Preface. You are advised to solve the problems yourself instead of using this book.

Transcription

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2 Preface It gives us immense pleasure to present Solutions To Concepts Of Physics. This book contains solutions to all the exercise problems from Concepts Of Physics and. The problems have been illustrated in detail with diagrams. You are advised to solve the problems yourself instead of using this book. The book is not written by any of our members and is not meant for sale. - Indrajeet Patil (dmin) - niket Panse (Mod Head)

3 . a) Linear momentum : mv [MLT ] b) Frequency : T [M L T ] Force [MLT ] c) Pressure : [ML T ] rea [L ]. a) ngular speed /t [M L T ] SOLUTIONS TO CONCEPTS CHPTE M L T b) ngular acceleration [M L T ] t T c) Torque F r [MLT ] [L] [ML T ] d) Moment of inertia Mr [M] [L ] [ML T ]. a) Electric field E F/q MLT [MLT I ] [IT] F MLT b) Magnetic field B [MT I ] qv [IT][LT ] B a MT I ] [L] c) Magnetic permeability [MLT I ] I [I] 4. a) Electric dipole moment P qi [IT] [L] [LTI] b) Magnetic dipole moment M I [I] [L ] [L I] 5. E h where E energy and frequency. E [ML T ] h [ML T ] [T ] Q [ML T ] 6. a) Specific heat capacity C [L T K ] mt [M][K] L L [L] b) Coefficient of linear expansion [K ] L T [L][] PV [ML T ][L ] c) Gas constant [ML T K (mol) ] nt [(mol)][k] 7. Taking force, length and time as fundamental quantity m (force/acceleration) [F /LT ] F 4 a) Density [FL T ] V Volume 4 [L ] L T b) Pressure F/ F/L [FL ] c) Momentum mv (Force / acceleration) Velocity [F / LT ] [LT ] [FT] Force d) Energy mv (velocity) acceleration F F [LT ] [L T ] [FL] LT LT ] metre 8. g 6 5 cm/min sec 9. The average speed of a snail is. mile/hr..6 Converting to S.I. units, m/sec [ mile.6 km 6 m].89 ms 6 The average speed of leopard 7 miles/hr In SI units 7 miles/hour 7.6 m/s 6.

4 . Height h 75 cm, Density of mercury 6 kg/m, g 9.8 ms then Pressure hfg 4 N/m (approximately) In C.G.S. Units, P 5 dyne/cm. In S.I. unit watt Joule/sec In C.G.S. Unit 9 erg/sec. micro century 4 years min So, min 5 / microcentury. Surface tension of water 7 dyne/cm In S.I. Unit, 7 dyne/cm.7 N/m 4. K ki a b where k Kinetic energy of rotating body and k dimensionless constant Dimensions of left side are, K [ML T ] Dimensions of right side are, I a [ML ] a, b [T ] b ccording to principle of homogeneity of dimension, [ML T ] [ML T ] [T ] b Equating the dimension of both sides, a and b a and b 5. Let energy E M a C b where M Mass, C speed of light E KM a C b (K proportionality constant) Dimension of left side E [ML T ] Dimension of right side M a [M] a, [C] b [LT ] b [ML T ] [M] a [LT ] b a ; b So, the relation is E KMC 6. Dimensional formulae of [ML T I ] Dimensional formulae of V [ML T I ] Dimensional formulae of I [I] [ML T I ] [ML T I ] [I] V I 7. Frequency f KL a F b M c M Mass/unit length, L length, F tension (force) Dimension of f [T ] Dimension of right side, L a [L a ], F b [MLT ] b, M c [ML ] c [T ] K[L] a [MLT ] b [ML ] c M L T KM b+c L a+b c T b Equating the dimensions of both sides, b + c () c + a + b () b () Solving the equations we get, a, b / and c / So, frequency f KL F / M / K / / F L M K L. F M Chapter-I

5 SCos 8. a) h rg LHS [L] MLT Surface tension S F/I L Density M/V [ML T ] adius r [L], g [LT ] HS [MT ] Scos [MT ] rg [ML T ][L][LT ] LHS HS So, the relation is correct [M L T ] [L] Chapter-I b) v p where v velocity LHS Dimension of v [LT ] Dimension of p F/ [ML T ] Dimension of m/v [ML ] HS p [ML T ] [L T ] [ML ] So, the relation is correct. c) V (pr 4 t) / (8l) LHS Dimension of V [L ] / [LT ] Dimension of p [ML T ], r 4 [L 4 ], t [T] Coefficient of viscosity [ML T ] HS 4 4 pr t [ML T ][L ][T] 8l [ML T ][L] So, the relation is correct. d) v (mgl /I) LHS dimension of v [T ] HS ( mgl /I) LHS HS [M][LT So, the relation is correct. 9. Dimension of the left side Dimension of the right side [ML ] ][L] [T ] dx L (a x ) (L L ) [L ] sin a dx So, the dimension of (a x ) a [L ] x sin a So, the equation is dimensionally incorrect. a x.

6 . Important Dimensions and Units : Physical quantity Dimension SI unit Force (F) [M L T ] newton Work (W) Power (P) Gravitational constant (G) ngular velocity () ngular momentum (L) Moment of inertia (I) Torque () Young s modulus (Y) Surface Tension (S) Coefficient of viscosity () Pressure (p) Intensity of wave (I) Specific heat capacity (c) Stefan s constant () Thermal conductivity (k) Current density (j) Electrical conductivity () Electric dipole moment (p) Electric field (E) Electrical potential (V) Electric flux () Capacitance (C) Permittivity () Permeability () Magnetic dipole moment (M) Magnetic flux () Magnetic field (B) Inductance (L) esistance () [M L T [M L T [M L T [T ] ] ] [M L T [M L ] [M L T ] [M L T ] [M T [M L ] [M L T T ] ] [M T [L T ] K ] [M T K ] ] 4 [M L T [I L K ] ] [I T M [LI T ] [M LI T L [M L I T ] [M T I [I T M L L ] ] 4 4 [I T M L [M LI [I L ] [M L I [M I T T T ] ] ] [M L I T [M L I T ] ] ] ] ] ] joule watt N-m /kg radian/s kg-m /s kg-m N-m N/m N/m N-s/m N/m (Pascal) watt/m J/kg-K watt/m -k 4 watt/m-k ampere/m m C-m V/m volt volt/m farad (F) C /N-m Newton/ N-m/T Weber (Wb) tesla henry ohm () Chapter-I * * * *.4

7 SOLUTIONS TO CONCEPTS CHPTE. s shown in the figure, The angle between and B 9 and B 4m esultant B Bcos 5 m Let be the angle between and 4 sin9 tan tan (4/) 5 4cos9 B y x esultant vector makes angle (5 + ) 7 with x-axis.. ngle between and B is 6 and B unit...cos 9. be the angle between and tan sin tan cos tan (.6795) 5 esultant makes angle with x-axis.. x component of cos 45 / unit x component of B cos 5 / x component of C cos 5 / esultant x component / / + / / y component of sin 45 / unit y component of B sin 5 / y component of C sin 5 / esultant y component / + / / / esultant Tan y component x component tan () 45 The resultant is unit at 45 with x-axis. 4. a 4 i j, b i 4 j a) a 4 5 b) b c) a b 7 i 7 j 7 d) a b ( 4)i ˆ ( 4 ) ˆj ˆi ˆj a b ( ).. 5 y B 6 45 x 5

