SOLUTIONS TO CONCEPTS CHAPTER 5


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1 1. m k S 10m Let, ccelertion, Initil velocity u 0. S ut + 1/ t 10 ½ ( ) 10 5 m/s orce: m 5 10N (ns) u 40 km/hr m/s m 000 k ; v 0 ; s 4m v u ccelertion s SOLUIONS O CONCEPS CHPE m/s (decelertion) So, brkin force m N (ns) 3. Initil velocity u 0 (neliible) v m/s. s 1cm 1 10 m. 4. ccelertion v u s ms m N. 1 0.k 0.3k fi fi 0.3k 0.k 0. fi m/s N 1 (0. + ) N ension in the two strins re 5N & 3N respectively. S i 1 m i i 3 + m 0 + m + / 6. m k s shown in the fiure, from (i) D 15 Slope of O nθ 5 m/s OD 3 So, t t sec ccelertion is 5m/s orce m N lon the motion 5.1 m 0 m (i) v(m/s) D E 6 C
2 Chpter5 t t 4 sec slope of 0, ccelertion 0 [ tn 0 0] orce 0 t t 6 sec, ccelertion slope of C. E 15 In EC tn θ 5. EC 3 Slope of C tn (180 θ) tn θ 5 m/s (decelertion) orce m N. Opposite to the motion. 7. Let, contct force between m & m. nd, f force exerted by experimenter. s f m 1 m m m i 1 + m f 0 m f 0 f m.(i) m...(ii) rom eqn (i) nd eqn (ii) f m m f m + m f (m + m ). f m (m + m ) 1 [becuse /m ] m m m he force exerted by the experimenter is 1 m 8. r 1mm 10 3 m k s 10 3 m. v 0 u 30 m/s. So, v u s m/s (decelertin) 3 10 kin nitude only decelertion is m/s So, force N 9. x 0 cm 0.m, k 15 N/m, m 0.3k. ccelertion m kx x m i m/s (decelertion) So, the ccelertion is 10 m/s opposite to the direction of motion 10. Let, the block m towrds left throuh displcement x. 1 k 1 x (compressed) k x (expnded) hey re in sme direction. esultnt 1 + k 1 x + k x x(k 1 + k ) x(k So, ccelertion 1 k) opposite to the displcement. m m 11. m 5 k of block. m 10 N 0.m 10N 10/5 m/s. s there is no friction between &, when the block moves, lock remins t rest in its position. m i 3 K 1 1 x m K
3 Initil velocity of u 0. Distnce to cover so tht seprte out s 0. m. ccelertion m/s s ut + ½ t (½) t t 0. t 0.44 sec t 0.45 sec. 1. ) t ny depth let the ropes mke nle θ with the verticl rom the free body dirm cos θ + cos θ 0 cos θ cos s the mn moves up. θ increses i.e. cos decreses. hus increses. b) hen the mn is t depth h cos orce h (d/ ) h d h 4 h 4h d 4h 13. rom the free body dirm w 0 w (10 ) 4N. So, the force exerted by the block on the block, is 4N. m d/ d m/s Chpter5 w 0.5 s i h 10N d/ ) he tension in the strin is found out for the different conditions from the free body dirm s shown below. ( ) m/s 0.55 N. b) N. c) hen the elevtor mkes uniform motion N 0 i5 Uniform velocity i i6 1.m/s i4 d) N. i7 1.m/s i8 e) ( ) N i9 1.m/s
