Preview prep guide to BITSAT 2020 by arihant publications

Preview prep guide to BITSAT 2020 by arihant publications Preview prep guide to BITSAT 2020 by arihant publications Preview prep guide to BITSAT 2020 by arihant publications Preview prep guide to BITSAT 2020 by arihant publications Preview prep guide to BITSAT 2020 by arihant publications Preview prep guide to BITSAT 2020 by arihant publications Preview prep guide to BITSAT 2020 by arihant publications

THE #1 BEST SELLING THE #1 BEST SELLING ARIHANT PUBLICATIONS (INDIA) LIMITED THE #1 BEST SELLING Arihant Publications (India) Ltd All Rights Reserved © Publisher No part of this publication may be re-produced, stored in a retrieval system or distributed in any form or by any means, electronic, mechanical, photocopying, recording, scanning, web or otherwise without the written permission of the publisher Arihant has obtained all the information in this book from the sources believed to be reliable and true However, Arihant or its editors or authors or illustrators don’t take any responsibility for the absolute accuracy of any information published and the damages or loss suffered there upon All disputes subject to Meerut (UP) jurisdiction only Administrative & Production Offices Regd Office ‘Ramchhaya’ 4577/15, Agarwal Road, Darya Ganj, New Delhi -110002 Tele: 011- 47630600, 43518550; Fax: 011- 23280316 Head Office Kalindi, TP Nagar, Meerut (UP) - 250002 Tele: 0121-2401479, 2512970, 4004199; Fax: 0121-2401648 Sales & Support Offices Agra, Ahmedabad, Bengaluru, Bareilly, Chennai, Delhi, Guwahati, Hyderabad, Jaipur, Jhansi, Kolkata, Lucknow, Meerut, Nagpur & Pune ISBN : 978-93-13196-77-8 Published by Arihant Publications (India) Ltd For further information about the books published by Arihant log on to www.arihantbooks.com or email to info@arihantbooks.com /arihantpub /@arihantpub Arihant Publications /arihantpub THE #1 BEST SELLING CONTENTS PHYSICS Units, Measurements and Dimensions 3-14 Scalars and Vectors 15-23 Motion in 1, & Dimensions and Projectile Motion 24-44 Newton's Laws of Motion and Friction 45-62 Circular Motion 63-72 Work, Energy and Power 73-83 Centre of Mass, Momentum and Collision 84-94 Rotational Motion of Rigid Body Gravitation 95-106 107-118 10 Simple Harmonic Motion 119-131 11 Fluid Mechanics 132-146 12 Elasticity 147-155 13 Waves Motion 156-164 14 Sound Wave 165-174 15 Heat, Temperature and Calorimetry 175-184 16 Physics for Gaseous State 185-194 17 Laws of Thermodynamics 195-205 18 Transmission 206-214 19 Ray Optics 215-233 20 Waves Optics 234-242 21 Electric Charge 243-251 22 Gauss's Law and Electric Potential Theory 252-262 23 Electric Capacitor 263-270 24 Current Electricity 271-288 25 Magnetic Field 289-300 THE #1 BEST SELLING 26 Magnetostatics 301-310 27 Electromagnetic Induction (EMI) 311-318 28 Alternating Current and EM Wave 319-330 29 Cathode Rays, Photoelectric Effect of Light and X-Rays 331-340 30 Atomic Structure 341-348 31 Nucleus 349-357 32 Semiconductor Devices and Logic Gates 358-375 33 Universe 376-380 CHEMISTRY Some Basic Concepts of Chemistry 383-394 Atomic Structure 395-408 Nuclear Chemistry 409-416 Chemical Bonding 417-431 Periodic Properties 432-441 States of Matter 442-454 Chemical Thermodynamics 455-466 Chemical and lonic Equilibria 467-480 Chemical Kinetics 481-491 10 Solution 492-501 11 Adsorption and Colloidal System 502-510 12 Redox Reactions 511-517 13 Electro-chemistry 518-529 14 Hydrogen 530-537 15 s-Block Elements 538-549 16 Metallurgy 550-558 17 p-Block Elements - I (Group 13 & 14) 559-567 18 p-Block Elements - II 568-586 19 d - and f-Block Elements 587-597 THE #1 BEST SELLING 20 Coordination Compounds and Organometallics 598-607 21 General Organic Chemistry 608-625 22 Purification and Estimation of Organic Compounds 626-631 23 Hydrocarbons 632-650 24 Halogen Derivatives of Hydrocarbons 651-663 25 Alcohol, Phenol and Ether 664-678 26 Aldehyde and Ketones 679-695 27 Carboxylic Acid and Its Derivatives 696-706 28 Nitrogen Containing Compounds 707-720 29 Polymers, Biomolecules and Chemistry in Action 721-738 30 Qualitative Analysis 739-743 31 Stereochemistry 744-756 MATHEMATICS Complex Number 759-769 Quadratic Equation 770-779 Sequences and Series 780-792 Exponential and Logarithmic Series 793-800 Permutations and Combinations 801-811 Binomial Theorem and Mathematical Induction 812-823 Matrices 824-832 Determinant 833-846 Sets, Relations and Functions 847-860 10 Linear Inequality 861-866 11 Trigonometry 867-897 12 Rectangular Coordinates and Straight Line 898-913 13 The Circle 914-928 14 Conic Sections 929-952 15 Three Dimensional Geometry 953-967 THE #1 BEST SELLING 16 Limits, Continuity and Differentiability 968-986 17 Differential Coefficients 987-998 18 Application of Derivatives 999-1013 19 Indefinite Integral 1014-1024 20 Definite Integral and Its Applications 1025-1039 21 Differential Equations 1040-1053 22 Probability 1054-1069 23 Vector Algebra 1070-1085 24 Statistics 1086-1096 25 Linear Programming 1097-1102 Ÿ Ÿ Ÿ Ÿ Ÿ Ÿ 1105-1143 1147-1191 1195-1251 1-29 1-34 1-31 English Proficiency Logical Reasoning Practice Sets (1-5) Solved Paper 2017 Solved Paper 2018 Solved Paper 2019 THE #1 BEST SELLING SYLLABUS PART I PHYSICS Units & Measurement 1.1 1.2 1.3 1.4 Units (Different systems of units, SI units, fundamental and derived units) Dimensional Analysis Precision and significant figures Fundamental measurements in Physics (Vernier calipers, screw gauge, Physical balance etc.) Kinematics 2.1 2.2 2.3 2.4 2.5 2.6 Properties of vectors Position, velocity and acceleration vectors Motion with constant acceleration Projectile motion Uniform circular motion Relative motion Rotational Motion 6.1 6.2 6.3 6.4 6.5 6.6 Gravitation 7.1 7.2 7.3 Newton’s Laws of Motion 3.1 3.2 3.3 3.4 3.5 Newton’s laws (free body diagram, resolution of forces) Motion on an inclined plane Motion of blocks with pulley systems Circular motion – centripetal force Inertial and