Work Power And Energy - Alfa Tutorialsalfatutorials.in/wp-content/uploads/2018/07/Work_theory.pdfWork Power And Energy XI Class ALFA PHYSICS CLASSES Sanjay Chopra 86, Chotti Baradari

Work Power And Energy XI Class

ALFA PHYSICS CLASSES

Sanjay Chopra 86, Chotti Baradari Part-1[ Near Medical College] Jalandhar 98152-15362

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Work, Power and Energy

Table of Contents Work:................................................................................................................................................................ 2

Nature of work.................................................................................................................................................. 3

Work Done by the variable force: ...................................................................................................................... 4

Power ............................................................................................................................................................... 5

Energy .............................................................................................................................................................. 6

Kinetic energy ................................................................................................................................................... 6

Potential Energy And Gravitational Potential Energy ......................................................................................... 9

Conservative force ..................................................................................................................................... 10

Energy stored in Spring ................................................................................................................................... 11

Mechanical Energy and its conservation:......................................................................................................... 13

Different forms of energy................................................................................................................................ 15

Einstein’s Mass Energy equivalence ................................................................................................................ 16

Collisions ........................................................................................................................................................ 17

Elastic Collision in two dimensions: ................................................................................................................. 18

Elastic Collision in One dimension: .................................................................................................................. 19

Conceptual Questions ..................................................................................................................................... 20

Work [Assignment No:1] ................................................................................................................................. 29

Work, Power And Energy [Assignment No:2] ................................................................................................... 30

Work, Power and energy [ Assignment No:3] .................................................................................................. 31

Collisions [Assignment No:4] ........................................................................................................................... 32

Work, Power And Energy [Assignment No:5] ................................................................................................... 35




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Work:

Work is said to be done by a force acting on the body, if force acting on the body produces some

displacement in the body in the direction of force or displacement of the body has some component

in the direction of the force. Thus, if displacement is perpendicular to the force acting on the body

then no work is done by the force.

Sup-

pose a force acting on the body, produces a dis-

placement s

, in the body, then the work done by

the force is

W = SF

. = [F Cos ]S

i.e mathematically work is defined as the product of the component of force in the direction of dis-

placement and the displacement of the body. Here, is the angle which the force vector makes with

the displacement vector. If force produces displacement in the direction of displacement, then =0,

W = FS Cos 0= FS

In co-ordinate form, if force and displacement produced by the force are given by

kFjFiFF zyxˆˆˆ

kzjyixS ˆˆˆ

Then, work done by the force is

W = SF

. =( kFjFiF zyxˆˆˆ ).( kzjyix ˆˆˆ )

W= Fxx + Fyy + Fzz

As, work is dot product of two vectors, thus work is a scalar quantity.

Units of Work:[a] Absolute Units: In SI the absolute unit of work is joule [J]. Work done on the body is

said, to be 1 J if force of 1N acting on the body produces a displacement of 1m in the body in the di-

rection of the force.

1 J = 1N x 1m= 1N-m

In cgs system, the absolute unit of work is erg. Work done by the force is said to be 1 erg, if the force

of 1 dyne acting on the body displaces the body through 1cm in the direction of the force.




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I erg = 1 dyne x 1cm = 1dyne-cm

Relation between joule and erg:

1 J = 1N-m= 105 dyne x 100cm

1 J=107 dyne-cm

1 J=107erg

Gravitational Units: In SI the gravitational unit of work is kg-m. The work done is said to be 1kg-m if

the force of 1kgf displaces a body through 1m in the direction of the force.

1kg-m = 1kgf x 1m = 9.8N-m

In cgs, the gravitational unit of work is g-cm. The work done by the force is said to be 1g-cm, if force

of 1 gf displaces the body through a distance of 1 cm in the direction of the force.

1g-cm = 1g f x 1cm = 980 erg

Nature of work:

Work done by the force can be classified into 3 categories:

[1] Positive work: The work done by the force is said to be positive if the angle between force and

displacement is acute. W = FS cos , where is angle less than 900

for e.g. 1. If we pull a heavy box using string and displace it through certain distance, then the work

done by the pulling force is positive work.

[b] When a body is falling under the effect of the gravity, then the work done by gravitational force is

positive as both force and displacement both are acting vertically downwards and angle between

force and displacement is 00.

[2] Negative work; The work done by the force is said to be negative, if angle between force and dis-

placement is obtuse [ > 900]. W = FS cos , where is obtuse angle.

For e.g. 1. When we pull a heavy box using the string, the frictional force tends to act opposite to the

displacement to oppose relative motion between box and the surface. The work done by frictional

force on the body is negative.




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2. When body is moved upwards from the ground, then the work done by the gravitational force on

the body is negative work because displacement is acting at an angle of 1800 with the gravitational

force.

[3] Zero Work: Work done by the force is said to be zero, if force and displacement are perpendicular

to each other or displacement of the body is zero. I.e. W = FS cos 90=0

For e.g. 1. When body is moving with constant speed in the circular path, then the centripetal force

acts perpendicular to the displacement of the body. The work done by the centripetal force on the

body is zero work.

2. When coolie moves on the horizontal railway platform with load on his head, then the work done

by the gravitational force is zero as gravitational force and displacement are perpendicular.

3. We try to push a heavy box by applying the force, but unable to displace it, in that case work done

is zero as displacement is zero.

4. When simple pendulum oscillates, the tension in the string is always perpendicular to the dis-

placement of the body, thus work done by the tension in the string is zero.

Work Done by the variable force:

Variable force means which changes its magnitude with the

change in position or the force which keeps on changing as the

body is displaced from one point to another. For e.g. if two bodies

are moving towards each other under mutually attractive gravita-

tional force, then the force which acts between the two bodies

goes on increasing as they approach each other.

Consider a variable force acting on the object that moves along straight line along x axis, as shown in

figure. Suppose the x component of force varies with position as shown in the graph, when body

moves from x=a to x=b.

The total displacement of the body is divided into large number of small displacements x1, x2,

x3……xn. In each small displacement we assume the force to be approximately constant. Thus, total

work done to move through from x=a to x=b is the sum of work done to move the body through n

small segments.

W= Fx1x1 + Fx2x2+………………….Fxn xn




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Also, Fx1 x1,…….Fxnxn are the areas of rectangles as shown in the figure. Thus, work done is the sum

of area of all the rectangles.

Thus, we conclude that when the force acting on the body varies with the distance, then the work

done by the force is given by area under the force-displacement graph.

Power

When we calculate the work, we never considered the time taken to do that work, a body is lifted

through a distance of 5m the work done is same if it is lifted in 1 second or 1minute. But in many

cases it is also useful to determine the rate at which the given work is done.

The time rate at which work is done by the force is called

power.

P = dt

dW

If force F

acting on the body displaces the body through S

, then work done is given by W= SF

. ,

then power can be written as

P = dt

SFd

.