4 f) hen the elevtor oes down with uniform velocity ccelertion N. 15. hen the elevtor is ccelertin upwrds, mximum weiht will be recorded. ( + m ) 0 + m m( + ) mx.wt. hen decelertin upwrds, mximum weiht will be recorded. + m 0 m m( ) So, m( + ) m( ) (1) () Now, + m m m K 9.9 So, the true weiht of the mn is 66 k. in, to find the ccelertion, + m m/s Let the ccelertion of the 3 k mss reltive to the elevtor is in the downwrd direction. s, shown in the free body dirm (/10) from fiure (1) nd, 3 3(/10) from fiure () (/10) nd 3 + 3(/10) 3 (i) (ii) Eqution (i) 3 + 3(/10) + 3 Eqution (ii) (/10) 3 Subtrctin the bove two equtions we et, 6 Subtrctin 6 in eqution (ii) (/10) (9.8) N cut is 1 shown in sprin. Mss wt k Given, m k, k 100 N/m rom the free body dirm, kl 0 kl l k m Suppose further elontion when 1 k block is dded be x, hen k(1 + x) 3 kx N 9. 8 x m 1 m (/10) 1.5 Uniform velocity Chpter5 m i m 3 3(/10) 3 kl 5.4
5 Chpter m/s kl ( + ) 0 kl l m 0.4 m kl hen 1 k body is dded totl mss ( + 1)k 3k. elontion be l 1 kl l urther elontion l 1 l m. 19. Let, the ir resistnce force is nd uoynt force is. Given tht v, where v velocity kv, where k proportionlity constnt. hen the blloon is movin downwrd, + kv (i) M kv kl m/s 3 M or the blloon to rise with constnt velocity v, (upwrd) let the mss be m Here, ( + kv) 0 (ii) + kv m kw So, mount of mss tht should be removed M m. kv kv kv kv kv (M ) {M (/)} G kv v kv i v 0. hen the box is ccelertin upwrd, U m(/6) 0 U + /6 m{ + (/6)} 7 /7 (i) m 6U/7. hen it is ccelertin downwrd, let the required mss be M. U M + M/6 0 U 6M M 6 5M 6 M 6U 5 V /6 /6 Mss to be dded M m 6U 1 U from (i) 6U 5 /5 m. he mss to be dded is m/5. 6U 7 6U V /6 /6 i 5.5
6 Chpter5 1. Given tht, u nd ct on the prticle. y or the prticle to move undeflected with constnt velocity, net force should be zero. ( u ) m x 0 ( u ) Z y ecuse, (u ) is perpendiculr to the plne continin u nd, u should be in the xzplne. in, u sin u sin u will be minimum, when sin u min lon Zxis. m 1 m m 1 m 1 m m m k, m 0.6 k (m 1 + m 1 ) 0 (i) m 1 + m 1 + m m 0 (ii) m m rom eqution (i) nd eqution (ii) m 1 + m 1 + m m 0, from (i) (m 1 + m ) (m m 1 ) m m f ms. m1 m ) t sec ccelertion 3.66 ms Initil velocity u 0 So, distnce trvelled by the body is, S ut + 1/ t 0 + ½(3.66) 6.5 m b) rom (i) m 1 ( + ) 0.3 ( ) 3.9 N c) he force exerted by the clmp on the pully is iven by N m/s 3.9 N fter sec mss m 1 the velocity V u + t m/s upwrd. t this time m is movin 6.5 m/s downwrd. t time sec, m stops for moment. ut m 1 is movin upwrd with velocity 6.5 m/s. It will continue to move till finl velocity (t hihest point) becuse zero. Here, v 0 ; u m/s [movin up wrd m 1 ] V u + t ( 9.8)t t 6.5/ /3 sec. Durin this period /3 sec, m mss lso strts movin downwrd. So the strin becomes tiht in fter time of /3 sec. 0.3k m 1 m 0.6k 5.6
