# User manual | bridge course - Narayana Schools

```VII - CLASS
BRIDGE COURSE
SUBJECT : PHYSICS
INDEX
Day
1
2,3
4
5
6,7
8
Name of the topic
Measurement: Introduction Physical quantity and Units
& Solve Worksheet – 1
Measurement: Multiples and Sub multiple factors
& Solve Worksheet – 2
Measurement: Density, Relative density area and volume
& Solve worksheet – 3
Kinematics:
Introduction to Rest, motion and its types
& Solve worksheet – 1
Kinematics:
Speed, velocity and its types
& Solve Worksheet – 2
11 to 13
Kinematics:
Acceleration and equations of motion for a body moving
with uniform acceleration in a straight line:
& Solve Worksheet – 3
13 to 15
Newton’s Laws of motion:
Force – Its effects and types, Newton’s First
law of motion, Inertia and its types
& Solve Worksheet - 1
Newton’s Laws of motion:
Linear momentum and Newton’s second law
of motion.
& Solve Worksheet – 2
16 to 18
Impulse, Mass and Weight, Newton’s Third
law and Law of conservation of momentum
& Solve Worksheet – 3
18 to 20
9
10,11
Newton’s Laws of motion:
12
Work, Power and Energy:
Work, its types and work done against
gravity
& solve worksheet – 1
Work, Power and Energy:
Energy and its types and relation between
kinetic energy and momentum
& Solve Worksheet – 2
13
14
Work, Power and Energy:
15
Law of conservation energy, power and its
units
& Solve Worksheet – 3
Key to worksheets
21 to 23
VII - Class _ Physics
Bridge Programme
DAY – 1 (SYNOPSIS)
Physical Quantity : The quantities which are
measurable in physics are called physical quantities.
unit of length is foot.
The unit of mass is pound. The unit of time is
second.
Ex : Length, mass, time, speed, etc.
ii. French or C.G.S system : In C.G.S system, the
unit of length is centimetre.
Unit : Unit is a standard which is used for the
measurement of physical quantity.
The unit of mass is gram. The unit of time is second.
Ex : (i) unit of length is metre
(ii) unit of time is second
Numerical value of physical quantity : The
number of times a unit is present in a given physical
quantity is called Numerical value of physical
quantity.
Relation among Physical quantity, Numerical
value and Unit :Physical quantity = Numerical
value × Unit
Ex : Let length of table = 3 metre
Here 3 is the Numerical value and metre is the
standard unit.
Fundamental quantities : The physical quantity
which does not depend upon (or independent of)
other physical quantities are called fundamental
quantities.
iii. M.K.S system or metric system : In M.K.S
system, the unit of length is metre.
The unit of mass is kilogram.The unit of time is
second.
iv. International system or S.I : S.I. system has
seven basic units and two supplementary units.
Quantity
length
mass
time
temperature
luminous intensity
electric current
amount of
substance
Plane angle
solid angle
Unit
Symbol
metre
kilogram
second
kelvin
candela
ampere
mole
m
kg
s
K
cd
A
mol
sr
Ex : Length, mass and time etc.
Derived physical quantities : The physical
quantities which depends on fundamental physical
quantities are called derived physical quantities.
Ex : Area, Volume, Speed, etc.
Fundamental units : The units used for measuring
fundamental quantities are called Fundamental units.
These are independent of other units.
Ex: The fundamental unit of length is metre.
Derived units : Derived units are the units of
derived physical quantities which are expressed in
terms of fundamental units.
Ex: The derived unit of speed is ms–1 ( read as
metre per second).
Systems of Units : There are four system of units:
i. British or F.P.S. system : In F.P.S. system, the
Narayana Group of Schools
DAY – 1 (WORKSHEET - 1)
1. Pick the odd man out.
1] Length
3] Time
2] Mass
4] Area
2. F.P.S stands for
1] Foot, pound, second
2] France, Paris, Spain
3] Force, pressure, second
4] Foot, Pace, Second
3. C.G.S stands for
1] Centimetre, Gravitation, second
2] Centisecond, gram, second
3] Centimetre, gram, second
25
VII - Class _ Physics
Bridge Programme
4] None of these
(c) Amount of substance
1] a  e; b  f; c  g
4. Ampere is the unit of
1] Length
3] Luminous intensity
2] Temperature
4] Current
5. Number of fundamental physical quantities in
M.K.S system are
1] Two
3] Seven
2] Three
4] Six
6. The temperature standard metre rod made of
platinum - Iridium alloy kept in the archives
of serves near Paris is
1] 0°C
3] 100° C
(g) Candela
2] 27°C
4] None
7. Statement I : Numerical value = Physical
quantity × unit
Statement II : Current strength is a
fundamental physical quantity according to
S.I system.
1) Statement I is true ; Statement II is true.
2) Statement I is true ; Statement II is false.
2] a  f; b  g; c  e
3] a  g; b  f; c  e
4] a  g; b  e; c  f
10) M.K.S. stands for
1) Meter, Kilometer, Second
2) Meter, Kilogram, Second
3) Mile, Kilogram, Second
4) None of these
11) The S.I. unit of luminous intensity
1) Ampere
2) Kelvin
3) Mole
4) Candela
12) The S.I. unit of plane angle
2) Steratian
3) Meter
4) Second
13) The S.I. unit of solid angle
3) Statement I is false ; Statement II is true.
4) Statement I is false ; Statement II is false.
3) Meter
4) Second
8. Observe the following
14) S.I. unit of thermodynamic temperature
a) Length
b) Mass
1) Ampere
2) Kelvin
c) Current strength
d) Temperature
3) Mole
4) Candela
pick the correct statement
1] Length is the odd man out.
2] All are Fundamental Physical Quantities
according to M.K.S system.
3] All are Fundamental Physical Quantities
according to S.I system.
4] Current strength is a derived quantity.
9.
Match the following.
List - A
List - B
(a) Temperature
(e) Mole
(b) Luminous intensity
(f) Kelvin
Narayana Group of Schools
26
VII - Class _ Physics
Bridge Programme
vi.
DAY – 2,3 (SYNOPSIS)
Multiples & submultiples of unit of mass :
Multiple and sub multiple factors :
i.
Quintal (qt) = 100 kg
ii.
Tonne = 1000 kg
Symbol
iii.
gram (g) = 10–3kg
T
G
M
k
h
da
d
c
m

n
p
f
a
iv.
milligram (mg) = 10–3g = 10–6kg
v.
micro gram (  g) = 10–6g = 10–9kg
Multiple and sub multiple factors
Multiplication factor Name
1012
109
106
103
102
10
10-1
10-2
10-3
10-6
10-9
10-12
10-15
10-18
pico metre (pm) = 10–12m
tera
giga
mega
kilo
hecto
deca
deci
centi
milli
micro
nano
pico
femto
atto
Multiples and Submultiples of Metre :
Multiples of metre
Multiples and Submultiples of second :
Submultiples of second
i.
ii.
milli second = 10–3s
micro second = 10–6s
Multiples of second
i.
1 minute = 60 second
ii.
1 Hour = 60 min = 3600s
iii.
1 Day = 24 hour = 24 × 60 × 60 second =
86400s Multiples & Submultiples of unit of
Area :
1cm2 = 10–4m2
1hectare = 104m2
ii.1m2 = 104cm2
i. decametre (dam) = 101m= 10m
i.
iii.
ii. hectometre (hm) =102m = 100 m
Multiples & Submultiples of unit of volume
iii. kilometre (km) = 103m = 1000 m
i.
1 cm3 = 10–6m3
iv. Astronomical unit (A.U): It is the mean distance
of the earth from the sun.1astronomical unit = 1.496
× 1011 m
ii.
1 m3 = 106cm3
iii.
1 litre = 103cm3 = 10–3m3
v. Light year (ly) : The distance travelled by the light
in one year. 1light year = 9.46×1015m.
vi. Parallactic second (Parsec) : It is the biggest unit
of distance. 1 parsec = 3.26 light year
Submultiples of metre
i.
deci metre (dm) = 10–1m
ii.
centi metre (cm) = 10–2m
iii.
milli metre (mm) =10–3m
DAY – 2,3 (WORKSHEET - 2)
1. 1cm2 = _________m2
1) 104
3) 10–3
2) 10–2
4) 10–4
2. 1m2 = _______cm2
1) 104
2) 10–2
3) 10–3
4) 10–4
3. 1cm3 = ______m3
iv.
micrometre  m  = 10–6m
v.
nano metre (nm) = 10–9m
Narayana Group of Schools
1) 10–6
2) 10–3
3) 106
4) 109
27
VII - Class _ Physics
Bridge Programme
4. 1m3 = ______cm3
11) 1 Decade = ________ years
1) 102
2) 104
1) 10
2) 100
3) 106
4) 108
3) 1000
4) 10,000
5. 1litre = ____cm3
12) 1 mean solar day = _________ seconds
1) 100
2) 1000
1) 60
2) 3600
3) 500
4) 10
3) 1440
4) 86,400
6. 1day = _____________ seconds.
1) 3600
2) 86400
3) 4200
4) 12300
7. Statement I: A thread is enough to measure
a curved line.
Statement II: A scale is enough to measure a
curved line.
1) Statement I is true ; Statement II is true.
2) Statement I is true ; Statement II is false.
8.
13) 1 light year = ______
1) 9.46×1012 m
2) 9.46×1013 m
3) 9.46×1014 m
4) 9.46×1015 m
14) 1 nano meter = __________
1) 10-3 m
2) 10-6 m
3) 10-9 m
4) 1012 m
15) 1 mega second = _________
3) Statement I is false ; Statement II is true.
1) 103 sec
2) 106 sec
4) Statement I is false ; Statement II is false.
3) 109 sec
4) 1012 sec
1
th of a mean solar day is called
24
16) 1 shake = __________ sec
1) Minute
2) Second
1) 102
2) 104
3) Hour
4) Day
3) 106
4) 108
9. One millennium is equal to how many decades
1) 10
2) 100
17) 1 tonn = __________ kg
3) 1000
4) 10,000
1) 10
2) 100
10. Match the following.
List A
List B
3) 1000
4) 10,000
i) 365
1
days
4
a) 1 Century
b) 3600 sec
1
iii)
the part of a
1440
mean solar day
iv) 1 hour
1) i  a, ii  b, iii 
2) i  a, ii  b, iii 
3) i  b, ii  a, iii 
c) 1 year
d) 1 minute
c, iv  d
d, iv  c
c, iv  d
4) i  c, ii  a, iii  d, iv  b
DAY – 4 (SYNOPSIS)
Density : The density of a substance is defined as
the mass per unit volume of the substance.