8 Chapter- 5. x component of O cos x component of BC.5 cos.75 x component of DE cos 7 y component of O sin y component of BC.5 sin. y component of DE sin 7 x x component of resultant m y resultant y component +.. m So, esultant.6 m If it makes and angle with positive x-axis Tan y component x component tan. 6. a m b 4. y m.5m 6 O 9 D B m E x a) If unit cos 8 b) cos 5 9 c) cos 7 ngle between them is. 7. D î.5ĵ 4Kˆ 6i ˆ.5ˆj C D E DE 6. KM Tan DE / E / tan (/) m B 6m 4m.5 km The displacement of the car is 6. km along the distance tan (/) with positive x-axis. D E.5 km 8. In BC, tan x/ and in DCE, tan ( x)/4 tan (x/) ( x)/4 4x 4 x 4x 6x 4 x / ft a) In BC, C B BC ft b) In CDE, DE (/) 4/ ft CD 4 ft. So, CE a) magnitude of displacement 74 ft b) the components of the displacement vector are 7 ft, 4 ft and ft.. CD DE 4 ft c) In GE, E G GE ft. 9. Here the displacement vector r 7î 4 ĵ kˆ C x B x G D z Y F E BC ft F ft DE x r

9 . a is a vector of magnitude 4.5 unit due north. a) a a is along north having magnitude.5 units. b) 4 a unit 4 a is a vector of magnitude 6 unit due south.. a m, b m angle between them 6 a) a b a b cos 6 / m b) a b a b sin 6 / m.. We know that according to polygon law of vector addition, the resultant of these six vectors is zero. Here B C D E F (magnitude) So, x cos + cos / + cos / + cos / + cos 4/4 + cos 5/5 [s resultant is zero. X component of resultant x ] cos + cos / + cos / + cos / + cos 4/ + cos 5/ Note : Similarly it can be proved that, sin + sin / + sin / + sin / + sin 4/ + sin 5/. a i j 4k; b i 4 j 5k a b a b ab cos cos ab cos cos ( B) (claim) s, B B sin nˆ Chapter- B sin nˆ is a vector which is perpendicular to the plane containing and B, this implies that it is also perpendicular to. s dot product of two perpendicular vector is zero. Thus ( B). 5. i ˆ j ˆ 4kˆ, B 4i ˆ ˆj kˆ ˆi ˆj kˆ B 4 4 ˆ i(6 ) ˆ j(4 6) k(6 ˆ ) 6i ˆ ˆ j 6k ˆ / 6. Given that, B and C are mutually perpendicular B is a vector which direction is perpendicular to the plane containing and B. lso C is perpendicular to and B B ngle between C and B is or 8 (fig.) So, C ( B ) The converse is not true. For example, if two of the vector are parallel, (fig.), then also C ( B ) So, they need not be mutually perpendicular. C ( B) C B.

10 7. The particle moves on the straight line PP at speed v. From the figure, OP v (OP)v sin nˆ v(op) sin ˆn v(oq) ˆn It can be seen from the figure, OQ OP sin OP sin So, whatever may be the position of the particle, the magnitude and direction of OP v remain constant. OP v is independent of the position P. 8. Give F qe q(v B) E (v B) So, the direction of v B should be opposite to the direction of E. Hence, v should be in the positive yz-plane. gain, E vb sin v E B sin For v to be minimum, 9 and so v min F/B Chapter- So, the particle must be projected at a minimum speed of E/B along +ve z-axis ( 9 ) as shown in the figure, so that the force is zero. 9. For example, as shown in the figure, B B along west B C along south C along north B B B C B C But B C. The graph y x should be drawn by the student on a graph paper for exact results. To find slope at any point, draw a tangent at the point and extend the line to meet x-axis. Then find tan as shown in the figure. It can be checked that, dy d Slope tan (x ) 4x dx dx Where x the x-coordinate of the point where the slope is to be measured.. y sinx So, y + y sin (x + x) y sin (x + x) sin x sin.57.. Given that, i i e t / C ate of change of current When a) t, di i dt C b) when t C, c) when t C, di d i / C d t / C ie i e dt dt dt di i dt Ce di dt i Ce i e C t / C Q B O y V y B P V E B P yx y y sinx x C x x.4

11 . Equation i i e t / C i, 6 5, C.5 6 F 5 7 F a) i e e amp. e b) di i e dt C t / C c) t t. sec, i e 4. y x + 6x + 7 when t. sec (. /.) 5.8 mp. e di (. /.) e mp / sec dt. e rea bounded by the curve, x axis with coordinates with x 5 and x is given by, y rea dy y 5. rea (x x x 6x 7)dx 5 7x dy sin xdx [cos x] y y sinx 5 sq.units. y Chapter- y x + 6x + 7 x 5 x 6. The given function is y e x y When x, y e x increases, y value deceases and only at x, y. So, the required area can be found out by integrating the function from to. So, rea mass 7. a bx length x x e dx [e ]. y x 8. a) S.I. unit of a kg/m and SI unit of b kg/m (from principle of homogeneity of dimensions) b) Let us consider a small element of length dx at a distance x from the origin as shown in the figure. dm mass of the element dx (a + bx) dx So, mass of the rod m dp dt ( N) + ( N/S)t momentum is zero at t momentum at t sec will be dp [( N) + Ns t]dt p dp dt (tdt) t L dm (a bx) dx t kg m/s. L bx bl ax al O x x.5

12 9. The change in a function of y and the independent variable x are related as dy x dx Taking integration of both sides, dy x dx y x c y as a function of x is represented by y. The number significant digits a) No.of significant digits 4 b). No.of significant digits 4 c). No.of significant digits 5 d). No.of significant digits 4 x c.. The metre scale is graduated at every millimeter. m mm dy. dx x Chapter- The minimum no.of significant digit may be (e.g. for measurements like 5 mm, 7 mm etc) and the maximum no.of significant digits may be 4 (e.g. mm) So, the no.of significant digits may be,, or 4.. a) In the value 47, after the digit 4, 7 is present. Its value is greater than 5. So, the next two digits are neglected and the value of 4 is increased by. value becomes 5 b) value 84 c).6 d) value is 8.. Given that, for the cylinder Length l 4.54 cm, radius r.75 cm Volume r l (4.54) (.75) Since, the minimum no.of significant digits on a particular term is, the result should have significant digits and others rounded off. So, volume V r l (.4) (.75) (.75) (4.54) cm Since, it is to be rounded off to significant digits, V 4.7 cm. 4. We know that, verage thickness mm ounding off to significant digits, average thickness.7 mm. 5. s shown in the figure, ctual effective length (9. +.) cm But, in the measurement 9. cm, the no. of significant digits is only. So, the addition must be done by considering only significant digits of each measurement. So, effective length cm. r l 9cm.cm * * * *.6