7 Chpter5 4. Mss per unit lenth 3/30 k/cm 0.10 k/cm. Mss of 10 cm prt m 1 1 k Mss of 0 cm prt m k. Let, contct force between them. rom the free body dirm (i) nd, 3 0 (ii) rom eq (i) nd (ii) / 3 4 m/s Contct force N k 1k 3m 4m 5m 1 i 1 0N m 0N m 1 0m m 3N 3N Sin 1 4/5 sin 1 ( + ) 0 sin 0 sin 3/5 sin 1 + (i) sin + (ii) + sin rom eqn (i) nd (ii), sin + + sin sin 1 sin / rom the bove ree body dirm M m 1 + (i) m 1 ccelertion of mss m 1 is m 1 m m rom the bove free body dirm + m 1 m(m 1 + ) 0 m 1 m 1 i m 1 m (m m 1 m 1 i ) m m rom the bove ree body dirm m + m 0.(ii) m + m 1 + m 0 (from (i)) (m 1 + m ) + m / m 0 {becuse f m /} (m 1 + m ) m 0 (m 1 + m ) m / towrds riht. m m rom the free body dirm (m + + m )0 m (m m 1 ) 5.7
8 Chpter5 m 1 + m (i) m + + m (ii) rom the eqn (i) nd eqn (ii) m/s [ 9.8m/s ] 5 ) ccelertion of block is 4. m/s b) fter the strin breks m 1 move downwrd with force ctin down wrd. m 1 + m 1 (1 + 5) 5( + 0.) orce 5( So, ccelertion mss 50.) ( + 0.) m/s 5 1N orce 1N, ccelertion 1/5 0.m/s. 8. / 3( 1+ ) 1 m 1 m / m 1 m m 3 l m 3 ( 1+ ) l 1 i ( 1 ) 3 i4 9. Let the block m+1+ moves upwrd with ccelertion, nd the two blocks m n m 3 hve reltive ccelertion due to the difference of weiht between them. So, the ctul ccelertion t the blocks m 1, m nd m 3 will be 1. ( 1 ) nd ( 1 + ) s shown (i) from fi () / ( 1 ) 0...(ii) from fi (3) / 3 3( 1 + ) 0...(iii) from fi (4) rom eqn (i) nd eqn (ii), elimintin we et, ( 1 + ) (iv) rom eqn (ii) nd eqn (iii), we et + ( 1 ) 3 3( 1 ) (v) Solvin (iv) nd (v) 1 9 So, nd So, ccelertion of m 1, m, m 3 e respectively. in, for m 1, u 0, s 0cm0.m nd 19 9 S ut + ½ t t t 0.5sec. 9 / [ 10m/s ] / (up) 9 9 (don) 1 (down) m 1 m m 3 i m i4 5.8
9 Chpter5 m 1 should be t rest. m 1 0 / 1 0 / m 1 (i) (ii) (iii) rom eqn (ii) & (iii) we et / Puttin yhe vlue of eqn (i) we et, m 1 4.8k k 1 1k 1 1 i (i) rom eqn (i) nd (ii), we et (ii) m/s 31. rom (ii) 1 5N. / m M m(/) i m M 0 + M M 0 M M /. M/ + m M. (becuse M/) 3 M M /3 ) ccelertion of mss M is /3. M M M b) ension 3 3 c) Let, 1 resultnt of tensions force exerted by the clmp on the pulley 1 M 3 M 3 in, n So, it is M t n nle of 45 with horizontl. 3 M 30 M m i m 5.9
10 33. M + M sin 0 + M M 0 M + M sin (i) (M + M sin ) + M M 0 [rom (i)] 4M + Msin + M M 0 6M + M sin30 M 0 6M M /6. ccelertion of mss M is s /6 /3 up the plne. M M D1 M D D3 s the block m does not slinover M, ct will hve sme ccelertion s tht of M rom the freebody dirms. + M M 0...(i) (rom D 1) M sin 0...(ii) (rom D ) sin m 0...(iii) (rom D 3) cos 0...(iv) (rom D 4) 4 t/ (ii) [rom D 4] in, /13 downwrd. (from D 3) ccelertion of mss () k is 10/13 () & 5k () is 5/13. c) m M D4 M m Elimintin, nd from the bove eqution, we et M cot ) (i) (rom D1) 5 in (1/) (ii) (from D) rom equn (i) nd (ii), we et /7 So, ccelertion of 5 k mss is /7 upwrd nd tht of 4 k mss is /7 (downwrd). 5 b) D1 k 4 / 5 5 5k D3 D4 M 8 / 4 D Chpter5 5k D3 4k 1k k / / C 1 1 D5 4 D (i) [rom D 5] in, (ii) [rom D 6] [rom (i)] 5.10