If force acting on the body is constant, then

P = vFdt

SdF

..

Thus, instantaneous power is dot product of the force and velocity vector at that instant. Force and

velocity are vector quantities but their dot product is a scalar quantity.

Average power: It is the ratio of the total work done to the total time taken to do that work. If work

done is W and the time taken is t, then

Pav = t

W

Units of power: In SI the unit of power is watt or joule/sec. Power is said to be 1W, if 1J of work is

done in a time of 1 second. Other commonly used unit of work is H.P. [horse power]

1 H.P. = 746W

The dimensional formula for power is [M1L2T-3].




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Energy

Energy is defined as the ability of the body to do work and is measured by the total quantity of work

done it can do. The units of work and energy are thus also identical. The different forms of energy

are: kinetic energy, potential energy, chemical energy, electrical energy etc. But in mechanics we

mainly deal with the kinetic and potential energy associated with the body.

Kinetic energy

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

Consider a body of mass m, which is initially at rest on a

perfectly frictionless surface. When a constant force acts on it, the body starts moving with accelera-

tion ‘a’. If v

, denotes the velocity of the body after it displaces through S, then

S

va

2

2

Work done in displacing the body through distance S

is

W = SF

. = FS Cos0

W = maS =mS

v

2

2

S= 2

2

1mv

As, we have not wasted the energy for the body is moving on frictionless surface, whole of this work

done gets converted into the kinetic energy of the body.

Ek =2

2

1mv

Thus, kinetic energy is one half of the product of mass and square of the speed of the particle.

If force acting varies with position: But if the force acting on the body is not constant, then we divide

the total displacement into number of small displacements Sd

each. Work done by the force for one

such displacement in the direction of force is

FdSCosFdSSdFdW 0.

dW= [ma]dS

dW = mdt

dv dS




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dW = m dvdt

dS

dW = mvdv

Thus, the total work done if the velocity of the body increases from 0 to v is

W = v

mvdvdW0

W = m

v

v

0

2

2

W= 2

2

1mv

This, work done gets stored in the body in the form of its kinetic energy.

Thus

Ek = 2

2

1mv

Ek = m

p

2

2

Variation of kinetic energy with momentum: If the mass of the body is constant then the kinet-

ic energy varies with momentum as Ek p2

Variation of kinetic energy with mass: if the momentum of the

body is constant, then kinetic energy is inversely proportional to mass of

the body.




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Work Energy Theorem

According to work energy theorem, the work done by force in displacing the body measures the

change in kinetic energy of the body. Increase in kinetic energy of the body means positive work

done on the body and decrease in kinetic energy signifies negative work done on the body.

Consider a body of mass m moving with initial velocity

‘u’ when force F

is applied in the direction of velocity of the body. If Sd

is the small displacement

produced when force F acts, then work done on the body is given by

dW = SdF

.

dW= FdS cos 0

dW= madS

dW= mdt

dv dS

dW = m dvdt

dS

dW = mvdv

Total work done to increase the velocity of the body from u to v is

W =

v

u

v

u

vmmvdv

2

2

W= 22

2

1uvm

W= final KE- initial KE

W= Change in kinetic energy of the body.

In this we have assumed that whole of the work done is used in increasing or changing the kinetic

energy of the body only, but work done may also be stored in the form of PE of the body. Also, work

energy principle holds good for any system of particles in the presence of both internal as well as ex-

ternal forces and both conservative and non conservative forces.




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Potential Energy And Gravitational Potential Energy

The potential energy is energy possessed by the body by virtue of its position is some field. Potential

energy of a body can be defined only if the forces or field acting on it are conservative. Potential en-

ergy is due to configuration of the bodies in a given system and if the configuration changes potential

energy will change. In this chapter two potential energy under consideration are [a] gravitational po-

tential energy and [b] potential energy stored in spring or elastic potential energy.

Gravitational Potential energy:

The energy potential energy possessed by the body by virtue of its position in the gravitational field

of earth is called gravitational potential energy.

Whenever we try to move the body above the surface of earth the gravitational force will ex-

ert downward force on the body. Thus, to move the body upward we have to do work against the

gravitational field of earth, this work will be stored in the body in the form of gravitational potential

energy of the body. Thus, larger the displacement larger will be the potential energy of the body.

To calculate the gravitational potential energy of the body, consider a body of mass m on the surface

of earth. Force acting on a body due to gravitational force of earth is mg and in downward direction,

if we assume that the value of g is constant for small height then this force is also constant. An equal

and opposite force mg is to be applied on the body to make it move upward. The work done by this

force to move the body from ground to height h above the surface of earth is

W = mg x h= mgh

This work gets stored in the body in the form of gravitational potential energy of the body. Thus

U = mgh

If we find the work done by gravitational force on the body when it is moved from ground to height

h, then force and displacement are in opposite direction and work done will be negative

Wg = -mgh

Thus , potential energy is negative of the work done by conservative force acting on the body. If h is

variable then the rate of change of potential energy with distance is

−𝑑𝑈

𝑑ℎ = −

𝑑

𝑑ℎ(𝑚𝑔ℎ) = 𝑚𝑔

Which is same as the gravitational force acting on the body. Thus, in general for conservative forces




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F=−𝑑𝑈

𝑑𝑥

Note: [a] Potential energy is dependent on the frame of reference selected for measurement where-

as change in potential energy is independent of the reference frame.

[b] It is always calculated or defined when the forces or fields doing work are conservative on nature,

it can’t be defined for non conservative field.

[c] The relation U = mgh is gravitational potential energy of body of mass m for small height h near

the surface of earth, this relation can’t be used for large heights as ‘g’ will be variable if height is large

as compared to the radius of the earth.

Conservative force

A force is said to be conservative if the work done in moving the body from one point to another doesn’t de-

pend on the path followed but only on initial and final position of the particle.

Or

Force acting is said to be conservative if the work done in moving the body around the closed loop is zero. For

e.g. if body moves from A to B via path 1 and work done is W1 and it moves from B to A via path 2 and work

done is W2 , then for force to be conservative

W1 + W2 =0

Gravitational force is an example of conservative force, for e.g. if body moves straight from A to B upwards a

height of h, the work done to climb height h is

W = mgh

Similarly, the work done in moving same mass up the same

height through inclined plane of inclination θ is

W = �⃗�. 𝑆 = 𝐹 cos 𝜃 × 𝑆

Where F cos θ is the component of force parallel to the

inclined plane and S is the length of the inclined plane.




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W = mg sin θ x ℎ

𝑠𝑖𝑛𝜃=mgh

Some other examples of conservative forces are elastic forces in spring or the electrical force be-

tween the charges or the magnetic force.

Energy stored in Spring:

Springs are the elastic bodies which tends to regain their shape once external force acting on them is

removed. They can be of different shapes like helical spring in watch or spiral spring in shockers of

the vehicles.