13 SOLUTIONS TO CONCEPTS CHPTE. a) Distance travelled m b) F B BF B DC 5 M His displacement is D D F DF 4 5m In ED tan DE/E /4 /4 tan (/4) His displacement from his house to the field is 5 m, tan (/4) north to east.. O Starting point origin. i) Distance travelled m ii) Displacement is only OB m in the negative direction. Displacement Distance between final and initial position.. a) V ave of plane (Distance/Time) 6/.5 5 km/hr. b) V ave of bus /8 4 km/hr. c) plane goes in straight path velocity V ave 6/.5 5 km/hr. d) Straight path distance between plane to anchi is equal to the displacement of bus. Velocity V ave 6/8.5 km/hr. 4. a) Total distance covered km in hours. Speed 64/ km/h b) s he returns to his house, the displacement is zero. Velocity (displacement/time) (zero). 5. Initial velocity u ( starts from rest) Final velocity v 8 km/hr 5 sec (i.e. max velocity) Time interval t sec. v u 5 cceleration a ave.5 m/s. t 6. In the interval 8 sec the velocity changes from to m/s. verage acceleration /8.5 m/s change in velocity time Distance travelled S ut + / at + /(.5)8 8 m. 7. In st sec S ut + / at + (/ 5 ) 5 ft. t sec v u + at ft/sec. From to sec (t sec) it moves with uniform velocity 5 ft/sec,. W N S E 5 m B 4 m 4 m C m D Initial point (starting point) B ( m, ) 4 8 Time in sec 75 S (in ft) 5 O Y m E X ( m, ) Initial velocity u t (sec)

14 Chapter- Distance S 5 5 ft Between sec to sec acceleration is constant i.e. 5 ft/s. t sec velocity is 5 ft/sec. t s S ut + / at 5 + (/)( 5)() 5 m Total distance travelled is sec S + S + S ft. 8. a) Initial velocity u m/s. final velocity v 8 m/s time sec, 8 v u 8 acceleration.6 m/s 6 4 ta b) v u as t v u Distance S m. a.6 c) Displacement is same as distance travelled. Displacement 5 m. 9. a) Displacement in to sec is m. time sec. V ave s/t / m/s. 5 b) t sec it is moving with uniform velocity 5/.5 m/s. at sec. V inst m/s t t 5 sec it is at rest. (slope of the graph at t sec) V inst zero. t 8 sec it is moving with uniform velocity m/s V inst m/s t sec velocity is negative as it move towards initial position. V inst m/s.. Distance in first 4 sec is, OB + BCD m. verage velocity is as the displacement is zero. 5 m/s O t B D 4 t (sec). Consider the point B, at t sec t t ; s m and t sec s m So for time interval to sec Change in displacement is zero. So, average velocity displacement/ time The time is sec.. t position B instantaneous velocity has direction along BC. For average velocity between and B. V ave displacement / time ( B / t) t time C B y 4 B C. 4 6 x

15 We can see that B is along BC i.e. they are in same direction. The point is B (5m, m).. u 4 m/s, a. m/s, t 5 sec Distance s ut at 4(5) + / (.)5 5 m. 4. Initial velocity u 4. km/hr m/s u m/s, v a 6 m/s (deceleration) Distance S v u m ( 6) Chapter-.

16 5. Initial velocity u cceleration a m/s. Let final velocity be v (before applying breaks) t sec v u + at + 6 m/s a) S ut at 9 m when breaks are applied u 6 m/s v, t 6 sec ( min) Declaration a (v u)/t ( 6)/6 m/s. v u S 8 m a Total S S + S m.7 km. b) The maximum speed attained by train v 6 m/s c) Half the maximum speed 6/ m/s.4 Chapter- v u Distance S 5 m from starting point a When it accelerates the distance travelled is 9 m. Then again declarates and attain m/s. u 6 m/s, v m/s, a m/s v u Distance a 6 ( ) 5 m Position is km from starting point. 6. u 6 m/s (initial), v, s.4 m. v u Deceleration a m/s. s v u 6 Time t.5 sec. a 7. u 5 m/s, s 5 cm.5 m, v v u (5) Deceleration a s.5 Deceleration is. 5 m/s. 8. u, v 8 km/hr 5 m/s, t 5 sec a v u 5 m/s. t 5. 5 m/s. s ut at.5 m a) verage velocity V ave (.5)/5.5 m/s. b) Distance travelled is.5 m. 9. In reaction time the body moves with the speed 54 km/hr 5 m/sec (constant speed) Distance travelled in this time is S 5. m. When brakes are applied, u 5 m/s, v, a 6 m/s (deceleration)

17 Chapter- S v u a m ( 6) Total distance s s + s m..5

18 . (deceleration on hard braking 6 m/s ) B (deceleration on hard braking 7.5 m/s ) a 5 ( 6) 9 m Driver X eaction time.5 Speed 54 km/h Braking distance a 9 m Total stopping distance b m Speed 54 km/h Braking distance e 5 m Total stopping distance f 8 m.6 Driver Y eaction time.5 Chapter- Speed 7 km/h Braking distance c m Total stopping distance d 9 m. Speed 7 km/h Braking distance g 7 m Total stopping distance h m. So, b m Similarly other can be calculated. Braking distance : Distance travelled when brakes are applied. Total stopping distance Braking distance + distance travelled in reaction time.. V P 9 km/h 5 m/s. V C 7 km/h m/s. Police t In sec culprit reaches at point B from. Distance converted by culprit S vt m. culprit t time t sec the police jeep is m behind the culprit. Time s/v / 5 4 s. (elative velocity is considered). In 4 s the police jeep will move from to a distance S, where S vt 5 4 m. km away. The jeep will catch up with the bike, km far from the turning.. v 6 km/hr 6.6 m/s. v 4 km/h.6 m/s. elative velocity between the cars (6.6.6) 5 m/s. Distance to be travelled by first car is 5 + t m. Time t s/v /5 sec to cross the nd V car. In sec the st 5 m car moved 6.6. m H also covered its own length 5 m. Total road distance used for the overtake m.. u 5 m/s, g m/s when moving upward, v (at highest point). a) S v u a 5 5 m ( ) maximum height reached 5 m b) t (v u)/a ( 5)/ 5 sec c) s 5/ 6.5 m, u 5 m/s, a m/s, V 5 m Before crossing V t sec B V m fter crossing P C

19 v u as v (u as) 5 ( )(6.5) 5 m/s. 4. Initially the ball is going upward u 7 m/s, s 6 m, a g m/s s ut at 6 7t + / t 5t 7t 6 t ( 6) taking positive sign t 4. sec ( t ve) Therefore, the ball will take 4. sec to reach the ground. 5. u 8 m/s, v, a g 9.8 m/s a) S v u a 8 (9.8) 4 m v u 8 b) time t.85 a 9.8 t Chapter- v u + at 8 (9.8) (.85) 9.87 m/s. The velocity is 9.87 m/s. c) No it will not change. s after one second velocity becomes zero for any initial velocity and deceleration is g 9.8 m/s remains same. Fro initial velocity more than 8 m/s max height increases. 6. For every ball, u, a g 9.8 m/s 4 th ball move for sec, 5 th ball sec and rd ball sec when 6 th ball is being dropped. For rd ball t sec S ut at + / (9.8) 4.9 m below the top. For 4 th ball, t sec S + / gt / (9.8) 9.6 m below the top (u ) For 5 th ball, t sec S ut + / at + / (9.8)t 4.98 m below the top. 7. t point B (i.e. over.8 m from ground) the kid should be catched. For kid initial velocity u cceleration 9.8 m/s Distance S.8.8 m S ut at + / (9.8)t t.4 t.4. In this time the man has to reach at the bottom of the building. Velocity s/t 7/ m/s. 8. Let the true of fall be t initial velocity u.8 m 7m.8m 6 th 5 th 4 th rd