11 (/3) downwrd. ccelertion of mss 1k(b) is /3 (up) ccelertion of mss k() is /3 (downwrd). 35. m k m k m k (i) (ii) (iii) rom equn (ii) (iv) rom equn (i) (v) Equn (iii) becomes [rom (iv) nd (v)] downwrd ccelertion of 500m block is 8/13 downwrd. 36. m 15 k of monkey. 1 m/s. rom the free body dirm [ (1)] 0 15 (10 + 1) N. he monkey should pply 165N force to the rope. Initil velocity u 0 ; ccelertion 1m/s ; s 5m. s ut + ½ t (1/)1 t t 5 t 10 sec. ime required is 10 sec. Chpter Suppose the monkey ccelertes upwrd with ccelertion & the block, ccelerte downwrd with ccelertion 1. Let orce exerted by monkey is equl to rom the free body dirm of monkey m 0...(i) + m. + m + m 1 0 [rom (i)] m m 1 1. ccelertion downwrd i.e. upwrd. he block & the monkey move in the sme direction with equl ccelertion. in, from the D of the block, m 1 0. m m 1 If initilly they re rest (no force is exertied by monkey) no motion of monkey of block occurs s they hve sme weiht (sme mss). heir seprtion will not chne s time psses. 38. Suppose move upwrd with ccelertion, such tht in the til of mximum tension 30N produced. 0.5 D N 0.5 m 3 50 D 100 m D N 5 i (i) (ii) (5 5) 105 N (mx) m/s So, cn pply mximum force of 105 N in the rope to crry the monkey with it. 5.11
12 Chpter5 or minimum force there is no ccelertion of monkey nd. 0 Now eqution (ii) is N (wt. of monkey ) Eqution (i) is [s 1 0 N] N. he monkey should pply force between 70 N nd 105 N to crry the monkey with it. 39. (i) Given, Mss of mn 60 k. Let pprent weiht of mn in this cse. Now, [rom D of mn] 60...(i) (ii) [ rom D of box] [ rom (i)] 15 he weiht shown by the mchine is 15k (ii) o et his correct weiht suppose the pplied force is nd so, cclertes upwrd with. In this cse, iven tht correct weiht 60, where cc n due to rvity rom the D of the mn rom the D of the box [ 60] (i) (ii) rom eqn (i) nd eqn (ii) we et N. So, he should exert 1800 N force on the rope to et correct redin. 40. he drivin force on the block which n the body to move sown the plne is sin, So, ccelertion sin Initil velocity of block u 0. s l, sin Now, S ut + ½ t l 0 + ½ ( sin ) t t sin ime tken is sin sin 41. Suppose pendulum mkes nle with the verticl. Let, m mss of the pendulum. rom the free body dirm v m cos 0 m sin 0 cos m sin cos...(i) t m sin...(ii) 5.1
13 rom (i) & (ii) m tn cos sin he nle is n 1 (/) with verticl. (ii) m mss of block. Suppose the nle of incline is rom the dirm m cos sin 0 m cos sin tn 1 sin cos Chpter5 tn / tn 1 (/). 4. ecuse, the elevtor is movin downwrd with n ccelertion 1 m/s (>), the bodyets seprted. So, body moves with ccelertion 10 m/s [freely fllin body] nd the elevtor move with ccelertion 1 m/s Now, the block hs ccelertion 10 m/s u 0 t 0. sec So, the distnce trvelled by the block is iven by. s ut + ½ t 0 + (½) 10 (0.) m 0 cm. he displcement of body is 0 cm durin first 0. sec. 1 m/s 10 m/s * * * * m m 5.13
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