Title 05. NEWTONS LAWS OF MOTION Hard Ans.pdf ✅

1. (b) Let forces are F and 2F and angle between them is  and resultant makes an angle  with the force F. tan 90 2 cos 2 sin tan = + =    F F F =   F + 2Fcos = 0  cos = −1 / 2 or  = 120  2. (b) g m m m m a         + − = 1 2 2 1 = 8 g ; by solving 9 / 7 1 2 = m m 3. (b) g m m m a 1 2 2         + = 2 10 3 / 7 3 3  = m s      + = 4. (c) g m m m m a 1 2 1 2 + − = 10 10 6 10 6       + − = 2 = 2.5 m / s 5. (d) Since the mass m2 travels double distance in comparison to mass m1 therefore its acceleration will be double i.e. 2a 6. (a) When the block m2 moves downward with acceleration a, the acceleration of mass m1 will be 2a because it covers double distance in the same time in comparison to m2 . Let T is the tension in the string. By drawing the free body diagram of A and B T = m1 2a ........(i) m2g − 2T = m2a ........(ii) by solving (i) and (ii) ( ) 1 2 2 4m m m g a + = 7. (c) Two frictional force will work on block B. F = fAB + fBG =  ABma g +  BG (mA + mB )g = 0.2 × 100 × 10 + 0.3 (300) × 10 = 200 + 900 = 1100N. (This is the required minimum force) 8. (a) Coefficient of static friction 0.38 20 9.8 75 =  = = R Fl S . 9. (d) When the lift moves down ward with acceleration 'a' then effective acceleration due to gravity g' = g – a g' = g − g = 0 [As the lift falls freely, so a = g] So force of friction = mg ' = 0 10. (a) Here the given angle is called the angle of repose So, 3 1 = tan 30 = o  11. (d) For upper half by the equation of motion v u 2as 2 2 = + 0 2( sin ) / 2 2 2 v = + g  l = glsin [As u = 0, s = l / 2, a = g sin ] For lower half 0 2 (sin cos ) 2 = u + g  −   l /2 [As v = 0,s = l / 2,a = g(sin −  cos) ]  0 = gl sin  + gl(sin −  cos) [As final velocity of upper half will be equal to the initial velocity of lower half]  2 sin =  cos   = 2 tan  12. (a) Frictional force f = R  F cos 60 = (W + F sin 60)  ( 3 sin 60 ) 2 3 1 F cos 60 = g + F  F = 20 N . R F cos 60° f W+F sin 60° Smooth Rough fAB A B F fBG Ground 2T m2 m2g m2a m1 T m1(2a)
13. (d) For pulling of block P  f  mg sin  R  mg sin   (mg − mg cos)  sin   (1 − cos)         2 2 sin 2 cos 2 2 sin 2               2 cot 14. (a) Limiting friction between the block B and the surface FBS =  BS .R = 0.5(m + M ) g = 0.5(2 + 8)10 = 50N but the applied force is 25 N so the lower block will not move i.e. there is no pseudo force on upper block A. Hence there will be no force of friction between A and B. 15. (a) From the above expression, for the equilibrium R = mg cos and F = mg sin . Substituting these value in F = R we get tan  =  or 3 1 cot = =   . 16. (b) By comparing the given condition with general expression 1 2 m M M +  =   2 1 M m + M =  1 2 M M m = −  17. (c) Let the acceleration of the system is a From the F.B.D. of m2 T − F = m 2 a  T − m2 g = m2a  T − 0.03  20 10 = 20a  T − 6 = 20a .....(i) From the FBD of m1 m g T m a 1 − = 1  4 10 − T = 4a  40 − T = 4a ...(ii) Solving (i) and (ii) 1.4 / . 2 a = m s 18. (a) From the expression l l         + = 1 '   l      + = 0.25 1 0.25 [As  = 0.25]  l l 1.25 0.25 ' = 5 l = = 20% of the length of the chain. 19. (c) For vertical equilibrium T W o 1 sin 45 =  o W T sin 45 1 = For horizontal equilibrium o o W cos 45 sin 45 = = W and for the critical condition T2 = F  W = T2 = F = 450 N 20. (a) case (i) Without friction d = g/2 sin 45 t2 ..... (i) case (ii) With friction d = g/2 [sin 45 - k cos 45] t2 n 2 .... (ii) From (i) and (ii) k = 1 - 2 n 1 . 21. (c) mgs sin  = mg cos  s/2 or  = 2 tna . 22. (d) ma cos  = mg sin  or a = g tan  23. (d) +    2 M mcos mg sin cos W T1 45o T1 cos 45o T2 F T1 sin 45o T F m1 m1g m1 a F T m2 m2a 8kg 2kg 25 N A B Surface R+mg cos mg sin  =p f mg
