aid of a linked video.] we *Q.6. For the study of any kind of motion, never use Newton’s first law of motion directly. Why should it be studied? aAns: i, iil. Newton’s first law shows an equivalence between the ‘state of rest’ and ‘state of uniform motion along 4 straight line.’ Newton’s first law of motion defines force as a physical quantity that brings about a change in ‘state of rest’ or ‘state of uniform motion along a straight line’ of a body. Newton’s first law of motion defines inertia as a fundamental property of every physical object by which the object resists any change in its state of rest or of uniform motion along a straight line. Due to all these r should be studied. easons, Newton’s first law ~@Shou ym entally gees separated out cronaut neokd Che colorating In inter of his small spaceship: # " stellar space att constant rate of 100 m 8%, What is the neceler of the astronaut the instant after he is outside the spaceship? (Assume that there are nO nearby stars to rt gravitational force on him.) (NCERT) Q.7. exe! Ans: in the vicinity, the exerted on astronaut is Assuming absence of 8 ravitational for aceship. But this force is negligible. aut is out of the spaceship m can be taken Hence, once astron net external force acting on hin as zero. he first law of motion, iv. From the acceleration of astronaut is Zero. *Q.8. In real life, objects mever travel with uniform velocity, even on a horizontal surface, unless something is done? Why is it so? What is to be done? Ans: ig According to Newton’s first law, for a body to achieve uniform velocity, the net force acting onit should be zero. ii, In real life, a body in motion is constantly being acted upon by resistive or opposing force like friction, in the direction opposite to that of the motion. iii, To overcome these opposing forces, an additional external force is required. Thus, the net force is not maintained at zero, making it hard to achieve uniform velocity. iv. For an object to travel with uniform velocity, the surface has to be frictionless ie., the motion has to be free of resistive or opposing forces. eeAns: iil. IV. Vv. the Magnitude and force act on: a drop of rain fallin speed, a cork of mass 10 g floating on water. a kite skilfully hela stationary in the sky. a car moving with a constant velocity of 30 kmh™ ona rough road. a high Speed electron in Space far from all gravitating objects, and free of electric and magnetic fields, (NCERT) irection of the net & down with a constant The drop of rain falls down with a constant speed, hence according to the first law of motion, the net force on the drop of rain is zero, Since the 10 g cork is floating on water, its weight is balanced by the up thrust due to water. Therefore, net force on the cork is zero. As the kite is skilfully held stationary in the sky, in accordance with first law of motion, the net force on the kite is zero. As the car is moving with a constant velocity of 30 km/h on a road, the net force on the car is zero. As the high-speed electron in space is far from all material objects, and free of electric and magnetic fields, it doesn’t accelerate and moves with constant velocity. Hence, net force acting on the electron is zero.a. M1. Explain why a cricketer moves his hands Ans: il. ili. backwards while holding a catch. (NCERT) In the act of catching the ball, by drawing hands backward, cricketer allows longer time for his hands to stop the ball. By Newton’s second law of motion, force applied depends on the rate of change of momentum. Taking longer time to stop the ball ensures smaller rate of change of momentum. Due to this the cricketer can stop the ball by applying smaller amount of force and thereby not hurting his hands.*Q.20. A truck of mass 5 ton is travelling on a horizontal road with 36 km hr’, stops on traveling 1 km after its engine fails suddenly. What fraction of its weight is the frictional force exerted by the road? If we assume that the story repeats for a car of mass 1 ton ice., car moving with same speed stops at similar distance, how much will the fraction be? Solution: Given: Merck = 5 ton = 5000 kg, Mear = 1 ton = 1000 kg, u = 36 km/hr = 10 m/s, v=0m/s,s=1km=1000m To find: Ratio of force of friction to the weight of vehicle Formulae: i. v =w + 2as ii. F=ma Calculation: From formula (i), 2x ame Seva 2 x auc x 1000 = 0 — 10° 2000€trck = — 100 track = — 0.05 m/s”Negative li velocity is © sign indic: i oorea SB" indicates that velocity i From formula (i), Fears ‘ewsk = Mine Bin ~ 5000 x a 000 x 0.05 x B50 a Weighty, 500010 200 From formula (i), 2x darxs=y 2 aur x 1000 2000. =— 100 Bear =~ 0.05 revs? From formula (ii), B= marta = 5000 x 0.0: Fe io weights, 1000310 2 The frictional force acting on both the truck 1 and the caris of their weight, Q.21. A constant force acting on a body of mass 3 kg changes its speed from 2 m s to 3.5 m/s in 25 5. The direction of motion of the body remains unchanged. What is the magnitude and direction of the force? Solution: (NCERT) Given. u=2ms'm Wes Sans To find: Force (F) Formula; F=ma Calculation: Since, v= w+ at 3.5=2+a%25 _ 35-2 = = 7 0.06ms From formula, F=3x0.06=0.18N Since, the applied forve increases the speed of the body, it acts in the direction of the motion Ans: The applied force is 0.18 N along the direction of motion. Q.22. A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms”. How long does the body take to stop? (NCERT) Solution: Given m=20kg,u=15ms7,y=0, =50N (retarding force) To find: Time (1) Formula: v=utat Calculation: Since, F = ma Chapter From formula, 0=15+(-2:5)xt t=6s Ans: Time taken to stop the body is 6 s. : Laws of Motion +Q.23. A hose pipe used for gardening is ejecting Water horizontally at the rate of 0.5 m/s. Area of the bore of the pipe is 10 cm’. Calculate the force to be applied by the gardener to hold the pipe horizontally stationary. Solution: Let ejecting water horizontally be considered as the action force on the water, then the water exerts a backward force (called recoil force) ‘on the pipe as the reaction force f= 4p _ dmv) _ dm >, | dv OR aGEE a ce a the velocity of ejected water v is constant, pooty at Since, the force is in the direction of velocity (horizontal), direction can be ignored dm FE v where, SS the rate at which mass of water is ejected by the pipe dm _ Vp) _ (Als) _ | al a dt ee a Where, V ~ volume of water ejected ‘A = area of cross section of bore 10 cm? p= density of water = 1 glee = length of the water ejected in time t Apy a == velocity of water ejected =0.5 m/s = 50 cm/s dm 2 dt =(Apy)v =Apv’ 10x 1x50" 7 = 25000 dyne = 0.25 N ‘Ans: The gardener must apply an equal and opposite force of 0.25 N. 44 Inertial and non-inertial frames coe ' Q.24. What docs the term frame of reference mean? ‘Ans: A system of co-ordinate axes with reference to which the position or motion of an object is described is called a frame of reference.