Density of a substance (d) =
mass of the subs tance(M)
M
d
volume of the subs tance(V)
V
Unit : C.G.S unit : g cm–3
S.I. unit : kg m–3
Relationship between S.I. & C.G.S units of
density : 1g/cm3 = 1000 kg/m3
Narayana Group of Schools
28
VII - Class _ Physics
Bridge Programme
Relative density : It is the ratio of the density of
the substance to the density of water at 40C.
* Area of Triangle =
1
× base× height
2
Thus, Relative density (R.D.)

h
density the subs tance
base
density of water at 40 C
Units of relative density : As the relative density is * Area of circle = π ×Square of radius
the ratio of two similar quantities hence it is a pure
= π × r2
number and therefore it has no units.
r
Relationship between density and relative
density :
1) Density of a solid in S.I. Unit = R.D. of the solid
× Density of water (in S.I. Unit) = R.D. of the
solid × 1000 kg/m3
* The area of an irregular flat surface is measured by
Graph paper
2) Density of a solid in C.G.S.unit = R.D. of the solid Volume: The space occupied by a
substance (solid, liquid (or) gas) is called volume.
× Density of water (in C.G.S. unit) = R.D. of the
3
S.I. unit of volume is m3 .
solid × 1g/cm
Area: The amount of surface occupied by an object Sub multiple of Volume:
or a place is called Area.
1) cm 3 = cm × cm × cm
* S.I. Unit of Area is Metre 2 (or) Square metre.
= 10-2 m ×10-2 m ×10-2 m
2
* Symbol of S.I. unit of area is m .
= 10-6 m 3
Submultiple of standard unit of area:
2) mm3 = mm× mm× mm
1) cm 2 = c.m × c.m = 10-2 m ×10 -2 m = 10 -4 m 2
= 10-3 m ×10-3 m ×10-3 m
2
-3
-3
-6
2
2) m.m = m.m × m.m. = 10 m ×10 m = 10 m
= 10-9 m 3
Multiple of standard unit of area :
Multiples of volume:
1) Hectare = 100m ×100m = 104 m 2
2) Acre = 10m ×10m = 10 m
2
3) 1 Hectare = 100 Acress
Area of regular bodies:
1) km3 = 1km×1km×1km
2
s
= 103 m ×103 m ×103 m
= 109 m 3
s
s
s
* Area of square = Side X Side
=sXs
= s2
2) Volume of liquids is measured in liters and
milliliters.
3) 1liter = 1000ml = 1000cc = 1000cm3
4) 1m3 = 1000 liters
5) Volume of liquids measured by measuring jar
Volume of regular bodies:
* Area of rectangle = l×b
Narayana Group of Schools
29
VII - Class _ Physics
Bridge Programme
shown in fig. It is observed that the beaker
filled with water have more mass than the
beaker filled with kerosene oil. Because
1) Volume of cube = s× s ×s = s 3
2) Volume of cubiod = l  b  h
3) Volume of cylinder = πr 2 h
4) Volume of cone =
1 2
πr h
3
5) The volume of irregular shaped body (like stone)
measured by measuring jar.
DAY – 4 (WORKSHEET - 3)
1. 5 litres of alcohol has a mass of 4 kg. The
density of alcohol is
1) 0.8
3) 800
kg
kg
m3
2) 80
m3
4) 8000
kg
m3
kg
m3
2. A 50kg mass of a body is immersed in water
then we observe the mark on measuring
cylinder as 70cm3. If the body is taken out
from the cylinder, the mark observed on
cylinder is 50cm3. Then the density of that
body is [The graduations marked in cm3]
3
3
1) 2500 g/cm
2) 250 g/cm
3) 2.5 g/cm3
4) 0.025×103g/cm3
g
3. The density of lead is 11.6 cm3 and that of
wood is 800
kg
m3
. what do you understand
by these statements ?
1) The matter is only densely packed in lead
than wood.
2) The matter is only densely packed in wood
3) The density of lead is greater than the
density of wood.
4) Both (1) and (3).
4. Take two identical 100cm3 beakers. Fill one
beaker completely with water and the other
with kerosene oil. Place the beakers in the
scale pans of an ordinary beam balance as
Narayana Group of Schools
1) The matter in water is more densely
packed than in kerosene oil.
2) The matter in kerosene oil is more densely
packed than in water.
3) Water and kerosene oil are occupied the
same volume.
4) None of these.
5. Mass of liquid = 72 g Initial volume of water
in measuring cylinder = 24 cm3 Final volume
of water + solid in measuring cylinder = 42
cm2 From the above data the density of solid
is ___________
1) 4000 kg/m3
3)
72 g
3
42 cm
2) 3.0 g/cm3
4) 4.0 kg/m3
6. If relative density of gold is 19.3. Then the
density of gold is __________times greater
than the density of water.
1) 8.9
2) 19.3
3) 1.29
4) 0.8
7. The density of copper is 8.9 × 103 kg/m3. The
relative density of copper is____
1) 8.9 × 103
2) 8.9 × 102
3) 8.9 × 101
4) 8.9 × 100
8. If the length of the wooden cube is 4m and
30
VII - Class _ Physics
Bridge Programme
the mass is
1
kg,
8
then the relative density of
the wooden cube is ______________
1) 19.53 × 10–7
2)
1
512
1
51200
4)
1
32
3)
9. An iron cylinder of radius 1.4 cm and length 8
cm is found to weigh 369.6 g. The relative
density of iron cylinder is _______Take
volume of cylinder = r 2 .
1) 7.2
2) 7.5
3) 8
4) 8.2
10. Calculate the mass of a body whose volume
is 2m3 and relative density is 0.52
1) 1040 kg
2) 1000 kg
3) 950 kg
4) 750 kg
3) 1000
16) The unit of melative density is
1) gm / cm3
2) kg / m3
3) gm / c.c
4) No units
17) R.d of solid is
1) Density of solid
2) Density of solid X Density of water at 40 c
3) Density of solid + Density of water at 40 c
4) None of these
18) The amount of surface occupied by a body
is called
1] Area
3] Density
2) Area
3) Volume
4) Speed
12) Density = ________
1) Mass× Volume
2) Mass + Volume
3) Mass / Volume
4) None of these
13) C.G.S. unit of density
1) gm / cm3
2) gm / cm 2
3) gm / cm
4) gm / m3
1] m
3] m3
4) None
20) 1 Hectare = _________ m 2
1) 10
2) 102
3) 103
4) 104
21) 1 Acre = ___________
1) 10
2) 102
3) 103
4) 104
22) 1 Hectare = __________ acres
1) 10
2) 102
3) 103
4) 104
1) gm / cm3
2) kg / m3
1) s×s
3) kg / m 2
4) kg / m
3)
15) 1 gm / cm3 = ___________ kg / m3
Narayana Group of Schools
2] m 2
23) The area of circle is
14) S.I. unit of density
1) 10
2] Volume
4] Speed
19) The S.I. unit of area
11) The mass per unit volume is
1) Density
4) 10,000
2) 100
1
b h
2
2) l×b
4) πr 2
24) The area of irregular flat surface is
measured by
31
VII - Class _ Physics
Bridge Programme
1) Graph paper
2) Scale
3) Tape
4) None of these
25) The space occupied by a substance is called
1) Area
2) Volume
3) Density
4) None of these
26) The S.I. unit of volume
1) m
2) m 2
3) m3
4) None of these
27) 1 litre = _________ cm3
1) 10
2) 102
3) 103
4) 104
28) 1 m3 = ___________ lit
1) 10
2) 102
3) 103
4) 104
29) Volume of liquids is measured by
1) Measuring jar
3) Pipette
4) All of these
30) The volume of cylinder
1) s×s×s
2) l  b  h
3) πr 2 h
4)
1 2
πr h
3
31) The volume of cone
1) s×s×s
2) l  b  h
3) πr 2 h
4)
1 2
πr h
3
DAY – 5 (SYNOPSIS)
Rest : A body is said to be at rest if it does not change
its position with respect to the reference point.
Ex : A chair lying in a room is in the state of rest,
because it does not change its position with
respect to the surroundings of the room.
Narayana Group of Schools
Motion : A body is said to be in motion if it changes
its position with respect to the surroundings with
the passage of time. All moving things are said
to be in motion.
Ex : A car is changing its position w.r.t trees, houses
etc. is in the state of motion.
Rest and motion are relative terms : Rest and
motion are relative terms. A body can be at rest
as well as in motion at the same time.
Ex : A person sitting in the compartment of a moving
train is in the state of rest, with respect to the
surroundings of compartment. Yet he is in the
state of motion, if he compares himself with
surroundings outside the compartment.
Types of motion :
a)
Rectilinear motion : When an object moves
along a straight line, its motion is called rectilinear
motion.
Ex : A car or a train moving on a straight road.
b)
Rotatory motion : A motion in which a body
moves about a fixed axis without changing its
position (from the fixed axis) is called the rotatory
motion.
Ex : (i) The motion of the blades of a fan. (ii) The
wheel of a sewing machine.
c)
Oscillatory motion : The to-and-fro motion is
called oscillatory or vibratory motion.
A motion in which the body as a whole moves
to-and-fro about its mean position is called
oscillatory motion.
Ex : The needle of a sewing machine moving up and
down.
Distance : It is defined as the actual path followed by
a body between the points between which its
moves.
Unit : C.G.S unit : cm S.I unit : m
Note: The distance travelled by body is always
positive.
Displacement : It is the shortest distance between
initial and final point in a definite direction.
Unit : C.G.S unit : cm S.I unit : m
Note : (i) For a moving body displacement can be
positive, negative or zero. (ii) If initial point and
final points are same then displacement is zero.
32
VII - Class _ Physics
Bridge Programme
1) Vector quantity
Scalar quantities : A physical quantity which is
2) Scalar quantity
described completely by its magnitude is called
3) Either vector quantity or scalar quantity
a scalar quantity. It has only magnitude and no
4) Neither vector quantity nor scalar quantity
specific direction.