a = ( ) ms 2 10 0.3 15 0.2 m kx − = = . 24. (c) Apply Lami’s theorem sin120 10 sin120 T sin120 T1 2 = =  T1 = T2 = 10 N 25. (d) For equilibrium f = Mg and F = N. for maintaining rotational equilibrium N will shift downward. Hence, torque due to friction about COM = Torque due to Normal reaction about COM. 26. (c) mg sin  = ma cos  or a = g tan  N = ma sin  + mg cos  = mg tan  sin  + mg cos  = cos  mg (sin2  + cos2 ) = cos  mg 27. (a) An object released from a moving body acquires its velocity also. 28. (a) 2 2.4t 0.6t dt dx = − and 2.4 1.2t 0 dt d x 2 2 = − = or t = 2s. 29. (a) mg = kv or v = mg/k 30. (a) m2a = T1 – T2 – m2g sin 37 - m2g cos 37 or a = 2 1 2 m T − T - g sin 37 - 4 1 g cos 37 or T1 – T2 – 0.8g = a ........ (ii) T2 – 0.5 g = 0.5a ......... (iii) m1 g – T1 = m1a or 2g – T1 = 2a .... .... (i) Adding (i), (ii) and (iii) 3.5 a = 0.7 g or a = g/5. 31. (a) v 2 = (2g sin )s or sin  = 0.20 2.4 0.5 2 10 1.5 2.5 2.5 2gs v 2 = =    = or  = 120 . 32. (b) 450 = 60(g - a) or 60 a = 150a = 2.5 ms-2 downwards. 33. (c) 2F = W or F = W/2. 34. (c) N = mg + F sin 53; N = F cos 53 F = sin53 cos53 mg −  = ( ) 16.9N 0.8 0.15 0.6 12 = −  . 35. (c) 1/2 kv dt mdv = − or   = − t 0 v v 1/2 m kdt v dv 0 or m kt v = v0 − 36. (c) at cos  = 2m at cos or v dt mdv 2  = At the break off point mg = at sin  or t = a sin mg . v =    =        2 2 2 2a sin mg cos a sin mg 2m a cos . 37. (c) ma = mg sin  + ma0 sin  or a = g sin  + a0cos ... (i) N = mg cos  - ma0 sin  ..... (ii) N sin  = Ma0 .... (iii) From (ii) and (iii) sin Ma0 = mg cos  - ma0 sin  or a0 = +    2 M msin mg cos sin
38. (c) T = mg = 10 (9.8) = 98 N 39. (d) keq = 2k + 2k = 4k        l 1  k . 40. (c) p = ∫ Fdt = ∫ kt(τ − t)dt = kτ 3 2 τ 0 − kτ 3 3 = kτ 3 6 . 41. (c) F = -kv2 , a = 2 v m k dt dx . dx dv − = or   − = h 0 v v m kdx v dv 0 . or loge h.........( 1) m k v v0 = t.........( 2) m k v 1 v 1 dt m k v dv 0 t 0 v v 2 0  − = − =   form (1) and (2) t = ( ) v v v v log v v h 0 0 e 0 − 42. (d) The force mg is vertical, 2mv perpendicular to vertical plane and mr 2 outward along the diameter. The resultant force is F = ( ) 2 2 4 2 m g + r  + 2v 43. (b) force acting along the tangent to the radius r = R sin . mg sin  - mr  2 cos  = 0 or mg sin           − cos g r 1 2 = 0 or  = cos-1        2 R g 44. (b) N = m(g cos  + a sin ) ma cos  = mg sin  + N ma cos  = mg sin  +  mg cos  +  ma sin  ma (cos   sin ) = mg sin + mg cos  or a = ( )  −    +   cos sin g sin cos 45. (b) m1g – T = m1a1 [from Fig. (ii)] m2g – T = m2a2 [from Fig. (iii)] m0a = 2T [from Fig. (i)] Solving we get a1 =  ( ) ( ) 1 2 0 1 2 1 2 0 1 2 4m m m m m 4m m m m m g + + + − 46. (c) F = 2mv  sin 600 = 2 × 2 × 106 × 2 3 24 60 60 15 2      . = 3.6 × 103N. 47. (d) F = R mv2 x = R mv cos 2 2 0  48. (d) Pseudo force will act in noninertial frame. 49. (b) Resultant of forces be zero. 50. (b) T = M m 2Mmg + . As M > > m × T ~ 2mg 51. (a) If N1 and N2 are the normal reactions at one foot and the other foot, then N1 + N2 = mg ........... (i) For equilibrium, (ma) H = dN1 – dN2 = d(N1 – N2) (taking moments about centre of mass) or N1 – N2 = (ma) d H ........... (ii) From (i) and (ii)