20 cceleration a 9.8 m/s Distance S / m S ut at. + / (9.8) t t..46 t.57 sec 4.9 For cadet velocity 6 km/hr.66 m/sec Distance vt m. The cadet,.6 m away from tree will receive the berry on his uniform. 9. For last 6 m distance travelled s 6 m, u? t. sec, a g 9.8 m/s S ut at 6 u(.) u 5.8/. 9 m/s. For distance x, u, v 9 m/s, a g 9.8 m/s v u 9 S 4.5 m a 9.8 Total distance m.. Consider the motion of ball form to B. B just above the sand (just to penetrate) u, a 9.8 m/s, s 5 m S ut at 5 + / (9.8)t t 5/4.9. t.. velocity at B, v u + at 9.8. (u ) 9.89 m/s. From motion of ball in sand u 9.89 m/s, v, a?, s cm. m. v u (9.89) a 49 m/s s. The retardation in sand is 49 m/s.. For elevator and coin u s the elevator descends downward with acceleration a (say) The coin has to move more distance than.8 m to strike the floor. Time taken t sec. S c ut at + / g() / g S e ut at u + / a() / a Total distance covered by coin is given by.8 + / a / g.8 +a/ 9.8/ 4.9 a 6. m/s ft/s.. It is a case of projectile fired horizontally from a height..8 a x m 6m Chapter-.66 m/s.6m t. sec 6 m B cm C 6ft.8m /a

21 h m, g 9.8 m/s a) Time taken to reach the ground t (h / g) Chapter- 4.5 sec. 9.8 b) Horizontal range x ut m. c) Horizontal velocity remains constant through out the motion. t, V m/s V y u + at m/s. esultant velocity V r ( 44.) 48.4 m/s. Tan V y V x tan (.5) 6. The ball strikes the ground with a velocity 48.4 m/s at an angle 66 with horizontal.. u 4 m/s, a g 9.8 m/s, 6 ngle of projection. a) Maximum height h u sin g 4 (sin 6) 6 m m b) Horizontal range X (u sin ) / g (4 sin (6 )) / 8 m. m/s V y V x V r.9

22 4. g 9.8 m/s,. ft/s ; 4 yd ft horizontal range x ft, u 64 ft/s, 45 We know that horizontal range X u cos t t x.65 sec. ucos 64 cos 45 y u sin (t) / gt 64 (.)(.65) (.65). Chapter- 7.8 ft which is less than the height of goal post. In time.65, the ball travels horizontal distance ft (4 yd) and vertical height 7.8 ft which is less than ft. The ball will reach the goal post. 5. The goli move like a projectile. Here h.96 m Horizontal distance X m cceleration g 9.8 m/s. u Time to reach the ground i.e. t h.96. sec g 9.8 Horizontal velocity with which it is projected be u. x ut u x t. m/s. 6. Horizontal range X ft covered by te bike. g 9.8 m/s. ft/s. gx sec y x tan u To find, minimum speed for just crossing, the ditch y ( is on the x axis) gx sec x tan gx sec gx gx u u x tan sincos sin u (.)(6.7) / (because sin /) u.79 ft/s ft/s. 7. tan 7/8 tan (7/8) The motion of projectile (i.e. the packed) is from. Taken reference axis at. 7 as u is below x-axis. u 5 ft/s, g. ft/s, y 7 ft x gsec y x tan u 7 x (.756) x g(.568) (5).5x x 7 x 5.78 ft (can be calculated).96m 7ft 5ft y ft.7ft m 5 5 u 8ft 5ft ft x

23 Horizontal range covered by the packet is 5.78 ft. So, the packet will fall ft short of his friend. Chapter-.

24 8. Here u 5 m/s, 6, g 9.8 m/s Horizontal range X u sin (5) sin( 6) 9.88 m g 9.8. Chapter- In first case the wall is 5 m away from projection point, so it is in the horizontal range of projectile. So the ball will hit the wall. In second case ( m away) wall is not within the horizontal range. So the ball would not hit the wall. usin 9. Total of flight T g B change in displacement H/ H H/ verage velocity time From the figure, it can be said B is horizontal. So there is no effect of vertical component of the velocity during this displacement. So because the body moves at a constant speed of u cos in horizontal direction. The average velocity during this displacement will be u cos in the horizontal direction. 4. During the motion of bomb its horizontal velocity u remains constant and is same as that of aeroplane at every point of its path. Suppose the bomb explode i.e. reach the ground in time t. Distance travelled in horizontal direction by bomb ut the distance travelled by aeroplane. So bomb explode vertically below the aeroplane. Suppose the aeroplane move making angle with horizontal. For both bomb and aeroplane, horizontal distance is u cos t. t is time for bomb to reach the ground. So in this case also, the bomb will explode vertically below aeroplane. 4. Let the velocity of car be u when the ball is thrown. Initial velocity of car is Horizontal velocity of ball. Distance travelled by ball B S b ut (in horizontal direction) nd by car S c ut + / at 9.8 m/s where t time of flight of ball in air. Car has travelled extra distance S c S b / at. Ball can be considered as a projectile having 9. u sin 9.8 t sec. g 9.8 S c S b / at m The ball will drop m behind the boy. 4. t minimum velocity it will move just touching point E reaching the ground. is origin of reference coordinate. If u is the minimum speed. X 4, Y, x sec Y x tan g u cm/s ) 4 x tan u (because g m/s u m/s cm cm cm C E cm cm

25 Chapter- u cm/s m/s. The minimum horizontal velocity is m/s. 4. a) s seen from the truck the ball moves vertically upward comes back. Time taken time taken by truck to cover 58.8 m. s 58.8 time 4 sec. (V 4.7 m/s of truck) v 4.7 u?, v, g 9.8 m/s (going upward), t 4/ sec. v u + at u 9.8 u 9.6 m/s. (vertical upward velocity). b) From road it seems to be projectile motion. Total time of flight 4 sec In this time horizontal range covered 58.8 m x X u cos t u cos 4.7 () Taking vertical component of velocity into consideration. y (9.6) ( 9.8) 9.6 m [from (a)] y u sin t / gt 9.6 u sin () / (9.8) u sin 9.6 u sin 9.6 (ii) u sin ucos 9. 6 tan tan (.) 5 gain u cos u 4.4 m/s. ucos 5 The speed of ball is 4.4 m/s at an angle 5 with horizontal as seen from the road , so cos 5 /5 Sec 5 m/s 5/9 and tan 4/ 5 Suppose the ball lands on nth bench So, y (n ) () [ball starting point m above ground] gx sec gain y x tan u ( y) (5 / 9) y ( + y)(4/) ( y) y 8 5 From the equation, y can be calculated. y 5 [x + n + y] n 5 n 6. The ball will drop in sixth bench. 45. When the apple just touches the end B of the boat. x 5 m, u m/s, g m/s,?. 5 y

26 u sin x g sin 5 5 sin sin / sin or sin 5 5 or 75 Similarly for end C, x 6 m Then sin (gx/u ) sin (.6) 8 or 7. So, for a successful shot, may very from 5 to 8 or 7 to a) Here the boat moves with the resultant velocity. But the vertical component m/s takes him to the opposite shore. Tan / /5 Velocity m/s distance 4 m Time 4/ 4 sec. b) The boat will reach at point C. In BC, tan BC B BC 4 BC 4/5 8 m. 47. a) The vertical component sin takes him to opposite side. Distance.5 km, velocity sin km/h Time Distance.5 hr Velocity sin 5 /sin min. b) Here vertical component of velocity i.e. km/hr takes him to opposite side. Distance.5 Time. 6 hr Velocity.6 hr minute. 48. Velocity of man V m km/hr BD horizontal distance for resultant velocity. X-component of resultant x 5 + cos t.5 / sin which is same for horizontal component of velocity. H BD (5 + cos ) (.5 / sin ) For H to be min (dh/d) d 5 cos d 6 sin 8 (sin + cos ) cos 5 cos 6 sin cos 8 cos 8 / /5.4 5km/h 4m sin 5m m/s 5 m m m/s 5km/h km/h 5km/h Chapter- / m / m B m/s W km/h B C N S 5km/h 5km/h sin km/h E D

