{"id":38089,"date":"2022-12-03T11:36:21","date_gmt":"2022-12-03T06:06:21","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=38089"},"modified":"2022-12-03T20:22:35","modified_gmt":"2022-12-03T14:52:35","slug":"ts-inter-1st-year-physics-study-material-chapter-6","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-study-material-chapter-6\/","title":{"rendered":"TS Inter 1st Year Physics Study Material Chapter 6 Work, Energy and Power"},"content":{"rendered":"
Telangana TSBIE\u00a0TS Inter 1st Year Physics Study Material<\/a> 6th Lesson Work, Energy and Power Textbook Questions and Answers.<\/p>\n Very Short Answer Type Questions<\/span><\/p>\n Question 1. Question 2. Question 3. Power : Energy : Question 4. <\/p>\n Question 5. b) Work done by gravitational force is positive.<\/p>\n Question 6. b) Work done by gravitational force when a body is sliding down is positive.<\/p>\n Question 7. b) In pendulum a\u221d – y. So work done by air resistance to bring it to rest is considered as positive.<\/p>\n Question 8. b) Gravitational forces are conservative forces. Work done by conservative force around a closed path is zero.<\/p>\n Question 9. In inelastic collision : Question 10. Question 11. Short Answer Questions<\/span><\/p>\n Question 1. Equation for potential energy : Work done against gravity W = m.g.h. <\/p>\n Question 2. So lighter body (car) will travel longer distance when P, F are same. Question 3. Example:<\/p>\n ii) Non-conservative forces : Question 4. Let two bodies of masses m1<\/sub>, m2<\/sub> are moving with velocities u1<\/sub>, u2<\/sub> along the straight line in same direction collided elastically.<\/p>\n Let their velocities after collision be v1<\/sub> and v2<\/sub>. According to the law of conservation of linear momentum According to law of conservation of kinetic energy Question 5. Let the first body moving with initial velocity ‘u’ collides with the second body at rest. In elastic collision, momentum is conserved. So, conservation of momentum along X-axis yields. From eq. (3) and (4) 2v1<\/sub>v2<\/sub> cos(\u03b81<\/sub> + \u03b82<\/sub>) = 0 <\/p>\n Question 6. Velocity with which it strikes the plate u1<\/sub> = \\(\\sqrt{2gh}\\) The velocity of plate before and after collision is zero i.e., u2<\/sub> = 0, v2<\/sub> = 0 For 3rd rebound it goes to a height For nth rebound height attained Question 7. Explanation : Since the universe may be considered as an isolated system, the total energy of the universe is constant.<\/p>\n Long Answer Questions<\/span><\/p>\n Question 1. Kinetic energy : The kinetic energy of an object is a measure of the work that an object can do by the virtue of its motion.<\/p>\n Kinetic energy can be measured with equation K = \\(\\frac{1}{2}\\)mv\u00b2 Work energy theorem (For variable force): Proof : When force is conservative force F = F(x) i.e., work done by a conservative force is equal to change in kinetic energy of the body. When Force (\\(\\overline{\\mathrm{F}}\\)) and displacement (\\(\\overline{\\mathrm{S}}\\)) are perpendicular work done is zero, i.e., when <\/p>\n Question 2. Collisions are two types :<\/p>\n To show relative velocity of approach before collision is equal to relative velocity of separation after collision.<\/p>\n Let two bodies of masses m1<\/sub>, m2<\/sub> are moving with velocities u1<\/sub>, u2<\/sub> along the same line in same direction collided elastically.<\/p>\n Let their velocities after collision are v1<\/sub> and v2<\/sub>. According to the law of conservation of linear momentum According to law of conservation of kinetic energy i.e., In elastic collisions relative velocity of approach of the two bodies before collision = relative velocity of separation of the two bodies after collision.