Ex: Length, distance, area, volume, mass, time and 5. Which is a vector quantity among these ?
1) My mass is 20 kg
energy are examples of scalar quantities.
2) Himalayas are in the northern India
Vector quantities : A physical quantity which is
3) The sun rises in the east.
described completely by its magnitude as well
4) 200 m towards north is Ramoji film city
as specific direction is called vector quantity. It
from my house.
has both magnitude and direction.
Ex : Displacement, velocity, acceleration, force and 6. Three girls skating on a circular ice ground
of radius 200 m start from a point P on the
weight are examples of vector quantities.
edge of the ground and reach a point ‘Q’
diametrically opposite to ‘P’ following
DAY – 5 (WORKSHEET - 1)
different paths as shown in figure.
1. Ravi and Raju are sitting in the moving train
compartment. Rama has come to see them off.
He is standing on the plat form. In this situation,
which is correct?
Q
B
A
C
1) Ravi and Raju are at rest with respect to Earth.
2) Ravi and Raju are at rest with respect to Rama
P
3) Rama is at rest with respect to Ravi and Raju
4) For Rama, Raju and Ravi both are in motion.
2. Imagine that you are travelling by a bus. Which
is correct in these situation ?
What is the magnitude of the displacement
vector for each? For which girl is this equal
to the actual length of path skated?
1) 200, C
2) 200, A
3) 400, B
4) 400, A
1) With respect to the surroundings outside the
bus, you are in motion.
7. A cyclist moves from a certain point X and
goes round a circle of radius ‘r’ and reaches
2) With respect to the surroundings inside the
Y, exactly at the other side of the point X, as
bus, you are at rest.
3) With respect to you, you are in motion
shown in figure.
4) Both (1) and (2)
3. Statement I : Blackboard in your classroom is
in the state of rest with respect to you.
Statement II : You are in motion in relation to
your friend while both travelling in the same
school bus.
Which is correct among these ?
1) Statement-I is correct, Statement-II is also
correct
r
X
The displacement of the cyclist would be
__________
1)  r
2) 2 r
3) 2r
4)
2) Statement-I is correct, Statement-II is wrong
3) Statement-I is wrong, Statement-II is correct
4) Statement-I is wrong, Statement-II is also
wrong
4. Displacement is a ________________
Narayana Group of Schools
Y
O
2
r
8. In the above problem, the distance covered by
the cyclist would be
1)  r
2) 2 r
33
VII - Class _ Physics
Bridge Programme
3) 2r
4)
2
r
9. A man walks 8m towards East and then 6m
towards north. His magnitude of displacement
is _________________
1) 10 m
2) 14 m
3) 2 m
4) zero
17) A particle complete half of circle radius R
then distance and displacement is
1) 2πR, πR
2) πR, 2R
3) 2R, πR
4) πR, 2πR
DAY – 6, 7 (SYNOPSIS)
10. A player completes a circular path of radius ‘r’ Speed: The rate of change of motion is called speed.
in 40s. At the end of 2 minutes 20 seconds,
The speed can be found by dividing the distance
displacement will be
covered by the time in which the distance is covered.
1) 2r
2) 2 r
Dis tance travelled
Formula : Speed 
3)
4) Zero
7 r
Time taken
11) Spinning top is example of
1) Rectilinear
2) Rotatory
3) Vibratory
4) Curvilinear
12) Given below which is vector quantity
1) Energy
2) Distance
3) Force
4) Time
13) Distance is _________ quantity
1) Scalar
2) Vector
3) Both (1) & (2)
4) None of these
14) A particle moving straight line in that case
distance and displacement are
1) Both zero
2) Distance > Displacement
3) Distance = Displacement
4) Distance < Displacement
15) An object is complete one revolution in
circle path circle readius R X then distance
and displacement
1) 0, 2πR
2) 2πR, 0
3) 2πR, 2πR
4) 0, 0
16) The relation of distance and displacement
1) Always > 1
2) Always < 1
3) Always  1
4) Always  1
Narayana Group of Schools
Units : C.G.S unit : cm/s
S.I unit : m/s
Nature : Scalar
Kinds of Speed :
a)
Uniform speed : When a body covers equal
distances in equal intervals of time (however
small the time intervals may be), the body is said
to be moving with a uniform speed.
Ex : A rotating fan, a rocket moving in space, etc.,
have uniform speeds.
b)
Non-Uniform Speed : When a body covers
unequal distances in equal intervals of time, the
body is said to be moving with a nonuniform
speed.
Ex : A train starting from a station, a dog chasing a
cat, have variable speeds.
c)
Average Speed:- When a body is moving with
a variable speed, then the average speed of the
body is defined as the ratio of total distance
travelled by the body to the total time taken i.e.,
 Average speed

Total dis tance cov ered
Total time taken to cov er the dis tance
Velocity : Velocity is the rate of change of motion in a
specified direction.
Formula : Velocity =
displacement
time
Units : C.G.S. unit : cm/s
S.I. unit : m/s
34
VII - Class _ Physics
Bridge Programme
distance with speed 2m/s and the last one
Nature : Vector
third distance with speed 3m/s then average
Kinds of velocity :
speed is ____________
a)
Uniform velocity :- When a body covers equal
1) 2 m/s
2) 1.79 m/s
distances in equal intervals of time in a specified
3) 2.66 m/s
4) 1.64 m/s
direction, (howsoever short, the time intervals
may be) the body is said to be moving with a 5) A body goes from A to B with a velocity of 20
m/s and comes back from B to A with a
uniform velocity.
velocity of 30 m/s. The average velocity of
Ex : Imagine a car is moving along a straight road
the body during the whole journey is
towards east, such that in every one second it
1) 25 m/s
2) 24 m/s
covers a distance of 5m then the car is said to
be moving with uniform velocity.
3) zero
4) none of these
b)
Variable Velocity:- When a body covers 6) An insect crawls a distance of 4m along north
unequal distances in equal intervals of time in a
in 10 seconds and then a distance of 3m
along east in 5 seconds. The average
specified direction or equal distances in equal
velocity of the insect is ______
intervals of time, but its direction
changes, then the body is said to be moving with
7
1
1) m/sec
2) m/sec
a variable velocity.
5
5
Ex : Now imagine the car is moving along a circular
5
path, such that it is covering 5m in every one
m/sec
3)
4) None of these
15
second but as the direction of the car is changing
at every instant, we say the car is moving with 7) A car travels half the distance with constant
velocity 50 km/h, and another half with a
variable velocity.
constant velocity of 40 km/h along a straight
c)
Average velocity:- It is the ratio of total
line. The average velocity of the car in km/h
displacement to total time taken.
is ___________
Average velocity 
Total displacement
Total time taken
DAY – 6, 7 (WORKSHEET - 2)
1) A horse runs a distance of 1200m in 3 min
and 20 s. The speed of the horse is_______
1) 60 ms–1
2) 65 ms–1
3) 40 ms–1
4) 6ms–1
2) A car is moving at a speed of 15 ms–1. In how
much time will it cover a distance of 1.2 km ?
1) 70 s
2) 80 s
3) 18 s
4) 45 s
3) A bus is moving at 20ms–1. How much
distance in kilometres will the bus cover in
25 minutes ?
1) 30 km
2) 20 km
3) 40 km
4) 50 km
4) When a body covers first one third distance
with speed 1m/s, the second one third
Narayana Group of Schools
1) 45
3) 0
2) 44.4
4)  50  40 
8) If a motorist completes half a revolution in a
circular track of radius 100 m in one minute,
then what will be his average speed and
average velocity ? ______
1) 314 metre/minute, 0
2) 0, 314 metre/minute
3) 314 metre/minute, 200 metre/minute
4) 200 metre/minute , 314 metre/minute
9) Speed and velocity are
1) Vector, Scalar
2) Scalar, Vector
3) Vector, Vector
4) Scalar, Scalar
10) The body move in uniform speed then
average speed, intial speed are
1) Average speed and intial speed is not equal
to uniform speed
35
VII - Class _ Physics
Bridge Programme
2) Average speed and intial speed are equal to
uniform speed
3) Average is equal to uniform but
instantneous is not equal to uniform speed
4) None of these
11) A particle move in straight line in uniformly
therefor uniform speed and uniform velocity
are
1) Both are diffferent
2) Both are zero
3) Both are equal
4) All of these
12) A body move in circular path in uniform
motion which is constant
1) Speed
2) Velocity
3) Acceleration
4) All of these
13) A particle half the distance travel in v1 next
half the distance travel v 2 therefore
average speed is
v1 + v 2
2
2v1 v 2
2) v + v
1
2
v -v
3) 1 2
2
2v12
4)
v1 + v 2
1)
14) A particle half of the time travel speed v1
next half of time time travel speed v 2
average speed is
2v1 v 2
2) v + v
1
2
v + v2
1) 1
2
v -v
3) 1 2
2
4) All of these
DAY – 8 (SYNOPSIS)
Acceleration : The rate of change of velocity of a
body is called acceleration.
Formula : Acceleration =
Change in Velocity
Time taken
Units : C.G.S. unit : cm/s2
S.I. unit : m/s2
Narayana Group of Schools
Nature : Vector
Uniform Acceleration: When a body describes
equal changes in velocity in equal intervals of
time (however small may be the time intervals)
it is said to be moving with uniform acceleration.