non-inertial frames Impulse and Momentum 4.1 4.2 4.3 4.4 4.5 Definition of impulse and momentum Conservation of momentum Collisions Momentum of a system of particles Center of mass Work and Energy 5.1 5.2 5.3 5.4 5.5 Work done by a force Kinetic energy and work-energy theorem Power Conservative forces and potential energy Conservation of mechanical energy Description of rotation (angular displacement, angular velocity and angular acceleration) Rotational motion with constant angular acceleration Moment of inertia, Parallel and perpendicular axes theorems, rotational kinetic energy Torque and angular momentum Conservation of angular momentum Rolling motion Newton’s law of gravitation Gravitational potential energy, Escape velocity Motion of planets – Kepler’s laws, satellite motion Mechanics of Solids and Fluids 8.1 8.2 8.3 8.4 Elasticity Pressure, density and Archimedes’ principle Viscosity and Surface Tension Bernoulli’s theorem Oscillations 9.1 9.2 9.3 Kinematics of simple harmonic motion Spring mass system, simple and compound pendulum Forced & damped oscillations, resonance 10 Waves 10.1 10.2 10.3 10.4 Progressive sinusoidal waves Standing waves in strings and pipes Superposition of waves, beats Doppler Effect 11 Heat and Thermodynamics 11.1 Kinetic theory of gases 11.2 Thermal equilibrium and temperature 80 SELF STUDY GUIDE BITSAT Since, F < f So, both blocks move together The acceleration of system is F 0.5 a= = = = m/s2 1+ 30 f kg For lower block, 1 = N So, the force of friction due to lower block on upper block is N in the opposite direction of displacement of the block ∴ Work done by lower block on upper block is W = − fs = − × = − 1J a a (c)W1 = ∫ F dx = ∫ −k (y $i + x$j )$i dx f = 2a = × 0 a = ∫ −k ( 0$i + x$j )$i dx = zero a a W2 = ∫ F ⋅ dy = ∫ −k (y $i + x$j )$j dy =∫ a 0 a −k (a $i + a$j )$j dy = − ka ∫ dy = − ka Total work done, W = W1 + W2 = − ka = − ka (d) Since, work done by this force depends upon path followed by the particle between two points Hence, the force must be non-conservative ∆U ( −Mgh ) Mgh 10 (a) Here, F = − =− = x x ∆x 11 (c)U = k (x + y + z ) = kr 2 F =− 15 (c) The spring loses elastic potential energy which appears as energy of molecules of acid 16 (c) Initially potential energy = kx 2 2U ⇒ U = kx or 2U = kx ⇒ k = 2 x When it is stretched to nx cm 1 2U PE = kx12 = × × n 2x = n 2U ∴ 2 x ∴ Potential energy stored in the spring = n 2U 17 (d) (a) This statement is correct only when forces are of conservative nature (b) In the absence of non-conservative force, mechanical energy and non-mechanical energy are separately constant (c) The work done by internal forces may not be zero 18 (c) Here, mg(h + x ) = kx 2 2mg(h + x) ⇒ kx = 2mg(h + x ) ⇒ k = x2 19 (e) If particle moves on a curved path, then there are two components of force (i) Tangential Component This component provides tangential acceleration Due to this, speed increases (ii) Radial Force This component provides centripetal acceleration v 90 dU − d = (kr ) dr dr ^ ^ a b − x 12 x At the stable equilibrium, 1/ 12a 6b dU  2a  = ⇒ − 13 + = ⇒ x =   b dx x x ∂U $ ∂U $ 13 (d) F = − i− j = 7$i − 24$j ∂x ∂y Fy ∴ ax = = = 1.4 m/s2 m Fy 24 ay = =− m = 4.8 m/s2 along negative y -axis 12 (d)Q U (x ) = and ∴ 14 (a) F = − v x = a x t = 14 ×2 = 2.8 m/s along negative y -axis v y = 4.8 × = 9.6 m/s v = v x2 + v y2 = 10 m/s ∂U $ ∂U $ i− j = 7$i − 24 $j ∂x ∂y ∴ | F | = ( 7)2 + ( −24)2 = 25 units ° F ^ −2 kr = − kr = − 2k ( i x + j y + k z ) ∴ Ft θ< Fr The resultant force is F = Fr + Ft tan θ = Fr Ft But Ft is not zero ∴ tan θ < ∞ ⇒ θ < 90° When speed increases, p = mv increases F F 20 (d) a = = = F m/s2 m v = u + at v = v + (F × 1) ⇒ v = v + F where,v = final velocity s = ut + at 2 F s =v0 + pi (initial momentum) = mv − v pf (final momentum) = mv = v + F Change in momentum, ∆p = pf − pi = F WORK, ENERGY AND POWER 21 (c) Here, u = 100 m/s,v = 0,s = 1cm = 10−2 m ∴ Tension at A = Weight of half the chain v = u − 2as 2 + Weight of 100 × 100 u = × 105 m/s2 = 2s × × 10−2 ∴ a= ∴ F = ma = 10 × 10−3 × × 105 = 5000 N = 1 mv − mu 2 = − × 10 × 10−3 × 10000 or −F × 10−2 = − × 10 × 10−3 × 104 ∴ F = 5000 N 60 22 (a) m = 60 g = = 0.06 kg 1000 s = 200 mm = 0.2 m u = 600 m/s v =0 Using v = u + 2as B A x l/2 B x/2 l O u=o Tmax v mg …(ii) mv m = 2gl = 2mg l l From Eqs (i), (ii) and (iii), Tmax = mg + 2mg = 3mg Q …(iii) 26 (c) L cosθ L L h 23 (b) From force diagram as shown in T figure, mg − T = ma mg 3mg ∴ T = mg − ma = mg − = 4 QWT = work done by tension s 3mgd mg = T ⋅ s = Ts cos 180° = − m w 24 (c) Mass per unit length, λ = = l lg x  Velocity,v = 2g   orv = gx  2 A 3x   1+  l  x length + F 2 ( 0)2 = ( 600)2 + × a × 0.2 −600 × 600 × 10 a= 0.4 Average retarding force 0.06 × 600 × 600 × 10 F = ma = F = 54 kN Change in momentum when an element dx w gx dx lg dp w dx w F = gx = = x dt lg dt l w wx wx w + + = 2 2l l 25 (c) The tension in string is maximum at the lowest position of the ball mv ∴ Tmax − mg = l mv …(i) ∴ Tmax = mg + l The loss in potential energy = gain in kinetic energy or mgl = mv 2 ∴ v = 2gl Alternate method W = ∆K or − Fs = 81 a = g/4 falls is  dx  Q = gx  dt  W = ∆K or WT + Wg + WF = [Since, change in kinetic energy is zero] Here,WT = work done by tension = Wg = work done by force of gravity = − mgh = − mgL (1 − cos θ ) ∴ WF = − Wg = mgL (1 − cos θ ) 27 (d) Elastic force in string is conservative in nature ∴ W = − ∆U where,W = work done by elastic force of string ∆U = change in elastic potential energy Q W = − (Uf − Ui ) = Ui − Uf 1 or W = kx − k (x + y )2 2 or W = kx − k (x + y + 2xy ) 2 2 = kx − kx − ky − k ( 2xy ) 2 2 = − kxy − ky = ky ( −2x − y ) 2 The work done against elastic force is ky Wexternal = − W = ( 2x + y ) SELF STUDY GUIDE 28 (a) F = ma = BITSAT v − 0  Qa = T   mv T mv at T mv v mv = t= t T T T Instantaneous power = Fv = mav = 29 (c) The final speeds of each ball are same ∴ Loss in PE by ball A = gain in KE by the system 1 or mgr = mv + mv 2 or mgr = mv ∴ v = gr = 9.8 × 0.5 = 4.9  49  =   = 2.21 m/s  10  ( 60 + 20) × (10) × ( 20 × 0.2) 30 (a) Power = = 160 W 20 31 (a) Hight of liquid column, h = 125 mm = 0.125 m Density of blood, ρ = 103 × 103 kg/m Accleraction due to gravity, g = 9.8 m/s2 Volume of blood,V = litre = × 10− m3 Time, t = 1minute = 60 second Work done,W = force × distance = pressure × area × distance = pressure × volume = hρgV W hρgV Power, P = = t t =  d 2x  33 (d )QP = Fv = (ma ) v = m    dt  Since, power is constant  d 2x   dx  ∴  2   =k  dt   dt  or d  dx   dx    = k or   = k1t  dt  dt  dt  dx = k1t = k 2t 1/ or x = k 3t 3/ dt dx Hence, ∝ t 1/ ∝ x 1/ dt or 34 (b) ac = v2 = k 2rt r Q v = krt The tangential acceleration is dv d (krt ) a1 = = = kr dt dt The work done by centripetal force will be zero So, power is delivered to the particle by only tangential force which acts in the same direction of instantaneous velocity Power = Fv ∴ t = mat krt = m(kr )(krt ) = mk 2r 2t 35 (b) Mass per unit length = = 112.76 × 10− HP where, F is the resistive force andv is the initial speed Case II Let the further distance travelled by the bullet before coming to rest iss ∴ −F ( + s ) = Kf − Ki = − mv 02 1 − mv 02( + s ) = − mv 02 ⇒ or (3 + s ) = s or + =1 4 or s = cm M = = kg m–1 L 1.4 m 0.125 × 103 × 103 × 9.8 × × 10− 60 32 (d) According to the work-energy theorem, W = ∆K v  Case I −F × = m   − mv 02   2  dx     dt  0.6 m 82 The mass of 0.6 m of chain = 0.6 × = 12 kg The height of the centre of mass of the hanging part 0.6 + h= = 0.3 m Hence, work done in pulling the chain on the table = work done against gravity the force of gravity i.e W = mgh = 1.2 × 10 × 0.3 = 3.6 J 36 (c) Pinst = F ⋅ v = (10$i + 10$j + 20k$ ) ⋅ (5$i − 3$j + 6k$ ) = 50 − 30 + 120 = 140 J/s WORK, ENERGY AND POWER 83 BITSAT Archives (c) According to figure a a T T 0.36 kg v (velocity of the larger fragment) = m/s Kinetic energy = mv 2 = × 12 × ( 4)2 = 96 J 1 2 (a) mv = kx 2 v = 0.72 kg 0.72 g Net pulling force Total mass 0.72g − 0.36g g = = 0.72 + 0.36 Distance,s = ut + at g g = + × × (1)2 = So, T − mg = ma T − 0.36 g = 0.36 a T = 0.48 g Now, work done by string on the block WT = Ts cos 0° (on 0.36 kg of mass) g = ( 0.48 g )   (1) = 0.08 g  6 = 0.08 × (10) = J 2 at s = + × × 10 × 10 s = 100 m Hence, work done W = F ×s W = ma × s ∴ W = 50 × × 100 W = 10000 = 104 J (b) Now, s = ut + (b) In stable equilibrium, the centre of gravity of object, lies at minimum height from ground As the object is given a slight push, its centre of gravity rises because it comes in unstable equilibrium (a) Total mass of the shell = 20 kg Ratio of the masses of the fragments = : ∴ Masses of the fragments are kg and 12 kg Now, according to the conservation of momentum m1v1 = m2v ∴ × = 12 × v 50 (10 × 10−2 ) 20 × 10−3 = So, acceleration = R x m = 50 × 10−1 = m/s (c) Let the velocity with which stone is pushed = u We have, r mu + mgH = mg 2 r  u = 2g  − H 2  or u = g (r − 2H ) (c) Given, velocity of river, (v ) = m/s Density of water ρ = 12 g/cc Mass of each cubic metre 1.2 × 10−3 m= (10−2 )3 = 1.2 × 103 kg ∴ mv 2 = × 1.2 × 103 × ( 2)2 = 2.4 kJ Kinetic energy = (c) When the ball is released from the top of tower, then ratio of distance covered by the ball in first, second and third seconds is [because hn ∝ ( 2n − 1)] hI : hII : hIII = 1: : ∴ Ratio of work done mghI : mghII : mghIII = 1: : F (c)W = 2k If both springs are stretched by same force, thenW ∝ K As K1 > K Therefore,W1 > W2 i.e more work is done in case of second spring Centre of Mass, Momentum and Collision Centre of Mass The centre of mass of a body is a point where the whole mass of the body is supposed to be concentrated for describing its translatory motion In a uniform gravitational field, the centre of mass and the centre of gravity of a system are coincident Centre of Mass of Two Particles System Consider the two particles system in which two masses m1 and m2 are separated by a distance d The Centre of Mass (CM) divides internally the line joining two particles (or masses) in inverse ratio of their masses m2 d m1 d and QR = PR = m1 + m2 m1 + m2 m1 P R CM Q m2 d Centre of mass of two particles system Centre of Mass of N-Particles System If there are N-particles distributed in space whose masses are m1 , m2 , m3 ,…,mN and position vectors r1 , r2 , r3 , K, rN respectively, then the coordinates of the centre of mass are given as below : XCM = m1 x + m2 x + m3 x3 +…+ mN x N m y + m2 y + m3 y3 +…+ mN y N , YCM = 1 m1 + m2 + m3 +…+ mN m1 + m2 + m3 +…+ mN ZCM = m1 z + m2 z + m3 z3 +…+ mN z N m1 + m2 + m3 +…+ mN CENTRE OF MASS, MOMENTUM AND COLLISION Coordinates of centre of mass of a rigid body are 1 XCM = x dm, YCM = y dm M∫ M∫ and ZCM = z dm M∫ We can also write as, XCM = 1 Σ mi xi ; YCM = Σ mi yi ; ZCM = Σ mi zi M M M N-particles Y 85 mi (xi, yi, zi) where, M is total mass of the rigid body and x , y and z are the coordinates of elementary mass dm CM (xCM, yCM, zCM) X Centre of Mass of Some Rigid Bodies Z Centre of mass of -particles system Centre of mass of a uniform rectangular, square or circular plate lies at its centre where, M = m1 + m2 + m3 +…+ mN The position vector of centre of mass of N-particles system is given by m r + m2 r2 + m3 r3 +…+ mN rN rCM = 1 m1 + m2 + m3 +…+ mN Important Formulae of Centre of Mass ● Velocity of centre of mass of N-particles system is given by n vCM = ● Σ mi vi Centre of mass of a uniform semi-circular ring lies at a 2R distance of h = from its centre on the axis of π symmetry where R is the radius of the ring i =1 M Similarly, acceleration of centre of mass is given by CM R n a CM = ● ● Σ mi a i i =1 n FCM = Σ Fi i =1 This is called as equation of motion of centre of mass In pure translation, every particle of the body moves with the same velocity at any instant of time Rigid Body It is a continuous solid body of finite size in which deformation is not possible, i.e the relative distance between the two given points within the rigid body is always constant whatever the force is