<\/p>\n Velocities of two bodies after elastic collision: Question 3. Proof : In case of a freely fidling body : Forces due to gravitational field are conservative forces, so total mechanical energy (E = P.E + K.E.) is constant i.e., neither destroyed nor created. The conversion of potential energy to kinetic energy for a ball of mass ra dropped from a height H<\/p>\n 1. At point H : Velocity of body v = 0 2. At point 0 : Proof: 3. At any point h: Conditions to apply law of conservation of energy:<\/p>\n When the above two conditions are satisfied then total mechanical energy of a system will remain constant.<\/p>\n Problems<\/span><\/p>\n Question 1. Question 2. <\/p>\n Question 3. Question 4. Question 5. Question 6. Question 7. <\/p>\n Question 8. Amount of work done by the force to displace the particle from x = -a to x = +2a is, Question 9. Velocity of approach, Question 10. Question 11. <\/p>\n Question 12. Telangana TSBIE\u00a0TS Inter 1st Year Physics Study Material 6th Lesson Work, Energy and Power Textbook Questions and Answers. TS Inter 1st Year Physics Study Material 6th Lesson Work, Energy and Power Very Short Answer Type Questions Question 1. If a bomb at rest explodes into two pieces, the pieces must travel in opposite directions. Explain. … Read more<\/a><\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[27],"tags":[],"yoast_head":"\nTS Inter 1st Year Physics Study Material 6th Lesson Work, Energy and Power<\/h2>\n
\nIf a bomb at rest explodes into two pieces, the pieces must travel in opposite directions. Explain.
\nAnswer:
\nExplosion is due to internal forces. In law of conservation of linear momentum internal forces cannot change the momentum of the system. So after explosion m1<\/sub>v1<\/sub> + m2<\/sub>v2<\/sub> = 0 or m1<\/sub>v1<\/sub> = – m2<\/sub>v2<\/sub> \u21d2 they will fly in opposite directions.<\/p>\n
\nState the conditions under which a force does no work.
\nAnswer:<\/p>\n\n
\n\u2235 W = \\(\\overline{\\mathrm{F}}.\\overline{\\mathrm{S}}\\) = |F| |S| cos \u03b8 when \u03b8 = 90\u00b0 work W = 0<\/li>\n
\nDefine Work, Power and Energy. State their S.I. units.
\nAnswer:
\nWork :
\nThe product of force and displacement along the direction of force is called work.
\nWork done W = \\(\\overline{\\mathrm{F}}.\\overline{\\mathrm{S}}\\)
\n= \\(|\\overline{\\mathrm{F}}||\\overline{\\mathrm{S}}|\\) cos \u03b8
\nS.I. unit of work is Joule.
\nDimensional formula : ML\u00b2T-2<\/sup>.<\/p>\n
\nThe rate of doing work is called power.
\n
\nS.I. unit: Watt;
\nD.F. : ML\u00b2T-3<\/sup><\/p>\n
\nIt is the capacity or ability of the body to do work. By spending energy we can do work or by doing work energy contentment of the body will increase.
\nS.I. unit: Joule ; D.F. : MML\u00b2T-2<\/sup><\/p>\n
\nState the relation between the kinetic energy and momentum of a body.
\nAnswer:
\nKinetic energy K.E = \\(\\frac{1}{2}\\)mv\u00b2 ;
\nmomentum \\(\\overline{\\mathrm{p}}\\) = mv
\nRelation between P and KE is
\nK.E = p\u00b2\/2m. \u21d2 P = \\(\\sqrt{K.E.2m}\\)<\/p>\n
\nState the sign of work done by a force in the following.
\na) Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
\nb) Work done by gravitational force in the above case.
\nAnswer:
\na) When a bucket is lifted out of well work is done against gravity so work done is negative.<\/p>\n
\nState the sign of work done by a force in the following.
\na) work done by friction on a body sliding down an inclined plane.
\nb) work done gravitational force in the above case.
\nAnswer:
\na) Work done by friction while sliding down is negative. Because it opposes downward motion of the body.<\/p>\n
\nState the sign of work done by a force in the following.
\na) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity.