We always assume the spring

to be massless, thus we can assume

tension [restoring force] to be same

everywhere in the spring. To produce

either extension or compression in

the spring equal and opposite force

must acts from the two ends of the

spring. For e.g. in bullworker we exert

force from both sides to compress

the spring or if we stretch the spring

whose other end is fixed in a hook,

then one force is exerted by us on the

spring and equal and opposite force

is exerted by the hook on the spring.

In equilibrium the restoring force is

equal to the force applied on either

end.

Now, whenever spring is

compressed or stretched restoring

force is developed in the spring as the

length changes. Some work has to be

done against this restoring force to

change the length of the spring. This

work done will be stored in the form




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of the potential energy of the spring. To find the work done in changing the length of the spring, con-

sider a spring of force constant k. As shown in figure [a] when the spring is neither stretched nor

compressed restoring force Fs in the spring is zero. But in fig. [b] When the spring is stretched the re-

storing force Fs will be negative. When the x is negative as in figure [c] the force will be positive. The

restoring force Fs will obey Hooke’s law within elastic limits. Hooke’s law states that the restoring

force developed in a spring is directly proportional to the change in length of the spring.

Fs α x

Fs = -kx

To compress the spring or stretch the spring we have to exert equal and opposite force on the spring.

Thus force exerted on the spring is kx and work done to compress the spring further by distance dx is

dW = Fdx

dW = kx dx

Total work done to change the length from 0 to X is

W= ∫ 𝑑𝑊 = ∫ 𝑘𝑥 𝑑𝑥𝑋

0

W= 1

2𝑘[𝑥2]0

𝑋 = 1

2𝑘𝑋2

This work done will be stored in the form of potential energy of the spring

U = 1

2𝑘𝑋2

We can also find the work done by restoring force from area under the force displacement graph as

shown in the figure. Work done by restoring force is

W = area under the graph OAB

W = - 1

2− 𝑘𝑥𝑚 × 𝑥𝑚 = −

1

2𝑘𝑥𝑚

2

Note: The spring constant of the spring depends on length, radius and nature of the material of

which the spring is formed. Experimentally it was found that spring constant of the spring is inverse-

ly proportional to the length of the spring.

K α 1

𝑙




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[b] The concept of energy stored in the spring is used in number of problems like shooting of missile

with spring or catapult. If we stretch a catapult to project a missile the potential energy is stored in it

which gets converted into kinetic energy of the missile. Thus larger we stretch larger will be potential

energy stored in string and larger will be the kinetic energy and velocity of the missile.

Mechanical Energy and its conservation:

The sum of kinetic and potential energy of the body is called its mechanical energy. If K and U de-

notes the kinetic and potential energy of the body respectively then the mechanical energy is given

by

K + U = E

Conservation of mechanical energy implies that if the forces doing work on the body are conservative

then the total mechanical energy of the body is constant.

We assume the case of one dimensional motion where force F is exerted on the body to dis-

place it through small distance Δx, the work done is given to the body in the form of change in kinetic

energy

Change in kinetic energy = F Δ x

Δ K = F Δ x

Also, as the force is conservative, the change in potential energy is given by

Δ U = - F Δx

Thus, total change in mechanical energy is

Δ K + Δ U = F Δx + (-F Δx) =0

Δ ( K + U) = 0

Or K + U = constant

This relation implies that if work is done by conservative forces on the body, the kinetic and potential

energy of the body can change but total energy will remain constant. Mechanical energy is depend-

ent on the frame of reference selected. The magnitude of the mechanical energy is different in differ-

ent reference frames.




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Conservation of mechanical energy [In freely falling body]

Consider a body of mass m which is dropped from height h above the surface of earth. The forces

acting on the body are conservative and we assume no effect of resistive forces on the body. As the

body falls we calculate the total mechanical energy at three points in the path, if the total mechanical

energy is same at all three points we can say that the energy is conserved.

When Body is at A Initially the body is at A at height h above the surface of earth and is at rest.

Kinetic energy at A, KA=0

Potential energy at A, UA= mgh

Total energy at A

EA= KA+UA= mgh [1]

When body is at B: In moving from A to B the body falls through distance x in the gravitational field

of earth, thus the body acquires some kinetic energy and also as the height decreases the potential

energy will decrease. Using kinematics

𝑣2 = 𝑢2 + 2𝑎𝑆

In this case, u=0 , a=g and S =x

V2 = 2gx

Thus kinetic energy at B

Kb= 1

2𝑚(2𝑔𝑥)= mgx

Potential energy at B

UB = mg ( h-x)

Total energy at B

Eb= UB+ KB

EB= mgx + mg (h – x) = mgh [2]

When body is at C: As the body has reached the surface of earth, therefore the gravitational poten-

tial energy will be zero. The velocity at point C can be calculated using

𝑣2 = 𝑢2 + 2𝑎𝑆




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If we assume initial point as A, then

U =0, a =g, S =h

𝑣𝑐2 = 0 + 2𝑔ℎ

Thus,kinetic energy at C is

Kc = 1

2𝑚𝑣𝑐

2 = 𝑚𝑔ℎ

Total energy at C

EC= Uc + Kc

Ec = mgh = mgh [3]

From equation [1], [2], and [3] we can say that the total mechanical energy at all three points is same

thus energy is conserved as the body falls in the gravitational field of earth.

Different forms of energy

The various important forms of energies present in this world are

[a] Heat Energy: Heat energy is the energy in transition it implies that if two bodies are at different

temperature then energy which is transferred from higher to lower temperature is called heat ener-

gy. It is never the energy possessed by the body. In mechanics also, if a body is moving on rough hor-

izontal surface then work is done against the frictional force results in rise in temperature of the

body. This happens because of the kinetic energy which is transferred to the body in the form of heat

energy.

[b] Internal energy: It is the sum of kinetic and potential energy possessed by the molecules of the

system. The potential energy arises because of the intermolecular interactions and kinetic energy

because of the random motion of the molecules.

[c] Electrical energy: It is energy which is used to make the electrical charge carriers move in a

particular direction. For e.g. when we switch on the fan the electrical energy gets converted into the

rotational kinetic energy of the fan or turbines in dams convert the potential energy into electrical

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[d] Nuclear Energy: It is the energy which is obtained from the nucleus of the atom in some kind of

nuclear transition. There are two main processes [a] nuclear fission [b] Nuclear fusion.

In nuclear fusion reaction, smaller nuclei combine to form heavier nuclei with the result of

release of large amount energy being released. In nuclear fission reaction heavier nuclei splits into

smaller nuclei with the result of release of large amount of energy. In both the processes, the energy

release is in accordance with the Einstein’s mass energy equivalence concept.