Equations of motion for a body moving with
uniform acceleration in a straight line:
a)
First Equations of Motion : It gives the
velocity acquired by a body in time t which is v
= u + at where, v = Final velocity of the body
[velocity after time (t) seconds] u = Initial velocity
of the body [velocity at time (t) = 0 second] a =
Acceleration (uniform) t = Time taken
b)
Second Equation of Motion : It gives the
displacement of the body in a time t, which is
1
s = ut + at 2
2
where, s = displacement of the
body in time t
c)
Third Equation of motion : It gives the velocity
acquired by a body in displacement ‘s’ which is
v2 – u2 = 2as
Points to remember :
i)
If a body starts from rest, its initial velocity, u =
0
ii)
If a body comes to rest (it stops),its final velocity,
v=0
iii) If a body moves with uniform velocity, its
acceleration, a = 0
Distance travelled in nth second : It gives the
distance travelled by the body in nth second
which is sn = u +
a
 2n - 1 where, sn = distance
2
travelled by the body in nth second
DAY – 8 (WORKSHEET - 3)
1. The velocity of car changes from 18 km/h to
72 km/h in 30s. What will be its acceleration
in km/h2 and in m/s2
1) 6480 km/h2, 0.50 m/s2
2) 6450 km/h2, 0.40 m/s2
3) 6580 km/h2, 0.30 m/s2
4) 6840 km/h2, 0.50 m/s2
36
VII - Class _ Physics
Bridge Programme
2. A car is moving with uniform velocity of 72
km/h for 15s on a National Highway. Its
acceleration and the distance travelled are
2
distance of 5m. The retardation produced by
the brakes is
1) 90 m/s2
2) 80 m/s2
3) 95 m/s2
4) 85 m/s2
2
1) 0m/s , 200m
2) 0m/s , 300m
3) 20m/s2, 300m
4) 2m/s2, 400m
9. A body originally at rest is subjected to
3. A motor bike is moving with a velocity of 5m/s.
uniform acceleration of 4ms–2. The distance
It is accelerated at a rate of 0.6m/s2 for 20s.
travelled by it in 5th second is
Then the final velocity of motor bike is
1) 10 m
2) 15 m
1) 16m/s
2) 18 m/s
3) 18 m
4) 20 m
3) 15m/s
4) 17m/s
10) Acceleration is a ________ quantity
4. If a bus starts from rest and attains a speed
1) Scalar
2) Vector
of 36 km/h in 10 minutes while moving with
uniform acceleration, then the acceleration of
3) Both (1) & (2)
4) None of these
the bus is
11) The rate continue increase the change of
velocity is called _________
1
1
1)
m/s2
2)
m/s2
40
60
1) Uniform velocity
2) Acceleration
3)
1
m/s2
50
4)
1
m/s2
30
5. A girl running in a race accelerates at 2.5 m/
s2 for the first 4s of the race. How far does
she travel in this time? (Assume a girl has
started from rest)
1) 10m
2) 15m
3) 20m
4) 25m
6. A body starting from rest travels a distance
of 200m in 10 seconds, then the value of
acceleration is
–2
4) All of these
12) A body has an acceleration of -3m / s 2 what
is retardation is
1) 3m / s 2
2) 4 m / s 2
3) 0.5 m / s 2
4) 5m / s 2
13) The acceleration of the body has the
direction of
1) Displacement
2) Velocity
3) Change in velocity
4) All of these
–2
1) 2ms
2) 4ms
3) 5ms–2
4) 3ms2
7. A scooter moving at a speed of 10 m/s is
stopped by applying brakes which produce a
uniform acceleration of, –0.5 m/s2. How much
distance will be covered by the scooter
before it stops?
1) 200 m
2) 300 m
3) 100 m
4) 250 m
8. A car moving with a speed of 30ms–1 upon the
application of brakes comes to rest within a
Narayana Group of Schools
3) Constant velocity
14) If a particle is in uniform velocity along the
straight line then its acceleration is
1) Increase
2) Decrease
3) Constant
4) Zero
15) A body intial velocity of 10 m / s and it
attain a velocity 20 m / s in 5 sec then
accleration ________ m / s 2
1) 3
2) 4
3) 2
4) 5
37
VII - Class _ Physics
Bridge Programme
16) C.G.S. unit of acceleration is
1) m / s 2
2) cm / s 2
3) km / n 2
4) None of these
DAY – 9 (SYNOPSIS)
Force : Push or pull is called force. The cause of
motion is force.
Effects of force :
A force can cause a motion in stationary object
A force can stop the moving objects or slow
them down
A force can make a moving object move faster
Force can change direction of moving objects
Force can change the shape of objects.
From the above examples, we are in position to
define force.
Force is an external agent which changes or
tends to change the state of rest or uniform
motion of a body or changes its direction or
shape.
Types of forces :
a)
Muscular forces : The force applied by the
muscles of our body is called muscular force or
biological force.
Ex : Lifting of heavy weight pulling of wheel cart,
pushing a lawn roller etc. involves muscular
forces.
b)
Mechanical forces :The forces generated by a
machine are called mechanical forces.
Ex : The force used to run a motor car engine is
produced by using the energy of petrol. The force
used to run steam engine, is produced by using
the energy of coal.
c)
Gravitational force : The force of attraction
exerted by the earth on all the objects is called
the force of gravity or gravitational force.
Ex : A stone falls downwards due to gravitational force.
It is the gravitational force of the sun that keeps
the planets in their orbits.
d) Electrostatic force : The force exerted by
electrostatic charge is called electrostatic force.
Ex : Charged comb attracts small pieces of paper.
Narayana Group of Schools
e)
Magnetic force : The force by which a magnet
attracts or repels objects of iron, steel, nickel and
cobalt is called magnetic force.
Newton’s first law of motion :
A body at rest will remain at rest and a body in
motion will remain in uniform motion, unless it is
compelled by an external force to change its state
of rest or of uniform motion.
Ex : A book lying on a table is in the state of rest w.r.t.
the table. It will remain at rest unless some one
picks it up or moves it from one position to
another.
Inertia : Inertia of a body may be defined as the
tendency of a body to oppose any change in its
state of rest or uniform motion.
Ex : A book lying on a table will remain placed at its
place unless it is displaced.
Measure of inertia : Mass is a measure of inertia.
Ex : If a body has a mass of 1 kilogram and another
body has a mass of 20 kg, then the body having
20 kg mass will have more inertia since its mass is
more.
Types of inertia : Inertia can be divided into three
types.
1) Inertia of rest
2) Inertia of motion
3) Inertia of direction
a)
Inertia of rest : The tendency of a body by
virtue of which it cannot change its state of rest
by itself is called inertia of rest.
Ex : A passenger in a bus tends to fall backward when
the bus starts suddenly
b)
Inertia of motion : The tendency of a body by
virtue of which it cannot change its state of
motion by itself is called inertia of motion.
Ex : A passenger in a moving bus tends to fall forward
when the bus stops suddenly
c)
Inertia of direction : The tendency of a body
to oppose any change in its direction of motion
by itself is known as inertia of direction.
Ex : A stone tied to a string and whirled along a circular
path flies off tangentially due to inertia of
direction, if the string breaks.
DAY – 9 (WORKSHEET - 1)
38
VII - Class _ Physics
Bridge Programme
1. Identity the situations where a pull is
involved.
7. Inertia is that property of a body by virtue of
which the body is
a) Man sitting on a chair
1) Unable to change by itself its state of rest.
b) ball falling to the ground
2) Unable to change by itself its state of
uniform motion.
c) Woman drawing water from a well
d) tube light fixed on the wall
1) both ‘a’ and ‘b’
2) both ‘b’ and ‘c’
3) both ‘c’ and ‘d’
4) both ‘a’ and ‘d’
2. A force can
3) Unable to change by itself its direction of
motion.
4) All the above.
8. An athlete runs some distance before taking
a long jump, because
a) move a body from rest.
1) It helps him to gain energy
b) change the direction of a moving body.
2) It helps to apply large force
c) increases the mass of a body.
3) It gives himself large amount of inertia of
motion
1) only ‘a’ is true
2) only ‘b’ is true
3) both ‘a’ and ‘b’ are ture
9. A rider on a horse back falls forward, when
the horse suddenly stops, this is due to
4) ‘b’ and ‘c’ are true
3. Apple is falling from the tree towards the
ground due to
1) Magnetic force
2) Gravitational
force
1) inertia of the horse
2) inertia of the rider
3) large weight of the horse
4) losing the balance
3) Mechanical force
4) Electrostatic force
4. A boy used ________ force to kick a football.
1) Muscular force
2) Gravitational force
3) Mechanical force
10. A passenger sitting in a bus gets a
backward jerk when the bus starts suddenly
due to the
1) inertia of rest
2) inertia of motion
3) inertia of direction
4) none of these
11) _________ is the measure of inertia
4) Electrostatic force
5. Magnetic force can cause
1) only attraction
2) only repulsion
3) both attraction and repulsion
4) none of these
6. A tailor cuts a piece of cloth using a pair of
scissors. The force involved here is
1) electrostatic force
2) mechanical force
3) magnetic force
4) none of these
Narayana Group of Schools
4) None of these
1) Volume
2) Mass
3) Area
4) Length
DAY – 10,11 (SYNOPSIS)
Linear momentum : The total quantity of motion
contained in a body is called linear momentum.
Formula : Momentum of a body is equal to the product
of the mass (m) of the body and the velocity


by P .Momentum
 v  of the body. It isdenoted


= mass × velocity ( P = m v )
Units : C.G.S. unit: g cms–1 S.I. unit: kg ms–1
39
VII - Class _ Physics
Bridge Programme
Nature : Vector , The direction of momentum of a
1) 6 m/s
2) 5 m/s
body is same as that of the direction of the
3) 4 m/s
4) 3 m/s
velocity of the body.
3. A cricket ball of mass 100 g is moving with
Newton’s Second Law of motion :
velocity 25 m/s. Then the momentum of the
The magnitude of the resultant force acting on a
ball is
body is proportional to the product of the mass
of the body and its acceleration. The direction
1) 7.5 kg.m/s
2) 3 kg.m/s
of the force is the same as that of the
3) 2 .5 kg.m/s
4) 4 kg.m/s
acceleration.
Newton’s second law gives the quantitative 4. Two bodies A and B of same mass are moving
with velocities V and 3 V respectively, then
definition of force in other words it measures
the ratio of their momentum will be
force. i.e. Fnet = ma, where, F = Force, m =
Mass, a = Acceleration
1) 1 : 2
2) 2 : 1
So, Force acting on the body = mass of the
3) 3 : 1
4) 1 : 3
body × acceleration produced in the body.
Newton’s 2nd Law in terms of momentum : The 5. A cricket ball of mass 100g strikes the hand
rate of change of momentum of an object is
of a player with a velocity of 20 m/s and is
proportional to the net force applied on the
brought to rest in 0.01 s, then the force
object. The direction of the change of momentum
applied by the hands of the player is
is the same as the direction of the net force.