applied on it Practically, there is no existence of it In order to assume a body to be completely rigid, deformation would be neglected For a rigid body which is not under any external force, centre of mass is always a fixed point while in case of body which is not rigid, the centre of mass can vary Y Rigid body dm (x,y,z) O Z 2R π O M In accordance with Newton’s second law of motion FCM = F1 + F2 + K + Fn or CM CM CM X Centre of mass of a uniform semi-circular disc of radius R lies at a 4R distance of h = 3π CM 4R R from the centre on 3π the axis of symmetry O as shown in figure Centre of mass of a hemispherical shell of radius R lies R at a distance of h = from CM R R its centre on the axis of symmetry as shown in figure O Centre of mass of a solid hemisphere of radius R lies at 3R from its centre on the axis of a distance of h = symmetry If some mass or area is removed from a CM rigid body, then the R 3R position of centre of mass of the remaining portion is O obtained from the following formula r CM = m1 r − m2 r A r − A2 r or r CM = 1 m1 − m2 A1 − A2 86 BITSAT SELF STUDY GUIDE In rotation about a fixed axis, every particle of the rigid body moves in a circle which is in a plane perpendicular to the axis and has its centre on the axis Perfectly Elastic Collision in a Plane In a two-dimensional (or oblique) collision between two bodies, momentum remains conserved v1 Collision m1 It is defined as an isolated event in which two or more colliding bodies exert relatively strong forces on each other for a relatively short time, as a result of which the kinetic energy and momentum of the colliding particles change A A m1 B θ m2 u2 φ u1 Before collision B m2 Types of Collision v2 There are two basic types of collision in one dimension as given below : (i) Elastic collision The collision is said to be elastic, if both the bodies regain their original shape and size completely, after they are separated Also, the total linear momentum as well as kinetic energy of bodies remain conserved m2 v2 m1 v1 m2 (a) Before collision v2′ m1 v1 ′ (b) After collision Let the two balls of mass m1 and m2 collide each other elastically with their respective velocities v1 and v2 in the same direction as shown in figure Their velocities become v1 ′ and v2 ′ after the collision along the same line Applying conservation of linear momentum, we get m1v1 + m2v2 = m1v1 ′+m2v2 ′ Solving Eqs (i) and (ii) for v1 ′ and v2 ′, we get  2m2   m − m2  v1 ′ =   v2  v1 +  m m +  m1 + m2   2 …(iii)  2m1   m − m1  v2 ′ =   v1  v2 +   m1 + m2   m1 + m2  …(iv) (ii) Inelastic collision The collision is said to be inelastic, if the bodies not return to their original form after the collision In this type of collision, the total linear momentum of the system of bodies will remain conserved but kinetic energy of the system is not conserved i.e p initial = p final ⇒ ∆p = But, ⇒ K initial ≠ K final ∆K ≠ ∴ Along the x-axis m1u + m2u = m1v1 cos θ + m2v2 cos φ … (i) and along the y-axis = m1v1 sin θ − m2v2 sin φ … (ii) As the total kinetic energy remains unchanged 1 1 Hence, m1u 12 + m2u 22 = m1v12 + m2v22 2 2 …(iii) We can solve these equations provided that either the value of θ or φ is known to us Rebounding of a Ball on Collision with the Floor Speed of the ball after the nth rebound = e nv0 = e n gh0 …(i) In an elastic collision, kinetic energy before and after collision is also conserved Hence, 1 1 …(ii) m1v12 + m2v22 = m1v1 ′ + m2v2 ′ 2 2 and After collision Height covered by the ball after the n th rebound hn = e nh0 Total distance (vertical) ball before it stops bouncing covered by the 1 + e2  H = h0 + 2h + 2h + 2h + = h0   1 − e  Total time taken by the ball before it stops bouncing T = t0 + t + t + t3 + = h0 h1 h2 h0  + e  +2 +2 + =   g g g g 1 − e Coefficient of Restitution (e) For a collision, it is defined as the ratio of relative velocity of separation to the relative velocity of approach v − v1 Thus, coefficient of restitution, e = u1 − u2 For a perfectly elastic collision, e = CENTRE OF MASS, MOMENTUM AND COLLISION If < e < 1, the collision is said to be partially elastic But if, e = 0, the collision is said to be perfectly inelastic ● In a perfectly inelastic collision, e = which means that v2 − v1 = or v2 = v1 ● It can be shown that for an inelastic collision, the final velocities of the colliding bodies are given by  m − em2  (1 + e )m2 v1 =  u2  u1 + (m1 + m2 )  m1 + m2  and ● v2 =  m − em1  (1 + e )m1 u1 +   u2 (m1 + m2 )  m1 + m2  If a particle of mass m, moving with velocity u , hits an identical stationary target inelastically, then final velocities of projectile and target are correlated as i.e m1 = m2 = m v 1−e and u2 = ; = v2 + e Important Concepts Regarding the Collision (a) Two bodies of mass m1 (heavy) and mass m2 (light) are moving with same kinetic energy If they are stopped by the same retarding force, then (i) The bodies cover the same distance before coming to rest (ii) The time taken to come to rest is lesser for m2 as it p has less momentum, i.e t = F (iii) The time taken to come to rest is more for m1 as it has greater momentum 87 (b) Two bodies A and B having masses m1 and m2 have equal kinetic energies If they have velocities v1 and v2 , then m  v1 m2 p and =   = v2 m1 p2  m2  where, p1 and p2 are their momenta (c) If a bullet of mass m1 moving with a velocity u, strikes a mass m2 which is free to move in the direction of the bullet and is embedded in it, then the loss of kinetic  m m u2  energy is    m1 + m2  (d) A shell of mass m1 is ejected from a gun of mass m2 by an explosion which generates kinetic energy equals to E Then, the initial velocity of the shell is   2m2 E m (m + m )  1  (e) A gun of mass m2 fires a shell of mass m1 horizontally and the energy of explosion is such as would be sufficient to project the shell vertically to a height h Then, the velocity or recoil of the gun is  2m12 gh    m2 (m1 + m2 ) (f) A bullet of mass m1 penetrates a thickness of a