SOLUTIONS TO CONCEPTS CHAPTER

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1 1. m = kg S = 10m Let, ccelertion =, Initil velocity u = 0. S= ut + 1/ t 10 = ½ ( ) 10 = = 5 m/s orce: = = 5 = 10N (ns) SOLUIONS O CONCEPS CHPE u = 40 km/hr = = m/s m = 000 kg ; v = 0 ; s = 4m v u ccelertion = s = = = 15.4 m/s (decelertion) 8 So, brking force = = = = = N (ns) 3. Initil velocity u = 0 (negligible) v = m/s. s = 1cm = 1 10 m ccelertion = v u s = 110 = = ms = = = = N. 0.kg 0.3kg fig 1 0.3g fig 0.3kg 0.kg fig 3 0.g g = 10m/s 0.3g = 0 = 0.3g = = 3 N 1 (0.g + ) =0 1 = 0.g + = = 5N ension in the two strings re 5N & 3N respectively. S ig 1 1 ig ig 3 + = 0 = + = + = = / 6. m = 50g = 5 10 kg s shown in the figure, from (i) D 15 Slope of O = nθ = = 5 m/s OD 3 So, t t = sec ccelertion is 5m/s orce = = = 0.5N long the motion 5.1 = 0 = (i) v(m/s) D E 6 C

2 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 = = = 0.5 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 ig 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 = 4 = kg s = 10 3 m. v = 0 u = 30 m/s. So, = v u s 3030 = = m/s (decelerting) 3 10 king gnitude only decelertion is m/s So, force = = 1.8 N 9. x = 0 cm = 0.m, k = 15 N/m, m = 0.3kg. ccelertion = m = kx x = m g ig = = 10m/s (decelertion) So, the ccelertion is 10 m/s opposite to the direction of motion 10. Let, the block m towrds left through 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 kg of block. = 10 N 0.m 10N 10/5 = m/s. s there is no friction between &, when the block moves, lock reins t rest in its position. m g ig 3 K 1 1 x m K 5.

3 Initil velocity of = u = 0. Distnce to cover so tht seprte out s = 0. m. ccelertion = m/s s= ut + ½ t 0. = 0 + (½) t t = 0. t = 0.44 sec t= 0.45 sec. 1. ) t ny depth let the ropes ke ngle θ with the verticl rom the free body digrm cos θ + cos θ = 0 cos θ = = cos s the n moves up. θ increses i.e. cos decreses. hus increses. b) hen the n is t depth h cos = orce = h (d/ ) h d h 4 h 4h d 4h 13. rom the free body digrm w = 0 = w 0.5 = 0.5 (10 ) = 4N. So, the force exerted by the block on the block, is 4N. 14. ) he tension in the string is found out for the different conditions from the free body digrm s shown below. ( ) = 0 = m/s = 0.55 N ig-1 ig b) = 0 = = 0.43 N. 1.m/s ig-3 ig-4 c) hen the elevtor kes uniform motion = 0 = = = 0.49 N =0 ig-5 Uniform velocity ig-6 d d/ ig-1 m/s w 0.5 s ig- h 10N d/ == d) = 0 = = 0.43 N. ig-7 =1.m/s ig-8 e) ( ) = 0 = = 0.55 N ig-9 1.m/s ig

4 f) hen the elevtor goes down with uniform velocity ccelertion = 0 = 0 = = = 0.49 N. 15. hen the elevtor is ccelerting upwrds, ximum weight will be recorded. ( + ) = 0 = + = m(g + ) x.wt. hen decelerting upwrds, ximum weight will be recorded. + = 0 = = m(g ) So, m(g + ) = m(g ) = (1) () Now, + = = = m = = 66 Kg 9.9 So, the true weight of the n is 66 kg. gin, to find the ccelertion, + = = 0. 9 m/s Let the ccelertion of the 3 kg ss reltive to the elevtor is in the downwrd direction. s, shown in the free body digrm 1.5 g 1.5(g/10) 1.5 = 0 from figure (1) nd, 3g 3(g/10) + 3 = 0 from figure () = 1.5 g + 1.5(g/10) nd = 3g + 3(g/10) 3 (i) (ii) Eqution (i) 3g + 3(g/10) + 3 = Eqution (ii) 1 3g + 3(g/10) 3 = Subtrcting the bove two equtions we get, = 6 Subtrcting = 6 in eqution (ii) 6 = 3g + 3(g/10) 3. 33g (9.8)33 9 = = = 3.59 = 6 = = = = 1.55 = 43.1 N cut is 1 shown in spring. Mss = wt 43.1 = 4.39 = 4.4 kg g Given, m = kg, k = 100 N/m rom the free body digrm, kl g = 0 kl = g l = g k = = 0. m Suppose further elongtion when 1 kg block is dded be x, hen k(1 + x) = 3g kx = 3g g = g = 9.8 N 9. 8 x = 100 = = 0.1 m ig g 1.5(g/10) ig Uniform velocity ig-1 ig- m 3g 3(g/10) 3 kl g 5.4