[e] Chemical energy: Chemical energy arises from the fact that the molecules participating in

chemical reactions have different binding energies. A stable chemical compound has less energy than

the separated parts. If total energy of reactants is more than the total energy of products then heat

is released and the reaction is called exothermic and if the total reaction of products is more than the

toal energy of reactants then the energy is absorbed and reaction is called endothermic. Chemical

energy is associated with forces that binds the atoms and molecules together. These forces bind at-

oms and molecules to form polymeric chains and it released on break up. For e.g. the energy re-

leased in the combustion of coal.

Einstein’s Mass Energy equivalence:

According to Einstein mass energy equivalence concept, the mass and energy are inter convertible

whenever certain amount of mass disappears an equivalent amount of energy is produced and vice

versa. For e.g. as we say the universe is expanding the stars are moving away out of the universe but

new stars are created from the energy present in the universe so that the number of stars are con-

stant or in nuclear reactions the difference in mass of parent nuclei and the daughter nuclei is con-

verted into energy in both fission and fusion. Similarly in electron positron annihilation, all of the

mass gets converted into energy in the form of gamma radiation. According to Einstein if mass ∆m

disappears the equivalent energy produced is

E = ∆ m c2

According to Einstein relativistic mass equation, the mass of the body varies with velocity ac-

cording to the relation

m = 𝑚0

√1−𝑣2

𝑐2

where m0 is the mass of the particle when it is at rest and m is the mass when it is moving with veloc-

ity v. c denotes the velocity of light.




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m = m0(1 −𝑣2

𝑐2)−1/2

m=m0(1 +𝑣2

2𝑐2)

m- m0 = 𝑚0𝑣2

2𝑐2

1

2𝑚0𝑣2 = ∆𝑚𝑐2

Ek = ∆m c2

Collisions

Any kind of interaction between bodies resulting in change in momentum or kinetic energy of body

or bodies is called collision. Broadly collision can be classified into 2 categories

[a] Direct Collision: In direct collision the two bodies came in actual contact with each other i.e.

there is physical contact between the two bodies. For e.g. the collision between bat and ball is an ex-

ample of direct collision or the collision between carom coins.

[b] Indirect Collision: In indirect collision there is no direct contact between the bodies and they

interact with each other from some distance. This happens because of interaction between the field

of two bodies. For e.g. if alpha particle approaches gold nucleus it gets deflected from some distance

without direct contact. Such collisions are called indirect collisions.

On the basis of conservation laws collisions can be classified into 3 categories:

[a] Elastic Collision : In elastic collision of two or more than two bodies, the momentum as well as

kinetic energy of the system both are conserved. The kinetic energy and momentum of individual

bodies can vary. For e.g. the collision between gas molecules can be considered to be elastic colli-

sion.

[b] Inelastic Collision: In these types of collisions the momentum is conserved but kinetic energy

of the system is not conserved which implies that initial and final momentum of the system are equal

but initial and final kinetic energy of the system are not equal. For e.g. collision between carom coins

or two billiard balls.




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[c] Perfectly Inelastic Collision: In perfectly inelastic collision, the momentum of the system is

conserved but kinetic energy is not conserved and two bodies stick to each other after the collision.

For e.g. if bullet is fired into a wooden pendulum, the bullet gets embedded in the pendulum and

they move together after the collision. In perfectly inelastic collision the relative velocity of the bod-

ies after the collision is zero.

Elastic Collision in two dimensions:

Consider the case of elastic collision of two bodies of mass m1 and m2. Initially m1 is moving with ve-

locity v1i and m2 is at rest. After the collision m1 moves with velocity v1f and m2 with velocity v2f. the

angle made by these two velocities relative to initial direction of m1 is θ1 and θ2 respectively as

shown in the figure.

Initial momentum of the system before the collision in horizontal direction = m1vii

Final momentum of the system in horizontal direction = m2v2f cos θ2 + m1v1f cos θ1

Using the principle of conservation of momentum, the equation can be written as

m1vi = m1v1f cosθ1 + m2v2f cosθ2 [1]

Similarly the initial momentum of the system in vertical direction is zero, the final momentum will

also be zero in the vertical direction.

m1v1f sinθ1 – m2v2 sin θ2 = 0 [2]

Also, as the collision is also elastic the initial and final kinetic energy of the system is also same.

1

2𝑚𝑣1𝑖

2 =1

2𝑚𝑣1𝑓

2 +1

2𝑚𝑣2𝑓

2 [3]

Using these three equations we can find the unknown variables in the problem.




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Elastic Collision in One dimension:

Consider two masses m1 and m2 moving with velocity u1 and u2 in the same direction. They collide

with each other and after the collisions their velocity becomes v1 and v2 respectively. We assume

that all the particles are moving in the same direction before and after the collision.

Initial momentum of the system before the collision

p1= m1u1+ m2u2

Final momentum of the system after the collision

P2 = m2v2 + m1v1

Using conservation of linear momentum, the momentum before and after the collision should be

same

miu1 + m2u2 = m1v1 = m2v2

m1( u1 – v1) = m2 ( v2 – u2) [1]

Also using the conservation of kinetic energy before and after the collision

1

2𝑚1𝑢1

2 +1

2𝑚2𝑢2

2 =1

2𝑚1𝑣1

2 +1

2𝑚2𝑣2

2

𝑚1𝑢12 + 𝑚2𝑢2

2 = 𝑚1𝑣12 + 𝑚2𝑣2

2

𝑚1(𝑢12 − 𝑣1

2) = 𝑚2(𝑣22 − 𝑢2

2) [2]

Dividing eqn [2] by eqn [1]

u1 + v1 = u2 + v2

v1 = u2 + v2 – u1 [3]

v2 = u1 + v1 – u2 [4]

Substitute [3] in [1] and solve for the value of v2 and similarly substitute [4] in [1] and solve for value

of v1, the following results can be obtained

v1 = 𝑚1−𝑚2

𝑚1+𝑚1𝑢1 +

2𝑚2𝑢2

𝑚1+𝑚2

and




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v2 = 𝑚2−𝑚1

𝑚1+𝑚1𝑢2 +

2𝑚1𝑢1

𝑚1+𝑚2

Special Cases: [a] If mass of both the bodies is same, in that case m1=m2, substitute these two val-

ues in equation for final velocities we get

v1 = u2 and v2 = u1

This implies that the velocities of bodies gets interchanged when two bodies of equal mass collides

elastically. There can be special case where second body is initially at rest i.e. u2 = 0, it implies that

v1 = 0, v2 = u1

which implies that if a body of mass m moving with certain velocity collides with another body of

same mass at rest then after the collision their velocities gets interchanged i.e. the first body will

come to rest and second body begins to move.

Conceptual Questions

Q.1 Will water at the foot of waterfall at higher temperature as compared to water at the top? Explain.

A.1 As water falls down its gravitational potential energy is converted into the kinetic energy of water molecules. This kinetic energy is further converted into heat energy. Thus water the lower end of the waterfall is at higher temperature as compared to water at the top.




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Q.2 How is kinetic energy related to direction of motion of the particle? Can kinetic energy of the body be negative?

