Tiêu đề 05. NEWTONS LAWS OF MOTION EASY Ans.pdf ✅

24. (a) Force on the block F N dt dm u  = 5 1 = 5      =  Acceleration of block a 2 2.5 / 2 5 m s m F = = = 25. (a) Opposing force N dt dm F u  = 2 0.5 = 1      =       = dt du As,F u So same amount of force is required to keep the belt moving at 2 m/s 26. (d) Resultant force is w + 3w = 4w 27. (c) Acceleration 2 5 / 10 50 Mass Force m s k g N = = = From v = u + at = 0 + 5  4 = 20 m /s 28. (c) Thrust       = dt dm F u N 4 6 = 5 10  40 = 2 10 29. (d) In stationary lift man weighs 40 kg i.e. 400 N. When lift accelerates upward it's apparent weight = m(g + a) = 40(10 + 2) = 480 N i.e. 48 kg For the clarity of concepts in this problem kg-wt can be used in place of kg. 30. (d) As the apparent weight increase therefore we can say that acceleration of the lift is in upward direction. R = m(g + a)  4.8 g = 4(g + a)  2 a = 0.2g = 1.96 m /s 31. (d) T = m (g + a) = 6000 (10 + 5) = 90000 N 32. (a) F = ma N t m v 40 0.1 0.2 20 =  =   = 33. (a) N dt dv F m 5000 0.1 100 5 =   =      = 34. (d) 35. (b) F m g a N 3 4 = ( + ) = 20 10 (10 + 4) = 28 10 36. (b) 2 3 ( ) = m g − a mg  a = g / 3 37. (a) T = m (g + a) = 500 (10 + 2) = 6000 N 38. (a)       = dt dm F u  kg s u F dt dm 0.7 / 300 210 = = = 39. (d) R = m (g + a) = m (g + g) = 2mg 40. (a) 2 3 2 1 ( ) 1 g g T m g a g  =      = + =  + 2 2 2 ( ) 1 g g T m g a g  =      = − =  −  1 3 2 1 = T T 41. (b) m(g a) dt udm F = = +  u m g a dt dm ( + ) = 800 5000 (10 + 20) = = 187 .5 kg/s 42. (c) Initially due to upward acceleration apparent weight of the body increases but then it decreases due to decrease in gravity. 43. (b) g l T = 2 and 4 /3 ' 2 g l T =  3 4 3 [ ' g g As g = g + a = g + = ]  T' = T 2 3 44. (b) Density of cork = d, Density of water =  Resultant upward force on cork = V( − d) g This causes elongation in the spring. When the lift moves down with acceleration a, the resultant upward force on cork = V( − d)(g − a) which is less than the previous value. So the elongation decreases. 45. (d) When trolley are released then they posses same linear momentum but in opposite direction. Kinetic energy acquired by any trolley will dissipate against friction.  m p mg s 2 2  =  2 1 m s  [As P and u are constants]  1 9 1 3 2 2 1 2 2 1  =      =         = m m s s 46. (b) Apparent weight = m(g − a) = 50 (9.8 − 9.8) = 0 47. (b) Opposite force causes retardation. 48. (a) T = m(g − a) = 10(980 – 400) = 5800 dyne 49. (d) g l T = 2 . T will decrease, If g increases. It is possible when rocket moves up with uniform acceleration. Cork
50. (c) We know that in the given condition 2 1 m s   2 2 1 1 2         = m m s s  1 2 2 1 2 s m m s          = 51. (a) 200 1 6 8 10 2 2 2 = + + = = a F m = 10 2 kg 52. (b) In the absence of external force, position of centre of mass remain same therefore they will meet at their centre of mass. 53. (d)Force N dt dv m 25 20 500 0.01 0.25 [(10) ( 10)] =  =  − −  =      = 54. (d) T mg 50 10 10 0.5 N 3 = =   = − 55. (a) N dt dm F u 16.66 60 50  = 20  =      = 56. (d) u = 250 m / s , v = 0 , s = 0.12 metre 2 0.12 20 10 (250 ) 2 2 2 3 2    =         − = = − s u v F ma m  F N 3 = 5.210 57. (a) N t v u F m 12.5 0.2 5(65 15) 10 2 = −   =      − = − 58. (c) t m s m F v u 10 15 / 1000 1000 500 10  =      − = + = + 59. (b) N t m u v F ma 4 4 ( ) 2 (8 0) =  − = − = = 60. (d) R = m(g + a) = 10 (9.8 + 2) = 118 N 61. (d) N S m u v F 1800 2 12 10 30 10 (120 ) 2 ( ) 2 2 2 3 2 =     = − = − − 62. (b) dp = Fdt = 10 10 = 100 k g m/s 63. (d) R = m (g − a) = m (10 −10) = zero 64. (b) Force exerted by the ball  N dt dv F m 30 0.1 20  = 0.15  =      = 65. (d) If rope of lift breaks suddenly, acceleration becomes equal to g so that tension, T = m(g − g) = 0 66. (d) R = m(g + a) = 50 (10 + 2) = 600 N = 60 k gwt 67. (b) N dt dm F u 500 50 10 25 3  =   =      = − 68. (a) SHorizontal= ut = 1.5 4 = 6 m t m m F S at 1 16 8 2 1 2 1 2 1 2 2 Vertical = = =   = S 6 8 10 m 2 2 Net = + = 69. (c) T = m (g + a) = 1000 (9.8 +1) = 10800 N 70. (d) The effective acceleration of ball observed by observer on earth = (a – a0) As , a0  a hence net acceleration is in downward direction. 