fixed plate of mass m2 If m2 is free to move and the resistance is supposed to be uniform, then the m2s thickness penetrated is m1 + m2 Practice Exercise Two homogeneous spheres A and B of masses m and 2m having radii 2a and a respectively, are placed in touch The distance of centre of mass from first sphere is a a c 3a b 2a d None of these A circular hole of radius cm is cut-off from a disc of radius cm The centre of hole is m from the centre of the disc Find the position of centre of mass of the remaining disc from the centre of disc a − cm 35 cm c 10 cm b 35 d None of these A non-uniform thin rod of length L is placed along X -axis as such its one of end is at the origin The linear mass density of rod is λ = λ x Find the distance of centre of mass of rod from the origin a L / b 2L / c L / d L / Which of the following has centre of mass not situated in the material of body? a A rod bent in the form of a circle b Football c Hand ring d All of the above 88 SELF STUDY GUIDE BITSAT A ball kept in a closed box moves in the box making collisions with the walls The box is kept on a smooth surface The centre of mass a of the box remains constant b of the box plus the ball system remains constant c of the ball remains constant d of the ball relative to the box remains constant A man of mass M stands at one end of a plank of length L which lies at rest on a frictionless surface The man walks to the other end of the plank If the mass of plank is M / 3, then the distance that the mass moves relative to the ground is a 3L / b L / c 4L / d L / The centre of mass of a system cannot change its state of motion, unless there is an external force acting on it Yet the internal force of the brakes can bring a car to rest Then a the brakes stop the wheels b the friction between the brake pads and the wheel stops the car c the car is stopped by the road d the car is stopped by the driver pressing the pedal Two blocks A and B are connected by a massless string (shown in figure) A force of 30 N is applied on block B Find the B distance travelled by 10 kg 20 kg F=30 N centre of mass in Smooth s starting from rest a m c m b m d None of these A loaded spring gun of mass M fires a shot of mass m with a velocity v at an angle of elevation θ The gun is initially at rest on a horizontal frictionless surface After firing, the centre of mass of the gun-shot system m M vm b moves with a velocity cos θ in the horizontal M direction c remains at rest v (M − m ) d moves with a velocity in the horizontal (M + m ) direction a moves with a velocityv 10 Two bodies A and B of masses m1 and m respectively are connected by a massless spring of force constant k A constant force F k m2 m1 F starts acting on the body A at t = Then, A B a at every instant, the acceleration of centre of mass is F m1 + m2 b at t = 0, acceleration of B is zero but that of A is maximum c the acceleration of A decreases continuously and finally becomes equal d All of the above 11 In the given figure, the mass m starts with velocity v and moves with constant velocity on the surface During motion, the normal reaction between the horizontal surface and fixed triangular block m1 is N Then, during motion a N = (m1 + m2 ) g c N < (m1 + m2 ) g m2 m1 θ A b N = m1g d N > (m1 + m2 ) g 12 Four particles of masses kg, kg, kg and kg are placed at the corners A, B , C and D respectively of a square ABCD of edge m If point A is taken as origin, edge AB is taken along X -axis and edge AD is taken along Y -axis Find the coordinates of centre of mass in SI a (1, 1) c (0.5, 0.7) b (5, 7) d None of these 13 Three bodies having masses kg, kg and kg are moving at the speed of m/ s, m/s and m/s, respectively along X-axis The magnitude of velocity of CM is a 1.0 m/s c 0.9 m/s b m/s d 1.3 m/s 14 Figure shows the situation involving two blocks of masses m1 and m The heavier block of mass m1 is pulled by a force F The acceleration of CM is (all the surfaces are smooth) a F m2 b F m1 c m2 m1 F m1 + m2 d F 2F m1m2 15 A particle is projected such that its horizontal range would be R At the highest point, the particle breaks into two identical parts P and Q If P comes to rest, the horizontal distance of point from the point of projection (i.e origin) where the particle Q lands on the ground is a 1.5 R c 2.5 R b R d 2R 16 A string of negligible thickness is wrapped several times around a cylinder kept on a rough horizontal surface A boy standing at a distance l from the cylinder holds one end of the string and pulls the cylinder towards him Assuming no slipping, the length of the thread passed through the hands of the man is a l b l c 2l d 3l 17 If momentum of a body remains constant Then mass-speed graph of body is a circle c rectangular hyperbola b straight line d parabola CENTRE OF MASS, MOMENTUM AND COLLISION 18 Two bodies of masses m and 4m are moving with equal linear momentum The ratio of their kinetic energies is a : c : b : d : 19 A man is sitting in a moving train, then a his momentum must not be zero b his kinetic energy is zero c his kinetic energy is not zero d his kinetic energy may be zero 20 When a meteorite burns in the atmosphere Then, a the momentum conservation principle is applicable to the meteorite system b the energy of meteorite remains constant c the conservation principle of momentum is applicable to a system consists of meteorites, earth and air molecules d the meteorite momentum remains constant 21 If a bullet is fired from a gun, then a the mechanical energy of bullet-gun system remains constant b the mechanical energy is converted into non-mechanical energy c the mechanical energy may be conserved d the non-mechanical energy is converted into mechanical energy 22 A nucleus moving with a velocity n emits an α-particle Let the velocities of the α-particle and the remaining nucleus be n1 and n2 and their masses be m1 and m 2, then a n, n1and n2 must be parallel to each other b none of the two of n, n1, n2 should be parallel