(Sit eS SS Ans: frames of reference. *Q.26. Distinguish between inertia 1 and non-inertial body moves _ | The body moves with | The 1 |. constant velocity with variable _(can be zero). velocity. ne . | Newton’s laws are Newton’s laws are 1 | obeyed. _ not obeyed. | . |The body does not The body undergoes Nl. | accelerate. acceleration. Iv. In this frame, force acting on a body is a real force. The acceleration of the frame gives rise to a pseudo force. Example: A rocket in inter-galactic space | just starts its motion (gravity free space | from rest, then between galaxies) with | during the time of all its engine shut. acceleration the car ; will be in a non- inertial frame of reference. zl Example: If a car*Q.28. Are there any situations in which we cannot Ans: 1 iil. apply Newton’s laws of motion? Is there any alternative for it? Limitation: Newton’s laws of motion cannot be applied for objects moving in non-inertial (accelerated) frame of reference. Alternative solution: For non-inertial (accelerated) frame of reference, pseudo force needs to be considered along with all the other forces. Limitation: Newton’s laws of motion cannot be applied for objects moving with speeds comparable to that of light. Alternative solution: Einstein’s special theory of relativity has to be used. Limitation: Newton’s laws of motion cannot be applied for studying the behaviour and interactions of objects having atomic or molecular sizes. Alternative solution: Quantum mechanics has to be used. :*Q.32. You are sitting next to your friend on Ans: ii. lil, iv. ground. Is there any gravitational force of attraction between you two? If so, why are you not coming together naturally? Is any force other than the gravitational force of the earth coming in picture? Yes, there exists a gravitational force between me and my friend sitting beside each other. The gravitational force between any two objects is given by, F= Goa Where, G = universal gravitational constant, m, and m) = mass of the two objects, r = distance between centres of the two objects Thus, me and my friend attract each other. But due to our small masses, we exert a force on each other, which is too small as compared to the gravitational force of the earth. Hence, me and my friend don’t move towards each other. Apart from gravitational force of the earth, there is the normal force and frictional force acting on both me and my friend.*Q.34, Among the four fundament: Ans: a ii, iii. Thus, al forces, only one force governs your daily life almost entirely. Justify the Statement by stating that force. Electromagnetic force is the attractive and repulsive force between electrically charged Particles, Since electromagnetic force is much stronger than the gravitational force, it dominates all the phenomena on atomic and molecular scales. Majority of the forces experienced in our daily life like friction, normal reaction, tension in strings, elastic forces, viscosity etc. are electromagnetic in nature. The structure of atoms and molecules, the dynamics of chemical reactions etc. are governed by electromagnetic forces. out of the four fundamental forces, electromagnetic force governs our daily life almost entirely,a 36, Distinguish between contact ang non-contact forces Ans: (PPV Contact forces | Non-contact forces | i. | The forces | The forces experienced by a experienced by ody due to physical | body without any contact are called | physical contact are contact forces alled non-contact | | forces. ii, | Example: Example: Frictional gravitational force, | force, fore exerted electrostatic foree, | due 0 collision, | | magnetostatic force | normal reaction ete, “| | otc | +Q.37. Find the odd man out: (i) Force responsible for a string to become taut on stretching (ii) Weight of an object (iii) The force due to which we can hold an object in hand. Ans: Weight of an object. Reason: Weight of an object (force due to gravity) is a non-contact force while force responsible for a string to become taut (tension force) and force due to which we can hold an ‘object in hand (normal force) are contact forces. +Q.38, Distinguish between real and pseudo force. Ans: A force which | A pseudo force produced due to | one_ which aes | [interaction between due to the [the objects is called | acceleration of the real force. observer's frame of reference. | ii, | Real forees obey | Pseudo forces do Newton's laws _of'| not obey Newton’s | |__| motion. laws of motion. _ iii, | Real forces are one of | Pseudo forces are | the four fundamental | not among any of | forces. the four | fundamental forces. Example: The earth | Example: Bus is | revolves around the sun moving with an. in circular path due to | acceleration (a) on a gravitational force of | straight road in | attraction between the | forward direction, a sun and the earth. person of mass ‘m’ experiences a backward pseudo force of eee ‘ma’. — ae eeii, iit iv, ‘ { I { 1 : posi | equations like x? — Gy pen Ve Solution set of ' ' | \ i { Solution: Chapter 4: Laws of Motion A Enrich Your icnowtedge In mathematics we q, lefine a its square is sen. SURE 8 number to be Feat if o x + 10 = 0 do 4 real number. § complex numbers. wi long with some rea) non-real number the criterion to be eS not satisfy uch numbers hich include | = part. eed not be im, aT It means every laginary, pened during that time? complete numerical Explain H analys sis. The weight recorded by always apparent weight reaction force actin; Weighing machine is and a measure of Weight on the weighing machine is recorded as 50 kg-wt mg = 50 kg-wt inside the lift mg This weight acts on the weighing machine which offers @ reaction R given by the reading of the weighing machine 9 R=45 kg-wt= 2 mg 8 10 The forces acting on person inside lift are as follows: a. Weight mg downward (exerted by the earth) b. Normal reaction (R) upward (exerted by the floor) As, R < mg, the net force is in downward direction and given as, mg-R=ma 9 But R= mg. 9 g——mg=ma mg— 75 ms { Reading between the lines astm ( g=1 mit) ‘Therefore, the elevator must be accelerated downwards with an acceleration of 1 m/s? at that time, s The lift can be accelerating in downward direction: i When it has jusi started moving in downward direction. i, When it is about to stop at a higher floor *Q.40, Distinguish between conservative and non- "Conservative force conservative forces. Non-conservative f force i [If work done by or | If work done by or | | against a force is | against a force is | independent of the | dependent of the | actual path, the force | actual path, the is said to be a force is said to be a | conservative force. non- conservative | | force ii, | During work done by a | During work done | conservative force, the by a non | mechanical energy is | conservative force, | conserved the mechanical | | energy may not be conserved. Work done is | Work done is not | completely | recoverable, | recoverable. | Example: gravitational | Exampl | foree, magnetic force | Frictional force, air | | ete. drag ete. nsevation of mechanical energy The total mechanical energy of a system is said to be conserved if the forees doing work on it are conservative. AK +AV=0 AK = Kinetic energy AV = Potential energyi. XI Sel: Perfect Physics Qui. In the following table, every entry left coh ‘an mateh with any 0) entries on the right side, Pick up all those and write respectively against (1, (i), (iD and (iy). Name of the forge | TYe of he ree | i. Force due 10 | p. | EM force tension ina sti Normal force q._| Reaction force ii | Frictional force |r. | Conservative force iv. | Resistive force | s. | Nonconservative offered by ait or force water for objects | moving through it | Ans: i Force duc to tension in string: Electromagnetic (EM) force, reaction foree, non-conservative Foree |. Normal force: Electromagnetic (EM) force, non-conservative force, Reaction force {. Frictional force: Electromagnetic (EM) force, reaction