71. (c) Due to relative motion, acceleration of ball observed by observer in lift = (g – a) and for man on earth the acceleration remains g. 72. (c) For accelerated upward motion R = m (g + a) = 80 (10 + 5) = 1200 N 73. (c) Tension the string = m(g + a) = Breaking force  20(g + a) = 25  g  2 a = g / 4 = 2.5 m /s 74. (b) Rate of flow will be more when lift will move in upward direction with some acceleration because the net downward pull will be more and vice-versa. ( ) Fupward = m g + a and ( ) Fdownward = m g − a 75. (c) Initial thrust must be m g a N 4 5 [ + ] = 3.5 10 (10 +10) = 7 10 76. (b) When the lift is stationary W = mg  49 = m9.8  m = 5 kg. When the lift is moving downward with an acceleration R = m (9.8 − a) = 5[9.8 − 5] = 24 N 77. (a) When car moves towards right with acceleration a then due to pseudo force the plumb line will tilt in backward direction making an angle  with vertical. From the figure, tan  = a / g  tan ( / ) 1 a g −  = 78. (a) R = m(g − a) = 0 79. (b) Displacement of body in 4 sec along OE  a  g a
sx = vx t = 3  4 = 12 m Force along OF (perpendicular to OE) = 4 N  2 2 / 2 4 m s m F ay = = = Displacement of body in 4 sec along OF  2 2 1 s u t a t y = y + y 2 (4) 16 m 2 1 2 =   = [As uy = 0 ]  Net displacement s sx sy (12) (16) 20 m 2 2 2 2 = + = + = 80. (d) When the whole system is accelerated towards left then pseudo force (ma) works on a block towards right. For the condition of equilibrium mg sin = ma cos     cos g sin a =  Force exerted by the wedge on the block R = mg cos  + ma sin R     sin cos sin cos       = + g mg m    cos (cos sin ) 2 2 + = mg R cos mg = 81. (d) u = velocity of bullet = dt dm Mass fired per second by the gun dt dm = Mass of bullet (mB) × Bullets fired per sec (N) Maximum force that man can exert       = dt dm F u  F = u mB  N  3 40 10 1200 144 3 =   =  = − m u F N B 82. (d) The stopping distance, 2 S  u ( 2 ) 2 2 v = u − as  4 60 120 2 2 1 2 1 2  =      =         = u u S S  S2 = 4 S1 = 4  20 = 80 m 83. (d) The apparent weight, R = m(g + a) = 75(10 + 5) = 1125 N 84. (c) By drawing the free body diagram of point B Let the tension in the section BC and BF are T1 and T2 respectively. From Lami's theorem  =  = sin120  sin120 sin120 T1 T2 T  10 . T = T1 = T2 = N 85. (d) a bt bt dt d dt dp F ( ) 2 2 = = + =  F  t 86. (a) When the lift moves upwards, the apparent weight, = m(g + a) . Hence reading of spring balance increases. 87. (c) When lift is at rest, T = 2 l / g If acceleration becomes g/4 then T g l g l T = = = 2 4 2 / 4 2  88. (b) The apparent weight of man, R = m(g + a) = 80(10 + 6) = 1280 N 89. (b) 10 200 /sec 5 100 0 t cm m F v u at  =       =      = + = + 90. (b) 91. (a) p p p mv mv mv  = i − f = −(− ) = 2 92. (d) In the condition of free fall apparent weight becomes zero. 93. (a) Total mass of bullets = Nm, time n N t = Momentum of the bullets striking the wall = Nmv Rate of change of momentum (Force) = t Nmv = nmv. 94. (b) 95. (c) If man slides down with some acceleration then its apparent weight decreases. For critical condition rope can bear only 2/3 of his weight. If a is the minimum acceleration then, Tension in the rope = m(g − a) = Breaking strength  m g a mg 3 2 ( − ) =  3 3 2g g a = g − = 96. (a) For exerted by ball on wall = rate of change in momentum of ball F = 4N vx = 3m/s O E F ux = 0 mg sin  ma cos ma mg mg cos + ma sin  a  R   T1 T2 120° A C F 120° 120° T=10N B