to each other c n1 + n2 must be parallel to n d m1 n1 + m2 n2 must be parallel to n 23 Two blocks of mass m1 and m are connected by a massless spring and placed at smooth surface The spring initially stretched and released Then, a the momentum of each particle remains constant separately b the momentum of both bodies are same to each other c the magnitude of momentum of both bodies are same to each other d the mechanical energy of system remains constant e Both (c) and (d) are correct 24 When two blocks A and B coupled by a spring on a frictionless table are stretched and then released, then a kinetic energy of body at any instant after releasing is inversely proportional to their masses b kinetic energy of body at any instant may or may not be inversely proportional to their masses KE of A mass of B c , when spring is massless = KE of B mass of A d Both (b) and (c) are correct 25 Two bodies are projected from roof with same speed in different directions If air resistance is not taken into account Then, 89 a they reach at ground with same magnitude of momenta if bodies have same masses b they reach at ground with same kinetic energy c they reach at ground with same speed d both (a) and (c) are correct 26 If a ball is dropped from rest It bounces from the floor respectively The coefficient of restitution is 0.5 and the speed just before the first bounce is m/s The total time taken by the ball to come to the rest is a s b s c 0.5 s d 0.25 s 27 Three identical blocks A, B and C are placed on horizontal frictionless surface The blocks B and C are in rest But A is approaching towards B with a speed of 10 m/s The coefficient A B C of restitution for all collision is 0.5 The speed of the block C just after collision is a 5.6 m/s b m/s c m/s d 10 m/s 28 A thin uniform bar lies on a frictionless horizontal surface and is free to move in any way on the surface Its mass is 0.16 kg and length is 1.7 m Two particles each of mass 0.08 kg are moving on the same surface and towards the bar in the direction perpendicular to the bar, one with a velocity of 10 m/s and other with velocity of m/s If collision between particles and bar is completely inelastic Both particles strike with the bar simultaneously The velocity of centre of mass after collision is a m/s b m/s c 10 m/s d 16 m/s 29 When two bodies collide elastically The force of interaction between them is a conservative b non-conservative c either conservative or non-conservative d zero 30 A body of mass M moving with a speed u has a head-on collision with a body of mass m originally at rest If M > > m , then the speed of the body of mass m after collision will be nearly a um M b uM m c u d 2u 31 A ball moving with a certain velocity hits another identical ball at rest If the plane is frictionless and collision is elastic, then the angle between the directions in which the balls move after collision, will be a 30° b 60° c 90° d 120° 32 A shell is fired from a cannon with a velocity v at an angle θ with the horizontal direction At the highest point in its path, it explodes into two pieces, one retraces its path to the cannon and the speed of the other pieces immediately after the explosion is a 3v cos θ  3 c   v cos θ  2 b 2v cos θ d v cos θ 90 SELF STUDY GUIDE BITSAT 33 A smooth steel ball strikes a fixed smooth steel plate at an angle θ with the vertical If the coefficient of restitution is e, then the angle at which the rebounce will take place is a θ c e tan θ  tan θ  b tan−1  e   e  d tan−1   tan θ  b N d 12 N 35 A machine gun fires a steady stream of bullets at the rate of n per minute into a stationary target in which the bullets get embedded If each bullet has a mass m and arrive at the target with a velocity v , the average force on the target is c mnv 60 60v mn mv d 60n b b 3v m I m rest II u=0 c v d at a speed of m/s collides with another body of kg at rest and sticks together The loss of kinetic energy would be v a 30 J b 40 J µ=0 v m⇒ M µ=0 c 33 J d 32 J 39 A bullet collides with a block which was initially at rest and embedded in it Upto what height from the horizontal, the block bullet system will rise? M 2v 2 (m + M )2g mv c (m + M ) a m 2v 2 (m + M )2g Mv d (m + M ) b 40 A heavy sheet of 36 Two boys of masses 10 kg and kg are moving along a vertical rope, the former climbing up with acceleration of m/s while later coming down with uniform velocity of m/s Then, tension in rope at fixed support will be (Take g = 10 m/s 2) a 200 N c 180 N v e=1/2 38 A kg body moving of each bullet is 10 g and the muzzle velocity is 800 m/s, then the average recoil force on the machine gun is a 60mnv v identical ball which was initially at rest The speed of the second ball after collision is a 34 A machine gun fires 120 shots per minute If the mass a.120 N c 16 N 37 A ball collides with another b 120 N d 160 N 2v wood and a light ball is moving towards each other as shown in the figure What will be the speed of the ball after collision? a 3v b 2v v m M>>>m M e = 1/3 c v Wall d v /2 BITSAT Archives In completely inelastic collision, [2013] a the complete KE of the medium must lost b the linear momentum of the system must remain conserved during collision c Both (a) and (b) are correct d Both (a) and (b) are incorrect A shell of mass 20 kg at rest explodes into two fragments whose masses are in the ratio : The smaller fragment moves with a velocity of ms −1 The kinetic energy of the larger fragment is [2012] a 96 J c 144 J b 216 J d 360 J In non-elastic collision, a b c d [2012] A sphere of mass m moving with a constant velocity u hits another staionary sphere of the same mass If e is the coefficient of restitution, then the ratio of the velocity of two spheres after collision will be [2010] 1− e 1+ e e+1 c e −1 a 1+ e 1− e e −1 d t e +1 b A 600 kg rocket is set for a vertical firing If the momentum is conserved energy is conserved momentum and energy are conserved momentum and energy are non-conserved exhaust speed