force, non-conservative force. iy Resistive force offered by air or water for objects moving through it: Electromagnetic (EM) force, non-conservative force. State the formula for calculating work done by a force. Are there any conditions or limitations in using it directly? If so, state those clearly, Is there any mathematical way out for it? Expla Ans: i. Suppose a constant force F acting on a body produces @ displacement s in the body along the positive X-direction. Then the work done by the force is given as, W=F.scos@ Where 0 is the angle between the applied force and displacement. ii, If displacement is in the direction of the force applied, @ = 0° W=F.s Conditionslimitations for application of work formul: The formula for work done is applicable only if both force F and displacement § are constant and finite i.e., it cannot be applied when the force is variable. ‘The formula is not applicable in several real- life situations like lifting an object through | several thousand kilometres since the gravitational force is not constant. It is not icable to viscous forces like fluid iv. v. vie vii. viii. resistance as they depend upon speed and thus are often not constant with time Tine method of integration has to be applied to find the work done by a variable force Integral method to find work done BY variable force: Force Displacement —> Figure (a) Let the force vary non-linearly in magnitude between the points A and B as shown in figure @). In order to calculate the total work done during the displacement from s; to s2, we need to use integration. For integration, we need to divide the displacement into large numbers of infinitesimal (infinitely small) displacements. Let at Pi, the magnitude of force be F = PiPy' Due to this force, the body displaces through infinitesimally small displacement ds, in the direction of force. | It moves from P) to P2 | ae \ ds = BP. | But direction of foree and displacement are same, we have | ds =PyPY. \ dis so small that the force F is practically \ constant, the area of the strip F .d's is the work done dW for this displacement. Hence, small work done between P) to P, is dW and is given by dW = F.ds = PiPy’ x PyPy. = Area of the strip P)P2P2'P)'. ‘The total work done can be found out by dividing the portion AB into small strips like P\P:P2'P\' and taking sum of all the areas of the strips. w eas = Area ABB'A’ constant for the displacement. As the force is | ‘Method of integration is applicable if the exact way of variation in Fand § is known and that function is integrable.variable force: The work done force is: represe portion of fore iS Non-linear variable by th nted ‘by the area earn under the Sin pla ‘raph, ‘ the Able force (traper alison: (figure (b)) ork done W Displacement 5 Figure (b) icthod to find work done by a Suppose a body under the actior shown in figure, Let *O” be the reference point. are magnitudes of initial ei erect displacements of the body at points A and B with respect to ‘0 respectively. Suppose at some instant of time, the body is at Py and the force acting at P, is is moving from A to B n of variable force as F. Now, due to the application of force the body moves ‘ ° SAP ee ds” $+ Body under variable force through infinitesimally small distance ds and reaches P,, RR= & Though the force acting on the body is varying, for infinitesimal small displacement ds it can be assumed to be constant. Hence, small amount of work done dW a by the foree F between the points Py and P) is given by, Chapter 4: Laws of Motion vi. The total work done in moving the body from A to B can be obtained by integrating equation (1) within proper limits of integration. We [aw = fids 1 \ 1 1 \ ' 1 1 If s and s; be the magnitudes of displacements of the body at points A | and B wart, point O then, t we faa : Vii If 0 is the angle between the force acting | on the body along the path and | displacement, then the work done is given | as, 1 W= frdscoso Q) : Vili, Equation (2) represents expression for | work done in moving a body fiom one | point to another point under the action of | variable force. \ Case 1: i If F and ds are along the same direction ' 7 a 1 then, W= | Fdscos0 = [Fds i 5 5 1 [is cos0°=1] | Cases 2: 4 If F and & are perpendicular to each | other then, ' 1 1 i w- frasxo-o [> cos 90°= 0) Q.43.Power is rate of doing work or the rate at which energy is supplied to the system. A constant force F is applied to a body of mass m. Power delivered by the force at time t from the start is proportional to 2 (A) t (B) ¢ © ve () Derive the expression for power in terms of F,mandt. Ans: @ t i, A constant force F is applied to a body of mass (rn) initially at rest (a ~ 0). ii, We have, v=utat 2 v=O+at v=at+Q.46. a we mass 15 ton is running at 72 kmph a ‘aight horizontal road. On turnii th engine off, it stops in 20 seconds. Prnife running at the same speed, on the eanie al the driver observes. an accident 50 m in front of him. He immediately applies the brakes and just manages to stop the car at the accident spot. Calculate the braking force. Solution: i Given: m= 1.5 ton= 1500kg, cae : w= 72 kmph = 72. = mvs = 20m s! : (on turning engine off), v=0,t=20s,s=50m To find: Braking force (F) Formula: 1 v=utat ii, vw? = 2as iii,” F=ma Calculation: On turning the engine off, From formula (i), a 0 20 inst This is frictional retardation (negative acceleration). After seeing the accident, From formula (ii), aie Org =4ms? 2(50) This retardation is the combined effect of braking and friction, braking retardation =4-1=3 ms From formula (iii), the braking force, F = 1500 x3 = 4500 N Ans: The braking force is 4500 N.+Q.48. Over a given region, a force (in newton) varies as F = 3x" — 2x + 1. In this region, an object is displaced from x; = 20 cm to x2 = 40 cm by the given force. Calculate the amount of work done. Solution: Given: F = 3x’ — 2x +1, x1 =20 cem=02 m, X2= 40 cm= 0.4 m. To find: Work done (W) Bees Formula: W= JFds A Calculation: From formula, 2 o w= ese =e Xe aoe [0.43 — 0.4? + 0.4] — [0.23 — 0.27 + 0.2] = 0.304 — 0.168 = 0.136 J Ans: The work done is 0.136 J. *Q.49.Variation of a force in a certain region is given by F = 6x" — 4x — 8. It displaces an object from x = 1 m to x = 2 m in this region. Calculate the amount of work done. Solution: ‘(ox —4x-8) = i se 6x'dx— | 4x7dx - x=l xel 6x* 4x =| 8x ere”: =(16-2)—(8-2)—(16-8)=0 Ans: The work done is zero. a = W anS Ne ce ee *Q.51. While decreasing linearly ‘ force displaces an object rom 3 eanunce the ai done by this force during this displacement. Solution: ot 5 For a variable force, work done is giveD ey area under the curve of force v/s displacement graph. From given data, graph can be plotted as follows: Force (F) in N Li, Dissent eet aes © Displacement (s) in m — ee Work done, W = Area of 0 ABCD. 2 | W2=A (A AEB) +A (CG EBCD) 1 = [}xEBxAE] + (DE x EB) = [5x22] +(3x2) =8J Ans: Work done is 8 J. [Note: According to the definition of work done, S.1. unit of work done is joule (J)] ae*Q.52. 