# 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 Radian steradian rad 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 1) Radian 2) Steratian 3) Meter 4) Second 13) The S.I. unit of solid angle 3) Statement I is false ; Statement II is true. 1) Radian 2) Steradian 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) 1012 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) 102 2) 104 3) Hour 4) Day 3) 106 4) 108 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 ii) 10 decades 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 measuring flask, pipette, burette. 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 than lead. 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 2) Measuring flask 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 isdenoted = 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) 105 3) 105 4) 107 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 brings about motion (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 The density of lead = 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. 25 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 sign indicates that it is moving in opposite 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 applied which brings about motion. 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 2mE 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

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