is 1000 m/s, the mass of the gas ejected per second to supply the thrust needed to overcome the weight of rocket is [2009] Which of the following is not an example of perfectly inelastic collision? a A bullet fired into a block, if bullet gets embedded into block b Capture of an electron by an atom c A man jumping onto a moving boat d A ball bearing striking another ball bearing [2011] a 117.6 kg s−1 c kg s−1 b 58.6 kg s−1 d 76.4 kg s−1 CENTRE OF MASS, MOMENTUM AND COLLISION If gas molecules undergo inelastic collision with the walls of the container, a b c d [2009] temperature of gas will increase temperature of gas will fall pressure of the gas will increase neither the temperature nor the pressure will change A ball is dropped from a height h on a floor of coefficient of restitution e The total distance covered by the ball just before second hit is [2008] a h(1 − 2e ) b h(1 + 2e ) c h(1 + e ) instance when the speed of A is v and speed of B is [2008] 2v , the speed of centre of mass (CM) is a zero each other, under mutual force of attraction At an c 2.5v b v d 4v 10 Consider the following statement When jumping from some height, you should bend your knees as you come to rest instead of keeping your legs stiff Which of the following relations can be useful in explaining the statement? [2007] b ∆E = − ∆( PE + KE) = d ∆ x ∝ ∆ F a ∆ p1 = − ∆ p2 c F ∆t = m∆ v d he Two particles A and B initially at rest, move towards 91 where, symbols have their usual meanings Answer with Solutions Practice Exercise (b) We assume that origin is situated at the O1 ∴ x1 = 0, m1 = m x = 3a, m2 = 2m m x + m2x x CM = 1 ∴ m1 + m2 m × + 2m × 3a = m + 2m 6ma = = 2a 3m L A B O1 O2 m C 2m O1 O2 λ0 (d) (a) If a rod is bent in the form of a circle Its centre of mass is at the centre of curvature (b) Football is in the shape of hollow sphere So, its centre of mass is at the centre of sphere (c) The reason is same as (a) In all the above geometrical shapes, no material is present at the centre of mass (b) Since, no external force is present on the system So, the centre of mass of system will not be changed O O where, σ = Surface mass density The mass of cutting portion is m2 = π (1)2 σ = πσ m x − m2x x CM = 1 m1 + m2 (b) Since, external force on system is zero So, no change takes place in the centre of mass m x + m2x x CM = 1 m1 + m2 m1 ∆x1 + m2 ∆x or ∆x CM = m1 + m2 Here, ∆x CM = M ∆x = M ∆x … (i) ∆x = − ∆x = − 3M x rel = ∆x1 − ∆x or L = ∆x1 − ∆x ∆x [from Eq (i)] L=− − ∆x = − ∆x 3 ∆x = − L ∆x L −3L ∆x = − =− =− 4×3 ∴ m1 ∆x1 + m2 ∆x = or M ∆x1 + Taking origin at the centre of disc, x1 = 0, x = cm 36πσ × − πσ × x CM = 36πσ − πσ −3πσ cm = =− 35πσ 35 L3 = 2L = = L L2 x  λ0 λ0   2  0 3a (a) For the calculation of the position of centre of mass, cut-off mass is taken as negative The mass of disc is m1 = πr12σ = π 62 σ = 36πσ (b) The mass of considered element is dm = λdx = λ 0xdx O L L ( λ ) xdm x xdx ∫ ∫ = L ∴ x CM = dm ∫ ∫ λ xdx x  λ0    0 ∴ But, or Q dm dx x ∴ Negative sign indicates that both move in opposite direction 92 SELF STUDY GUIDE BITSAT (c) The car is stopped by the contact force exerted due to road (b) The acceleration of centre of mass is F 30 a CM = = = 1m/s2 mA + mB 10 + 20 1 ∴ s = a CM t = × × 22 = m 2 (c) If we consider a system consists of gun plus mass m Then, no external force is present on the system So, the acceleration of centre of mass of system will be zero Since, initially the system is in rest So, centre of mass will always remains at rest 10 (d) (a) Since, external force on the system is F F ∴ a CM = m1 + m2 (b) At t = 0, elongation in spring is zero So, force on m2 is zero but force on m1 is only in forward direction, i.e F F ∴ a1 = , at t = m1 This is maximum acceleration of m1 11 (c) When mass m2 moves downwards, the centre of mass of system (m1 + m2 ) moves downwards It means the acceleration is found in centre of mass in downward direction during motion of m2 This is possible only when net downward force is greater than that of upward force Mathematically, m1g + m2g > N ∴ N < m1g + m2g m x + mBxB + mCxC + mDxD 12 (c) x CM = A A mA + mB + mC + mD 1× + × 1+ × 1+ × 1+ + + 2+ = = = 0.5 m 10 Similarly, m y + mByB + mCyC + mDyD y CM = A A mA + mB + mC + mD 1× + × + × 1+ × = 1+ + + = = 0.7 m 10 m v + m2v + m3v 13 (b) As, v CM = 1 m1 + m2 + m3 ×5 + × + × = 5+ 4+ 25 + 16 + = 11 45 v CM = ≈ 4.09 ≈ m/s ∴ 11 = N m2 g m 1g N2 14 (c) FBD of two blocks F For m1, a1 = m1 m1 m and N2 + m1g = N1 For m2, a = 0,N2 = m2g m a + m2 a2 ∴ aCM = 1 N2 m2g m1 + m2 F m1 × +0 F m1 = = m1 + m2 m1 + m2 15 (a) Here, m1 = m2 = m V (suppose), when P comes to rest, it falls vertically downward X1 R O X1 = Centre of mass, m X + m2X X CM = 1 m1 + m2 R m + mX 2 R= 2m ⇒ X = 1.5 R F N1 m1g m1 m 2Q R X2 [Q X CM = R] 16 (c) If velocity of centre of mass isv, then velocity of contact is and that of the top is v, hence when centre of mass covers a distance l, thread covers a distance 2l 2v v p = mv p m= ∴ v Hence, m-v graph will be rectangular hyperbola 17 (c) y D (0, 1) C (1, 1) 18 (b) A (0, 0) B (1, 0) x Ek1 Ek2 p12 4m m 2m = 21 = = = :1 m1 m p2 2m2 [Q p1 = p 2] 19 (d) The momentum and kinetic energy depend upon reference frame In the frame of train, the man is in rest So, the momentum and kinetic energy of the man in the frame of train are zero But in the frame of ground, velocity of the man is non-zero So, in the frame of ground, momentum and kinetic energy are not zero 20 (c) Since, gravitational force is present So, conservation principle is only applicable to a system consists of air molecules + earth + meteorite 21 (d) (a) During firing of the bullet from gun, the chemical energy of powder of bullet is converted into the kinetic energy of the system So, final mechanical energy of system is greater than initial kinetic energy of the system 93 CENTRE OF MASS, MOMENTUM AND COLLISION (b) In this case, chemical energy (non-mechanical) is converted into mechanical energy of the system (c) The same