40000 litre of oil of density 0.9 g/ce is pumped from an oil tanker ship into a storage tank at 10 m higher level than the ship in half an hour. What should be the power of the pump? Solution: Given: h=10m, p=0.9 g/cc = 900 kg/m’, g=10 m/s’, ¥ = 40000 litre = 40000 10° x 10° m3 =40 m* T =30 min = 1800s To find: Power(P) Formula: P= ~ = hes Calculation: From formula, = 10 x 900 x 10 x 40 1800 P =2000 W P=2kW Ans: The power of the pump is 2 kW. Work done = force x distance = pressure x area x distance = pressure x volume Work done =hpgv Power, P = Wenpe. ee Re eas*Q.55. Justify the statement, “Work and energy Ans: i il. ill. Ve are the two sides of a coin.” Work and energy both are scalar quantities. Work and Chere, both have the same dimensions i.e., [M Teale] Work and energy both have the same units em SI unit: joule and CGS unit: erg. Energy refers to the total amount of work a body can do. A body capable of doing more work possesses more energy and vice versa. Thus, work and energy are the two sides of the same coin.Tr Chapter 4: Laws of Motion *Q.56. From the terrace of a building of height 1 Ans: iii. iv. Vi. you dropped a ball of mass m. Tt reached the ground with speed y. Is the relation 1 mgh = gm’ applicable exactly? If not, how can you account for the difference? Will the ball bounce to the same height from where it was dropped? Let the ball dropped from the terrace of a building of height h have mass m. During free fall, the ball is acted upon by gravity (accelerating conservative force). While coming down, the work that is done is equal to the decrease in the potential energy. This work done however is not entirely converted into kinetic energy but some part of it is used in overcoming the air resistance (retarding non-conservative force). This part of energy appears in some other forms such as heat, sound, etc. Thus, in this case of an accelerating conservative force along with a retarding non-conservative force, the work-energy theorem is given as, Decrease in the gravitational P.E. = Increase in the kinetic energy + work done against non-conservative forces. Thus, the relation mgh =; my’ is not applicable when non-conservative forces are considered. The part of the energy converted to heat, sound etc also needs to be added to the equation. The ball will not bounce to the same height from where it was dropped due to the loss in kinetic energy during the collision making it an inelastic collision. :he gravitational force iS, 3= 9.8 J work done byt 0x 10° x 9.8% 10 = I We mgh= 1. w,= 9:8 a From formula, wr= 4K. — Ww, =1.25—- 9.8 wr=- 8.55 J unknown force is 8.55 J. ns: Work done by the A body of mass 0.5 kg travels in a straight = ax??, where with velocity Vo ~— he work done by the A Q.58- line a=5 m2 57. What is t net force during its displacement from x = tox=2m? (NCERT) Solution: Given: M=0.5 kg, v= ax’, 172,-1 wherea=5m “S Let v; and v2 be the velocities of the body, when x = 0 and x = 2m respectively. Then, vy, =5x 0°? =0, v2=5 X20 10/2 m Work done (W) To find: Work done = Increase in kinetic energy Formula: I aie la 5 M(¥%2=%1) Calculation: From formula, 1 W = 5 x05 x [(ov2)" -0"} Z A W=50J ns: Work done by the net force on the body is 50 J——~—=—_—~?~Z[_—_—SSoeo"™"S—s—asesettettit™s(C ‘RE *Q.60. State the law of conservation of linear Ans: iii. momentum. It is a consequence of which law? Give an example from our daily life for conservation of momentum. Does it hold good during burst of a cracker? Statement: The total momentum of an isolated system is conserved during any interaction. The law of conservation of linear momentum is a consequence of Newton’s second law of motion. (in combination with Newton’s third law) Example: When a nail is driven into a wall by striking it with a hammer, the hammer is seen to rebound after striking the nail. This is because the hammer imparts a certain amount of momentum to the nail and the nail imparts an equal and opposite amount of momentum to the hammer. Linear momentum conservation during the burst of a cracker: a. The law of conservation of linear momentum holds good during bursting of a cracker. b. When a cracker is at rest before explosion, the linear momentum of the cracker is zero. c. When cracker explodes into number of pieces, scattered in different directions, the vector sum of linear momentum of these pieces is also zero. This is as per the law of conservation of linear momentum.‘Enrich Yi There are no external forces acting in an isolated system. A system refers to a set of ace colliding objects, exploding Shigcis a Interiction refers to collision, explosion ete. | During any interaction among such objects the total linear momentum of the entire system of these particles/objects is constant. Remember. forces during collision or during explosion are internal forces for that entire system. During collision of two particles, the two ! particles exert forces on each other. If these particles are discussed independently, these are | external forcés. However, for the system of the | two particles together, these forces are internal | forces. ; } *Q.61. A lighter object A and a heavier object B are initially at rest. Both are imparted with the same linear momentum. Which will start with greater kinetic energy: A or B or both will start with the same energy?ii. iii. iv. Let m, and m be the masses of light object A and heavy object B and vy, and v2 be their respective velocities. Since both are imparted with the same linear momentum, Mm) Vi = M2 V2 Kinetic energy of the lighter object A 1 =K.E.= 5 m,v; Kinetic energy of the heavier object B 1 2 Kp ay, B ae din 2 KB ae (m,v,) /m, —m,V> a 2%2 eae .-e[°° My V1. = M2 V2] KE As m, < mp, therefore K.E.4 > K.E.p i.e, the lighter body A has more kinetic energy. aeSolved Examples wee eee @ *Q.63.Ten identical masses (m each) are connected one below the other with 10 strings. Holding the topmost string, the system is accelerated upwards with acceleration g/2. What is the tension in the 6th string from the top (Topmost string being the first string)?Solution: Consider the 6” sting from the top. The number of masses below the 6" string is 5, Thus, FBD for the 6 mass is given in figure (b). 5 mg Figure (b) The force equation for that mass is, 5 ma=T-—S5mg As, a= & 2 Smg T= ome Figure (a) ~ 5 me(51] pene 2 = 7.5mg ‘Ans: Tension in the 6" string is 7.5mg. [Note: The answer given above is modified considering the correct textual concepts. ]consiaerins s a fixed pulley. A massless +Q.64. nen oe string with masses mj and m, > ™ attached to its two ends is Passing over the pulley. Such an arrangement is called an Atwood machine. Calculate accelerations of the masses and force due to the tension along the string assuming axle of the pulley to be frictionless. T T Zt aw mg mg Solution: Method I: As m2 > mj, mass m2 is moving downwards and mass m, is moving upwards. Net downward force = F = (mz) g — (mj) g = (m2 —m)) g the string being inextensible, both the masses travel the same distance in the same time. Thus, their accelerations are equal in epee (one upward, other downward). Let ee I a.Chapter 4: Laws of Motion to tension, upward acceleration T=m(g +a) Using equation (i), we pey 8. Thus, T. T= mj a+ m,) UC Am, Fim, Method 11; m, i mig Z Body A Body B tom the free body equatis quation fe re ee a the first body, From the fre equation fi fee ee body equation for the second meg —T=ma (ii) Adding (i) and (ii), we get, (iii) Solving equations. (i) and (iii) for T, we get, 2m,m, Ee "Q.65. Figure below shows a block of mass 35 kg resting on a table. The table is so rough that it offers a self adjusting resistive force 10% of the weight of the block for its sliding motion along the table. A 20 kg wt load is attached to the block and is passed over a pulley to hang freely on the left side. On