1) 200 N
2) 300 N
Force  Rate of change of momentum. i.e.,
3) 400 N
4) 500 N
Force 
Change of momentum
Time
6. Two bodies have masses in the ratio 3 : 4.
When a force is applied on first body, it
Absolute Units of Force : C.G.S unit : g cm/s2 or
moves with an acceleration of 6 m/s2. How
dyne.
much acceleration the same force will
Definition of dyne : The force is said to be 1 dyne if
produce in the other body ?
it produces 1 cm/s2 acceleration in a body of 1g
mass.
1) 5.5 m/s2
2) 3.5 m/s2
S.I unit of force : kg m/s2 or newton(N).
3) 4.5 m/s2
4) 2.5 m/s2
Definition of newton (N) : 1 newton is that much
7. A force of 200 dyne acts on a body of mass
force which produces an acceleration of 1 m/s2
10 g for 5 sec. What will be the final velocity
in a body of mass 1 kg.
of body if it starts from rest ?
Relation between newton and dyne :1 newton (N)
1) 50 cm/s
2) 80 cm/s
= 105 dyne
DAY – 10,11 (WORKSHEET - 2)
1. What will be the momentum of a toy car of
mass 200 g moving with a speed of 5 m/s ?
1) 1 kg m/s
2) 2 kg m/s
3) 3 kg m/s
4) 4 kg m/s
2. A body of mass 25 kg has a momentum of
125 kg m/s what is its velocity?
Narayana Group of Schools
3) 100 cm/s
4) 200 cm/s
8. A force of 10 kg wt acting on a certain mass
for 2 second gave it a velocity 10 m/s. What is
the mass in kg ? (g = 9.8 m/s2)
1) 19.6
2) 9.8
3) 15
4) 5
9) Linear momentum 
P =?
40
VII - Class _ Physics
Bridge Programme

1) m× v
m
2)

v


2) m  v
4)
v

m
10) The total quantity of motion contained in a
body is called __________
1) Linear momentum
2) Velocity
3) Force
4) Mass
11) One newton = _____ dyne
1) 107
2) 105
3) 105
4) 107
12) According newtons second law Fnet = ____
1) m×a
Formula : Weight = mass × acceleration due to
gravity.
Unit : C.G.S unit : dyne S.I. unit : newton
Nature : Vector
Note: Mass remains constant whereas weight
changes from place to place.
Newton’s Third Law : ‘To every action, there is an
equal and opposite reaction’
Note : Action and reaction force are equal in
magnitude but opposite in direction.
i.e. Action = – Reaction
Ex : Jet aeroplanes and rockets works on the principle
of Newton’s third law of motion.
Note : Action and Reaction act simultaneously but
act on different bodies. So they donot cancel with
each other.
m
2)
a
Law of conservation of momentum : According
to this law, the total momentum of a system remains
constant if no net external force acts on the system.
4) m  a
That is, momentum of a system. p = constant, if net
external force acting on it is zero

a
3)
m

DAY – 12 (SYNOPSIS)
Impulsive Force : A large force which acts for a
small interval of time is called impulsive force.
Impulse : Impulse of a force is defined as the
change in momentum produced by the given force
and it is equal to the product of force and the time
for which it acts.
Formula : Impulse = Change in momentum =
Force × Time.
Unit : C.G.S unit : dyne second (or) g cm/s
S. I. unit : N s (or) kg m/s Nature : Vector
Mass : The quantity of matter contained in the body
is called its mass.
Unit : C.G.S unit : gram (g). S.I. unit : kilogram
(kg). Nature : Scalar
Weight : The weight of a body is the force with
which it attracted towards the centre of the earth.
Narayana Group of Schools
(i.e. Fexternal = 0)
DAY – 12 (WORKSHEET - 3)
1. A force of 50 N acts on a body for 10 s. What
will be the change in momentum ?
1) 200 Ns
2) 400 Ns
3) 500 Ns
4) 1000 Ns
2. A body of mass 100 kg moving straight line
with a velocity of 30 m/s, moves in opposite
direction with a velocity of 20 m/s after
hitting a wall. What is its magnitude of
impulse ?
1) 6000 Ns
2) 5000 Ns
3) 4000 Ns
4) 3000 Ns
3. How much would a 70 kg man weigh on the
moon ? What will be his mass on the earth
and on the moon ? [ g on moon = 1.7 m/s2]
41
VII - Class _ Physics
Bridge Programme
1) 119 N, 70 kg
2) 115 N, 68 kg
1) m1u1 + m2u2 = m1v – m2v
3) 116 N, 65 kg
4) 114 N, 55 kg
2) m2u1 – m1u2 = (m1 + m2)v
4. If the weight of man on earth surface 30 N.
What will be his weight on moon surface ?
(gmoon = (1/6) gearth)
1) 5 N
2) 4 N
3) 3 N
4) 2 N
5. In case of a book lying on a table.
1) action of book on table and reaction of
table on book are equal and opposite and
are inclined to vertical.
2) action and reaction are equal and opposite
and act perpendicular to the surfaces of
contact.
3) action and reaction are equal but act in the
same direction.
4) action and reaction are not equal but are in
opposite direction.
6. Whenever an object A exerts a force on
another object B, object B will exert a return
force back an object A. The two forces are
1) equal in magnitude and in same direction
2) equal in magnitude but opposite in direction
3) not equal and in opposite direction
4) not equal and in the same direction
7. When two bodies of masses m1 and m2
moving with velocities u1 and u2 in the same
direction collide with each other and v1 and
v2 are their velocities after collision in the
same direction, then
1) m1v1 + m2v2 = m2u2 – m1u1
3) m1u1 – m2u2 = (m1 – m2)v
4) v 
m1u1  m2u2
m1  m2
9. The car A of mass 1500 kg travelling at 25 m/
s collides with another car B of mass 1000 kg
travelling at 15 m/s in the same direction.
After collision the velocity of car A becomes
20 m/s. The velocity of car B after the
collision is
1) 12.2 m/s
2) 11.5 m/s
3) 22.5 m/s
4) 5.22 m/s
10) A large amount of force which acts for a
small interval of time is called _______
1) Gravitational force
2) Measure force
3) Impulsive force
4) No force
11) To every action, there is equal and opposive
reach on explains newtons ______ law
1) Ist law
2) IInd law
3) IIIrd law
4) None of these
12) Newton third law explains _________
1) Action = -Reaction
2) Action  -Reaction
3) Both (1) and (2)
4) None of these
13) According to ___________ law, the total
momentum of a system remain constant if
no external force acts on the system
2) m1v1 + m2v2 = m1u1 – m2u2
1) Law of conservation of energy
3) m2u2 + m2u1 = m2v1 + m1v2
2) Law of conservation of momentum
4) m1u1 + m2u2 = m1v1 + m2v2
3) Law of conservation of mass
8. When two bodies of masses m1 and m2
moving with velocities u 1 and u 2 in the
opposite direction collide with each other
and move together after collision in the
same direction with a common velocity v,
then
Narayana Group of Schools
4) None of these
14) Impulse = _________
1) Force X time
2)
force
time
42
VII - Class _ Physics
Bridge Programme
has positive work done by the gravitational
force.
Negative work done : The work done by a force on
15) Unit of Impulse
a body is said to be negative work done when
the body is displaced in a direction opposite to
1) N - S
2) N/S
the direction of the force.
2
2
3) N / s
4) N - s
Ex : Work done by frictional force as force of friction
and the displacement are opposite to each other.
16) Weight = ___________
Zero work : Work done is zero if
1) Mass X Velocity
i)
The displacement is zero.
2) Mass X time
Ex : When a person pushes a wall but fails to move the
wall, then work done by the force on the wall is
3) Mass X acceleration due to gravity
zero.
ii)
The force and t he displacement are
Velocity
4)
perpendicular to each other.
Mass
Ex : When a person carrying a suitcase in his hand or
17) Impulse is a ____________
on his head is walking horizontally, the work
done against gravity is zero.
1) Vector
2) Scalar
No work is done on a body when the body
3) Both (1) & (2)
4) None of these
moves along a circular path.
DAY – 13 (SYNOPSIS)
Work done against gravity :
Work : Work is said to be done when a force produces
Work done in lifting a body = weight of body ×
motion.
vertical distance = W = mg × h where, m = mass
Ex : When an engine moves a train along a railway line,
of body g = acceleration due to gravity at that
it is said to be doing work.
place h = height through which the body is lifted.
Mathematical Expression for work :
DAY – 13 (WORKSHEET - 1)
If a force F acts on a body and moves it a
distance S in the direction of the force then work 1. When a stone tied to a string is whirled in a
circle, the work done on it by the string is
done. i.e.,
W=F×S
1) Positive
2) Negative
Units of work : C.G.S. unit : g cm2 s–2 or erg.
3) Zero
4) None of these
S.I unit : kg m2 s–2 or joule(J).
Relation between Joule and erg : 1 J = 1 N × 1 m 2. A man with a box on his head is climbing up a
ladder. The work done by the man on the box
= 105 dyn × 100 cm = 107 dyn cm.
7
is
(or) 1 J = 10 erg.
1) Positive
2) Negative
Note : Work is a scalar quantity.
3)
time
force
4)
-force
time
3) Zero
4) Undefined
Types of Work : Work done can be positive, negative
or zero depending upon the direction of force 3. Work is said to be done if
and direction of motion. (displacement)
Statement A : a force is applied which
Positive work done : Work done by a force on a body
(or an object) is said to be positive work done
Statement B : a force is applied but no
when the body is displaced in the direction of
motion is produced
applied force.
1) only statement A is true
Ex : The body falling freely under the action of gravity
Narayana Group of Schools
43
VII - Class _ Physics
Bridge Programme
2) only statement B is true
1) A force acts on it
3) both the statement A and B are true
2) It moves through the certain distance
4) none of these
4. Work done is zero
1) When force and displacement of the body
are in the same direction
2) When force and displacement of the body
are in the opposite directions
3) When force acting on the body is
perpendicular to the direction of the
displacement of the body
4) None of these
5. How much work is done by a force of 10 N
is moving a body through a distance of 2 m
in its own direction ?
1) 20 J
2) 24 J
3) 26 J
4) 30 J
6. Calculate the work done by a passenger
standing on a platform holding a suitcase of
weight 10 kgwt.