5 18. = m/s kl (g + ) = 0 kl = g + = = l = 100 = 0.36 m = 0.4 m kl g hen 1 kg body is dded totl ss ( + 1)kg = 3kg. elongtion be l 1 kl 1 = 3g + 3 = l 1 = 100 = = 0.36 urther elongtion = 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 moving downwrd, + kv = (i) M = kv g or the blloon to rise with constnt velocity v, (upwrd) let the ss be m Here, ( + kv) = 0 (ii) = + kv m = kw g So, mount of ss tht should be removed = M m. kv kv kv kv kv = = g g g g (Mg ) = {M (/g)} G 0. hen the box is ccelerting upwrd, U m(g/6) = 0 U = + /6 = m{g + (g/6)} = 7 /7 (i) m = 6U/7g. hen it is ccelerting downwrd, let the required ss be M. U Mg + Mg/6 = 0 U = 6Mg Mg 6 5Mg 6 M = 6U 5g kv ig-1 v kl m/s 3g M v kv ig- V g/6 /6 ig-1 Mss to be dded = M m = 6U 1 U = g g = 35 6 g from (i) 6U 5g = /5 m. he ss to be dded is m/5. 6U 7g 6U 1 1 g 5 7 V g/6 /6 ig- 5.5

6 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 contining u nd, u should be in the xz-plne. gin, u sin = u = sin u will be minimum, when sin = 1 = 90. u min = m 1 m long Z-xis. m 1 = 0.3 kg, m = 0.6 kg (m 1 g + m 1 ) = 0 (i) = m 1 g + m 1 + m m g = 0 (ii) = m g m rom eqution (i) nd eqution (ii) m 1 g + m 1 + m m g = 0, from (i) (m 1 + m ) = g(m m 1 ) m = m f 9.8 = 3.66 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 (g + ) = 0.3 ( ) = 3.9 N c) he force exerted by the clmp on the pully is given by = 0 = = 3.9 = 7.8 N. 3. = 3.6 m/s = 3.9 N fter sec ss m 1 the velocity V = u + t = = 6.5 m/s upwrd. t this time m is moving 6.5 m/s downwrd. t time sec, m stops for moment. ut m 1 is moving upwrd with velocity 6.5 m/s. It will continue to move till finl velocity (t highest point) becuse zero. Here, v = 0 ; u = 6.5 m 1g m 1 = g = 9.8 m/s [moving up wrd m 1 ] V = u + t 0 = 6.5 +( 9.8)t t = 6.5/9.8 = 0.66 =/3 sec. m g m During this period /3 sec, m ss lso strts moving downwrd. So the string becomes tight gin fter time of /3 sec. 0.3kg m 1 m 0.6kg 5.6

7 4. Mss per unit length 3/30 kg/cm = 0.10 kg/cm. Mss of 10 cm prt = m 1 = 1 kg Mss of 0 cm prt = m = kg. Let, = contct force between them. rom the free body digrm 0 10 = 0 (i) nd, 3 = 0 (ii) rom eq (i) nd (ii) 3 1 = 0 = 1/ 3 = 4 m/s Contct force = = = 4 N kg 1kg 3m 4m Sin 1 = 4/5 g sin 1 ( + ) = 0 g sin = 0 sin = 3/5 g sing 1 = + (i) = g sin + (ii) + g sin rom eqn (i) nd (ii), g sin + + g sin 1 = = g sin 1 g sin = g = g / = g 1 g 5 10 rom the bove ree body digrm M 1 + = 0 = m 1 + (i) m 1 ccelertion of ss m 1 is m 1 5m ig-1 m ig-1 ig-1 m rom the bove free body digrm + m 1 m(m 1 g + ) = 0 m 1 m 1g ig- m 1 (m m 1 1g ig- m 1g ig- ) m g m ig-3 rom the bove ree body digrm m + m g =0.(ii) m + m 1 + m g = 0 (from (i)) (m 1 + m ) + m g/ m g = 0 {becuse f = m g/} (m 1 + m ) m g =0 (m 1 + m ) = m g/ = towrds right. 1 0N 1g 1 1g rom the free body digrm (m g + + m )=0 m m g ig-3 ig-3 1 0N m 1 (m m 1 10m ) 0m m g 3N 3N 5.7