the right side there is a 2 kg wt pan attached to the block and hung freely, Weights of 1 kg wt each, can be added to the pan. Minimum how many and maximum how many such weights can be added into the pan so that the block does not slide along T=m,(g-a) the table? @ 35 kg wt bk 20 ke. wt on’rough table Fei load Solution: ii, Frictional (resistive) force £= 10% (weight) iO 35 x10= 3 35410 35 N ff Hoe a 2 rn ne 7 mg = 200 N Figure (a) Figure (b) Consider FBD for 20 kg-wt load. Initially, the block kept on the table is moving towards left, because of the movement of block of mass 20 kg in downward direction, Thus, for block of mass 20 kg, ma=mg-T; a) Consider the forces acting on the block of mass 35 kg in horizontal direction only as shown in figure (b). Thus, the force equation for this block is, mja=T)-T,-f 2) To prevent the block from sliding across the table, mja=ma=0 T)=mg=200N {From (1)] Ti=Th+f [From (2)} Ta +f=200 T:~ 200-35 ~165N Thus, the total force acting on the block from right hand side should be 165 N. Total mass = 16.5 kg Minimum weight to be added = 16.5-2=14.5 kg = 15 weights of 1 kg each Now, considering motion of the block towards right, the force equations for the masses in the pin and the block of mass 35 kg can be determined from FBD shown a eae Scoay(i3Sica lap mm) | me f ma me ) Figure (a) From figure (c) ma=T2-T)-f ---iil) From figure (d), mya = mg —T> iv) To prevent the block of mass 35 kg from sliding across the table, mja=m,a=0Std. XI Sci.: Perfect Physics = = a = nee From equations (iii) and (iv), T> = T; ate f T2 = mg mog = 200 + 35 = 235 N The maximum mass required to stop the sliding = 23.52 =21.5 kg ~ 21 weights of 1 kg each : The minimum 15 weights and maximum 21 weights of 1 kg each are required to stop the block from sliding. EEE NT“Sooty epawaen aucr COMISION: *Q.71. Discuss the following as special cases of elastic collisions and obtain their exact or approximate final velocities in terms of their initial velocities. ib Colliding bodies are identical. ii. A very heavy object collides on a lighter object, initially at rest. iii, A very light object collides on a comparatively much massive object, initially at rest. Ans: The final velocities after a head-on elastic collision is given as, m,-m, 2m, v, = u,| + }+u, 2 m, +m, m, +m, 2m, m,—m, YA hh | SS | hl SS m,+m, m, +m, 1. Colliding bodies are identical If m; = mp, then v; = uw and v2 = uy rs Thus, objects will exchange their velocities after head on elastic collision.same speed while the massive object is unaffected. #Q.72.Are you aware of elasticity of materials? Is there any connection between elasticity of materials and elastic collisions? (Students should answer the question as per their understanding).terms of their masse “~~ : Refer Q.76. (i), (ii) and (iii) my travelling with 2 tationary wooden block of mass m2 and gets embedded into it. Determine the expression for loss 10 the kinetic energy of the system. Is this violating the principle of conservation of energy? If not, how can you account for this loss? A bullet of mass *Q.78 velocity u strikes aS Ans: A bullet of mass m; travelling with a velocity u, striking a stationary wooden block of mass im and getting embedded into it is a case of perfectly inelastic collision. In a perfectly inelastic collision, although there is a loss in kinetic energy, the principle of conservation of energy is not violated as the total energy of the system is conserved. Loss in the kinetic energy during a perfectly inelastic head on collision: Refer 0.76*Q.82. In the following table, every item on the left side can match with any number of items Elastic | collision Inelastic collision ‘| Perfectly inelastic collision Head collision on the right-hand side. Select all those. | agas. ‘A ball hit by a bat. Molecular collisions responsible, for pressure exerted by A stationary marble A is hit by marble B and the marble. B comes to rest. A blob of clay dropped on the ground sticks to the o1 ound.+Q.83.One marble collides head-on with another identical marble at rest. If the collision is partially inelastic, determine the ratio of their final velocities in terms of coefficient of restitution e. Solution: According to conservation of momentum, myU, + m2 U2 = M)V; + MeV2 As mj) = mp, we get, uy + up = Vv; + V2 If up = 0, we get, v) + v2 =u, (4) Coefficient of restitution, C= vi u, -U, ‘V2 — V1 = euy (ii) Dividing equation (i) by equation (ii), Vinivammes V,-vV, € Using componendo and dividendo, we get,’ Vv. l+e 2 V, l-eLs *Q.85. A marble of mass 2m travelling directly followed by ano! m with double speed. After ¢ heavier one travels with the ave speed of the two. Calculate the co! restitution. : Solution: Given: m, = 2m, mp) =m, w= 6 cm/s, Uy = 2u, = 12 cm/s, y= ne collision, To find: Coefficient of restitution (¢) Formulae: i. myu; + myU2 = MyV; + M2V2 Pe Vp =a ii. e= 2 u, —U> Calculation: From formula (i), at 6 cm/s is ther marble of mass rage initial efficient of [(2 m) x 6] + (m x 12) = (2 mx 9) + mv2 12% 12=18 + v2 = 6 cm/s From formula (ii), 6-9 -3 e= =—=0. 6-12 -6 is Ans: The coefficient of restitution is 0.5.si un move 8" Shows that the machin : eine dns 5 “HREOC the butt, HSN opposite to ie Calculate the during explosion, supplied Solution: ™m +m: ~0.5 kgm, om, l m= + kg gk 1 m= 1 a= 5 kg Initially, when the ball v=uyat=0+10 (0) v= 100 mV/s =u, =u, (m, + m)v = my, + mayp 's falling freely for 10s, =a 1 9.5 x 100 = = (60) + $Y 5p — 1 50=10+ Ly, dial 2 mi tym 5 (m,+m,) 1 (1) eee ss +(2) 60" +5 x5 * (120) $0.5 (100) = 300 + 2400-2500 K.E.=200J. Kinetic energy supplied is 200 J. \.A shell of mass 3 kg is dropped from some height. After falling freely for 2 seconds, it explodes into two fragments of masses 2 kg and I kg. Kinetic energy provided by the explosion is 300 J. Using g = 10 m/s’, calculate velocities of the fragments. Justify your answer if you have more than one options. Total mass =m, +m; =3 kg Initially, when the shell falls freely for 2 seconds, “a + at=0+10(2)=20 ms! = uy =u, According to momentum, conservation of linear Muy + mu = Mv, + myvp 3 20 =2v, + V2 V2 = 60 ~2v; (i) K.E, provided = Final K. E. ~ Initial K.E. KE Provided =m vis dmv (2vi)+ me 313007] Substituting equation (i, 2v} +(60-2v,)* = 1800 3600 —240y, + 6v? = 1800 vi ~ 40v, +300=0 vi =30 ms" or 10 ms and v2=0 or 40 ms! There are two possible answers since the Positions of two fragments can be different as explained below. Case 1: v) = 30 ms! and v2 = 0 with the lighter fragment 2 above, Case 2: v; = 10 ms” and vy = 40 ms with the lighter fragment 2 below, both moving downwards. +Q.91. Bullets of mass 40 g each, are fired from a machine gun at a rate of 5 per second towards a firmly fixed hard surface of area 10 em’, Each bullet hits normal to the surface at 400 m/s and rebounds in such a way that the coefficient of restitution for the collision between bullet and the surface is 0.75. Calculate average force and average Pressure experienced by the surface due to this firing. Solution: For the collision, u, =400 ms"! e=0.75 For the firmly fixed hard surface, up u,-u, 0-400 v; =-300 m/s. Negative sign indicates that the bullet rebounds in exactly opposite direction Change in momentum of each bullet m (vy; ~ uy) ‘The same momentum is transferred to the surface per collision in opposite direction.after second bout and after third bounce, ye e(eu) eu Butu= /3gh vse) x Jig =~ (0.8) x \In10%5 (b= 5m given) (0.8) «10 ~ 5.12 m/s iii, Impulse given by the ball during third bounce, is, J= Ap = my; — mv; m x (—e'u~ e*u) mxe'ux(e+1) 100 x 107 x (0.8) x 10 x (0.8 + 1) 1.152 Ns iv. Average force exerted in 250 ms, qd 1.152 t 250x107 =~ 4.608 N v. Average pressure for area 0.5 cm? =0.5 x 104 m? = F__ 4.608 NTE P= a ioe = 9.216 x 10° N/m’ Ans: i. Coefficient of restitution is 0.8. Speed after third bounce is 5.12 m/s. Impulse given by ball during third bounce is 1.152 Ns. iv. Average force exerted by ground is 4.608 N. v. Average pressure exerted on given area is 9.216 x 10* Nim’. [Note: Negative sign indicates ball is moving in upward direction] Q.93. A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m s“, what is the recoil speed of the gun? (NCERT) Solutior Given: m, = 0.02 kg, m: = 100 kg, vi=80 ms! To find. Recoil speed (v2) Formula: » mu, + mzu2 = mv; + m2v2, Calculation: Initially gun and shell are at rest. myuy +m2u2=0 Final momentum = m;v; ~mv2 Using formula, 0 =0.02 (80) - 100(v) = 902x80 - 9.016 ms? v= ‘Ans: The recoil speed of the gun is 0.016 ms”.Ans: ii, iv. Impulse is the mantity re momentum, — M*Mtity related tp change in Tega aaa a5 th momentum of ject when, @n objece acted upon by g pe Mhen interval. ah change of the object is 4 given time Need to define impulse: In cases when time fo, ime for force ae Which an appreciable it becomes difficult time independently, Example: Hitting a ball with, a bat, giving a kick to a foot-ball, hammering a nail bouncing a ball froma hard surface, ete, Impulse for a variable force; Consider the collision between a bat and ball. The variation of the force as a function of time 19 ‘Showa Bslowalcihe) ioreetexialla starting from zero, From the graph, it ean be seen that the force is zero before ‘the impact. It rises to a maximum during the impact and decreases to zero after the impact. The shaded area or the area under the curve of the force -time graph gives the product of force against the corresponding time (At) which is the impulse of the force. Area of ABCDE = F. At= impulse of force For a constant force, the area under the curve isa rectangle. In case of a softer tennis ball, the collision time becomes larger and the maximum force becomes less keeping the area under curve of the (F - t) graph same. Chapter 4: Laws of Motion Area of ABCDE = Area of PQRST time (t) Graphical representation of impulse of a force In Chapter 3, you have studied the concept of using area under the curve. Solved Examples J}- rd - Q.95. Two billiard balls each of mass 0.05 kg Moving in opposite directions with speed 6 ms" collide and rebound with the same speed. What is the impulse imparted to each ball due to the other? (NCERT) Solution: Given: m= 0.05 kg, u=6 m/s, v=~6 m/s To find: Impulse (J) Formula: J=m(y—u) Caleulation:Etom formula, J=0.05 (-6~ 6) =-0.6 kg ms Ans: Impulse received by each ball is 0.6 kg ms, 2.96. A bullet of mass 0.1 kg moving horizontally with a velocity of 20 nvs strikes a target and brought to rest in 0.1 s, Find the impulse and average force of impact. Solution: Given; To find: Formulae: i, 1s m=0.1 kg, u=20 mis, Impulse (J), Average force (F) J=my—mu ii. F=m Calculation: From formula (i), J=m(v=u)=0.1 (0-20) =-2Ns From formula (i), 2 =20N OL Ans: Magnitude of impulse is 2 Ns, average force of impact is 20 N.erfect ENysits Q.97. A batsman deflects a ball by a2 an: without changing its i : equal to 54 km/h, What is the im Std. XI Sci gle of 45° ich is imparted to the ball? (Mass of th 01s kg.) (NCERT) ae the position of bat. Let the point B represents The ball strikes the bat wil the path AB and gets deflecte velocity along BC, such that Z AB th velocity v along d with same c= 45° Initial momentum of the ball = mv cos (2) along NB Final momentum of the ball = mv cos (3) along BN Hence, Impulse = change in momentum ed ae! = mv cos| — |—| —mvcos| — 2 2 af 8 = 2mv cos ai (- v = 54 kiv/hr = 15 m/s) =2 x 0.15 x 15 x cos (22.5°) = 4.157 kg m st Thus, impulse imparted to the ball is 4.157kg ms".Solution: Be travelling with a velocity of 500 my, in exactly opposite direction, Caleutaig the impulse received by each of them dur, collision. Assuming that the collision lasts, oe I ms, how much is the average fora, experienced by each molecule? i t Let, m) = mo = 5.35 x 107" kg, m; = my = 4.65 107° kg. j u; = 400 ms! and u; = -500 ms" | taking direction of motion of oxygen moleculg as the positive direction. For an elastic collision, v= (BSB ae osana m, +m, m, +m, v: eee i et ?* (Gn, ¢m,)?"(m,+m, ) * 26 i (535-4. 5) cc x 400 (5.35+4.65)x10 | in (24.6510) (535+ 4.65) x10" = (0.07 x 400) — (0.93 x 500) =— 437 mis (4.65 —5.35) x10 * = eS a0) ¥2 = ((sq5calss) aioe ee = (— 500) 2x5.35x10% . (5.35 +4.65) x10 = 35 + 428 = 463 m/s Jo = Mo (vi — 4) | = 5.35 x 10° x (- 437-400) | =-4.478 x 10° Ns, | Jy = my (v2 — U2) = 4.65 x 10 x (463 + 500) | =+4.478 x 10 Ns, | Hence, the net impulse or net change in momentum is zero. | _ 4g Jo _ 4478x100” dt At 10° =-4.478 x 107° N Fxo =~ Fon = 4.478 x 107° N The average force experienced by the nitrogen molecule and the oxygen molecule are 4.478 x 107° N and 4.478 x 10° N.*Q.106.Why is the moment of a couple independent Ans: ili. of the axis of rotation even if the axis is fixed? , Consider a rectangular sheet free to rotate only about a fixed axis of rotation, perpendicular to the plane. A couple of forces F and eT is acting on the sheet at two different locations. Consider the torque of the couple as two torques due to individual forces causing rotation about the axis of rotation. Case 1: The axis of rotation is between the lines of action of the two forces constituting the couple. Let x and y be the perpendicular distances of the axis of rotation from the forces F and -F respectively.Ans: In this ca, anticlockwise direction of j forces is the same Figure (a) Sr Sction of beth th Side of the a same 5 © aXis Of rotat be the perpendicutay nfo axis Of rotation from the respectively, Case 2: Lin on the same and p » the rotation of +F is While that of top anticlockwise, (from the “Fis clockwise view) individual torques are orci deste oe | pe + -(2) From equation (1) and (2), it is ae that that torque of a coupl leis ind rotation “pendent of the axis of Figure (b) rotating an object or while opening a door or a water tap we apply a force or forces. Under which conditions is this Process easy for us? Why? Define the vector. quantity concerned. How does it differ for a single force and for two opposite forces with different lines of action? Opening a door can be done with ease if the force applied is: a. proportional to the mass of the object b. far away from the axis of rotation and the direction of force is perpendicular to the line joining the axis of rotation with the point of application of force. This is because, the rotational ability of a force depends not only upon the magnitude and direction of force but also on the point where the force acts with respect to the axis of rotation. Rotating an object like a water tap can be done with ease if the two forces are equal in magnitude but opposite in direction are applied along different lines of action. vi, Chapter 4: Laws of Motion The ability of « force to produce rotational ‘motion is measured by its “moment of force’ or “torque” However, a moment of couple or rotational Cffect of a couple is also called torque For differences in the two vector