A.2 Kinetic energy of body is independent of the direction of motion as it is scalar quantity. No kinetic energy of the body can never be negative.

Q.3 What is the work done by the coolie walking on the platform with load on his head?

A.3 The work done by the coolie against the gravitational force of earth is zero, as the coolie moves in horizontal direction and the gravitational force acts in vertical direction. If force and displacement are perpendicular work done is zero.

Q.4 The work done in lifting a box on to a platform does not depend upon how fast the box is lift-

ed. Why?

A.4 Work done in lifting a box depends only on the force needed to lift the box and displacement of the body. It is independent of the time in which body is lifted. Time is required if we need to calculate the power.

Q.5 Is it possible that the body is in accelerated motion under the force acting on the body, yet no

work is done by the force? Explain your answer giving suitable example.

A.5 Yes, it is possible if force and displacement are perpendicular to each other. For e.g. if body is moving in circular path due to centripetal force, then it is accelerated motion as direction of velocity is changing continuously. But work done is zero because centripetal force and dis-placement are perpendicular to each other.

Q.6 A body moves along the semicircular path of radius r. What is the work done in doing so?

A.6 If the body moves in semicircular path no work is done by the centripetal force acting on it but if tangential force is acting on the body then the work done by tangential force is equal to change in kinetic energy of the body.

Q.7 A man rowing a boat upstream is at rest with respect to the shore. Is he doing any work?

A.7 No net work is done on the boat because the displacement relative to the shore is zero, he is not moving relative to the earth.

Q.8 If block attached to a spring is pulled up by a distance ‘x’ and released, the amplitude of its

motion cannot exceed ‘x’. Why?

A.8 Amplitude can’t exceed ‘x’ because the work done in stretching by distance ‘x’ is converted into potential energy and body oscillates. To increase amplitude beyond ‘x’ some external work needs to be done on the body.

Q.9 Why a metal ball rebounds better than the rubber ball?




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A.9 Metal ball rebound better than the rubber ball because when it strikes the floor the time of impact is very small which results in large change in momentum of the ball and it rebounds higher.

Q.10 A person continues to push a rock for some time but fails to move it. What is the work done?

A.10 Work done by the man on the rock is zero because although he is applying force displacement of the rock is zero.

Q.11 A truck and car are moving with the same kinetic energy on a road. Which one will have

greater momentum?

A.11 Momentum and kinetic energy are related as p = kmE2 . Thus if the kinetic energy of the

two bodies are same, p m , thus as truck has larger mass it will have larger momentum.

Q.12 Two bodies of different masses have same kinetic energy. What is the ratio of the momentum of the two bodies?

A.12 As, p = kmE2 . Thus for two bodies having same kinetic energy

2

1

2

1

m

m

p

p

Q.13 Two bodies of different masses have same momentum. What is the ratio of the kinetic energy

of the two bodies?

A.13 The kinetic energy and momentum are related as E = m

p

2

2

, thus for two bodies having same

momentum 1

2

2

1

m

m

E

E

Q.14 Explain how fast moving neutrons can be slowed down on passing through heavy water?

A.14 heavy water contains deuterium atoms and their mass is same as that of neutron. So when neutrons strikes deuterium atoms, most of the kinetic energy of neutrons is given to them and neutrons gets slowed down.

Q.15 Can a body have energy without possessing momentum?

A.15 Yes, when spring is compressed energy is stored in it in the form of potential energy but its momentum is zero as it is at rest.

Q.16 A rocket explodes in mid air. How does it effect their momentum and kinetic energy?




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A.16 When rocket explodes in mid air its momentum remains constant [same as before explosion] because no external force acts on it and the momentum remains same. But kinetic energy will increase as the chemical energy gets converted into kinetic energy.

Q.17 Two ball bearings of mass m each, moving in opposite directions with equal speed v collide

head on with each other. Predict the outcome of the collision assuming it to be perfectly elas-

tic?

A.17 If two bodies of equal mass collide with each other their velocities gets interchanged.

Q.18 When an arrow is shot from the bow, it has kinetic energy. From where does it get the kinetic energy?

A.18 When we stretch the string of the bow, potential energy gets stored in it, the potential energy gets converted into kinetic energy when the arrow is shot.

Q.19 Is whole kinetic energy lost in the perfectly inelastic collision?

A.19 No whole kinetic energy may not be lost as when one body collides with the other, they stick and moves together after the collision.

Q.20 Explain, why throwing mud on the wall is an example of perfectly elastic collision?

A.20 When mud is thrown on the wall after the collision with wall mud sticks to the wall and there

is no relative motion between them.

Q.21 What should be the angle between the force and displacement for work to be maximum?

A.21 The body should displace in the direction of force for work done to be maximum, i.e. the an-

gle between the force and displacement should be 00.

Q.22 We wind our watch once a day, what happens to its energy?

A.22 When we wind our watch the potential energy gets stored in the spring, this potential energy

gets converted into rotational kinetic energy of the hands of the watch as the hands move.

Q.23 When a ball is thrown up, the magnitude of its momentum decreases and then increases.

Does this violate the conservation of momentum principle?




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A.23 No, it does not violate the principle of conservation momentum because momentum will re-

main conserved if no external force acts on the body but in this case gravitational force acts

on the body which results in change in momentum of the body.

Q.24 Is collision possible without even direct contact between the particles?

A.24 Yes, collision is possible without direct contact if bodies can interact from some distance. For

e.g. if the alpha particle approaches gold foil they can be deflected without direct contact be-

tween alpha particle and the gold nucleus. Such collisions are called indirect collision.

Q25 How will the momentum of the body change if its kinetic energy is doubled?

A.25 As, p = kmE2 , thus if kinetic energy is doubled the momentum of the body increases 2

times.

Q.26 What is the most common feature in all types of collisions?

A.26 Momentum remains conserved in all the collisions i.e. in elastic, inelastic, perfectly inelastic

and super elastic collisions.

Q.27 A spring is cut into two halves, how is the spring constant affected?

A.27 Spring constant is inversely proportional to the length of the spring. Thus if spring is cut into

two halves the spring constant will be doubled.

Q.28 Comets move around the sun in elliptical orbits and the force is not perpendicular to the ve-

locity of comet. Yet the work done in complete cycle is zero. Why?

A.28 The gravitational force is acting on the comet as it moves and gravitational force is conserva-

tive by nature. Thus, the work done in moving a comet around the close loop is zero.

Q.29 The sign of work done by a force on body is important to understand. State if the following

works are positive or negative:

[a] work done by man in lifting a bucket out of the well by rope tied to the bucket




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[b] work done by gravitational force in above case

[c] work done by frictional force on a body sliding down the inclined plane.

[d] work done by applied force on the body moving on rough horizontal plane with uniform

velocity.

[e] work done by resistive force of air on the vibrating pendulum in bringing it to rest.