\nb) work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
\nAnswer:
\na) Work done against the direction of motion of a body moving on a horizontal plane is negative.<\/p>\n
\nState if each of the following statements is true or false. Give reasons for your answer.
\na) Total energy of a system is always conserved, no matter what internal and external forces on the body are present
\nb) The work done by earth’s gravitational force in keeping the moon in its orbit for its one revolution is zero.
\nAnswer:
\na) Law of conservation of energy states that energy can be neither created nor destroyed. This rule is applicable to internal forces and also for external forces when they are conservative forces.<\/p>\n
\nWhich physical quantity remains constant (i) in an elastic collision (ii) in an inelastic collision?
\nAnswer:
\nIn elastic collision :
\nP and K.E. are conserved, (remains constant)<\/p>\n
\nonly momentum is conserved, (remains constant)<\/p>\n
\nA body freely falling from a certain height ‘h’, after striking a smooth floor rebounds and h rises to a height h\/2. What is the coefficient of restitution between the floor and the body?
\nAnswer:
\nGiven that, h1<\/sub> = h and h2<\/sub> = \\(\\frac{h}{2}\\)
\nWe know that coefficient of restitution.
\n<\/p>\n
\nWhat is the total displacement of freely falling body, after successive rebounds from the same place of ground, before it comes to stop? Assume that V is the coefficient of restitution between the body and the ground.
\nAnswer:
\nTotal displacement of a freely falling body after successive rebounds from the same place of ground, before it comes to stop is equal to height (h) from which the body is dropped.<\/p>\n
\nWhat is potential energy? Derive an expression for the gravitational potential energy.
\nAnswer:
\nPotential energy :
\nIt is the energy possessed by a body by the virtue of its position.
\nEx: Energy stored in water a over head tank, wound spring.
\n<\/p>\n
\nLet a body of mass m is lifted through a height ‘h’ above the ground. Where ground is taken as refe-rence. In this process we are doing some work.<\/p>\n
\ni. e., Force \u00d7 displacement along the direction of force applied. This work done is stored in the body in the form of potential energy. Because work and energy can be interchanged.
\n\u2234 Potential Energy P.E. = mgh.<\/p>\n
\nA lorry and a car moving with the same momentum are brought to rest by the application of brakes, which provide equal retarding forces. Which of them will come to rest in shorter time? Which will come to rest in less distance?
\nAnswer:
\nMomentum (\\(\\overline{\\mathrm{P}}\\) = mv) is same for both lorry and car.
\nWork done to stop a body = Kinetic energy stored
\n\u2234 W = F. S = \\(\\frac{1}{2}\\) mv\u00b2 = K.E. But force applied by brakes is same for lorry and car.
\nRelation between \\(\\overline{\\mathrm{P}}\\) on K.E. is
\n<\/p>\n
\nThen mS = constant.
\n\u2234 So car travels longer distance than lorry before it is stopped.<\/p>\n
\nDistinguish between conservative and non-conservative forces with one example each.
\nAnswer:
\ni) Conservative forces :
\nIf work done by the force around a closed path is zero and it is independent of the path then such forces are called conservative forces.<\/p>\n\n
\nSo gravitational forces are conservative forces.<\/li>\n
\nFor non-conservative forces work done by a force around a closed path is not equal to zero and it is dependent on the path.
\nEx: Work done to move a body against friction. While taking a body between two points say A & B. We have to do work to move the body from A to B and also work is done to move the body from B to A. As result, the work done in moving the body in a closed path is not equals to zero. So frictional forces are non-conservative forces.<\/p>\n
\nShow that in the case of one dimensional elastic collision, the relative velocity of approach of two colliding bodies before collision is equal to the relative velocity of separation after collision.