quantities Refer 0.105 ing effect called (442 Mechanical equ ibrium J *Q.108.Expiai ‘Ans: ii, iti, iv. balancing or mechanical equilibrium. Linear velocity of a rotating fan 8 a whole is generally zero. Is it in mechanical equilibrium? Justify your answer. The state in which the momentum of a system is Constant in the ‘absence of an external unbalanced force is called mechanical equilibrium. A’ particle is said to be in mechanical equilibrium, ifno net force is acting upon it In case of a system of bodies to be in mechanical equilibrium, the net force acting on any part of the system should be zero i.e., the velocity or linear momentum of all parts of the system must ‘be constant or zero. There should be no acceleration in any part of the system Mathematically, for a system in mechanical equilibrium, )F =0, In case of rotating fan, if linear velocity is zero, then the linear momentum is zero. That means there is no net force acting on the fen Hence, the fan is in mechanical equilibrium. Q.109. The figure below shows three situations of a Ans: ball at rest under the action of balanced forces. Is the ball in mechanical equilibrium? Explain how the three situations differ. © In all these cases, as the ball is at rest under the action of balanced forces i.e, there is no net force acting on it. Hence, it is in mechanical equilibrium. ; However, potential energy-wise, the three situations show the different states of mechanical equilibrium.Moment about pivot 5xS~0.25x $m 23.7SNm 2 Las 3m ee T Pivot Moment about pivot = $5630.35 4 7 =13.25Nm. Ans: i, The moment of ‘ent of couple at center is 23.75 N-m, ii, The moment of couy 1813.25 Nom, UME #t new position +Q.113. A uniform wooden ph supported symimeteteally ‘by nc KE identical cables; each can sustain a tent up 9 500 XN After tying, the cables = exactly vertical and are separat 7 boy of mass 50 kg, standing eae the plank, is interested in walking om th plank. How far can he walk? (g= 10 ni) i + 300N 500N Solution: Let T; and T; be the tensions along the cables, both acting vertically upwards, Weight of the plank is 30 x 10 = 300 N is acting vertically downwards through the centre, | m from either cable. Weight of the boy, 500 N (50 x 10) is vertically downwards at the point where he is standing. T) + Tz: = 300 + 500 = 800 N ‘Suppose that the boy is able to walk x m towards the right, Obviously, the tension in the right side cable goes on increasing as he walks towards the cable. Moments of 300 N and 500 N forces about left end A are clockwise, while that of T> is anticlockwise. As the cable can sustain 500 N, (T2)max = 100 N ‘Thus, for the equilibrium about A, we can write, 300 x 1 + 500 x (1 +x) = 500 x2 800 +500x = 1000 0.4m Thus, the boy can walk side of the centre. up to 40 em on either ie Chapter 4: Laws of Motion +Q.114, A Indder of negligible mass raving a cross bar Is resting on a frictionless horizontal floor with angle between its legs to be 40°. Each leg is 1 m long. Calculate the force experienced by the cross bar when a person Solution: Tension T' along, the Lot L be the length of 388 bar is horizontal. h leg, which is 1 m. As there is no friction, there is no horizontal Teaction at the floor. Reaction N given by the floor at the base of the ladder will then be only vertical. Thus, along the vertical, two such reactions balance weight W - mg of the person, Sf eye SOHIOR eo 2 a ‘At the left leg, about the upper end, the torque due to N is clockwise and that due to the tension T is anticlockwise. For equilibrium, these two torques should have same magnitude, T +(5) cos 20° 2 - NxLsin20°= Ntan20° =2x 250 x 0.364 182N *Q.115. A 2 m long wooden plank of mass 20 kg is pivoted (supported from below) at 0.5 m from either end. A person of mass 40 kg starts walking from one of these pivots to the farther end. How far can the person walk before the plank topples? Solution: Let the person starts walking from pivot P2 as shown in figure. 4 : 3 2 Wke ke +$—_—_—_—§—? m- :——— ee VOtION all the mass of an atom is concentrated in its nucleus, Song eteenteated in ts ncn cE cl X CMM x= STA cme yp Mm, = me) = 35.5my, X1 = 0, 127A Tofind: Centre of mass (Xc44) Formula: Xx = Subst mghy ™, +m, Caleulation: From formula, X_,, — Mu x0+35.Sm, 1.27 ou, = SSN OTE m,, +35.5m, = 35.Smyx1.27 36.5m,, = 1.235 A The location of centre of mass from the nucleus of hydrogen atom is 1.235 A Ans: *Q.125. Three thin walled uniform hollow spheres of radii 1 em, 2 em and 3 cm are so located that their centres are on the three vertices of an equilateral triangle ABC having each side 10 em. Determine centre of mass of the system. r=lem Solution: Mass of a thin walled uniform hollow sphere is proportional to its surface area, (as density is constant) hence proportional to r°. Thus, if mass of the sphere at A is my = m, then mp = 4m and mc = 9m. By symmetry of the spherical surface, their centres of mass are at their respective centres, i.e., at A, B and C. Let us choose the origin to be at C, where the largest mass 9m is located and the Point B with mass 4m on the positive x-axis. With this, the co-ordinates of C are (0, 0) and that of B are (10, 0). If A of mass m is taken in the first quadrant, its co-ordinates will be [558] rrrStd. xl Sci.: Pertect rnysics —Eeeeeeeeeeeeee Mm, X, + MypXy + Me Xe m, + my + Me A mx 5+ 4 Am x 104 4. 9m x0 _ wos m+ 4m + om 1A) _ My Ya + Mp Yo + Moo m, +My, + Mm, _mx5¥3+mx0+9mx0_ 5 m+4m+9m ai 10V3 i cm =[ 28 { \ | \ | For the ee triangle, AABC, \ AD? + BD’ x | oe = AB’ — BD* | \ = 100225 \ Ess 10 cm t | | BI { 5em D5cm )A fore of 6 N acts on a body of MASS 1 ky during this time, the body initially at rest and The time for attains a velocity of 10 avs Awhich the foree acts on a body is (A) 10 second (B) 8 second (©) 7 second (D) S second Consider following pair of forces of equal magnitude and opposite dir (P) Gravitational: f other by (wo po a dista (Q) Couple of forces used 10 rotate a water it masses separated by witational fore and normal force i) fenced by an object kept on q For which of these pait/pairs the two forces do NOT cancel each other's translational effect? (A) OnlyP (B) Only PandQ (©) OnlyR (D) Only QandR "7. Consider following forces: (ww) Force due to tension along a string, (x) Normal force given by a surface, (y) Force due to air resistance and (2) Buoyant force or up thrust given by a fluid. Which of these are. electromagnetic forces? (A) Only w, y and z (B) Only w, xandy (©) Only x,y andz (D) All four. 8. A bullet of mass 10 g is fired from a gun of mass 1 kg with recoil velocity of gun = 5 m/s. ‘The muzzle velocity will be (A) 30km/min (B) 60 km/min (© 30ms (D) 500 m/s 9. The velocity of rocket with respect to ground is vy; and velocity of gases ejecting from rocket with respect to ground is v2. Then velocity of gases with respect to rocket is given by (A) v2 @) uty © wx Ou {0, At a given instant three point masses m, 2m and 3m are equidistant from each. other. Consider only the gravitational forces between them. Select correct statement/s for this instance only: : (A) Massm experiences maximum force. ( ‘Mass 2m experiences maximum force. ass 3m experiences maximum force. masses experience force of same itude.emia tS LN *16. The rough surface of a horizontal table offers a definite maximum opposing | initi ; g force to initiate motion of a block alo ha 4 ng the table, which is proportional to the resultant