SOLUTIONS TO CONCEPTS CHAPTER 9

Transcription

1 SOUTIONS TO CONCEPTS CHPTER 9. kg, kg, kg, x 0, x, x / y 0, y 0, y / The position of centre of ass is C. x x x y y y, ( 0) ( ) ( / ) ( 0) ( 0) ( (, 7, fro the point B. / )). et be the origin of the syste In the aboe figure g, x ( )sin 5 y 0 g, x ( )sin 5 y 0 x 0 y ( ) cos 5 The position of centre of ass x x x y y y, 0 0 ( ) sin5 ( ) sin y, 6 8 0, 8 / cos5. et O (0,0) be the origin of the syste. Each brick is ass & length. Each brick is displaced w.r.t. one in contact by /0 The X coordinate of the centre of ass o Y (0, 0) B , 5 5 H Q H (0, 0) (, ) X C /0 X c et the centre of the bigger disc be the origin. R Radius of bigger disc R Radius of saller disc R T (R) I T where T Thickness of the two discs Density of the two discs The position of the centre of ass O O (0, 0) R (R, 0) X 9.

2 x x y y, x R y 0 x 0 y 0 R TR 0 0, R TR R,0 R T (R) T,0 5R T 5 t R/5 fro the centre of bigger disc towards the centre of saller disc. 5. et 0 be the origin of the syste. R radius of the saller disc R radius of the bigger disc The saller disc is cut out fro the bigger disc s fro the figure R T x R y 0 (R) T x 0 y 0 The position of C.. R TR 0 0 0, R T (R) TR R TR R, 0, 0 R T C.. is at R/ fro the centre of bigger disc away fro centre of the hole. 6. et be the ass per unit area. ass of the square plate d d ass of the circular disc 4 et the centre of the circular disc be the origin of the syste. Position of centre of ass d d (d / 4) 0 0 0, d 4d d (d / 4),0,0 d 4 4 4d The new centre of ass is right of the centre of circular disc kg..5 cos 7 î.55 sin 7 ĵ. î 0.9 ĵ.kg. 0.4 ĵ.5kg 0.8 î ĵ 4 0.5kg 4 î 5 kg 5.6 î. ĵ So, 44 5 c 4 (.î 0.9ĵ).(0.4 ĵ).5( 0.8î 0.6ĵ) 0.5(î) (.6î.ĵ) 5..î 0.9ĵ 4.8 ĵ.î.90 ĵ.5î.6î.ĵ î 0.7ĵ kg.5/s 7 05kg d/ d/ O (0, 0) (x, y ) 0.4/s.kg /s d/ d kg O (0, 0) (d, 0) (x, y ) /s Chapter 9 7 R (R, 0) 7.5kg /s

3 8. Two asses & are placed on the X-axis 0 kg, 0kg. The first ass is displaced by a distance of c x x 0 0x Xc 0 Chapter 9 0 0x x x x. The nd ass should be displaced by a distance c towards left so as to kept the position of centre of ass unchanged. 9. Two asses & are kept in a ertical line 0kg, 0kg The first block is raised through a height of 7 c. The centre of ass is raised by c. y y 0 7 0y y 70 +0y 40 0y 0 y. 40 The 0 kg body should be displaced c downward inorder to raise the centre of ass through c. 0. s the hall is graity free, after the ice elts, it would tend to acquire a spherical shape. But, there is no external force acting on the syste. So, the centre of ass of the syste would not oe.. The centre of ass of the blate will be on the syetrical axis. y c ( / )R 4 R / (R (R 4R R 4R R R ( / )R R ) 4 (R R)(R R RR ) (R R )(R R ) R RR ) aboe the centre. R R. 60kg, 40kg, 50kg, et be the origin of the syste. Initially r. Vera & r. athur are at extree position of the boat. The centre of ass will be at a distance fro When they coe to the id point of the boat the C lies at fro. 60kg The shift in C towards right. But as there is no external force in longitudinal direction their C would not shift. So, the boat oes 0. or c towards right.. et the bob fall at,. The ass of bob. The ass of cart. Initially their centre of ass will be at 0 Distance fro P When, the bob falls in the slot the C is at a distance O fro P. 40kg P R 0kg R B 9.