\nAnswer:
\nTo show relative velocity of approach of two colliding bodies before collision is equal to relative velocity of separation after collision.<\/p>\n
\n<\/p>\n
\nm1<\/sub>u1<\/sub> + m2<\/sub>u2<\/sub> = m1<\/sub>v1<\/sub> + m2<\/sub>v2<\/sub> or m1<\/sub>(u1<\/sub> – v1<\/sub>) = m2<\/sub>(v2<\/sub> – u2<\/sub>) ………… (1)<\/p>\n
\n\\(\\frac{1}{2}\\)m1<\/sub>u\u00b21<\/sub> + \\(\\frac{1}{2}\\)m2<\/sub>u\u00b22 = \\(\\frac{1}{2}\\)m1<\/sub>v\u00b22<\/sub> + \\(\\frac{1}{2}\\)m2<\/sub>v\u00b22<\/sub>
\nm1<\/sub>(u\u00b21<\/sub> – v\u00b21<\/sub>) = m2<\/sub>(v\u00b22<\/sub> – u\u00b22<\/sub>) ………. (2)
\nDividing eqn. (2) by (1)
\n
\nu1<\/sub> + v1<\/sub> = v2<\/sub> + u2<\/sub> \u21d2 u1<\/sub> – u2<\/sub> = v2<\/sub> – v1<\/sub> ……. (3)
\ni.e., relative velocity of approach of the two bodies before collision = relative velocity of separation of the two bodies after collision. So coefficient of restitution is equal to ‘1’.<\/p>\n
\nShow that two equal masses undergo oblique elastic collision will move at right angles after collision, if the second body initially at rest.
\nAnswer:
\nConsider two bodies possess equal mass (m) and they undergo oblique elastic collision.<\/p>\n
\n<\/p>\n
\nmu = mv1<\/sub> cos \u03b81<\/sub> + mv2<\/sub> cos \u03b82<\/sub>.
\n(i.e.) u = v1<\/sub> cos \u03b81<\/sub> + v2<\/sub> cos \u03b82<\/sub> ……. (1)
\nalong Y-axis
\n0 = v1<\/sub> sin \u03b81<\/sub> – v2<\/sub> sin \u03b82<\/sub> ……… (2)
\nsquaring and adding eq. (1) and (2) we get
\nu\u00b2 = v\u00b21<\/sub> + v\u00b22<\/sub> + 2v1<\/sub>v2<\/sub> cos (\u03b81<\/sub> + \u03b82<\/sub>) …. (3)
\nAs the collision is elastic,
\nKinetic Energy (K.E.) is also conserved.<\/p>\n
\nAs it is given that v1<\/sub> \u2260 0 and v2<\/sub> \u2260 0
\n\u2234 cos(\u03b81<\/sub> + \u03b82<\/sub>) = 0 or \u03b81<\/sub> + \u03b82<\/sub> = 90\u00b0.
\nThe two equal masses undergoing oblique elastic collision will move at right angles after collision, if the second body initially at rest.<\/p>\n
\nDerive an expression for the height attained by a freely falling body after ‘n’ number of rebounds from the floor.
\nAnswer:
\nLet a small ball be dropped from a height ‘h’ on a horizontal smooth plate. Let it rebounds to a height ‘h1<\/sub>‘.
\n<\/p>\n
\nVelocity with which it leaves the plate v1<\/sub> = \\(\\sqrt{2gh_1}\\)<\/p>\n
\nCoefficient of restitution,
\n
\nFor 2nd rebound it goes to a height
\nh2<\/sub> = e\u00b2h1<\/sub> = e\u00b2e\u00b2h = e4<\/sup>h<\/p>\n
\nh3<\/sub> = e\u00b2h2<\/sub> = e\u00b2e4<\/sup>h = e6<\/sup>h<\/p>\n
\nhn<\/sub> = e2n<\/sup>h.<\/p>\n
\nExplain the law of conservation of energy.
\nAnswer:
\nLaw of conservation of energy:
\nWien forces doing work on a system are conservative then total energy of the system is constant i.e., energy can neither be created nor destroyed.
\ni.e., Total energy = (K + u) = constant form.<\/p>\n
\nConsider a body undergoes small displacement \u2206x under the action of conservative force F. According to work energy theorem.