27 Sin H cos 4/5 5 cos 5 ( / 5) km. 6 sin 6 (4 / 5) 49. In resultant direction the plane reach the point B. Velocity of wind V w m/s Velocity of aeroplane V a 5 m/s In CD according to sine formula sin 5 sin sin 5 sin 5 sin (/5) a) The direction is sin (/5) east of the line B. b) sin (/5) (5) cos m/s. s 5 Time 994 sec 49 5 min. v Velocity of sound v, Velocity of air u, Distance between and B be x. In the first case, resultant velocity of sound v + u (v + u) t x v + u x/t () In the second case, resultant velocity of sound v u (v u) t x v u x/t () From () and () v v x t From (i) u t x t v Velocity of air V x t x t x t x t t x t x t x t nd velocity of wind u t x t t x t t 5. Velocity of sound v, velocity of air u Velocity of sound be in direction C so it can reach B with resultant velocity D. ngle between v and u is > /. esultant D (v u ).5 5 Chapter- Here time taken by light to reach B is neglected. So time lag between seeing and hearing time to here the drum sound. W N m/s D 5m/s S vy v C 5 x x x v D V w u m / s v B B

28 t t t. Displacement velocity x (v u)(v u) v x x u (x / t )(x / t ) [from question no. 5] Chapter- 5. The particles meet at the centroid O of the triangle. t any instant the particles will form an equilateral BC with the same centroid. Consider the motion of particle. t any instant its velocity makes angle. This component is the rate of decrease of the distance O. Initially O a a Therefore, the time taken for O to become zero. a / v cos a v a v a. B O C * * * *.6

29 SOLUTIONS TO CONCEPTS CHPTE 4. m gm / kg F N F Gm m r 6.67 (/ ) (/ ) r m gm r m gm r r metre. So, the separation between the particles is m.. man is standing on the surface of earth The force acting on the man mg (i) ssuming that, m mass of the man 5 kg nd g acceleration due to gravity on the surface of earth m/s W mg 5 5 N force acting on the man So, the man is also attracting the earth with a force of 5 N. The force of attraction between the two charges q q o r r The force of attraction is equal to the weight Mg r 9 r m 9 m 8 [Taking g m/s ] r 9 m 8 m 4 mt For example, ssuming m 64 kg, r mass 5 kg r cm. m 4 75 m mm FG G r Coulomb s force F C 4 o q q r q Since, F G F c q q q C. 4.

30 Chapter-4 5. The limb exerts a normal force 48 N and frictional force of N. esultant magnitude of the force, ( 48) () N 6. The body builder exerts a force 5 N. Compression x cm. m Total force exerted by the man f kx kx 5 F x 48N F 5 5 k 75 N/m. 7. Suppose the height is h. t earth station F GMm/ M mass of earth m mass of satellite adius of earth F GMm ( h) GMm ( + h) h h h + h H ± (.44) km 8. Two charged particle placed at a sehortion m. exert a force of m. F N. r cm F? r 5 cm Since, F 4 o q q r, F r F r F F F r r r 5 9. The force between the earth and the moon, F G N N. 5 5 m m c r m F N N. Charge on proton.6 9 F electrical q q 4 r o 9 mass of proton.7 7 kg 9 4 r

31 mm F gravity G r r Fe r F g r. The average separation between proton and electron of Hydrogen atom is r 5. m. a) Coulomb s force F 9 9 q q r N. Chapter-4 b) When the average distance between proton and electron becomes 4 times that of its ground state Coulomb s force F 4 o q q 4r N.. The geostationary orbit of earth is at a distance of about 6km. We know that, g GM / (+h) t h 6 km. g GM / (6+64) g` g g [ taking g 9.8 m/s at the surface of the earth] kg equipment placed in a geostationary satellite will have weight Mg` N * * * * 4.

32 . m kg S m Let, acceleration a, Initial velocity u. S ut + / at ½ a ( ) a a 5 m/s Force: F ma 5 N (ns) 4. u 4 km/hr. m/s. 6 m kg ; v ; s 4m v u acceleration a s SOLUTIONS TO CONCEPTS CHPTE m/s (deceleration) So, braking force F ma N (ns). Initial velocity u (negligible) v 5 6 m/s. s cm m. 4. acceleration a v u s ms F ma N. T.kg.kg fig.g fig.kg.kg.g T fig 5. g m/s T.g T.g. N T (.g + T) T.g + T. + 5N Tension in the two strings are 5N & N respectively. B S Fig T F ma B T mg Fig mg Fig T + ma F F T + ma F T + T T F T F / 6. m 5g 5 kg s shown in the figure, from (i) D 5 Slope of O Tanθ 5 m/s OD So, at t sec acceleration is 5m/s Force ma 5 5.5N along the motion 5. T ma T ma (i) v(m/s) 5 5 D B 8 4 E 6 C

33 Chapter-5 t t 4 sec slope of B, acceleration [ tan ] Force t t 6 sec, acceleration slope of BC. BE 5 In BEC tan θ 5. EC Slope of BC tan (8 θ) tan θ 5 m/s (deceleration) Force ma N. Opposite to the motion. 7. Let, F contact force between m & m B. nd, f force exerted by experimenter. s f m m m Ba F m Ba Fig F F + m a f m B a f F f m a.(i) F m B a...(ii) From eqn (i) and eqn (ii) f m a m B a f m B a + m a f a (m + m B ). f F m (mb + m ) F [because a F/m B ] mb m B m The force exerted by the experimenter is F m 8. r mm m 4mg 4 6 kg s m. v u m/s. So, a v u s m/s (decelerating) Taking magnitude only deceleration is m/s So, force F N 9. x cm.m, k 5 N/m, m.kg. cceleration a m F kx x m g Fig B m/s (deceleration) So, the acceleration is m/s opposite to the direction of motion. Let, the block m towards left through displacement x. F k x (compressed) F k x (expanded) They are in same direction. esultant F F + F F k x + k x F x(k + k ) F x(k So, a acceleration k) opposite to the displacement. m m. m 5 kg of block. B ma N.m N a /5 m/s. s there is no friction between & B, when the block moves, Block B remains at rest in its position. m Bg Fig K F x m K F

34 Initial velocity of u. Distance to cover so that B separate out s. m. cceleration a m/s s ut + ½ at. + (½) t t. t.44 sec t.45 sec.. a) at any depth let the ropes make angle θ with the vertical From the free body diagram F cos θ + F cos θ mg mg F cos θ mg F cos s the man moves up. θ increases i.e. cos decreases. Thus F increases. b) When the man is at depth h cos Force h (d/ ) mg h d h 4 h mg 4h d 4h. From the free body diagram +.5 w w.5.5 ( ) 4N. So, the force exerted by the block on the block B, is 4N. B F ma d/ F d Fig- m/s mg F Chapter-5 w.5 s mg Fig- h N d/ F Wmg.5 4. a) The tension in the string is found out for the different conditions from the free body diagram as shown below. T T (W +.6.) T m/s.55 N. b) T T N. c) When the elevator makes uniform motion T W T W N Fig- a Fig-5 Uniform velocity W.5. Fig- T W Fig-6 Fig- a.m/s.5. T W Fig-4 T d) T +.5. W T W.5..4 N. T Fig-7 a.m/s W.5. Fig-8 e) T (W +.5.) T W N Fig-9.m/s a W.5. Fig- 5.