8 = m 1 g + m 1 = 5g (i) = m g + + m = g (ii) rom the eqn (i) nd eqn (ii) 5g = g g 7 = 0 7 = 3g 3 g 9. 4 = = 7 7 = 4. m/s [ g= 9.8m/s ] 5 ) ccelertion of block is 4. m/s b) fter the string breks m 1 move downwrd with force cting down wrd. m 1 = + m 1 g = (1 + 5g) = 5(g + 0.) orce 5(g So, ccelertion = = ss 50.) = (g + 0.) m/s 5g =1N orce = 1N, ccelertion = 1/5= 0.m/s m 1 m m 3 ig-1 ( 1+ ) 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 weight between them. So, the ctul ccelertion t the blocks m 1, m nd m 3 will be 1. ( 1 ) nd ( 1 + ) s shown = 1g 1 = 0...(i) from fig () / g ( 1 ) = 0...(ii) from fig (3) / 3g 3( 1 + ) = 0...(iii) from fig (4) rom eqn (i) nd eqn (ii), eliminting we get, 1g + 1 = 4g + 4( 1 + ) = 3g (iv) rom eqn (ii) nd eqn (iii), we get g + ( 1 ) = 3g 3( 1 ) = (v) g Solving (iv) nd (v) 1 = 9 So, 1 g 19g 17g = g 19g nd = g 5 1 = g = g 19g 1g = So, ccelertion of m 1, m, m 3 e respectively. gin, for m 1, u = 0, s= 0cm=0.m nd = 19 g 9 S = ut + ½ t 1 19 = 0. gt t = 0.5sec. 9 m 1 lg l 1 ig- / m g ig-3 ( 1 ) / m 3 3g ig-4 [g = 10m/s ] 3( 1+ ) / / 19 g 17g (up) 9 9 (don) 1 g (down) 9 =0 1 m 1 m m 3 ig- m 1g g 1 ig-3 3g 3 1 ig-4 ig-1 5.8

9 m 1 should be t rest. m 1 g = 0 / g 1 = 0 / 3g 3 1 =0 = m 1 g (i) 4g 4 1 = 0 (ii) = 6g 6 1 (iii) rom eqn (ii) & (iii) we get 3 1g = 1g = 4g/5= 408g. Putting yhe vlue of eqn (i) we get, m 1 = 4.8kg kg 1 ig-1 1kg = 1g...(i) rom eqn (i) nd (ii), we get g = 1g = g = = = 5m/s rom (ii) = 1 = 5N. 1 =0 = 1 (ii) M = 0 + M Mg = 0 M = = M /. M/ + = Mg. (becuse = M/) 3 M = Mg = g/3 ) ccelertion of ss M is g/3. M M g Mg b) ension = = = = 3 3 c) Let, 1 = resultnt of tensions = force exerted by the clmp on the pulley 1 = / m = ig-1 Mg 3 Mg 3 gin, n = = 1 = 45. M 1g ig- m(/) ig- 1g ig-3 ig So, it is Mg t n ngle of 45 with horizontl. 3 M 30 ig-1 M ig- ig-3 5.9

10 33. M + Mg sin = 0 + M Mg = 0 = M + Mg sin (i) (M + Mg sin ) + M Mg = 0 [rom (i)] 4M + Mgsin + M Mg =0 6M + Mg sin30 Mg = 0 6M = Mg =g/6. ccelertion of ss M is = s g/6 = g/3 up the plne. M Mg D-1 Mg D- D-3 s the block m does not slinover M, ct will hve sme ccelertion s tht of M rom the freebody digrms. + M Mg = 0...(i) (rom D 1) M sin = 0...(ii) (rom D -) sin = 0...(iii) (rom D -3) cos =0...(iv) (rom D -4) D-4 M m Eliminting, nd from the bove eqution, we get M = cot ) 5 + 5g = 0 = 5g 5...(i) (rom D-1) 5 gin (1/) 4g 8 = 0 = 8g 16...(ii) (from D-) rom equn (i) nd (ii), we get 5g 5 = 8g = 3g = 1/7g So, ccelertion of 5 kg ss is g/7 upwrd nd tht of 4 kg ss is = g/7 (downwrd). 5g b) D-1 kg 4 / 5g 5 g 5kg D-3 D-4 M M 8 / 4g D- 5kg D-3 4kg 4 t/ = 0 8 = 0 = 8 (ii) [rom D -4] gin, + 5 5g = g = g = 0 = 5g/13 downwrd. (from D -3) ccelertion of ss () kg is = 10/13 (g) & 5kg () is 5g/13. c) 1kg kg / / C 1g 1 D-5 4 g D g = 0 = 1g 1 (i) [rom D 5] gin, g 4 = 0 4g 8 = 0 (ii) [rom D -6] 1g 1 4g 8 = 0 [rom (i)] 5.10