\nChange in K.E = work done
\n\u2206K = F(x)\u2206x ………….. (1)
\nbut Potential energy Au = -F(x)\u2206x ………….. (2)
\nfrom (1) and (2) = \u2206K = – \u2206u
\n\u21d2 \u2206(K + u) = 0
\nHence (K + u) = constant
\ni.e., sum of the kinetic energy and potential energy of the body is a constant<\/p>\n
\nDevelop the notions of work and kinetic energy and show that it leads to work- energy theorem. State the conditions under which a force does no work. [AP Mar. I 7, 15, May 1 7; TS Mar. 15]
\nAnswer:
\nWork :
\nThe product of component of force in the direction of displacement and the magnitude of displacement is called work.
\nW = \\(\\overline{\\mathrm{F}}.\\overline{\\mathrm{S}}\\)
\nWhen \\(\\overline{\\mathrm{F}}\\) and \\(\\overline{\\mathrm{S}}\\) are parallel W = \\(|\\overline{\\mathrm{F}}|\\times|\\overline{\\mathrm{S}}|\\)
\nWhen \\(\\overline{\\mathrm{F}}\\) and \\(\\overline{\\mathrm{S}}\\) has some angle 6 between them
\nW = \\(\\overline{\\mathrm{F}}.\\overline{\\mathrm{S}}\\) cos \u03b8
\n<\/p>\n
\nEnergy possessed by a moving body is called kinetic energy (k)<\/p>\n
\nEx : All moving bodies contain kinetic energy.<\/p>\n
\nWork done by a variable force is always equal to the change in kinetic energy of the body.
\nWork done W = \\(\\frac{1}{2}\\)mV\u00b2 – \\(\\frac{1}{2}\\)mV\u00b20<\/sub>? = Kf<\/sub> – Ki<\/sub><\/p>\n
\nKinetic energy of a body K = \\(\\frac{1}{2}\\)mv\u00b2
\nTime rate of change of kinetic energy is
\n<\/p>\n
\n\u2234 On integration over initial position (x1<\/sub>) and final position x2<\/sub>
\n<\/p>\n
\nCondition for Force not to do any work.<\/p>\n
\n\u03b8 = 90\u00b0 then W = \\(\\overline{\\mathrm{F}}.\\overline{\\mathrm{S}}\\) = 0<\/p>\n
\nWhat are collisions? Explain the possible types of collisions? Develop the theory of one dimensional elastic collision. [TS Mar.’ 18; AF Mar. 19. May 14]
\nAnswer:
\nA process in which the motion of a system of particles changes but keeping the total momentum conserved is called collision.<\/p>\n\n
\n<\/p>\n
\nm1<\/sub>u1<\/sub> + m2<\/sub>u2<\/sub> = m1<\/sub>v1<\/sub> + m2<\/sub>v2<\/sub>
\nor m1<\/sub> ( u1<\/sub> – v1<\/sub> ) = m2<\/sub> ( v2<\/sub> – u2<\/sub> ) ……… (1)<\/p>\n
\n<\/p>\n
\n<\/p>\n
\nState and prove law of conservation of energy in case of a freely falling body. [TS Mar. ’19, ’17, ’16, May ’18, ’17, ’16, June ’15; AP Mar. ’18, ’16, ’15, May ’18, ’16, June ’15, May ’13]
\nAnswer:
\nLaw of conservation of energy :
\nEnergy can neither be created nor destroyed. But it can be converted from one form into the another form so that the total energy will remains constant in a closed system.<\/p>\n
\nLet a body of mass is dropped from a height H’ at point A.<\/p>\n
\n<\/p>\n
\n\u21d2 K = 0
\nPotential energy (u) = mgH
\nwhere H=height above the ground
\nT.E = u + K = mgH (1)<\/p>\n
\ni.e., just before touching the ground :
\nA constant force is a special case of specially dependent force F(x) so mechanical energy is conserved.