35 f) When the elevator goes down with uniform velocity acceleration T W T W N. 5. When the elevator is accelerating upwards, maximum weight will be recorded. (W + ma ) W + ma m(g + a) max.wt. When decelerating upwards, maximum weight will be recorded. + ma W W ma m(g a) So, m(g + a) m(g a) () () Now, mg + ma mg ma mg 6.8 m Kg 9.9 So, the true weight of the man is 66 kg. gain, to find the acceleration, mg + ma a. 9 m/s Let the acceleration of the kg mass relative to the elevator is a in the downward direction. s, shown in the free body diagram T.5 g.5(g/).5 a from figure () and, T g (g/) + a from figure () T.5 g +.5(g/) +.5a nd T g + (g/) a (i) (ii) Equation (i) g + (g/) + a T Equation (ii) g + (g/) a T Subtracting the above two equations we get, T 6a Subtracting T 6a in equation (ii) 6a g + (g/) a. g (9.8) 9a a. 4 a.59 T 6a T T N cut is T shown in spring. Mass wt kg g Given, m kg, k N/m From the free body diagram, kl g kl g l g k m Suppose further elongation when kg block is added be x, Then k( + x) g kx g g g 9.8 N 9. 8 x.98. m Fig- T W ma a.5g.5(g/) Fig-.5a Uniform velocity Chapter-5 W ma a a T W Fig- T Fig- m g (g/) a kl g 5.4

36 Chapter-5 8. a m/s kl (g + a) kl g + a l.6 m.4 m a kl g a a When kg body is added total mass ( + )kg kg. elongation be l kl g + a l.4.6 Further elongation l l.6. m. 9. Let, the air resistance force is F and Buoyant force is B. Given that F a v, where v velocity F a kv, where k proportionality constant. When the balloon is moving downward, B + kv mg (i) M B kv g kl m/s g M For the balloon to rise with a constant velocity v, (upward) let the mass be m Here, B (mg + kv) (ii) B mg + kv m B kw g So, amount of mass that should be removed M m. B kv B kv B kv B kv g g g kv g (Mg B) {M (B/g)} G kv B mg Fig- v B mg kv Fig- v. When the box is accelerating upward, U mg m(g/6) U mg + mg/6 m{g + (g/6)} 7 mg/7 (i) m 6U/7g. When it is accelerating downward, let the required mass be M. U Mg + Mg/6 U 6Mg Mg 6 5Mg 6 M 6U 5g V g/6t mg mg/6 Fig- Mass to be added M m 6U U g 5 5 g 7mg 5 6 g from (i) 6U 5g /5 m. The mass to be added is m/5. 6U 7g 6U g 5 7 V g/6t mg mg/6 Fig- 5.5

37 Chapter-5. Given that, F u and mg act on the particle. y F For the particle to move undeflected with constant velocity, net force should be zero. ( u ) m x mg ( u ) Z y mg mg Because, (u ) is perpendicular to the plane containing u and, u should be in the xz-plane. gain, u sin mg u mg sin u will be minimum, when sin 9. u min mg along Z-axis. T T m m m g m a m g m a m. kg, m.6 kg T (m g + m a) (i) T m g + m a T + m a m g (ii) T m g m a From equation (i) and equation (ii) m g + m a + m a m g, from (i) a(m + m ) g(m m ) m a m.6. f ms. m m.6. a) t sec acceleration.66 ms Initial velocity u So, distance travelled by the body is, S ut + / at + ½(.66) 6.5 m b) From (i) T m (g + a). ( ).9 N c) The force exerted by the clamp on the pully is given by F T F T N.. a.6 m/s T.9 N fter sec mass m the velocity V u + at m/s upward. t this time m is moving 6.5 m/s downward. t time sec, m stops for a moment. But m is moving upward with velocity 6.5 m/s. It will continue to move till final velocity (at highest point) because zero. Here, v ; u 6.5 g 9.8 m/s [moving up ward m ] V u + at 6.5 +( 9.8)t t 6.5/ / sec. During this period / sec, m mass also starts moving downward. So the string becomes tight again after a time of / sec. a T F.kg m m T.6kg 5.6

38 Chapter-5 4. Mass per unit length / kg/cm. kg/cm. Mass of cm part m kg Mass of cm part m kg. Let, F contact force between them. From the free body diagram F (i) nd, F a (ii) From eqa (i) and (ii) a a / 4 m/s Contact force F + a N. 5. a a T kg kg m 4m 5m Fig- g Fig- a N g a T Fig- a g m N m a a F F m m g N N Sin 4/5 g sin (a + T) T g sin a sin /5 g sing a + T (i) T g sin + a (ii) T + a g sin From eqn (i) and (ii), g sin + a + a g sin 4 a g sin g sin g g / a g 5 g From the above Free body diagram M a + F T T m a + F (i) a m cceleration of mass m is m m Fig- Fig- a a m From the above free body diagram T + m a m(m g + F ) T m a m g F Fig- m a F mg (m m m g Fig- ) T m g F m a Fig- T a From the above Free body diagram m a + T m g.(ii) m a + m a + F m g (from (i)) a(m + m ) + m g/ m g {because f m g/} a(m + m ) m g a(m + m ) m g/ a towards right. m a T m g Fig- From the free body diagram T (m g + F + m a) a mg (m m ) 5.7

39 Chapter-5 T m g + F m a T 5g + 5a (i) T m g +F + m a T g + + a (ii) From the eqn (i) and eqn (ii) 5g + 5a g + + a g 7a 7a g g 9. 4 a m/s [ g 9.8m/s ] 5a a) acceleration of block is 4. m/s b) fter the string breaks m move downward with force F acting down ward. m a F + m g ( + 5g) 5(g +.) Force 5(g So, acceleration mass 5.) (g +.) m/s 5g FN Force N, acceleration /5.m/s. 8. T T/ (a +a ) a m a m T/ m m m lg g m (a +a ) la Fig- Fig- (a a ) g Fig-4 9. Fig- Let the block m++ moves upward with acceleration a, and the two blocks m an m have relative acceleration a due to the difference of weight between them. So, the actual acceleration at the blocks m, m and m will be a. (a a ) and (a + a ) as shown T g a...(i) from fig () T/ g (a a )...(ii) from fig () T/ g (a + a )...(iii) from fig (4) From eqn (i) and eqn (ii), eliminating T we get, g + a 4g + 4(a + a ) 5a 4a g (iv) From eqn (ii) and eqn (iii), we get g + (a a ) g (a a ) 5a + a (v) g Solving (iv) and (v) a 9 So, a a g 9g 7g and a g 5a g g 9 9g 9 a + a g 9g g So, acceleration of m, m, m ae respectively. gain, for m, u, s cm.m and a 9 g 9 S ut + ½ at 9. gt t.5sec. 9 T T/ [g m/s ] T/ 9 g 7g (up) 9 9 (doan) g (down) 9 a a m m m Fig- m g g a Fig- g a Fig-4 Fig- 5.8