11 = (g/3) downwrd. ccelertion of ss 1kg(b) is g/3 (up) ccelertion of ss kg() is g/3 (downwrd). 35. m 1 = 100g = 0.1kg m = 500g = 0.5kg m 3 = 50g = 0.05kg g = 0...(i) g =...(ii) g = 0...(iii) rom equn (ii) 1 = 0.05g (iv) rom equn (i) 1 = 0.5g (v) Equn (iii) becomes g = g g g = 0 [rom (iv) nd (v)] = 0.4g = = g = g downwrd D ccelertion of 500gm block is 8g/13g downwrd. 36. m = 15 kg of monkey. = 1 m/s. rom the free body digrm [15g + 15(1)] = 0 = 15 (10 + 1) = = 165 N. he monkey should pply 165N force to the rope. Initil velocity u = 0 ; ccelertion = 1m/s ; s = 5m. s = ut + ½ t 5 = 0 + (1/)1 t t = 5 t = 10 sec. ime required is 10 sec. 37. 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 digrm of monkey = 0...(i) = +. gin, from the D of the block, = 1 = = 0 [rom (i)] = 1 = 1. ccelertion downwrd i.e. upwrd. he block & the monkey move in the sme direction with equl ccelertion. If initilly they re rest (no force is exertied by monkey) no motion of monkey of block occurs s they hve sme weight (sme ss). heir seprtion will not chnge s time psses. 38. Suppose move upwrd with ccelertion, such tht in the til of ximum tension 30N produced g 1 = 30N 0.5 m 3 50g D- 100g m 0.5g g D g 15g 15 5g 1 = 30N 5 ig- ig-3 g 5g 30 5 = 0...(i) 30 g = 0...(ii) = (5 5) = 105 N (x) 30 0 = 0 = 5 m/s So, cn pply ximum force of 105 N in the rope to crry the monkey with it. 5.11

12 or minimum force there is no ccelertion of monkey nd. = 0 Now eqution (ii) is 1 g = 0 1 = 0 N (wt. of monkey ) Eqution (i) is 5g 0 = 0 [s 1 = 0 N] = 5g + 0 = = 70 N. he monkey should pply force between 70 N nd 105 N to crry the monkey with it. 39. (i) Given, Mss of n = 60 kg. Let = pprent weight of n in this cse. Now, + 60g = 0 [rom D of n] = 60g 30g = 0...(i)...(ii) [ rom D of box] 60g 30g = 0 [ rom (i)] = 15g he weight shown by the chine is 15kg. (ii) o get his correct weight suppose the pplied force is nd so, cclertes upwrd with. In this cse, given tht correct weight = = 60g, where g = cc n due to grvity 60 rom the D of the n rom the D of the box g 60 = g 30 = = 0 [ = 60g] 1 60g 30g 30 = 0 1 = 60...(i) 1 30 = 90g = = (ii) rom eqn (i) nd eqn (ii) we get 1 = = 1800N. So, he should exert 1800 N force on the rope to get correct reding. 40. he driving force on the block which n the body to move sown the plne is = sin, So, ccelertion = g sin Initil velocity of block u = 0. s = l, = g sin Now, S = ut + ½ t l = 0 + ½ (g sin ) t g = t = g sin ime tken is g sin g sin 41. Suppose pendulum kes ngle with the verticl. Let, m = ss of the pendulum. rom the free body digrm 1 60g 30g 30 60g 30g v cos = 0 sin =0 cos = = sin = cos...(i) t = sin...(ii) 5.1

13 rom (i) & (ii) tn = = cos sin g he ngle is n 1 (/g) with verticl. (ii) m ss of block. Suppose the ngle of incline is rom the digrm cos sin = 0 cos = sin tn 1 g sin cos tn = /g = tn 1 (/g). 4. ecuse, the elevtor is moving downwrd with n ccelertion 1 m/s (>g), the bodygets seprted. So, body moves with ccelertion g = 10 m/s [freely flling body] nd the elevtor move with ccelertion 1 m/s Now, the block hs ccelertion = g = 10 m/s u = 0 t = 0. sec So, the distnce trvelled by the block is given by. s = ut + ½ t = 0 + (½) 10 (0.) = = 0. m = 0 cm. he displcement of body is 0 cm during first 0. sec. 1 m/s 10 m/s * * * * g 5.13