\nSo energy at H = Energy at 0 = mgH<\/p>\n
\nAt point ‘0’ height h = 0 \u21d2
\n\u21d2 v = \\(\\sqrt{2gH}\\) ; u = 0
\nK0<\/sub> = \\(\\frac{1}{2}\\) mv\u00b2 = \\(\\frac{1}{2}\\) m2gH = mgH
\nTotal energy E = mgH + 0 = mgH ………….. (2)<\/p>\n
\nLet height above ground = h
\nu = mgh, Kh<\/sub> = \\(\\frac{1}{2}\\)mV\u00b2
\nwhere v = \\(\\sqrt{2g(h – x)}\\)
\n\u2234 Velocity of the body when it falls through a height (h – x) is \\(\\sqrt{2g(h – x)}\\)
\n\u2234 Total energy =mgh + \\(\\frac{1}{2}\\)m2g(H – h)
\n\u21d2 E = mgh + mgH – mgh = mgH ………… (3)
\nFrom eq. 1, 2 & 3 total energy at any point is constant.
\nHence, law of conservation of energy is proved.<\/p>\n\n
\nA test tube of mass 10 grams closed with a cork of mass 1 gram contains some ether. When the test tube is heated the cork flies out under the presssure of the ether gas. The test tube is suspended horizontally by a weight less rigid bar of length 5 cm. What is the minimum velocity with which the cork should fly out of the tube, so that test tube describing a full vertical circle about the point O. Neglect the mass of ether.
\nSolution:
\nLength of bar, L = 5 cm, = \\(\\frac{5}{100}\\), g = 10m\/s\u00b2
\nFor the cork not to come out minimum velocity at lowest point is, v = \\(\\sqrt{5gL}\\). At this condition centrifugal and centripetal forces are balanced.
\n<\/p>\n
\nA machine gun fires 360 bullets per minute and each bullet travels with a velocity of 600 ms-1<\/sup>. If the mass of each bullet is 5 gm, find the power of the machine gun? [AP May ’16, ’13, June ’15, Mar. ’14; AP Mar. ’18. ’16; TS May ’18]
\nSolution:
\nNumber of bullets, n = 360
\nTime, t = 1 minute = 60s
\nVelocty of the bullet, v = 600 ms-1<\/sup> ; Mass of each bullet, m = 5gm = 5 \u00d7 10-3<\/sup> kg
\n
\n\u21d2 P = 5400W = 5.4KW<\/p>\n
\nFind the useful power used in pumping 3425 m\u00b3 of water per hour from a well 8 m deep to the surface, supposing 40% of the horse power during pumping is wasted. What is the horse power of the engine?
\nSolution:
\nMass of water pumped, m = 3425 m\u00b3
\n= 3425 \u00d7 10\u00b3 kg.
\nMass of lm\u00b3 water = 1000 kg
\nDepth of well d = 8 m., Power wasted = 40%
\n\u2234 efficiency, \u03b7 = 60%
\ntime, t = 1 hour = 3600 sec.
\n<\/p>\n
\nA pump is required to lift 600 kg of water per minute from a well 25m deep and to eject it with a speed of 50 ms-1<\/sup>. Calculate the power required to perform the above task? (g = 10 m sec-2<\/sup>) [TS Mar. ’19, ’16; AP May 18, Mar. 15, June 15]
\nSolution:
\nMass of water m = 600 kg; depth = h = 25 m
\nSpeed of water v = 25 m\/s; g = 10 m\/s\u00b2, time t = 1 min = 60 sec.
\nPower of motor P = Power to lift water (P1<\/sub>) + Kinetic energy of water (K.E) per second.
\nPower to lift water
\n
\n\u2234 Power of motor P = 2500 + 3125 = 5625 watt 5.625 K.W.<\/p>\n
\nA block of mass 5 kg initially at rest at the origin is acted on by a force along the X-positive direction represented by F=(20 + 5x)N. Calculate the work done by the force during the displacement of the block from x = 0 to x = 4m.