4 Shift in C 0 towards left towards right. But there is no external force in horizontal direction. So the cart displaces a distance towards right. 4. Initially the onkey & balloon are at rest. So the C is at P When the onkey descends through a distance The C will shift 0 t o fro P So, the balloon descends through a distance 5. et the ass of the to particles be & respectiely kg, 4kg ccording to question ½ ½ Now, : / Chapter 9 6. s uraniu 8 nucleus eits a -particle with a speed of /sec. et be the speed of the residual nucleus thoriu /sec /sec so, the earth will recoil at a speed of.5 0 /sec. 8. ass of proton et V p be the elocity of proton a e p Gien oentu of electron kg /sec Gien oentu of antineutrino kg /sec a) The electron & the antineutrino are ejected in the sae direction. s the total oentu is consered the proton should be ejected in the opposite direction V p V p (0.4 /.67). /sec in the opposite direction. b) The electron & antineutrino are ejected r to each other. Total oentu of electron and antineutrino, e (4) (6.4) 0 7 kg /s kg /s Since, V p kg /s So V p 9. /s a g p

5 Chapter 9 9. ass of an, Initial elocity 0 ass of bad et the throws the bag towards left with a elocity towards left. So, there is no external force in the horizontal direction. The oentu will be consered. et he goes right with a elocity V V V..(i) h h et the total tie he will take to reach ground H / g t Hard ground pound et the total tie he will take to reach the height h t Then the tie of his flying t t H / g (H h) / g 9.5 (H h) / g / g H H h Within this tie he reaches the ground in the pond coering a horizontal distance x x V t V x / t x t g H H h s there is no external force in horizontal direction, the x-coordinate of C will reain at that position. (x) x 0 x x The bag will reach the botto at a distance (/) x towards left of the line it falls. 0. ass 50g 0.05kg cos 45 î sin 45 ĵ cos 45 î sin 45 ĵ a) change in oentu 0.05 ( cos 45 î sin 45 ĵ ) 0.05 ( cos 45 î sin 45 ĵ ) 0. cos 45 î 0. sin 45 ĵ +0. cos 45 î + 0. sin 45 ĵ 0. cos 45 î agnitude kg /s c) The change in agnitude of the oentu of the ball P i P f P incidence (h/) cos î (h/) sin ĵ P Reflected (h/) cos î (h/) sin ĵ The change in oentu will be only in the x-axis direction. i.e. P (h/) cos ((h/) cos ) (h/) cos. s the block is exploded only due to its internal energy. So net external force during this process is 0. So the centre ass will not change. et the body while exploded was at the origin of the co-ordinate syste. If the two bodies of equal ass is oing at a speed of 0/s in + x & +y axis direction respectiely, o cos 90 0 /s 45 w.r.t. + x axis x PR h/cos If the centre ass is at rest, then the third ass which hae equal ass with other two, will oe in the opposite direction (i.e. 5 w.r.t. + x- axis) of the resultant at the sae elocity.. Since the spaceship is reoed fro any aterial object & totally isolated fro surrounding, the issions by astronauts couldn t slip away fro the spaceship. So the total ass of the spaceship reain unchanged and also its elocity. y P R h/ x P h/cos P h/sin P R y x