1) 15
2) 10
3) 0
4) 5
7. Which of the following represents joule ?
1) Nm
2) dyn m
3) N/m
4) m/N
8. A person of mass 50 kg climbs a tower of
height 72 metre. The work done is [g = 9.8 m/
s2]
3) It experiences an increase in energy
through a external mechanical force &
dispalces
4) None of these
12. A force does not perform any work if
1) The displacement is parallel to the force
2) The displacement is perpendicular to the
force
3) The body is in motion 4) None of these
13. Under the action of a constant force a
particle is experiencing a constant
acceleration & displacement the work is
may be
1) Positive
3) Both (1) & (2)
2) Zero
4) None of these
14. The work done by a force on a body does
not depend upon
1) The mass depend upon
2) The displacement of the body
3) The intial velocity of the body
4) The angle b/n force & displacement
DAY – 14 (SYNOPSIS)
Energy : Energy is the ability to do work or the
3) 52380
4) 58320 J
capacity to do work.
9. How much is the mass of a man if he has to
Units : Unit of energy is same as that of the unit of work.
do 2500 joule of work in climbing a tree 5m
As work is a form of energy.
tall ? (g = 10 m/s2)
C.G.S unit of energy is erg. S.I. unit of energy
1) 30 kg
2) 40 kg
is Joule (J).
3) 50 kg
4) 45 kg
Nature : Energy is a scalar quantity.
10. An object of 100 kg is lifted to a height of 10 Mechanical energy (M.E) : The sum of kinetic
m vertically. What will be the work done? [g
energy (K.E) and potential energy (P.E) of a
= 9.8 m/s2]
body is known as mechanical energy. M.E
1) 9800 J
2) 9008 J
= K.E + P.E
Kinetic energy (K.E) : The word kinetic comes from
3) 9.8 J
4) 8.9 J
a Greek word which means motion.
11. Work is always done on a body when
1) 35280 J
Narayana Group of Schools
2) 32580 J
44
VII - Class _ Physics
Bridge Programme
The energy possessed by a body by virtue of its 3. A 1kg mass has a kinetic energy of 1 Joule
when its velocity is
motion is known as kinetic energy. Note : All
moving bodies possess kinetic energy.
1) 0.45 m/s
2) 1 m/s
Ex : A moving bus or a car or a train has kinetic energy.
3) 1.4 m/s
4) 4.4 m/s
Formula of kinetic energy:Kinetic energy, 4. If acceleration due to gravity is 10 m/s2, what
will be the potential energy of a body of mass
1
K.E  mv 2
1 kg kept at a height of 5 m ?
2
1) 20 J
2) 30 J
where, m = Mass of the body, v = Velocity of the
body.
3) 40 J
4) 50 J
Potential energy : (P.E) : The energy possessed by 5. An object of mass 1 kg has a potential
a body by virtue of its position is called potential
energy of 1 J relative to the ground, when it
energy.
is at a height of [g=10 m/s2]
Ex : Water stored in a dam has potential energy due to
1) 0.1 m
2) 1 m
its position.
3) 9.8 m
4) 32 m
Formula of potential energy :
6. A light and a heavy body have equal kinetic
Potential energy (P.E) of a body at a certain
energy. Which one has greater momentum ?
height = P.E = mgh
1) The lighter body has greater momentum
where, m = mass is the body, g = acceleration
2) The heavier body has greater momentum
due to gravity h = height from the ground.
3) both the bodies have same momentum
Examples of body possessing both the kinetic and
potential energies at the same time:
4) none of these
i) A flying aeroplane
7. What will be the momentum of a body of
ii) A bird flying in the sky
mass 100 g having kinetic energy of 20 J ?
Relation between kinetic energy and momentum
1) 2 kg m/s
2) 4 kg m/s
We know , P = mv  v = P/m and K.E. =
3) 5 kg m/s
4) 6 kg m/s
2
1 P
1
P2
P2
8.
Two
bodies
of
mass
1
kg
and 4 kg possess
2
mv  m   =
K.E 
equal momentum. The ratio of their
2
2 m
2m
2m
kinetic energies is
1
Note : For a body having same momentum, K.E 
1) 4 : 1
2) 1 : 4
m
For a body having same kinetic energy, P  m .
DAY – 14 (WORKSHEET - 2)
1. What will be the K.E of a body of mass 2 kg
moving with a velocity of 0.1 metre per
second ?
1) 0.1 J
2) 0.01 J
3) 0.001 J
4) 1 J
2. Two bodies of equal masses move with
uniform velocities v and 3v respectively. Find
the ratio of their kinetic energies.
1) 9 :1
2) 2 : 9
3) 1 : 9
4) 1 : 1
Narayana Group of Schools
3) 2 : 1
4) 1 : 2
9) A sphere of 4kg is dropped from a certain
height. After 5sec its K.E. is  g  10m / s 3 
1) 5 J
2) 50 J
3) 5 KJ
4) 50 KJ
10) A body is projected vertically upward from
the ground with 19.6 m/s i when it’s velocity
is 9.8 m/s. It’s height above the ground is
1) 14 m
2) 14.6 m
3) 18 m
4) None of these
11) A river is flowing with spend 4 m/s of the KE
45
VII - Class _ Physics
Bridge Programme
of cubic meter water is
Power 
1) 8 J
2) 800 J
energy consumed
time taken
Relationship between the S.I unit and C.G.S unit
of power :
12) A body of 1kg dropped from a height of
1W = 1Js–1 = 107 erg s–1
20m. After 2 seconds its KE is _____
Commercial Unit of Energy : The commercial unit of
1) 192.08 J
2) 140 J
energy is kilowatt hour(kWh).
Electric energy consumed in kWh = power in
3) 200 J
4) 500 J
kW × time in hrs
DAY – 15 (SYNOPSIS)
Relation between Kilowatt hour and Joule:
Law of conservation of energy : According to this
1 kWh = 1 kilowatt × 1 hour = 1000 watt × 1
law “Energy can neither be created nor be
hour = 1000 watt × 3600 second
destroyed, but can be changed from one form to
= 36 × 105 watt second or 1 kWh = 36 × 105
another form”.
J.
Ex :When a body falls from a certain height, its P.E
DAY – 15 (WORKSHEET - 3)
gradually changes into kinetic energy but the
total sum of both the energies remains the same. 1. A ball is thrown upwards from a point ‘A’. It
Note :
reaches up to the highest point ‘B’ and
returns then
i) For a freely falling body, potential energy
1) K.E at ‘A’ = K.E at ‘B’
changes into kinetic energy.
Hence, Loss of P.E = Gain of K.E
2) P.E at ‘A’ = P.E at ‘B’
ii) For a body projected vertically upwards,
3) P.E at ‘A’ = K.E at ‘B’
kinetic energy changes into potential
4) P.E at ‘B’ = K.E at ‘A’
energy. Hence, Loss of K.E = Gain of
2. A stone of mass ‘m’ is thrown vertically
P.E.
upwards with a velocity v. The K.E at the
Power : Power is defined as the rate of doing work.
highest point is
3) 600 J
i.e., P =
4) 8000 J
Workdone W

Time taken
t
Power in terms of force (F) and velocity (v) :
P=
Workdone W F  S
S


 F  = F × v
Time taken
t
t
t
 Power = Force × velocity..
Unit of Power:
2) zero
4) 2 mgh
3. A stone is thrown vertically upward. It comes
to rest momentarily at the highest point.
What happens to its kinetic energy ?
1) It converts into electrical energy
S.I unit of power
or watt(W)
1
mv2
2
1
2
3) 2  mv 
2

1)
Workdone
joule


= Js–1
time taken second
 1 watt = 1Js–1
C.G.S unit of power
Work done
erg


=
time taken second
erg s–1.
Note : In case of power of an electric bulb we use,
Narayana Group of Schools
2) It converts into potential energy
3) It converts into chemical energy
4) It is completely destroyed
4. The potential energy of a freely falling object
decreases continuously. What happens to the
loss of potential energy ?
1) It is continuously converted into sound
energy
46
VII - Class _ Physics
Bridge Programme
2) It is continuously converted into kinetic
energy
3) It is continuously converted into magnetic
energy
4) none of these
5. A body of mass 2kg moving up has potential
energy 400J and kinetic energy 580J at a
point ‘P’ in its path. The maximum height
reached by the body is (g = 10ms–2)
1) 49m
2) 98m
3) 196m
4) 392m
12. A water pump lift is 18  103 liters of water
to height 30m from depth in one hour the
power of pump is
1) 1.5 KW
2) 1.8 KW
3) 5 KW
4) 2 KW
6. A body is moving horizontally at a height of
10m has its P.E equal to K.E. Then velocity
of that body is (g = 9.8 m/s2)
1) 7ms–1
2) 14ms–1
3) 3.5ms–1
4) 2.8ms–1
7. A machine does 1920 J of work in 240 s, then
the power of the machine is
1) 8W
2) 10W
3) 20 W
4) 30 W
8. A person weighing 50 kg runs up a hill raising
himself vertically 10m in 20s, then the power
of the person is (Given g = 9.8ms–2)
1) 250W
2) 245W
3) 255W
4) 260W
9. A rickshaw puller pulls the rickshaw by
applying a force of 100N. If the rickshaw
moves with constant velocity of 9 kmh–1, then
the power of the rickshaw puller is
1) 250W
2) 245W
3) 255W
4) 260W
10. A mactin gun fires 360 bullets per minute. If
the mass of each bullet is 10gm & moves
with 400 m/s, power of the gun is
1) 4.8 KW
2) 4 KW
3) 3.6 KW
4) 5.4 KW
11. A motor can hoist 9000 kg of coal per hour
from a mine of 120 m deep the power of
motor is
1) 3000 W
2) 2000 W
3) 4000 W
4) 5000 W
Narayana Group of Schools
47
VII - Class _ Physics
Bridge Programme
cylinder = (V2) = 42 cm3
MEASUREMENT WORKSHEET - 1
1) 4
2) 1
3) 3
4) 4
8) 3
9) 2
10) 2 11) 4
5) 2
6) 1
Densit y
7) 2
MEASUREMENT WORKSHEET - 2
3) 1
4) 3
8) 3
9) 2
10) 4 11) 1
5) 2
6) 2
M
V2 -V1
=
72
72
=
=
= 4.0 g /cm3 4 g /cm3 = 4×1000
42- 24 18
6)
2) 1
solid
4
12) 1 13) 2
14) 2
1) 4
of
7)
7) 1
kg /m3
= 4000 kg/m3
The density of gold is 19.3 times greater than the
density of water.
