushing up a 30 plane by a force of 400 N acting

    parallel to the plane. If the initial velocity of the body is 1.5 m/sec and

    = 0.2, what velocity will the body have after moving 5 m?

    3 0

    4 00 N

    2 00 N

    Solution:

    Draw the FBD of the block.

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    h = 1 5 c m

    S

    3 0

    2 0 0 N

    200 sin3

    0

    200 c

    o s30

    R

    3 0

    0 .2R

    Fy

    = 0, R 200 cos 30 = 0

    R = 173.2 N

    Work done by force along the plane

    = (400 0.2 R 200 sin 30) distance

    = (400 0.2 173.2 200 sin 30) 5

    = 265.36 N-mChange in K.E

    ( )2 2

    1 wv u

    2 g

    =

    ( )2 2

    1 200v 1.5

    2 9.81

    =

    = 10.19 (v2 2.25)

    Apply work-energy equaiton,

    Work done = change in K.E

    265.36 = 10.19 (v2 2.25)

    = 28.29 m/sec

    v = 5.31 m/sec.

    7. A ball of mass 3 kg is dropped on to a spring of stiffness 300 N/m from a

    height of 15 cm. Find the maximum deflection of the spring.

    Solution:

    Given,

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    Mass of ball, m = 3 kg

    Weight of ball, w= 3 9.81 = 29.43 N

    Let s = maximum deflection of the spring.

    According to law of conservation of energy,

    Work done by the body= Work done by the spring.

    ( ) 2

    1w h s ks

    2+ =

    ( ) 2

    129.43 0.15 s 300 s

    2+ =

    4.414 + 29.43 s = 150 s2 or 150 s2 29.43 s 4.414 = 0

    2 b b 4 a cs

    2 a

    =

    ( )2

    29.43 29.43 4 150 4.414

    2 150

    + =

    29.43 59.28

    0.3 m or 0.1 m300

    = =

    Taking positive value, s = 0.3 m or 300 mm.

    SOLVED PROBLEMS ON IMPULSE-MOMENTUM METHOD

    1. A car of mass 200 kg is travelling on a horizontal track at 36 km/hr.

    Determine the time needed to stop the car. The co-efficient of friction between

    the tyres and the road is 0.45.

    Solution:Given,

    mass of car, m= 200 kg

    Weight of Car, w = 200 9.81 = 1962 N

    36 1000u 36 km / hr 10 m / sec

    3600

    = = =

    Applying Impulse-momentum equation,

    F t = m (u v) .... (i)

    To find F

    F = Force acting along the direction of motion

    = R

    F = 0.45 1962 = 882.9 N

    Substituting F value in (i)

    882.9 t = 200 (10 0) (v = 0 Since car is brought to rest)

    882.9 t = 2000

    t = 2.27 seconds

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    2. A mass of 20 kg block slides down from rest on an inclined plane inclined at

    30 with the horizontal. What will be the speed of the block at the end of 3

    seconds? Take the co-efficient of kinetic friction between the block and the

    plane is 0.25.

    Solution:

    3 0

    w

    w sin

    30

    w co

    s30

    R

    0 .25R

    mass of block,m = 20 kg

    Weight of block, w = 20 9.81 = 196.2 N

    Resolving the forces normal to the plane

    R w cos 30 = 0

    R = w cos 30 = 196.2 cos 30 = 169.91 N

    Let F = Net force acting along the plane

    = w sin 30 0.25 169.91 = 55.62 N

    Applying Impulse-momentum equation,

    F t = m (v u)

    where u = Initial velocity of block= 0

    t = Time = 3 sec (given)

    v = ?

    m = 20 kg

    55.62 3 = 20 (v 0) v = 8.34 m/sec.

    3. Two weights 50 N and 30 N are connected by thread and move along a

    rough horizontal plane under the action of a force 40 N applied to the first

    weight of 50 N as shown in figure. The of the weight and plane is 0.4.

    Determine the velocity of the system after 3 seconds. Also calculate the

    tension in the string using Impulse-momentum equation.

    3 0 N 5 0 N 4

    Solution:

    Draw the FBD of the body.

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    3 0 N 5 0 N

    ( 0 . 4 3 0 ) ( 0 . 4 5 0 )

    R = w = 3 0 N R = w = 3 0

    F B D o f t h e b o d

    4 0

    Let Fx

    = Net horizontal force acting along the direction of motion.

    = 40 (0.4 30) (0.4 50) = 8 N

    Applying Impulse momentum equation

    T

    R = w = 3 0 N

    3 0 N

    ( 0 . 4 3 0 ) N

    F t = m (v u)

    ( ) ( )30

    T 12 3 2.94 09.81

    =

    ( ) ( )

    30 508 3 v 0 v 0

    9.81 9.81 = +

    30 5024 v

    9.81

    + =

    v = 2.94 m/sec.

    To find T

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    Let T = Tension in the string.

    Consider any one of the block and apply Impulse momentum equation

    T

    R = w = 3 0

    3 0 N

    ( 0 . 4 3 0 ) N

    Consider 30 N block,

    [T (0.4 30)] t = m (v u)

    ( ) ( )

    30T 12 3 2.94 0

    9.81 =

    = 8.99

    8.99T 12

    3 = +

    = 15 N.

    4.Two block of weight 150 N and 50 N are connected by a string and passingover a frictionless pulley as shown in figure . Determine the velocity

    of 150 N block after 4 seconds.

    1 5 0 N

    5 0 N

    A

    B

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    Solution:

    Since the weight of block A is more than that of B, A will move down and B willmove up. If B moves up by y. A will move down by y/2. Hence the velocity of Block

    A will be half that of B. The tension in the string will be same throughout.