6 4. d c, 0 /s, u 0, 900 kg/ 0.9g/c olue (4/)r (4/) (0.5) 0.58c ass g ass of 000 hailstone Rate of change in oentu per unit area N/ Total force exerted N. 5. ball of ass is dropped onto a floor fro a certain height let h. gh, 0, gh & 0 Rate of change of elocity :- F t u + at gh gh, s h, 0 gh g t t Total tie F h g gh h t h g g 9.6 Chapter 9 6. railroad car of ass is at rest on frictionless rails when a an of ass starts oing on the car towards the engine. The car recoils with a speed backward on the rails. et the ass is oing with a elocity x w.r.t. the engine. The elocity of the ass w.r.t earth is (x ) towards right V c 0 (Initially at rest) 0 + (x ) (x ) x + x x 7. gun is ounted on a railroad car. The ass of the car, the gun, the shells and the operator is 50 where is the ass of one shell. The uzzle elocity of the shells is 00/s. Initial, V c V + 00 V /s /s towards left. 49 When another shell is fired, then the elocity of the car, with respect to the platfor is, 00 V` /s towards left. 49 When another shell is fired, then the elocity of the car, with respect to the platfor is, 00 ` /s towards left Velocity of the car w.r.t the earth is /s towards left Two persons each of ass are standing at the two extrees of a railroad car of ass resting on a sooth track. Case I et the elocity of the railroad car w.r.t the earth is V after the jup of the left an. 0 u + ( + ) V

7 Chapter 9 u V towards right Case II When the an on the right jups, the elocity of it w.r.t the car is u. 0 u u Ust e +e Und (V is the change is elocity of the platfor when platfor itself is taken as reference assuing the car to be at rest) So, net elocity towards left (i.e. the elocity of the car w.r.t. the earth) u u ( ) ( ) 9. sall block of ass which is started with a elocity V on the horizontal part of the bigger block of ass placed on a horizontal floor. Since the sall body of ass is started with a elocity V in the horizontal direction, so the total initial oentu at the initial position in the horizontal direction will reain sae as the total final oentu at the point on the bigger block in the horizontal direction. Fro.C.K. : + O ( + ) 0. ass of the boggli 00kg, V B 0 k/hour. ass of the boy.5kg & V Boy 4k/hour. If we take the boy & boggle as a syste then total oentu before the process of sitting will reain constant after the process of sitting. b V b boy V boy ( b + boy ) (00 +5) /sec 5. ass of the ball 0.5kg, elocity of the ball 5/s ass of the another ball kg et it s elocity /s Using law of conseration of oentu, Velocity of second ball is.5 /s opposite to the direction of otion of st ball.. ass of the an 60kg Speed of the an 0/s ass of the skater 40kg let its elocity /s loss in K.E. (/)60 (0) (/) J. Using law of conseration of oentu. u + u (t) + Where speed of nd particle during collision. u + u u + + (t/ t)( u ) + u t ( u) t t u ( u) t 4. ass of the bullet and speed ass of the ball frictional ass fro the ball. 9.7

8 Chapter 9 Using law of conseration of oentu, + 0 (+ ) + ( ) where final elocity of the bullet + frictional ass ( )V 5. ass of st ball and speed ass of nd ball et final elocities of st and nd ball are and respectiely Using law of conseration of oentu, ( + ). + () lso e () Gien that final K.E. ¾ Initial K.E. ½ + ½ ¾ ½ + ¾ e e e e 6. ass of block kg and speed /s ass of nd block kg. et final elocity of nd block using law of conseration of oentu. ( + ) /s oss in K.E. in inelastic collision (/) () (/) ( + ) () 4 J axiu loss b) ctual loss J (/) (/) + (/) 4 ( + ) ( e ) 4 4 ( + e ) + e e e 7. Final K.E. 0.J Initial K.E. ½ V + 0 ½ 0. u 0.05 u u Where and are final elocities of st and nd block respectiely. + u () ( ) + l (a u ) 0 la..() u 0, u u. dding Eq.() and Eq.() ( + l)u (u/)( + l) u u u ( l) u Gien (/) +(/) g 00 g u u 0 9.8