40 Chapter-5 m should be at rest. T m g T/ g a T/ g a T m g (i) T 4g 4a (ii) T 6g 6a (iii) From eqn (ii) & (iii) we get T g g T T 4g/5 48g. Putting yhe value of T eqn (i) we get, m 4.8kg.. kg a T Fig- B kg B T a g Fig- g Fig- T + a g...(i) From eqn (i) and (ii), we get T a T a (ii) g a + a g a g a 5m/s. From (ii) T a 5N. a/ B m Fig- M a m(a/) mg Fig- T T mg ma Fig- Ma T T + Ma Mg Ma T T Ma /. Ma/ + ma Mg. (because T Ma/) Ma Mg a g/ a) acceleration of mass M is g/. Ma M g Mg b) Tension T c) Let, resultant of tensions force exerted by the clamp on the pulley T T T Mg T Mg gain, Tan T T 45. T T T T. So, it is Mg at an angle of 45 with horizontal. T T a M Fig- M ma mg Fig- mg ma Fig- 5.9

41 . Ma + Mg sin T T + Ma Mg T Ma + Mg sin (i) (Ma + Mg sin ) + Ma Mg [From (i)] 4Ma + Mgsin + Ma Mg 6Ma + Mg sin Mg 6Ma Mg a g/6. cceleration of mass M is a s g/6 g/ up the plane. Ma T Mg FBD- Mg FBD- T FBD- s the block m does not slinover M, ct will have same acceleration as that of M From the freebody diagrams. T + Ma Mg...(i) (From FBD ) T Ma sin...(ii) (From FBD -) sin ma...(iii) (From FBD -) cos mg...(iv) (From FBD -4) 4a t/ 8a T T 8a (ii) [From FBD -4] gain, T + 5a 5g 8a + 5a 5g a 5g a 5g/ downward. (from FBD -) cceleration of mass () kg is a / (g) & 5kg (B) is 5g/. c) T T ma mg M FBD-4 M m Eliminating T, and a from the above equation, we get M cot 4. a) 5a + T 5g T 5g 5a...(i) (From FBD-) 5a gain (/) 4g 8a T 8g 6a...(ii) (from FBD-) T From equn (i) and (ii), we get 5g 5a 8g + 6a a g a /7g a So, acceleration of 5 kg mass is g/7 upward and that of 4 kg mass is a g/7 (downward). 5g b) T FBD- a kg 4a T/ a 5g 5a g B 5kg FBD- FBD-4 a a M a 8a T/ 4g FBD- a Chapter-5 a 5kg FBD- a 4kg a kg kg a T/ T/ C T B g a FBD-5 4a g FBD-6 T + a g T g a (i) [From FBD 5] gain, T g 4a T 4g 8a (ii) [From FBD -6] g a 4g 8a [From (i)] 5.

42 a (g/) downward. cceleration of mass kg(b) is g/ (up) cceleration of mass kg() is g/ (downward). 5. m g.kg m 5g.5kg m 5g.5kg. T +.5a.5g...(i) T.5a.5g a...(ii) T +.a T +.5g...(iii) From equn (ii) T.5g +.5a...(iv) From equn (i) T.5g.5a...(v) Equn (iii) becomes T +.a T +.5g.5g +.5a +.a.5g +.5a +.5g [From (iv) and (v)].65a.4g a g g downward 65 cceleration of 5gm block is 8g/g downward. 6. m 5 kg of monkey. a m/s. From the free body diagram T [5g + 5()] T 5 ( + ) T 5 T 65 N. The monkey should apply 65N force to the rope. Initial velocity u ; acceleration a m/s ; s 5m. s ut + ½ at 5 + (/) t t 5 t sec. Time required is sec. Chapter-5 7. Suppose the monkey accelerates upward with acceleration a & the block, accelerate downward with acceleration a. Let Force exerted by monkey is equal to T From the free body diagram of monkey T T mg ma...(i) T mg + ma. mg + ma + ma mg [From (i)] ma ma a a. cceleration a downward i.e. a upward. The block & the monkey move in the same direction with equal acceleration. gain, from the FBD of the block, mg T ma mg. ma ma a mg If initially they are rest (no force is exertied by monkey) no motion of monkey of block occurs as they have same weight (same mass). Their separation will not change as time passes. 8. Suppose move upward with acceleration a, such that in the tail of maximum tension N produced. T.5a T FBD-.5g a T N.5a m 5g FBD- a g m T a a T.5g.a 5g FBD-.g a T a 5g 5a a a B 5g T N 5a Fig- Fig- g a T 5g 5a...(i) g a...(ii) T 5 + +(5 5) 5 N (max) a a 5 m/s So, can apply a maximum force of 5 N in the rope to carry the monkey B with it. 5.

43 Chapter-5 For minimum force there is no acceleration of monkey and B. a Now equation (ii) is T g T N (wt. of monkey B) Equation (i) is T 5g [s T N] T 5g N. The monkey should apply force between 7 N and 5 N to carry the monkey B with it. 9. (i) Given, Mass of man 6 kg. Let apparent weight of man in this case. Now, + T 6g [From FBD of man] T 6g...(i) T g...(ii) [ From FBD of box] 6g g [ From (i)] 5g The weight shown by the machine is 5kg. T T 6g g (ii) To get his correct weight suppose the applied force is T and so, acclerates upward with a. In this case, given that correct weight 6g, where g acc n due to gravity a 6a T 6g T g a a From the FBD of the man From the FBD of the box T + 6g 6a T g a T 6a [ 6g] T 6g g a T 6a...(i) T a 9g 9 T a 9...(ii) From eqn (i) and eqn (ii) we get T T 8 T 8N. So, he should exert 8 N force on the rope to get correct reading. 4. The driving force on the block which n the body to move sown the plane is F mg sin, So, acceleration g sin Initial velocity of block u. s l, a g sin Now, S ut + ½ at l + ½ (g sin ) t g t g sin Time taken is g sin g sin 4. Suppose pendulum makes angle with the vertical. Let, m mass of the pendulum. From the free body diagram T a mg v ma mg T cos mg ma T sin T cos mg ma T sin T mg cos...(i) t ma sin...(ii) 5.

44 From (i) & (ii) mg ma a tan cos sin g The angle is Tan (a/g) with vertical. (ii) m mass of block. Suppose the angle of incline is From the diagram ma cos mg sin ma cos mg sin a tan g sin cos a g Chapter-5 tan a/g tan (a/g). 4. Because, the elevator is moving downward with an acceleration m/s (>g), the bodygets separated. So, body moves with acceleration g m/s [freely falling body] and the elevator move with acceleration m/s Now, the block has acceleration g m/s u t. sec So, the distance travelled by the block is given by. s ut + ½ at + (½) (.) 5.4. m cm. The displacement of body is cm during first. sec. m/s m/s * * * * ma ma a mg mg 5.

45 SOLUTIONS TO CONCEPTS CHPTE 6. Let m mass of the block From the freebody diagram, mg mg...() velocity gain ma ma mg (from ()) a g 4 g 4/g 4/.4 The co-efficient of kinetic friction between the block and the plane is.4. Due to friction the body will decelerate Let the deceleration be a mg mg...() velocity ma ma mg (from ()) a g. m/s. Initial velocity u m/s Final velocity v m/s a m/s (deceleration) S v u a ( ) 5m It will travel 5m before coming to rest.. Body is kept on the horizontal table. If no force is applied, no frictional force will be there f frictional force F pplied force From grap it can be seen that when applied force is zero, frictional force is zero. 4. From the free body diagram, mg cos mg cos..() For the block U, s 8m, t sec. s ut + ½ at 8 + ½ a a 4m/s gain, + ma mg sin mg cos + ma mg sin [from ()] m(g cos + a g sin ) cos g sin a a mg a ma ma p o mg a mg F ( / ) (/) 4 ma ( 5 / ) / ( 5 / ). Co-efficient of kinetic friction between the two is.. 5. From the free body diagram 4 4a + 4g sin () 4g cos...() 4g cos Putting the values of is & in equn. () 4 4a. 4g cos + 4g sin 4kg mg 4N 4 4a. 4 ( / ) + 4 (/) 4 4a.8 + a 5 m/s For the block u, t sec, a 5m/s Distance s ut + ½ at s + (/) 5 m The block will move m. ma mg 6.