The density of coper = 8.9 × 103 kg/m3 T he
density of water = 1000 kg /m3
12) 4 13) 4
The relative density of copper =
14) 3 15) 2 16) 4 17) 3
8.9  103
103
 8.9 = 8.9 × 100 = 8.9 × 1 = 8.9
8)
MEASUREMENT WORKSHEET - 3
1) 3
2) 1
3) 4
4) 1
8) 1
9) 2
10) 1 11) 1
5) 1
6) 2
7) 4
12) 3 13) 1
14) 2 15) 3 16) 4 17) 1 18) 1 19) 2
1
mass
1
 8 
kg /m3
volume 64 512
20) 4 21) 2 22) 3 23) 4 24) 1 25) 2
26) 3 27) 3 28) 3 29) 4 30) 3 31) 4
The relative density of wooden cube =
Density of wooden cube
Density of water
MEASUREMENT WORKSHEET - 3-HINTS
1)
Mass of alcohol (M) = 4 kg = 4000 g Volume
of alcohol(V)=5 litres=5000ml=5000 cm3
M
1
kg /m3
1

= 19.53 × 10-7
 512
3
512000
1000 kg /m
4000
3
 density of alcohol, (D)= V  5000 g/cm =
2)
0.8 g/cm3
density in kg/m3 = 1000 × 0.8 = 800 kg/m3
difference of two levels of water in a
measuring cylinder gives the volume of
immersed body. Volume of the body = 70
cm3 - 50 cm3 = 20 cm3
mass of the body = 50 kg
density =
50 kg
50000 g
mass
= 20 cm3 = 20 cm3
volume
-3
3)
4)
5)
= 2500 g cm
= 11.6 g/cm3 = 11.6 ×
3
1000 kg/m = 11600 kg/ m3
The density of wood
= 800 kg/m3
1
Mass of solid (M) = 72 gInitial volume of
water in measuring cylinder (v1) = 24 cm3
Final volume of water + solid in measuring
Narayana Group of Schools
Length of the wooden cube = 4m
Volume of wooden cube = (length)3 = (4m)3 = 64
m3
Densit y
of
wooden
cube
=
9)
Radius of an iron cylinder (r) = 1.4 cm
length (l) = 8 cm
Mass of an iron cylinder (m) = 369.6g
Volume of an iron cylinder (v) =
22
 r 2 l  7 1.4 1.4  8 = 49.28 cm3
Density of an iron cylinder 
mass
369.6

volume 49.28
= 7.5 g/cm3
Density of water = 1 g/cm3
Relative density of an iron cylinder

Density of an iron cylinder 7.5

= 7.5
Density of water
1
10) m = 2 × 0.52 × 1000 = 1040kg
48
VII - Class _ Physics
Bridge Programme
The magnitude of displacement = 10m
KINEMATICS WORKSHEET - 1
10) Whenever you have a problem involving motion
along a circular path remember that after every
complete turn the displacement is zero.
8) 1 9) 1 10) 1 11) 2 12) 3 13) 1
Here the player completes one round in 40
14) 3 15) 2 16) 3 17) 3
seconds. In the given time 2 minutes 20 seconds
(140 seconds), we have time to complete three
rounds (i.e., 3× 40=120 ) the displacement in the
KINEMATICS WORKSHEET - 1- HINTS
rest of the time (140-120) is the resultant

6) Displacement for each girl = PQ Magnitude
displacement. During this time the player can
of the displacement for each girl = PQ = diametre of
complete half a circle or his displacement is equal
circular ice ground = 2 × 200 = 400 m
to the diametre of the circle.
Diametre =
2r
For girl B, the magnitude of displacement is equal
to the actual length of path skated.
KINEMATICS WORKSHEET - 2
7) Displacement is the diametre of the circle along
1) 4 2) 2 3) 1 4) 4 5) 3 6) 3 7) 2
XOY.
diameter = 2 × radius = 2 × r = 2r
8) 3 9) 2 10) 2 11) 3 12) 1 13) 2
displacement = 2r
14) 1
8) The distance covered by the cyclist is equal to
the half the circumference.
 Total circumference of the circle = 2 r
KINEMATICS WORKSHEET - 2 - HINTS
1) 4
2) 4
3) 2
4) 1
5) 4
6) 3
Half the circumference of the circle 
7) 3
2 r
 r
2
1)
 the distance covered by the cyclist= r
9)
Distance covered by a horse = 1200 m
Time taken = 3 min and 20s = 180s+20s = 200
s
The speed of the horse =
North
2)
B
6 m/s
Speed
1200 m
6m
West
time =
East
O
8m
A
3)
South
Let the initial position of the man is at’O’. Let 4)
‘A’ be the where he stops on the East direction.
Let ‘B’ be the where he stops on the north
direction.
The shortest distance from ‘O’ to ‘B’ is OB.
 OA = 8 m ; AB = 6m
From the pythogorus theorem, in a  OAB,
(diagonal)2 = (side)2 + (side)2
(OB)2 = (OA)2 + (AB)2 = (8)2 + (6)2 = 64 + 36
= 100
OB = 10m
Narayana Group of Schools
= 15 m/s
dis tan ce 1200
time = 200 =
distance = 1.2 km =
dis tan ce 1200
time = 15 = 80 s
Speed
= 20 m/s; time = 25 minutes = 1500
s
distance = time × speed = 1500 × 20 = 30000
m = 30 km
Let ‘d’ be the total distance covered by the
object.
The body covers first one third distance is
speed is 1 m/s
d
3
and
The body covers second one third distance is
and speed is 2 m/s
The body covers last one third distance is
speed is 3 m/s
 Total time taken for these distances
d
3
d
3
and
49
VII - Class _ Physics
Bridge Programme
=
 d /3    d /3    d /3 
1
2
3
=
8) 1
d d d
  =
3 6 9
=
5)
6)
d
11d
18
=
10) 2 11) 2
12) 1 13) 3
14) 4 15) 3 16) 2
6d  3d  2d 11d
=
18
18

Average speed 
9) 3
KINEMATICS WORKSHEET - 3 - HINTS
1.
Change in velocity= (72 – 18) km/h = 54 km/h
Total dis tance travelled
Total time taken
 Change in velocity in m/s  54 
18d
= 1.64 m/s
11d
Here displacement = 0
m/s
Now
so, velocity = 0
in
km/h 2
change in velocity  54km/h
30
time
h
3600
=
net displacement
total time
Average Velocity =
acceleration
5
m/s = 15
18
= 54 × 120 km/h2 = 6480 km/h2
2

7)
2
1
4 3
25 5


m/s = m/s
3
10  5
15 15
time
Let the total distance travelled be (x + x) = 2 x
km
Time taken, t = t1 + t2 =
Average velocity =

8)
Acceleration in m/s2 = change in velocity
x
x
90 x
9x



50 40 2000 200

2.
2x
Total dis tance

9x
time
200
1
 2 r   r  3.14 100m = 314m
2
speed
t
3.
=
min.
Total displacement = diameter of a circle = 2 r =
2 × 100 = 200 m
4.
Average
velocity
Total displacement
200 m

= 200
Time take
1 min ute
metre / minute
3) 4
4) 2
Narayana Group of Schools
5) 3
6) 2
15s = 300m.
Initial velocity (u) = 5m/s;
Final velocity
(v) = ?
Time (t) = 20s;
Acceleration (a) = 0.6 m/s2
Applying the formula
v = u + at ;
v = 5 + 0.6 × 20m/s =
5 + 12 m/s = 17m/s
Initial velocity, u = 0
Final velocity,
v = 36 km/h  36 
5
m/s
18
= 10m/s
time, t = 10 minutes = 10 × 60s = 600s
Now, a 
KINEMATICS WORKSHEET - 3
2) 2
5
m/s = 20m/
18
We know that, V  s  s = V × t = 20m/s ×
dis tance travelled
314m

 314 metre/
time taken
1minute
1) 1
Velocity = 72 km/h  72 
s
Distance travelled = half of the perimeter =

(i)
Acceleration will be zero, as body is
moving with uniform velocity
acceleration should be zero.
(ii)
400 x
 44.4km/h
9x
Average
15m/s
= 0.50 m/s2
30s
v  u 10  0  m/s
1


m/s2
t
60
600s
7) 3
50
VII - Class _ Physics
Bridge Programme
Thus acceleration of the bus 
5.
a = 2.5 m/s2;
NEWTONS LAWS OF MOTION
WORKSHEET - 1
1
m/s2
60
s = ?, u = 0, t = 4s
1
2
2
s  ut  at  0 
1
× 2.5 × 4m × 4m
2
1) 2
2) 3
3) 2
4) 1
8) 3
9) 2
10) 1 11) 2
5) 3
6) 2
7) 4
NEWTONS LAWS OF MOTION
WORKSHEET - 1 - HINTS
7. Inertia is that property of a body by virtue
s = 200, u = 0,
t = 10s
of which the body is
1) Unable to change by itself its state of rest.
1 2
we know that, s  ut  at ;
2 0 0
2
2) Unable to change by itself its state of
uniform motion.
1
 0  × a × 10s × 10s
3) Unable to change by itself its direction of
2
motion.
200
8. An athlete runs some distance before taking a
200 = 50a  a 
= 4m/s2
50
long jump, because, it gives himself large amount
Initial speed, u = 10m/s;
Final
of inertia.
speed, v = 0
9. A rider on a horse back falls forward, when the
2
Acceleration, a = –0.5 m/s ;
Distance
horse suddenly stops. This is due to inertia of the
covered (s) = ?
rider.
2
2
From 3rd equation of motion;
v – u = 10. A passenger sitting in a bus gets a backward jerk
2
2
2as ;
2as = v – u
when the bus starts suddenly due to the inertia of
rest.
0  100
v2  u2

S 
m
NEWTONS LAWS OF MOTION
2   0.5 
2a
WORKSHEET - 1
100  10
m
=
100m

1) 1 2) 2 3) 3 4) 4 5) 1 6) 3 7) 3
1 25
 
×16m = 20m
2 10
6.
7.