9 u u u ( ) ( ) 4 ( ) 4 u For axiu alue of u, denoinator should be iniu, l 0. Chapter 9 u 8 u /s For iniu alue of u, denoinator should be axiu, l u 4 u /s 8. Two friends & B (each 40kg) are sitting on a frictionless platfor soe distance d apart rolls a ball of ass 4kg on the platfor towards B, which B catches. Then B rolls the ball towards and catches it. The ball keeps on oing back & forth between and B. The ball has a fixed elocity 5/s. a) Case I : Total oentu of the an & the ball will reain constant /s towards left b) Case II : When B catches the ball, the oentu between the B & the ball will reain constant (0/44) /s Case III : When B throws the ball, then applying.c.. 44 (0/44) /s (towards right) Case IV : When a Catches the ball, the applying.c ( 0.5) /s towards left. s/s o (core -) d e B +e c) Case V : When throws the ball, then applying.c (0/) V V 60/40 / /s towards left. Case VI : When B receies the ball, then applying.c /44 /s towards right. Case VII : When B throws the ball, then applying.c (66/44) V V 80/40 /s towards right. Case VIII : When catches the ball, then applying.c (/) 44 (80/44) (0/) /s towards left. Siilarly after 5 round trips The elocity of will be (50/) & elocity of B will be 5 /s. d) Since after 6 round trip, the elocity of is 60/ i.e. > 5/s. So, it can t catch the ball. So it can only roll the ball six. e) et the ball & the body at the initial position be at origin d 0 X C d u gh elocity on the ground when ball approaches the ground. d B u 9.8 elocity of ball when it separates fro the ground. u 0 u l K.E. of Nucleus (/) E (/) c Energy liited by Gaa photon E. E Decrease in internal energy E c E c linear oentu E/c V 9.9

10 Chapter 9 4. ass of each block and B kg. Initial elocity of the st block, (V) /s V /s, V B 0/s Spring constant of the spring 00 N/. The block strikes the spring with a elocity /s/ fter the collision, it s elocity decreases continuously and at a instant the whole syste (Block + the copound spring + Block B) oe together with a coon elocity. et that elocity be V. Using conseration of energy, (/) V + (/) B V B (/) + (/) B + (/)kx. (/) () + 0 (/) + (/) + (/) x 00 (Where x ax. copression of spring) + 50x () s there is no external force in the horizontal direction, the oentu should be consered. V + B V B ( + B )V. 4 V (/) /s. () /s Putting in eq.() kg kg (/4) + 50x++ (/) 50x B x /00 x (/0) 0. 0c. 4. ass of bullet 0.0kg. Initial elocity of bullet V 500/s 500 /s ass of block, 0kg. Initial elocity of block u 0. Final elocity of bullet 00 /s. et the final elocity of block when the bullet eerges out, if block. + u /s fter oing a distance 0. it stops. change in K.E. Work done 0 (/) 0 (0.8) The projected elocity u. The angle of projection. When the projectile hits the ground for the st tie, the elocity would be the sae i.e. u. Here the coponent of elocity parallel to ground, u cos should reain constant. But the ertical coponent of the projectile undergoes a change after the collision. u sin u e eu sin. Now for the nd u u sin projectile otion, U elocity of projection ( ucos ) (eu sin) u cos and ngle of projection tan eu sin tan (e tan ) a cos or tan e tan () gx sec Because, y x tan () u Here, y 0, tan e tan, sec + e tan nd u u cos + e sin Putting the aboe alues in the equation (), 9.0

11 x e tan x x eu eu u gx ( e (cos tan (cos g( e tan e tan cos g tan e ) sin ) ) sin ) eu sin g So, fro the starting point O, it will fall at a distance u sin eu g sin sin ( e) g g u 44. ngle inclination of the plane the body falls through a height of h, The striking elocity of the projectile with the indined plane Now, the projectile akes on angle (90 ) Velocity of projection u et B. So, x l cos, y l sin Fro equation of trajectory, y x tan gx sec u gh l sin l cos. tan (90 ) l sin l cos. cot So, l g g cos sec (90 gh cos cos ec 4gh cos cos ec sin + cos cot 4h cos 4h (sin + cos cot ) cos ec o ) gh 4h sin cos sin cos cos sin 4h 4 sin cos sin sin cos cos 6 h sin cos sin 45. h 5, 45, e (/4) Here the elocity with which it would strike g 5 0/sec cos 8h sin sin cos Chapter 9 fter collision, let it ake an angle with horizontal. The horizontal coponent of elocity 0 cos 45 will reain unchanged and the elocity in the perpendicular direction to the plane after wllisine. V y e 0 sin 45 l (/4) 0 (.75) /sec V x 0 cos 45 5 So, u x y /sec V V /sec ngle of reflection fro the wall tan.75 tan ngle of projection 90 ( + ) 90 ( ) 8 et the distance where it falls x cos, y sin ngle of projection () 8 l 9.