reason as the above (d) The same reason as (a) and (b) 22 (d) Since, no external force is present So, momentum conservation principle is completely applicable ∴ m n = m1n1 + m2n2 or (m1 + m2 ) n = m1n1 + m2n2 23 (e) Since, no external force is present on the system So, conservation principle of momentum is applicable ∴ pi = pf = p1 + p2 [∴ pi = 0] ∴ p1 = − p2 ∴ | p1| = | − p2| ∴ p1 = p2 From this point of view, it is clear that momenta of both particles are equal in magnitude but opposite in direction Also, friction is absent So, total mechanical energy of system remains conserved 24 (d) When spring is massless Then according to momentum conservation principle, pi = pf or = m1v1 + m2v ∴ m1v1 = − m2v ∴ m1v1 = m2v or p1 = p p2 p2 K1 = ⇒ K = Q 2m1 2m2 K1 m2 [Q p1 = p 2] ∴ = K m1 25 (d) In the case of projectile motion, if bodies are projected with same speed, they reaches at ground with same speeds So, if bodies have same masses, then momentum of body will be different for different directions of projection but magnitude of momenta must be same v −u v −v0 or a = 26 (c)Q Acceleration, a = t t v −v0 [considered average force] or g= t Q v =0 Speed before first bounce v = m/s ( ↑ ) v − v 0 − ( −5 ) ∴ t= = = = 0.5 s g 10 10 27 (a) For collision between blocks A and B, v − v A vB − v A vB − v A e= B = = u A − uB 10 − 10 ∴ vB − v A = 10e = 10 × 0.5 = From momentum conservation principle, mAu A + mBuB = mAv A + mB vB or m × 10 + = mv A + mvB ∴ v A + vB = 10 Adding Eqs (i) and (ii), we get vB = 7.5 m/s Similarly for collision between B and C, vC − vB = 7.5e = 7.5 × 0.5 = 3.75 ∴ vC − vB = 3.75 m/s and vC + vB = 7.5 Adding Eqs (iii) and (iv), we get 2vC = 1125 1125 vC = = 5.6 m/s ∴ 28 (b) Here, m = 0.08 kg 30 (d) v CM v2 31 (c) 33 (b) Since, no force is present v1 along the surface of plate So, momentum conservation principle for ball is applicable n along the surface of plate mv sin θ1 = mv1 sin θ or v sin θ1 = v1 sin θ e= ∴ ∴ ∴ Here, 35 (c) F = .(i) θ2 θ1 v v1 cos θ v1 cos θ = v cos θ1 v cos θ v1 cos θ = ev cos θ v1 sin θ v sin θ tan θ = = v1 cos θ ev cos θ e tan θ tan θ = e  tan θ  θ = tan− 1   e  ∴ ∴ …(ii) m0 32 (a) As we know that at the highest point, the shell has only the horizontal component of velocity which isv cosθ If u be the velocity of second exploded piece, then applying conservation of linear momentum along X-axis ∴ 2mv cos θ = − mv cos θ + m or u = 3v cosθ 34 (c) F = v … (i) v1 m m0 = 0.16 kg According to conservation principle of momentum mv1 + mv = ( 2m + m0 ) v CM mv1 + mv ∴ v CM = 2m + m0 m 0.08 × 16 = 0.16 + 0.16 128 128 = = = m/s 0.32 32 29 (a) …(iii) … (iv) ∆m ∆t ∆m nm 120 × 10 × 10− = = 60 ∆t t = 20 × 10− kg/s = 800 × 20 × 10− N = 16 N nm ∆p ∆m =v =v t ∆t ∆t n mvn F = vm = 60 60 θ1 Plate [Let θ1 = θ] (ii) 94 SELF STUDY GUIDE BITSAT 36 (a) Since, m2 moves with constant velocity 37 (a) ∴ f2 = m2g f2 = × 10 = 80 N Since, block m1 moves with acceleration a = m/ s2 in upward direction ∴ f1 − m1g = m1a ∴ f1 = m1g + m1a = 10 × 10 + 10 × = 120 N a=2 m/s2 m1=10 kg, m2=8 kg m1 T m2 v=2 m/s f1 m2 m 1g m2 g 39 (b) Initially from conservation of momentum theorem, for bullet and block system mv mv ′ = (m + M ) v ⇒ v ′ = (m + M ) Let the bullet-block system rises to height h, so from conservation of mechanical energy, (m + M ) v ′ = (m + M ) gh v′2 h= 2g m 2v × h= ⇒ (m + M )2 2g 40 (a) From definition, we can write coefficient of restitutions, v ′ − 2v 2v e= v + 2v v v ′ − 2v ⇒ = 3v f2 m1 38 (a) f1 f2 For equilibrium of rope, T = f1 + f2 = (120 + 80) N = 200 N After collision ⇒ 3v ′ = v ⇒ v ′ = 3v As wood sheet is heavier than ball, so its speed will not change BITSAT Archives dm dm Mg 600 × 9.8 = Mg or = = kgs−1 = dt 1000 dt u (b) In any type of collision, the linear momentum of the system remain conserved even during collision (c) Thrust = u (a) Total mass of the shell = 20 kg Ratio of the masses of the fragments = : ∴ Masses of the fragments are kg and 12 kg Now, according to the conservation of momentum m1v1 = m2v ∴ × = 12 × v v (velocity of the larger fragment) = m/s 1 Kinetic energy = mv = × 12 × ( 4)2 = 96 J 2 (d) Usually, temperature of the gas in a container is same as the temperature of the walls of the container Therefore, during collision, there is no exchange of energy Hence, whether collision is elastic or inelastic, molecules rebounce with same average speed Therefore, neither the temperature not the pressure will change (b) Total distance travelled by the ball before its second hit is (a) Momentum is conserved in non-elastic collision but kinetic energy is not conserved (d) A ball bearing striking another ball bearing is not an example of perfectly inelastic collision m m m (b) Since, if momentum is A B A B conserved, then v2 u = v1 u mu = mv1 + mv Before collision After collision ⇒ u = v1 + v …(i) As we know that coefficient of restitution can be given as v −v2 …(ii) e= u From Eqs (i) and (ii), we get v −v2 …(iii) e= v1 + v v 1+ e On solving Eq (iii), we get = v2 1– e h h1 H = h + 2h1 = h[1 + 2e 2] [Q h1 = he 2] (a) As initially both the particles were at rest, therefore velocity of centre of mass was zero and there is no external force on the system So, speed of centre of mass remains constant, i.e it should be equal to zero m∆v 10 (c) F ∆t = m∆v ⇒ F = ∆t By doing so, time of change in momentum increases and impulsive force on knees decreases ... Published by Arihant Publications (India) Ltd For further information about the books published by Arihant log on to www.arihantbooks.com or email to info@arihantbooks.com /arihantpub /@arihantpub Arihant. .. unit vector Basically, unit vector represents the direction of the given vector Consider a vector A This vector is represented as Vector = (Magnitude of the vector) × (Direction of the vector) i.e... vectors If two or more vectors are perpendicular to each other, then they are known as orthogonal vectors Unit vector A vector of unit magnitude and whose direction is same as the given vector