25
8.
u = 30m/s, v = 0, s = 5m, –a = ?;
know that v2 – u2 = 2as
 2as = v2 – u2 
a
9.
w
e
v2  u2
a
m/s2
2s
02  900
m/s2;
10
 –a = 90m/s2
a = –90 m/s2
8) 1
1.
2.
u = 0, a = 4m/s2 ; distance travel in 5th second?
3.
S5th  0 
4
[2 × 5 –1] = 2 [9] m = 18m
2
4.
Snth  u 
Narayana Group of Schools
10) 1 11) 3
12) 1
NEWTONS LAWS OF MOTION
WORKSHEET - 1 - HINTS
m = 200 g = 0.2 kg
v = 5 m/s
 p
= m × v = 0.2 × 5 kg m/s = 1 kg m/s
m = 25 kg, p = 125 kg m/s, v = ? we know
that, p = m v
 v
a
[2n – 1] ;
2
we know that
9) 1
p 125

m/s  5 m/s
m 25
m = 100 g = 0.1 kg, v = 25 m/s p = mv = 0.1
× 25 kg m/s = 2.5 kg m/s
A
B
mass = m
mass = m
velocity = v
velocity = 3v
as p = mv
51
VII - Class _ Physics
Bridge Programme
 p  v [as mass is constant] ;
so,
5.
p1
v
P
v
 1  1
 P 1 : P2 = 1 : 3
P2
v2
P2 3v
m = 100 g = 0.1 kg;
s
1.
v = 20 m/s, t = 0.01
v 20
m/s2 = 2000 m/s2
t 0.01
Now a  
2.
 F = m a = 0.1 × 2000 N = 200 N
6.
a1 = 6 m/s2
m1 : m2 = 3 : 4;
But, a 
1
(Since F is same)
m
ma
=

 m1a1  m2 a 2  cons tan t

=
7.
m1 a 2 3 a 2
2
  
m2 a1 4 6  4a2 = 18 m/s  a2
18
m/s2  a2 = 4.5 m/s2
4
F = 200 dyne, m = 10 g, t= 5s
a=
8.
3.
constant.
F 200

 20 cm/s2
m 10
As a body starts from rest, u = 0
4.
From 1st equation of motion
v = u + at  v = at
 v = 20 × 5 cm/s2
= 100 cm/s
F = 10 kg. wt = 10×9.8 N = 98 N;
t=2
s, v = 10 m/s;
m=?
Now a =
v 10

= 5 m/s2;
t 2
 m=
F
=
a
98
kg = 19.6 kg
5
1) 3
2) 2
3) 1
4) 1
8) 4
9) 3
10) 3 11) 3
14) 1 15) 1 16) 3 17) 1
Narayana Group of Schools
5) 2
6) 2
12) 1 13) 2
Therefore weight on moon =
1
th weight on
6
earth
weight of the man on earth surface = 30 N
 weight of the man on moon surface =
5.
NEWTONS LAWS OF MOTION
WORKSHEET - 3
NEWTONS LAWS OF MOTION
WORKSHEET - 3 - HINTS
Impulse = change in momentum.
Given
F = 50 N;
t = 10 s
change in momentum = ?
we know, change in momentum = Impulse = F
× t = 50 × 10 = 500 Ns
m = 100 kg, u = 30 m/s;
v = –20 m/s
impulse = m (v – u) = 100 [–20 – 30] = 100 × (–
50) = –5000 Ns
Therefore magnitude of impulse = 5000 Ns
Here, mass of the man, m = 7 kg
Acceleration due to gravity on moon.
gm =
2
1.7 m/s
Weight of the man on the moon,
W=
?
From relation,
W = mg
Putting values,we get, W = 70 × 1.7 = 119
i.e.
W = 119 N.
The weight of the man on the moon will be 119
N
The mass will remain same (70kg) on earth and
the moon
Since gmoon = 1/6 gearth
6.
7) 4
7.
1
× 30
6
N=5N
In case of a book lying on a table. action and reaction
are equal and opposite and act perpendicular to the
surfaces of contact.
Whenever an object A exerts a force on another
object B, object B will exert a return force back
an object A. The two forces are equal in
magnitude but opposite in direction.
When two bodies of masses m1 and m2 moving
with velocities u1and u2 in the same direction
collide with each other and v1, v2 are their
velocities after collision in the same direction, then
m1u1 + m2u2 = m1v1 + m2v2
52
VII - Class _ Physics
Bridge Programme
8.
Since initialy, the two bodies are moving in
opposite, therfore if velocity of one body is u1
and then velocity of other body is –u2. Negative
direction.
so, m1u1 – m2u2 = m1v + m2v = (m1+m2) v 
v
9.
m1u1  m2u2
m1  m2
m=?
we

10.
know
W
=
mgh
m
W
gh
2500
kg = 50 kg
10  5
m = 100 kg,
h = 10 m,
W=?
work done against gravity W = mgh = 100 ×
9.8 × 10 J = 9800 J
car A
car B
mA = 1500 kg
mB = 1000 kg
uA = 25 m/s
uB = 15 m/s
WORK, POWER & ENERGY WORKSHEET - 2
vA = 20 m/s
vB = ?
mAuA + mBuB = mAvA + mBvB
 1500 × 1) 2 2) 3 3) 3 4) 4 5) 1 6) 2 7) 1
25 + 1000 × 15 = 1500 × 20 + 1000 × vB
8) 1 9) 3 10) 2 11) 4 12) 1
 vB = 22.5 m/s
WORK, POWER & ENERGY WORKSHEET - 2 - HINTS
WORK, POWER & ENERGY WORKSHEET - 1
1) 3
2) 2
3) 1
4) 3
8) 1
9) 3
10) 1 11) 3
5) 1
6) 3
1.
1
1
2
mv2 =  2   0.1 = 0.01 J
2
2
7) 1
12) 2 13) 1
3.
WORK, POWER & ENERGY WORKSHEET - 1- HINTS
2.
3.
4.
5.
6.
7.
8.
When a stone tied to a string is whirled in a 4.
circle, the work done on it by the string is zero.
5.
Aman with a box on his head is climbing up a ladder.
The work done by the man on the box is negative
as force and displacement are opposite to each other.
The work is said to be done if a force is
6.
Work done is zero when force acting on the body
is perpendicular to the direction of the
displacement of the body.
F = 10 N,
S=2m
We know that W = F × S = 10 N × 2 m = 20
N m = 20 J
Work done is zero as force produces no motion.
1J=1N×1m
m = 50 kg, h = 72 m, g = 9.8 m/s2
W = F × S = mg × h = 50 × 9.8 × 72 J
 50 
9.
K.E = 1 J,
m = 1 kg,
We know that, K.E =
14) 3
1.
m = 2 kg, v = 0.1 m/s; Now K.E =
98
 72 J = 35280 J
10
W = 2500 J,
h = 5 m,
Narayana Group of Schools
g = 10 m/s2
 v2 = 2  v =
v=?
1
1
mv2  1 =  1  v2
2
2
2  v = 1.4 m/s
m = 1kg,
g = 10 m/s2, h = 5 m;
 P.E = mgh = 1 × 10 × 5 J = 50 J
P.E = 1 J,
m = 1 kg ;
=10 m/s2,
h=?
we know that P.E = mgh;
h
g
=
P.E
1

J = 0.1 m
mg 1  10
Let mass of heavy body = M;
of heavy body = V
mass of light body = m ;
of light body = v
Then for equal kinetic
velocity
velocity
energy
1
1
MV 2  mv2
2
2
V

v
Ratio
m
M
-------- (1)
of
t heir
P1
MV M V M




P2
mv
m v
m
moment um
m

M
M
m
Since M > m,
P1 > P 2
The heavier body has greater momentum.
or
53
VII - Class _ Physics
Bridge Programme
instant.
i.e., mgh = 400J + 580J = 980J  2 × 10 × h =
p2
We know, E =
 p2  2mE
2m
 p2  2mE  p  m E is same 
Thus, the heavier body has greater momentum.
7.
6.
1
kg ; K.E = E = 20 J,
m = 100 g =
10
P=?
we know that
2
8.
980  h 
1
 20 =
10
m1 = 1 kg,
P =
=
 v2 = 196  v  196 = 14m/s
P1 = P 2
P12
KE1
2m1
m
4


 2   4 :1
2
KE2
P2
m1 1
 P1  P2 
2m2
7.
3) 2
4) 2
8) 2
9) 1
10) 1 11) 1
5) 1
6) 2
Given W = 1920 J;
?
8.
7) 1
12) 1
WORK, POWER & ENERGY WORKSHEET - 3 - HINTS
2.
3.
4.
5.
A ball is thrown upwards from a point ‘A’. It
reaches up to the highest point ‘B’ and returns 9.
then P.E at ‘B’ = K.E at ‘A’
A stone is thrown vertically upwards with a
velocity v. The K.E at the highest point is zero.
As K.E 
1
mv 2 at the highest point, v = 0
2
 K.E 
1
m × 02 = 0 (zero)
2
W 1920J

 8Js 1 = 8
t
240s
W
Given, m = 50kg, h = 10m,
t = 20s,
g = 9.8ms–2,
we know that
P
1.
t = 240s, P =
we know that P 
WORK, POWER & ENERGY WORKSHEET - 3
2) 2
1
mv2  v2 = 2 × 10 × g  v2
2
= 2 × 10 × 9.8 [g = 9.8m/s2]
4 = 2 kg m/s
1) 4
Given that P.E of the body = K.E of the body.
 mg × 10 
2mE
m2 = 4 kg
980
 49m
20
P=?
W mgh 50  9.8  10
4900J



=
t
t
20
20s
245J s–1 = 245 W
Here,
F = 100N
v = 9 km h–1  9 
5
ms 1 = 2.5ms–1, P = ?
18
–1
–1
 P = Fv = 100N × 2.5ms = 250Js = 250W
A stone is thrown vertically upward. It comes to
rest momentarily at the highest point. Then its
kinetic energy is converted into gravitational
potential energy.
When the potential energy of a freely falling object
decreases continuously, then its loss of potential
energy is continuously converted into kinetic
energy.
m = 2kg, g = 10m/s2.
If the maximum height reached by the body is ‘h’.
On reaching it, its P.E = mgh
This should be equal to the total energy at any
Narayana Group of Schools
54
```