12 Chapter 9 Using equation of trajectory, y x tan l sin l cos tan 8 gx g cos sec u sec u 8 0 cos 45 sec 8 sin 45 cos 45 tan 8 ( ) (8.8) Soling the aboe equation we get, l ass of block Block of the particle 0g 0.kg. In the equilibriu condition, the spring is stretched by a distance x.00 c g K. x. K 0.0 K 00 N/. The elocity with which the particle will strike is gien by u /sec. So, after the collision, the elocity of the particle and the block is 0. 9 V /sec et the spring be stretched through an extra deflection of. 0 (/) 0. (8/64) 0. 0 ( / 00 ( + 0.) (/) 00 (0.0) Soling the aboe equation we get c 47. ass of bullet 5g 0.05kg. ass of pendulu 5kg. The ertical displaceent h 0c 0. et it strike the pendulu with a elocity u. et the final elocity be. u ( + ) u u u ( ) Using conseration of energy. 0 (/) ( + ). V ( + ) g h u 0 80 /sec. u (0) ass of bullet 0g 0.0kg. ass of wooden block 500g 0.5kg Velocity of the bullet with which it strikes u 00 /sec. et the bullet eerges out with elocity V and the elocity of block V s per law of conseration of oentu. u +.() gain applying work energy principle for the block after the collision, 0 (/) V gh (where h 0.) V gh V gh 0 0. /sec Substituting the alue of V in the equation (), we get\ V /sec. x

13 49. ass of the two blocks are,. Initially the spring is stretched by x 0 Spring constant K. For the blocks to coe to rest again, et the distance traelled by & Be x and x towards right and left respectiely. s o external forc acts in horizontal direction, x x () gain, the energy would be consered in the spring. (/) k x (/) k (x + x x 0 ) x o x + x x 0 x + x x 0 () x x 0 x siilarly x x0 (x 0 x ) x x 0 x x x x0 50. a) Velocity of centre of ass Chapter 9 b) The spring will attain axiu elongation when both elocity of two blocks will attain the elocity of centre of ass. d) x axiu elongation of spring. Change of kinetic energy Potential stored in spring. x x x (/) 0 0 (/) ( + ) ( (/) kx K 0 / 0 kx x 0 5. If both the blocks are pulled by soe force, they suddenly oe with soe acceleration and instantaneously stop at sae position where the elongation of spring is axiu. et x, x extension by block and Total work done Fx + Fx () Increase the potential energy of spring (/) K (x + x ) () Equating () and () F(x + x ) (/) K (x + x ) F (x + x ) K Since the net external force on the two blocks is zero thus sae force act on opposite direction. x x () F nd (x + x ) K x Substituting F + x K F K F F x x K Siilarly x F K F K 9.

14 Chapter 9 5. cceleration of ass F F Siilarly cceleration of ass F F Due to F and F block of ass and will experience different acceleration and experience an inertia force. Net force on F a F F F F F F F F F K F Siilarly Net force on F a F F F F F F F F F If displaces by a distance x and x by the axiu extension of the spring is x +. Work done by the blocks energy stored in the spring., F F F F x + x (/) K (x + x ) x + x K V p / V p V F F 5. ass of the an ( ) is 50 kg. ass of the pillow ( p ) is 5 kg. When the pillow is pushed by the an, the pillow will go down while the an goes up. It becoes the external force on the syste which is zero. acceleration of centre of ass is zero elocity of centre of ass is constant s the initial elocity of the syste is zero. V p V p () Gien the elocity of pillow is 80 ft/s. Which is relatie elocity of pillow w.r.t. an. Vp ( V ) V p + V V p V p / V Putting in equation () V p ( V V ) p / 50 V 5 (8 V ) pillow F 8 0 V 8 V V 0.77/s bsolute elocity of pillow ft/sec. S 8 Tie taken to reach the floor. sec. 7. s the ass of wall >>> then pillow The elocity of block before the collision elocity after the collision. Ties of ascent. sec. Total tie taken. +.. sec. 54. et the elocity of u. et the final elocity when reaching at B becoes collision. (/) (/)u gh h h u gh u gh () B When the block B reached at the upper an s head, the elocity of B is just zero. For B, block (/) 0 (/) gh gh 9.4