    Draw the FBD of the two blocks,

    Consider 50 N Block

    Applying Impulse- momentum equation.

    ( ) ( )50

    T 50 t v 09.81

    =

    (T 50) 4 = 5.097 v or T 50 = 1.274 v ....(i)

    Consider 150 N Block

    Applying Impulse-momentum equation.

    ( ) ( )

    150 v150 2T 4 029.81

    =

    150V150 2T 1.9113 v

    4 9.81 2 = =

    or 150 2T = 1.9113 v ....(ii)

    Solving the equation (i) and (ii)

    T 50 1.274 v 2

    150 2T 1.9113 v

    Add 50 4.46 v

    = =

    =

    v = 11.2 m/sec

    Velocity of 150 N block = v/2 = 11.2/2 = 5.6 m/sec.

    T = 1.274 v + 50 = 1.274 11.2 + 50

    = 64.29 N.

    TTT

    150N 50NFBD of A FBD of B

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    5. A boat of weight 4 kN is at rest in a river. A man weighing 60 kg jumps into

    the boat with a velocity of 1.2 m/sec. With what velocity the boat will move

    immediately after the man has jumped.

    Solution:

    Let mb

    = mass of boat

    4000407.74 kg.

    9.81= =

    mm

    = mass of man = 60 kg

    ub

    = Initial velocity of boat = 0

    um

    = Initial velocity of man = 1.2 m/sec

    vb

    = Final velocity of boat

    vm

    = Final velocity of man

    As soon as the man jumps into the boat both man and boat will have the samevelocity,

    vb

    = vm

    = v

    Apply law of conservation of momentum,

    mm

    um

    + mb

    ub

    = mm

    vm

    + mb

    vb

    mm

    um

    + mb

    ub

    = v (mm

    + mb)

    ( )m m b b

    m b

    m u m uv

    m m

    + =

    +

    ( )

    60 1.2 0

    60 407.74

    +=

    +

    = 0.153 m/sec.

    The positive sign for v indicates that the boat will move along the direction in

    which the man has jumped into the boat.

    6. A gun of mass 3500 kg fires horizontally a bullet of mass 40 kg with a velocity

    of 250 m/sec. What is the velocity with which the gun will recoil? Find also

    the force required to stop the gun in 50 cm.

    Solution:

    Let mg= mass of gun = 3500 kg

    mb = mass of bullet = 40 kgv

    b= velocity of bullet = 250 m/sec

    vb

    = velocity of gun (i.e., recoil velocity of gun)

    Applying law of conservation of momentum

    mb

    vb

    = mg

    vg

    sing is due to recoil of gun in opposite direction.

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    b bg

    g

    m v 40 250v 2.86 m / sec

    m 3500

    = = =

    Also, v2 = u2 + 2 a s

    0 = (2.86)2 + 2 a (0.5)

    a = 8.179 m/sec2

    Force required to stop the gun.

    = m a= 3500 8.179 = 28.62 kN.

    Review Questions

    1. Define work and energy.

    2. State the different forms of energy.

    3. Distinguish between potential energy and kinetic energy.

    4. State law of conservation of energy.

    5. State the principle of work and energy.

    6. What is impulsive force?

    7. State the equation for the law of conservation of momentum.

    8. Define potential energy.

    9. Define kinetic energy.

    10. Write the equation for the potential energy and kinetic energy of a particle.

    11. Define the term linear momentum?

    11. A body weighing 400 N is pushed up in a 25 plane by a 250 N force acting parallel

    to the plane. If the initial velocity of the body is 1.5 m/sec and co-efficient of kinetic

    friction is 0.2, What velocity will the body have after moving 6 m? (ANS: 2.17 m/sec)

    12. A block weighing 100 N is moving along a horizontal rough surface of friction co-

    efficient 0.2 with a velocity of 5 m/sec. A push of 80 N inclined at 30 to the horizontal

    acts on the block. Using work-energy principle, find the velocity of the block after it had

    moved through a distance of 20 m ? (ANS: 12 m/sec)

    13. A block of mass 5 kg resting on a 30 inclined plane is released from rest. The block

    after travelling a distance of 0.5 m along the inclined plane hits a spring of stiffness 1.5N/mm, placed along the inclined plane. Find the velocity with which the block hits the

    spring and also maximum compression of the spring. Assume the co-efficient of friction

    between block and plane as 0.2. (ANS: 115 mm )

    14.Two blocks A and B of masses 75 kg and 150 kg are connected by a string. If the

    system is released from rest, find the velocity of block A after it has moved a distance of

    0.75m. Assume the co-efficient of friction as 0.15. (ANS: 2.4 m/sec)

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    15. A 100 N block shown in fig, is released from rest and slides a distance S down the

    inclined plane. It strikes the spring which compresses 0.1m, before motion impends up

    the plane. Take = 0.25 and spring constant K = 3N/mm. Determine the value of S.

    (ANS: 0.429m)

    16. A machine gun fires 100bullets/min. The nozzle velocity of the bullet is 400

    m/sec. The mass of one bullet is 0.035kg. Determine the average reaction of the

    gun against the support if firing is continuous?( ANS: 23.33N)

    17. A pile driver of mass 3000kg falls through a height of 5m on a pile. Find the

    impulsive force of the blow if the hammer comes to rest in 0.01 sec. ( ANS:

    2970000N)

    18. A 15 KN car is moving at a speed of 60 Km/hr, when the brakes are fully appliedcausing all four wheels to skid. Determine the time required to stop the car? Take

    = 0.05. ( ANS : 33.84 sec)


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