\nSolution:
\nMass of block, m = 5 kg
\nForce acting on the block, F = (20 + 5x) N
\nIf ‘w’ is the total amount of work done to displace the block from x = 0 to x = 4m then,
\n<\/p>\n
\nA block of mass 5 kg is sliding down a smooth inclined plane as shown. The spring arranged near the bottom of the inclined plane has a force constant 600 N\/m. Find the compression in the spring at the moment the velocity of the block is maximum?
\n
\nSolution:
\nMass of the block, m = 5kg
\nForce constant, K = 600 N m-1<\/sup>
\nFrom figure, sin \u03b8 = \\(\\frac{3}{5}\\)
\nForce produced by the motion in the block,
\nF = mg sin0 \u21d2 F = 5 \u00d7 9.8 \u00d7 \\(\\frac{3}{5}\\) = 29.4 N
\nBut force constant K = \\(\\frac{F}{x}\\) x
\n\u2234 x = \\(\\frac{F}{K}=\\frac{29.5}{600}\\) = 0.05m = 5cm<\/p>\n
\nA force F = – \\(\\frac{K}{x^2}\\) (x \u2260 0) acts on a particle along the X-axis. Find the work done by the force in displacing the particle from x = + a to x = + 2a. Take K as a positive constant.
\nSolution:
\nForce acting on the particle, F = –\\(\\frac{K}{x^2}\\)
\nTotal amount of work done to displace the particle from x = + a to x = + 2a is,
\n<\/p>\n
\nA force F acting on a particle varies with the position x as shown in the graph. Find the work done by the force in displacing the particle from x = – a to x = + 2a?
\n
\nSolution:
\nAverage force acting on the particle, F = \\(\\frac{F}{K}\\)<\/p>\n
\n<\/p>\n
\nFrom a height of 20 m above a horizontal floor, a ball is thrown down with initial velocity 20 m\/s. After striking the floor, the ball bounces to the same height from which it was thrown. Find the coefficient of restitution for the collision between the ball and the floor? (g = 10 m\/s\u00b2)
\nSolution:
\nInitial velocity = u\u00b9 = 20 m\/s, h – 20 m,
\ng = 10 m\/s\u00b2<\/p>\n
\nu\u00b2 = u1\u00b2<\/sup> = u1\u00b2<\/sup> +2as = 400+ 2 \u00d7 10 \u00d7 20
\n\u21d2 u\u00b2 = 400 + 400 = 800 \u21d2 u = 20\u221a2
\nHeight of rebounce = h = 20 m.
\n\u2234 Velocity of separation
\n<\/p>\n
\nA ball falls from a height of 10 m on to a hard horizontal floor and repeatedly bounces. If the coefficient of restitution is \\(\\frac{1}{\\sqrt{2}\\), then what is the total distance travelled by the ball before it ceases to rebound?
\nSolution:
\nHeight from which the ball is allowed to fall, h = 10 m
\nCoefficient of restitution between the hard horizontal floor and the ball, e = \\(\\frac{1}{\\sqrt{2}\\)
\n\u2234 Total distance travelled by the ball before it ceases to rebound,
\n<\/p>\n
\nIn a ballistics demonstration, a police officer fires a bullet of mass 50g with speed 200 msr1 on soft plywood of thickness 2 cm. The bullet emerges with only 10% of its initial kinetic energy. What is the emergent speed of the bullet?
\nAnswer:
\nMass of bullet m = 50g = 0.05 kg
\nInitial velocity V0<\/sub> = 200 m\/s
\n<\/p>\n
\nFind the total energy of a body of 5 kg mass, which is at a height of 10 in from the earth and foiling downwards straightly with a velocity of 20 m\/s. (Take the acceleration due to gravity as 10 m\/s\u00b2) [TS May ’16]
\nAnswer:
\nMass m = 5 kg; Height h = 10 m ; g = 10 m\/s\u00b2
\nVelocity v = 20 m\/s.
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