15 Chapter 9 Before collision elocity of u, u B 0. fter collision elocity of (say) B gh Since it is an elastic collision the oentu and K.E. should be cosered gh gh lso, (/) + (/) I 0 (/) + (/) gh Diiding () by () ( )( ) ( ) dding () and () But gh + u u.5 gh gh () gh gh gh u 9 gh 4 gh gh gh gh gh + So the block will trael with a elocity greater than.5 Total distance (0. 0. cos 0 ) gh () 55. ass of block 490 g. ass of bullet 0 g. Since the bullet ebedded inside the block, it is an plastic collision. Initial elocity of bullet 50 Velocity of the block is 0. et Final elocity of both /s I 0 ( ) 0 V gh so awake the an by B. V 7 /s. When the block losses the contact at D the coponent g will act on it. VB g sin (V B ) gr sin () r Puttin work energy principle (/) (V B ) (/) (V ) g ( sin ) (/) gr sin (/) 7 g ( sin ).5 (/) sin ( + sin ) sin sin sin (/) 0 ngle of projection tie of reaching the ground h g (0. 0. sin0) 0.47 sec. 9.8 Distance traelled in horizontal direction. s V cos t V B /r gr sin t 9.8 (/ ) D O 90 - B C 490g 0g

16 Chapter et the elocity of reaching at lower end V Fro work energy principle. (/) V (/) 0 g l g. Siilarly elocity of heay block will be gh. V u(say) et the final elocity of and and respectiely. ccording to law of conseration of oentu. x + V + u u + + u () gain, (V V ) [u ( )] V () Subtracting. u g u Substituting in () u 5 - u u + u g - u g b) Putting the work energy principle (/) 0 (/) ( ) g h [ h height gone by heay ball] g (/) l h h 9 9 Siilarly, (/) 0 (/) g h [ height reached by sall ball] (/) 50g g h h Soeh is ore than l, the elocity at height point will not be zero. nd the will rise by a distance l. 57. et us consider a sall eleent at a distance x fro the floor of length dy. So, d dx So, the elocity with which the eleent will strike the floor is, So, the oentu transferred to the floor is, (d) dx gx [because the eleent coes to rest] So, the force exerted on the floor change in oentu is gien by, d dx F gx dt dt gx (Initial position) dx Because, dt gx (for the chain eleent) F gx gx gx gx gain, the force exerted due to x length of the chain on the floor due to its own weight is gien by, W gx (x) g So, the total forced exerted is gien by, gx gx gx F F + W dx x 9.6

17 58. V 0 /s V 0 V, elocity of CB after collision. a) If the edlision is perfectly elastic. V + V () gain, (u ) (0 0) 0 () Subtracting () fro () 0 0 /s. The deacceleration of B g Putting work energy principle (/) 0 (/) a h (/) 0 - g h 00 h b) If the collision perfectly in elastic. u + u ( + ) /s. The two blocks will oe together sticking to each other. Putting work energy principle. (/) 0 (/) g s 0 /s Chapter 9 B u s s et elocity of kg block on reaching the 4kg block before collision u. Gien, V 0 (elocity of 4kg block). Fro work energy principle, /s (/) u (/) ug s u u kg 4kg 5 6 u c u u u 6/s Since it is a perfectly elastic collision. et V, V elocity of kg & 4kg block after collision. V + V () gain, V V (u u ) (0.6 0) 0.6 () Subtracting () fro (). 0.4 /s /s Putting work energy principle for st kg block when coe to rest. (/) 0 (/) (0.) 0. 0 s (/) s S c. Putting work energy principle for 4kg block. (/) 4 0 (/) 4 (0.4) s s S 4 c. Distance between kg & 4kg block S + S c. 60. The block will slide down the inclined plane of ass with acceleration a g sin (relatie) to the inclined plane. The horizontal coponent of a will be, a x g sin cos, for which the block will accelerate towards left. et, the acceleration be a. ccording to the concept of centre of ass, (in the horizontal direction external force is zero). a x ( + ) a 9.7