A.29 [a]The work done by the tension in the string is positive as both tension and displacement are

in the same upward direction.

[b] the work done by gravitational force is negative as gravitational force acts in downward

direction and body displaces in opposite direction upward.

[c] frictional force will do negative work as body slides downward and frictional force acts in

upward direction.

[d] applied force and displacement they are in the same direction thus the work done by the

applied force is positive.

[e] the resistive force acts opposite to displacement of the body thus work done by the resis-

tive force is negative.

Q.30 A spring is cut into two halves. What is the spring constant of each portion?

A.30 A spring constant is inversely proportional to the length of the spring. Thus, if the length of

the spring is cut into two halves then the spring constant is doubled.

Q.31 The casing of a rocket in flight burns up due to friction. At whose expanse is the heat energy

required for burning obtained? The rocket or the atmosphere?

A.31 The rocket itself provides heat energy required for burning. The energy is obtained at the ex-

panse of mass of the rocket and its mechanical energy.

Q.32 An artificial satellite orbiting the earth in thin atmosphere loosing its energy, gradually due to

atmospheric resistance. Why then speed increases as I comes closer and closer to earth?

A.32 As the satellite comes closer to earth the potential energy of the satellite decreases. As the

total mechanical energy should remain constant, the kinetic energy and hence the velocity of

the satellite increases.




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Q.33 The force between pair of protons is repulsive. How will the potential energy change if they

are brought closer?

A.33 The potential energy of the system will increase and stability of the system decreases. The

work done in bringing them closer increases potential energy of the system.

Q.34 Why is potential energy said to be positive or negative?

A.34 The potential energy of system of particles is said to be positive if they are held at a position

due to repulsive force between them. Potential energy is said to be positive if the system of

particles is held at a position due to the attractive force between them.

Q.35 Chemical energy, gravitational energy and nuclear energies are nothing but different forms of

potential energy. Explain.

A.35 Potential energy is energy possessed by the system due to their position in some force or

field. Chemical energy results from bonding between atoms. Gravitational energy is because

of the position in gravitational field and nuclear energy is result of nuclear force between nu-

cleons.

Q.36 A lorry and car moving with same kinetic energy are brought to rest by the application of

equal retarding force. Which one of them will come to rest in shorter time?

A.36 As work done FS= change in kinetic energy. Thus S = change in kinetic energy/force. As

change in kinetic energy and force applied are same they will come to rest in same distance.

But stopping time t = mv/F i.e. t momentum and momentum is more for truck thus it will

take longer time to stop.

Q.37 Distinguish between head on and oblique collision.

A.37 If colliding bodies move along the straight line joining their centers, the collision is said to

head on collision and if the colliding bodies do not move along the straight line joining their

centers then collision is said to be oblique.

Q.38 Lead or heavy water which is preferred as moderator in nuclear reactor and why?




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A.38 Moderator is used to slow down the fast moving neutrons. When fast moving neutron col-

lides with heavy nuclei like lead it will rebound with almost same velocity and able to retain

its kinetic energy. But if it collides with heavy water molecules transfer of energy takes place

between them and neutrons will be slowed down.

Q.40 Select the correct alternative:

[a] When conservative force does positive work on a body, the potential energy of the body

increases/decreases/remains unaltered.

[b] Work done by the body against friction always results in loss of kinetic/potential energy.

[c] The rate of change of total momentum of a many particle system is proportional to the

external force/sum of internal forces in the system.

[d] in inelastic collision of two bodies, the quantities which do not change after the collision

are total kinetic energy/total linear momentum/total energy of the system.

A.40 A conservative force does positive work on the body displaces the body in direction of force.

Thus potential energy of the system will decrease.

[b] Work is done against friction by the kinetic energy thus kinetic energy of the system will

decrease.

[c] Total momentum can change only if some external force is applied on the body. Thus, rate

of change of linear momentum is equal to the external force acting on the body.

[d] In inelastic collision, the total linear momentum of the system and the total kinetic energy

of the system remains constant.

Q41 State which of the following statement is true or false

[a] In an inelastic collision of two bodies, the momentum and energy of each body is con-

served.

[b] Total energy of the system is always conserved no matter what internal or external forces

are acting on the body.

[c] Work done in motion of the body over a closed loop is zero for every force in nature.

[d] In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy.




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A.41 [a] False, as total momentum and total energy of the system is conserved and not for individ-

ual bodies.

[b] False, as external force acting on the system may increase as well as decrease the total

energy of the system.

[c] False, as work done for moving a body under the action of closed loop is zero only if the

force acting on the body is conservative in nature.

[d]True, this is generally a true statement as loss of kinetic energy takes place in an inelastic

collision.

Q.42 Kinetic energy of body is increased by 300%. What is the percentage change in linear momen-

tum?

A.42 300% increase means that kinetic energy becomes 4 times its initial value. As p = kmE2 ,

thus if kinetic energy increases 4 times the momentum is doubled or increased 100%.




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Work [Assignment No:1]

Q.1 Calculate the work done by applied force in lifting a mass of 8kg to a height of 2.5m. Will the work done be positive or negative? [Ans. 196J]

Q.2 A 1200kg car skids to rest in a distance of 25m. Assuming that coefficient of friction between the tyres and the road is 0.7, find the work done on the car by the force of friction.

[Ans. –2.06x105J]

Q.3 A man cycles up a hill whose slope is 1 in 20. If the mass of the man and the bicycle is 150kg and the man covers a distance of 100m along the incline. Calculate the work done by the man. [Ans. 7350J]

Q.4 A body is constrained to move along the z-axis of a coordinate system subjected to a force given by

kjiF ˆ3ˆ2ˆ

. What is the work done by the force in moving the body a distance of 4m along the

z-axis? [Ans. 12J]

Q.5 A force, which acts at angle of 370 to the horizontal, is pulling the box and the body moves at constant speed. The force of friction opposing the motion is 20N and the box has a mass of 30kg. Find the mag-nitude force and the work done by the force, as the box is moved 5m. [Ans. 100J]

Q.6 A worker pushes a lawn roller with a force of 180N at an angle of 240 downward from the horizontal. How much work does she do as she pushes the roller through a horizontal distance of 50m? [Ans. 8200J]

Q.7 A 52kg box rests on a floor. How much work is required to move it at constant speed [a] 12m along the floor against a frictional force of 150N and [b] 12m vertically upward?