normal force given by the table. Forces F; and F, act at the same angle 8 with the horizontal and both are just initiating the sliding motion of the block along the table. Force F, is a pulling force while the force F is a pushing force. F, > F,, because (A) Component of F2 adds up to weight to increase the normal reaction. (B) Component of F; adds up to weight to increase the normal reaction. (C) Component of F, adds up to the : opposing force. (D) Component of F; adds up to the opposing force. OE(C) 49 (D) zero *25. A uniform rod of mass 2m is held horizontal by two sturdy, practically inextensible vertical strings tied at its ends. A boy of mass 3m hangs himself at one third length of the rod. Ratio of the tension in the string close to the boy to that in the other string is (A) 2 (B) 1.5 (C) 4/3 (D) 5/3230! Sts K.E. increases by (A) 50% (B) 300% (C) 100% (D) 400% A mass 2m moving with some speed is directly approaching another mass m moving with double speed. After some time, they collide with coefficient of restitution 0.5. Ratio of their respective speeds after collision is (A) 28°") 372-2 (C) 2 (D) 1/2 In perfectly inelastic collision, which is conserved? (A) _ P.E. only (RR) K F onlv#38. *39% Nae Select. WRONG statement about Contre a mass: (A) Centre of mass of a ‘C’ shaped Dif, rod can never be a point on that rog, If the line of action of a force through the centre of mass, the ™ome, of that force is zero. (C) Centre of mass of our Earth is not atig geometrical centre. (D) While balancing an object on a Pita the line of action of the gr aVitationg| force of the earth passes through the centre of mass of the object. (B) For which of the following objects will the centre of mass NOT be at their geometrica| centre? (1) Anegg (II) acylindrical box full of rice (II) a cubical box containing assorted sweets (A) Only (1) s (B) Only (1) and (II) (C) Only (II) (D) All, (1), (ID and (111).Std. XI Sci.: Perfect Phystes. ‘As ceutre of mass of the full dise 8 at the origin, we can write ain x (X) 1: (Using negative mss): 1 of the centre of ‘M. Mass Let R_ be the ps mass of the uniform tion vector with centre of mass at post 2 i the centre of the disc be out O° from the Eraplete dise, Position veeter Of the centre 0} Fass ofthe remaining dise is then gIvc0 by MR= mr Mm (as if there is a negative Mass, ie,m=m) ‘Ag Gated in the previous method, M = 4m. m=m,R=Qandr=r = 3m +130. A uniform solid sphere of radius R has @ hole of radius R/2 drilled inside it. One end of the hole is at the centre of the sphere while the other is at the boundary. Locate centre of mass of the remaining sphere. Solution: Let the centre of the sphere be origin O. Then, 1 be the position vector of centre of mass of uniform solid sphere and r» be the position vector of centre of mass of the cut-out part of the sphere. Now, mass of the sphere is given as, M= (4:n)p eli) Hence, the mass of the cut part of the sphere will be, “ih _.(e sphere is uniform, p= constant) viz = 0,1 From figure, 1 or of centre of mass of Temaining Position vect | art, | pa MyM xb | fon ASN ee Ee ; egative sign indicates the distance ison lef side of the origin.) ‘tion of centre of mass of remaining sphere 4.14. Centre of gr: Q.131. Define centre of gravity of a body. Under ‘what conditions the centre of gravity and | centre of mass coincide? Ans: 2 Centre of gravity of a body is the point around ‘which the resultant torque due to force of gravity on the body is zero. ii. The centre of mass coincides with centre of gravity when the body is in a uniform gravitational field. *Q.132, Why do we need to know the centre of mass ‘of an object? For which objects, its position may differ from that of the centre of gravity? Ans: i. Centre of mass of an object allows us to apply Newton’s laws of motion to finite objects (objects of measurable size) by considering these objects as point objects. ii. For objects whose size is comparable to that of the Earth (size at least few thousand km), the position of centre of mass will differ than that of centre of gravity. Q.133. Explain how to find the location of centre of = ‘mass or centre of gravity of a laminar object. Ss? i. A laminar object is suspended from a rigi a support at two orientations. ii, Lines are drawn on the object parallel to th plumb line as shown in the figure, | |Using symmetry for ms, m z ms and my, there will be effective mass 3m at the origin (0, 0) Thus, effectively, 3m and 7m are separated by em along X-direction. Y-coordinate is not required. & ‘cading between the lines i \ 1 ' Altemately, for two point masses, the { iH centre of mass divides the distance | ' between them in the inverse ratio of their | ' masses. Hence, | em is divided inthe ratio. ' 1 1 1 =o x 1=0. ' ' 73 = 0.7 em from 3m, ic., i ne from the origin at ms. £ -Q.128. Four uniform solid cubes of edges 10 em, 20 cm, 30 cm and 40 cm are kept on the ground, touching each other in order. ocate centre of mass of their system. ¥: Solution S10 2030 45 60 «80100 The given cubes are arranged as shown in figure. Let one of the comers of smallest cube lie at the origin. ‘As the cubes are uniform, let their centre of masses lie at their respective centres. tm = (5, 5), tm = (20, 10), ro = (45, 15) and tp = (80, 20) Also, masses of the cubes are, my= 2.x p=10°p mp = (20)'p me = (30)'p mp = (40)'p ‘As the cubes are uniform, p is As 1p is same for all of For X ~ co-ordinate of centre of mass of the syste Dm Xem Lo MyXy + MyXy MK m, +m) +m-+m, __ [10°05] [20 p20] [30° p45] [40 ~pn0] 1p 20°xp+30"xp+40"xp _ 10°s(0154 8420+ 27% 45 4.6480) 10"s(1+ 842764) 6500 “100 = 65cm pa. Similarly, ‘YY — co-ordinate of centre of mass of system is, MY, _ MAY, +MgYq + Ye * MY M m, +m, #me+Mp 10° x(1x5 4810427154 64x20) 10'x( 14842764) =17.7 em ‘Ans: Centre of mass of the system is located at point (65 cm, 17.7 em). 40.129. A hole of radius r is cut from a uniform Gise of radius 2r. Centre of the hole is at a distance r from centre of the dise. Locate centre of mass of the remaining part of the disc. Solution: Before cutting the hole, em, of the full dise was at its centre, Let this be our origin O. Centre of mass of the cut portion is at its centre D. Thus, it is at a distance x; = r form the origin. Let C be the centre of mass of the remaining disc, which will be on the extension of the line DO at a distance x» = x from the origin, As the disc is uniform, mass of any of its partis proportional to the area of that part, =3m Thus, if'm is the mass of the cut disc, mass 0 the entire disc must be 4m and mass of thy remaining dise will be 3m. a i+Q.127. A letter ‘E’ is prepared from a uniform cardboard with shape and dimensions as shown in the figure. Locate its centre of mass. Solution: As the sheet is uniform, each square can be taken to be equivalent to mass m concentrated at its respective centre. These masses will then be at the points labelled with numbers 1 to 10, as shown in figure. Let us select the origin to be at the left central mass ms, as shown and all the co-ordinates to be in cm. By symmetry, the centre of mass of mj, My and m3 will be at m (1, 2) having effective mass 3m. Similarly, effective mass 3m due to mg, My and myo will be at mo (1, —2). Again, by symmetry, the centre of mass of these two (3m each) will have co-ordinates (1, 0). Mass mg is also having co-ordinates (1, 0). Thus, the effective mass at (1, 0) is 7m. Pai Ie. em