18 Chapter 9 6. a a x g sin cos () So, the absolute (Resultant) acceleration of on the block along the direction of the incline will be, a g sin - a cos g sin cos cos g sin g sin h a cos a g sin g sin sin So, a g sin () et, the tie taken by the block to reach the botto end be t. Now, S ut + (/) at h sin (/) at t a sin So, the elocity of the bigger block after tie t will be. V u + a t g sin cos h a sin g h sin cos ( ) Now, subtracting the alue of a fro equation () we get, V or V g h sin cos ( ) sin g hcos ( )( sin ) C ( ) gsin ( sin ) / / a sin a g sin cos g sin h h y B The ass is gien a elocity oer the larger ass. a) When the saller block is traelling on the ertical part, let the elocity of the bigger block be towards left. Fro law of conseration of oentu, (in the horizontal direction) ( + ) b) When the saller block breaks off, let its resultant elocity is. Fro law of conseration of energy, (/) (/) + (/) + gh gh..() ( ) gh ( ) ( ) gh / 9.8

19 e) Now, the ertical coponent of the elocity of ass is gien by, y ( ) ( ) [ y y ] gh ( ) ( ) gh ( ) () gh 9.9 Chapter 9 To find the axiu height (fro the ground), let us assue the body rises to a height h, oer and aboe h. Now, (/) y gh h So, Total height h + h h + [fro equation () and ()] H ( )g y g y g () h + h ( )g d) Because, the saller ass has also got a horizontal coponent of elocity at the tie it breaks off fro (which has a elocity ), the block will again land on the block (bigger one). et us find out the tie of flight of block after it breaks off. During the upward otion (BC), 0 y gt / y t gh (4) [fro equation ()] g g ( ) So, the tie for which the saller block was in its flight is gien by, / ( )gh T t g ( ) So, the distance traelled by the bigger block during this tie is, S T or S [ g [ g( ) / ( )gh] ( ) ( )gh] / / / 6. Gien h < < < R. G ass 6 I 0 4 kg. b 0 4 kg. et V e Velocity of earth V b elocity of the block. The two blocks are attracted by graitational force of attraction. The graitation potential energy stored will be the K.E. of two blocks. G pi (/) e e + (/) b b R (h / ) R h gain as the an internal force acts. bvb e V e b V b V e () e

20 Putting in equation () G e b (/) e R h R h b b e (/) b V b b e e + (/) b V b R h R h G (/) V b 0 (R h)(r h) 6 0 s h < < < R, if can be neglected 4 4 R Gh Rh h Chapter 9 (/) V b (/) G h (/) V gh b (/) V b R 6. Since it is not an head on collision, the two bodies oe in different diensions. et V, V elocities of the bodies ector collision. Since, the collision is elastic. pplying law of conseration of oentu on X-direction. u + xo cos + cos cos a + cos b u () u Putting law of conseration of oentu in y direction. 0 sin sin X sin sin () u gain ½ u + 0 ½ + ½ x u + () Squaring equation() u cos + cos + cos cos Equating () & () + cos + cos + cos cos sin + sin cos cos sin sin cos cos sin 64. sin sin cos cos cos cos sin sin 0 cos ( + ) 0 cos 90 ( + 90 cos sin r r r Earth R h Block et the ass of both the particle and the spherical body be. The particle elocity has two coponents, cos noral to the sphere and sin tangential to the sphere. fter the collision, they will exchange their elocities. So, the spherical body will hae a elocity cos and the particle will not hae any coponent of elocity in this direction. [The collision will due to the coponent cos in the noral direction. But, the tangential elocity, of the particle sin will be unaffected] So, elocity of the sphere cos nd elocity of the particle sin r r r 9.0 [fro (fig-)] * * * * *