[Ans. 1.8x103J , 6.9x103J]

Q.8 Find how much work is done in moving a 50kg block through a horizontal distance of 10m by applying a force of 100N, which makes an angle of 600 with the horizontal. [Ans. 500J]

Q.9 Calculate the work done in raising a stone of mass 6kg of relative density 2 immersed in water, from a depth of 4m to 1m below the surface. [Ans. 88.2J]

Q.10 an engine pulls a train 1km on the level track. Calculate the work done given that the frictional re-sistance is 5 x 105N. [Ans. 5 x 108J]




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Q.11 A rain drops of radius 2mm falls from a height of 500m above the ground. It falls with decreasing ac-celeration [due to viscous forces] until half of its original height, it attains its maximum [terminal] speed, and moves with constant speed thereafter. What is the work done by the gravitational force on this drop in first and second half of the journey? [Ans. 0.082J, 0.082J]

Q.12 a locomotive of mass m starts moving so that its velocity varies with the distance according to the law

v=ks where k is constant and s is the distance traveled. Find the work done by all the forces acting on the body during first t seconds after the beginning of the motion.

[Ans. 8

24tmk]

Work, Power And Energy [Assignment No:2]

Q.13 A force F =a +bx acts on a particle in the x direction where a and b are constants. Find the work done by this force during displacement from x=0 to x=d.

[Ans. dbd

a

2]

Q.14 A particle is projected from a 40m high cliff with an initial speed of 50m.s at an unknown angle. Find the speed when it hits the ground.

[Ans. 58m/s]

Q.15 Water falling from the 50m high falls is to be used for generating electrical energy. If 1.8 x 105kg of water falls per hour and half the gravitational potential energy can be converted into electrical energy, find how many 100W lamps can be lit.

[Ans. 122]

Q.16 Find the average frictional force needed to stop a car weighing 500kg in a distance of 25m if the initial speed of the car is 72km/h.

[Ans. 4000N]

Q.17 A particle of mass m moves in a straight line with its velocity varying with the distance traveled ac-cording to the equation v = a/x, where ‘a’ is constant. Find the total work done in displacement from x=0 to x=d.

[Ans. ma2d/2]

Q.18 A block of mass 2kg is being pushed down an inclined plane of inclination 370 with a force of 20N act-ing parallel to the incline. It is found that the block moves n the incline with an acceleration of 10m/s2. if the block started from rest, find the [a] work done by the applied force in first second, [a] the weight




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of the block in the first second and [c] the frictional force acting on the block in first second. Take g=10m/s2. [Ans. 100J, 60J, -60J]

Q.19 A 250kg block slides on a rough horizontal surface. Find the work done by the frictional force in bring-ing the block to rest if it is initially moving at a speed of 40cm/s. if the frictional coefficient between the block and the table is 0.1, how far does the block move before coming to rest.

[Ans. –0.02J, 8.2cm]

Q.20 A person painting his house walls. He stands on a ladder with the bucket containing the pain on one hand and brush in the other hand. Suddenly the bucket slips from his hands and falls down on the floor. If the bucket along with the paint has weight of 6kg and was at a height of 2m from the floor at the time it slipped, how much gravitational potential energy is lost along with the paint?

[Ans. 118J]

Q.21 A block of weight 100N is slowly slid up on a smooth incline of inclination 370 be a person. Calculate the work done by the person in moving the block through a distance of 2m if the driving force is [a] parallel to the incline and [b] in the horizontal direction.

[Ans. 120J, 120J]

Q.22 A water pumps lifts water from a level of 10m below the ground. Water is being pumped at a rate of 30kg/minute with negligible velocity. Calculate the minimum horsepower the engine should have to do this.

[Ans. 6.6x10-2HP]

Work, Power and energy [ Assignment No:3]

Q.23 A running man has kinetic energy half the kinetic energy of boy of half his mass> the man speeds up by 1m/s and then has the same kinetic energy as that of the boy. What are the original speeds of man and boy?

[Ans. 4.84m/s, 2.42m/s]

Q.24 A 30gm bullet initially traveling at 500m/s penetrates 12cm into the wooden block. What is the aver-age fore exerted on it/

[Ans. 3125N]

Q.25 A spring gun has spring constant of 80N/cm. The spring is compressed 12cm by a ball of mass 15gm> how much is potential energy of the spring? If the trigger is pulled, what will be the velocity of the ball?

[As. 57.6J, 87.6m/s]




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Q.26 An object of mass 5kg falls from rest through a vertical distance of 20m and reaches with velocity of 10m/s. How much work is done by the push of the air on the object?

[Ans. –730J]

Q.27 Find the horsepower used in pumping 2500kg of water per minute from a well 15m deep onto the surface. Supposing 40% of HP of the engine during the pumping is wasted, what is the horsepower of the engine/

[Ans. 13.6HP]

Q.28 An object is attached to the vertical spring and slowly lowered to its equilibrium position. This stretch-es the spring by ‘d’. If the same object is attached to the vertical spring but permitted to fall instead, through what distance will the spring stretch/

[Ans. 2d]

Q.29 A truck can move up a road having slope 1 in 50 with a speed of 24km/h, the resisting force is equal to 1/25th of the weight of the truck. How fast will the same truck move down the hill with the same pow-er?

[Ans. 72km/h]

Q.30 a 1kg block collides with horizontal weightless spring of force constant 2N/m. the block compresses the spring by 4m from the rest position. Assuming that the coefficient of friction between the block and the horizontal surface is 0.25, what was the speed of the block at the instant of collision?

[Ans. 7.18m/s]

Q.31 An engine is hauling a train of mass M kg on a level track at a constant speed of v m/s. The resistance due to friction is f N/kg. What power is the engine producing? What extra power must engine develop to maintain this speed? What is the new total power developed by the engine?

[Ans. Mfv watt,

s

ghfMv

s

Mghv, ]

Q.32 A body of mass 2kg slides down on inclined planes, which make an angle of 300 with the horizontal.

The coefficient of friction between the block and the surface is 3 /2. [a] What force should be applied to the block so that the block moves down without acceleration?[b] What force should be applied so that it moves up without any acceleration?

[Ans. 10.9W, 30.6W]

Q.33 A particle of mass 1gram executes an oscillatory motion in a concave surface of radius 2m placed on horizontal plane. If the motion of the particle begins from a height of 1cm from the horizontal plane and coefficient of friction is 0.01, find the total distance covered by the particle before it comes to rest.

[Ans. 100cm]

Collisions [Assignment No:4]




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Q.34 A 5gm bullet is fired horizontally into a 1.20kg wooden block resting on a horizontal surface. The coef-ficient of kinetic friction between the block and the surface is 0.2. The bullet remains embedded in the block, which is observed to slide 0.230m along the surface before stopping. What is initial speed of the block?

[Ans. 2.30x 102m/s]

Q.35 A 12gm rifle bullet is fired with a speed of 380m/s into a ballistic pendulum with a mass of 6kg sus-pended from a cord of length 70cm. Compute the vertical height through which the pendulum rises. [b] The initial kinetic energy of the bullet [c] the kinetic energy of the bullet and pendulum immediate-ly after the bullet becomes embedded into the pendulum.

[Ans. 2.93cm, 866J, 1.73J]

Q.36 Two frisky grasshoppers collide in mid air at the top of their trajectories and grab into each other, holding tight thereafter. One is robust 250gm beast initially moving south at 20cm/s and other is 150gm creature initially moving with 60cm/s. calculate the decrease in kinetic energy that results from the collision. What happens to the lost kinetic energy?