46 6. To make the block move up the incline, the force should be equal and opposite to the net force acting down the incline + g sin. (9.8) + I 9.8 (/) [from ()] N With this minimum force the body move up the incline with a constant velocity as net force on it is zero. b) Net force acting down the incline is given by, F g sin 9.8 (/).9 6.4N Due to F 6.4N the body will move down the incline with acceleration. No external force is required. Force required is zero. 7. From the free body diagram g m/s, m kg,,. mg cos - F sin mg cos + F sin...() nd mg sin + F cos mg sin + (mg cos + F sin ) F cos mg sin + mg cos + F sin F cos (mg sin mgcos) F ( sin cos) F (/ ). ( 8. m mass of child mg cos 45. (/ ) ( / ) / ) mg cos 45 mg /v...() N 7.5N.76 Net force acting on the boy due to which it slides down is mg sin 45 - mg sin 45 - mg cos 45 m (/ ).6 m (/ ) m [(5/ ).6 (5 / )] m( ) Force acceleration mass m( ) m m/s 9. Suppose, the body is accelerating down with acceleration a. From the free body diagram mg cos mg cos...() ma + mg sin mg(sin cos ) a g (sin cos ) m For the first half mt. u, s.5m, t.5 sec. So, v u + at + (.5)4 m/s S ut + ½ at.5 + ½ a (/5) a 4m/s...() For the next half metre u` m/s, a 4m/s, s.5..5 t + (/) 4 t t + t.5 6. ma Chapter 6 mg (body moving down) F mg (body moving us) mg mg mg 45 F

47 4 t + 4 t sec 4 8 Time taken to cover next half meter is.sec.. f applied force F i contact force F frictional force normal reaction tan F/ Chapter 6 When F, F is the limiting friction (max friction). When applied force increase, force of friction increase upto limiting friction () Before reaching limiting friction F < tan F. From the free body diagram tan tan T +.5a.5 g...() + a + T T...() + a T + a T...() From () & () + a T T T T T kg. B kg. Fi F.5kg a f Limiting Friction.5g.5g T T Equation () becomes + a + T T + a T T + a.g + a...(4) Equation () becomes T + /5a.5g T.5g.5a.5g.5a...(5) From (4) & (5).g + a.5g.5a T a g a g.5 a.4 I.4m/s.5 a) ccln of kg blocks each is.4m/s b) Tension T.g + a +.4.4N c) T.5g.5a N. From the free body diagram + 6 (g) + ( 5) 5/ g sin ( ) kg a.5 m/s 4kg Co-efficient of friction.75 &.6.5 6N 4g 4.5 6NT 6.

48 Chapter 6. a 5kg From the free body diagram T + 5a 5g T (T + 5a+ ) T 5g 5a T 5g 5 a...(i) T (5g + 5a + 5a + ) T 5g + 5a (iii) T 5g + a + (ii) From (i) & (ii) 5g 5a 5g + a +. (5g) 5a 9 a.6m/s Equation (ii) T N in the left string Equation (iii) T 5g + 5a N in the right string. 4. s 5m, 4/, g m/s u 6km/h m/s, v, v u a m/s s 5 From the freebody diagrams, mg cos ; g m/s mg cos.(i) ; 4/. gain, ma + mg sin - ma + mg sin mg cos a + g sin mg cos + sin - (4/) cos + sin 4 cos + sin 4 cos 4 cos - sin 4 B a a C 5kg sin + sin 6 ( sin ) sin + 8 sin 5a T 5g B T r5g 5g T 5a T 5g the max. angle velocity ma a mg sin 8 sin (.8) 6 8 4(5)( 7) [Taking +ve sign only] 5 Maximum incline is 6 5. to reach in minimum time, he has to move with maximum possible acceleration. Let, the maximum acceleration is a ma ma mg a g.9 9m/s a) Initial velocity u, t? a 9m/s, s 5m ma mg a s ut + ½ at 5 + (/) 9 t t sec. 9 b) fter overing 5m, velocity of the athelete is V u + at + 9 (/) m/s He has to stop in minimum time. So deceleration ia a 9m/s (max) a mg ma 6.4

49 Chapter 6 ma ma (max frictional force) a g 9m / s (Deceleration) u m/s, v v u t sec. a a a 6. Hardest brake means maximum force of friction is developed between car s type & road. Max frictional force From the free body diagram mg cos mg cos...(i) and + ma mg sin ) mg cos + ma mg sin g cos + a (/) (ii) a 5 { ( )} ( / ).5 m/s When, hardest brake is applied the car move with acceleration.5m/s S.8m, u 6m/s S, velocity at the end of incline mg a ma V u as 6 (.5)(.8) 6 64 m/s 6km/h Hence how hard the driver applies the brakes, that car reaches the bottom with least velocity 6km/h. 7. Let,, a maximum acceleration produced in car. ma [For more acceleration, the tyres will slip] ma mg a g m/s a For crossing the bridge in minimum time, it has to travel with maximum acceleration u, s 5m, a m/s s ut + ½ at 5 + (/) t t sec. mg If acceleration is less than m/s, time will be more than sec. So one can t drive through the bridge in less than sec. 8. From the free body diagram 4g cos 4 /...(i) + 4a P 4g sin. (4) cos + 4a P 4 sin (ii) P + a + g sin (iii) g cos /...(iv) Equn. (ii) 6 + 4a P Equn (iv) P + a + From Equn (ii) & (iv) 6 + 6a + 6a P 6.5 a.69.7m/s 4g g 6 b) can be solved. In this case, the 4 kg block will travel with more acceleration because, coefficient of friction is less than that of kg. So, they will move separately. Drawing the free body diagram of kg mass only, it can be found that, a.4m/s. 6.5 a P a kg 4kg a

50 Chapter 6 9. From the free body diagram M a M a M a M g cos M g cos...(i)...(ii) T + M g sin m a T M M a +...(iv)...(iii) Equn (iii) T + M g sin M a M g cos Equn (iv) T M g sin + M a + M g cos...(v) Equn (iv) & (v) g sin (M + M ) a(m + M ) g cos (M + M ) a (M + M ) g sin (M + M ) g cos (M + M ) a g(sin cos ) The blocks (system has acceleration g(sin cos ) The force exerted by the rod on one of the blocks is tension. Tension T M g sin + M a + M g sin T M g sin + M (g sin g cos ) + M g cos T. Let p be the force applied to at an angle From the free body diagram + P sin mg P sin + mg p cos...(ii)...(i) Equn. (i) is (mg P sin ) P cos mg sin P cos M p T mg sin cos M g T mg P pplied force P should be minimum, when sin + cos is maximum. gain, sin + cos is maximum when its derivative is zero. d/d ( sin + cos ) cos sin tan mg So, P sin cos mg sec mg ( tan mg / cos sin cos cos cos mg mgsec tan mgsec tan Minimum force is mg at an angle tan.. Let, the max force exerted by the man is T. From the free body diagram + T Mg Mg T mg + mg nd T...(i)...(ii) mg T m mg T 6.6