[Ans. 0.0300J]

Q.37 In Dallas, 1400kg automobile is going west at 35km/h collides with 2800kg truck going south at 50km/h. If they become coupled after the collision, what are the magnitude and direction of their ve-locity after colliding?

[Ans. 35.3m/h 70.70 south of west]

Q.38 A 0.150kg glider is moving to the right on a frictionless horizontal air track with a speed of 0.80m/s. It has a head on collision with 0.300kg glider that is moving to the left with a speed of 2.20m/s. Find the final velocity of each glider if the collision is elastic.

[Ans. 3.2m/s, 0.20m/s]

Q.39 A block A of mass 2kg and B of mass 10kg on a frictionless horizontal surface. Initially, block B is at rest and block A is moving towards it at speed of 2m/s. the blocks are equipped with ideal spring bumpers. The collision is head on, so all motion before and after the collision is along a straight line. [a] Find the maximum energy stored in the spring bumpers and the velocity of each block at that time [b] find the velocity of each block as they move apart.

[Ans. 3.33J, 0.67m/s]

Q.40 An object of mass m, initially at rest explodes into two fragments, one with mass ma and other with mb. [a] If energy Q is released in the explosion how much kinetic energy does each fragment have im-




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mediately after the collision? [b] What percentage of the total energy released does each fragment get when one is four times the mass of the other?

[Ans. 80% for ma and 20% for mb]

Q.41 A 8kg ball, hanging from the ceiling by a wire of 135cm long, is truck in an elastic collision by 2kg ball moving with speed v0 just before the collision. If the maximum tension that the wire can have without breaking is 1600N, what is the largest v0 can be without breaking?

[Ans. 40m/s]

Q.42 A 5gm bullet is shot through a 1kg wood block suspended on a string 2m long. The center of mass of the block rises by a distance of 0.45cm. Find the speed of the bullet as it emerges from the block if ini-tial speed is 450m/s?

[Ans. 390.6m/s]

Q.43 What percentage of the kinetic energy of the moving particle is transferred to the stationary particle, when it strikes the stationary particle of [a] 100 times its mass [b] equal mass/

[Ans. 3.92%, 100%]

Q.44 AA simple pendulum of length 1m has a wooden bob of mass 1kg. It is struck by a bullet of mass 10-2kg moving with a speed of 2x102m/s. the bullet gets embedded into the bob. Obtain the height to which the bob rises before swinging back.

[Ans. 0.2m]

Q.45 A moving ball of mass m undergoes head on collision with another stationary ball of mass 2m. Show that the colliding ball looses 8/9th of the energy after the collision.

Q.46 A bomb at rest explodes into two fragments of masses 3kg and 1kg. The total kinetic energy of the fragments is 6x104J. Calculate the [a] kinetic energy of the bigger fragment [b] momentum of the smaller fragment.

[Ans. 1.5x104J, -300kgm/s]

Q.47 A body of mass M at rest is struck by a moving body of mass m. Prove that the fraction of initial kinetic

energy of mass transferred to the struck body is 2

4

Mm

mM

Q.48 A railway carriage of mass 104kg is moving with a speed of 15m/s strikes a stationary charge of same mass. After the collision the carriages gets coupled and move together. What is their common speed after the collision?




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[Ans. 7.5m/s]

Q.49 A ball of 0.1kg makes an elastic head on collision with a ball of unknown mass that is initially at rest. If the 0.1kg ball rebounds at one third of its original speed, what is the mass of the other ball?

[Ans. m2=0.2kg]

Q.50 A bullet weighing 10gm is fired with a velocity of 800m/s. after passing through the mud wall 1m thick, its velocity decreases to 100m/s. find the average resistance offered by the mud wall.

[Ans. 3150N]

Q.51 A body falling on the ground from a height of 10 m rebound to a height of 2.5m. Calculate the [a]% of the kinetic energy lost during collision [b] The ratio of velocities before and after the collision.

[Ans. 75%, 2]

Work, Power And Energy [Assignment No:5] Q.52 A chain is held on a frictionless table with 1/nth of its length hanging over the edge. If the

chain has length L and mass M, how much work is done to pull the hanging part back on the table?

[Ans. 22n

MgL]

Q.53 A person decides to use his bath tub to generate electric power to run a 40W bulb. The bathtub is lo-

cated at a height of 10m from the ground and holds 200 liter of water. He installs a water wheel gen-

erator on the ground. At what rate should water drain from the bathtub to light the bulb? How long

can he keep the bulb on, if efficiency of generator is 90%?

[Ans. 0.453lt/s, 441second]

Q.54 An automobile of mass m accelerates starting from rest, while engine supplies constant power P,

show that [a] velocity is given as a function of time by V =

2/12

m

Pt. [b] position is given as a function

of time by S= 2/3

2/1

9

8t

m

P

.

Q.55 A massless platform is kept on a light elastic spring. When a body of mass 0.1kg mass is dropped from

height h of 0.24m the particle strikes the pan and spring is compressed by 0.01m. From what height

particle should by dropped to cause a compression of 0.04m?




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Q.56 The height of waterfall is 50m. Calculate the difference in temperature of water at the top and bottom

of the fall. [Ans.

0.11670C]

Q.57 A 50gm lead bullet of specific heat 0.02 is initially at 300C. It is fired with a speed of 840m/s and re-

turning to the starting level strikes a cake of ice at 00C. How much ice will melt? Assume that all the

energy is spend in melting of ice only. L=80cal/gm for ice.

[Ans. 52.875gm]

Q.58 Show that the speed v reached by a car of mass m driven with constant power P is given by

V=

3/13

m

xP

Where x is the distance traveled from rest.

Q.59 A block of wood is pressed against a wooden surface with a force os 10N and then rubbed on it to

sweep a distance of 10cm. This is done 30 times. If coefficient of friction is 0.4, calculate the heat pro-

duced. Can wood burn?

[Ans. 2.8cal, no this is small heat only]

Q.60 A particle slides along a tack with elevated ends and flat central part. The flat part has a length of 3m.

The curved portion of the track is frictionless. For the flat part the coefficient of friction is k=0.2, the

particle is released from A, which is at a height of 1.5m above the flat part of the track. Where does

will it come to rest.

[Ans. at the center of the flat panel]

Q.61 A bullet of mass 20gm strikes a target with a velocity of 150m/s and is brought to rest after piercing

10cm into it. Calculate the average force of resistance offered by the target.

[Ans. 2250N]

Documents

Work Power And Energy - Alfa Tutorialsalfatutorials.in/wp-content/uploads/2018/07/Work_theory.pdfWork Power And Energy XI Class ALFA PHYSICS CLASSES Sanjay Chopra 86, Chotti Baradari