Title Notes_Exercise_05.Netwons Laws Of Motion.pdf ✅

5.1 SOLUTIONS TO CONCEPTS CHAPTER – 5 1. m = 2kg S = 10m Let, acceleration = a, Initial velocity u = 0. S= ut + 1/2 at2  10 = 1⁄2 a (22 )  10 = 2a  a = 5 m/s2 Force: F = ma = 2 × 5 = 10N (Ans) 2. u = 40 km/hr = 3600 40000 = 11.11 m/s. m = 2000 kg ; v = 0 ; s = 4m acceleration ‘a’ = 2s v u 2 2  =   2 4 0 11.11 2 2   = 8 123.43  = –15.42 m/s2 (deceleration) So, braking force = F = ma = 2000 × 15.42 = 30840 = 3.08 104 N (Ans) 3. Initial velocity u = 0 (negligible) v = 5 × 106 m/s. s = 1cm = 1 × 10–2m. acceleration a = 2s v u 2 2  =   2 2 6 2 1 10 5 10 0      = 2 12 2 10 25 10    = 12.5 × 1014ms–2 F = ma = 9.1 × 10–31 × 12.5 × 1014 = 113.75 × 10–17 = 1.1 × 10–15 N. 4. g = 10m/s2 T – 0.3g = 0  T = 0.3g = 0.3 × 10 = 3 N T1 – (0.2g + T) =0  T1 = 0.2g + T = 0.2 × 10 + 3 = 5N Tension in the two strings are 5N & 3N respectively. 5. T + ma – F = 0 T – ma = 0  T = ma ............(i)  F= T + ma  F= T + T from (i)  2T = F  T = F / 2 6. m = 50g = 5 × 10–2 kg As shown in the figure, Slope of OA = Tanθ OD AD = 3 15 = 5 m/s2 So, at t = 2sec acceleration is 5m/s2 Force = ma = 5 × 10–2 × 5 = 0.25N along the motion fig 1 0.2kg 0.3kg 0.2kg fig 3 0.2g T1 T fig 2 0.3g 0.3kg A Fig 2 mg T R F B Fig 3 mg ma R T B Fig 1 S A 2 v(m/s) 180°–  A B D E C  4 6 5 15 10 Page 1 NETWON'S LAW OF MOTION
Chapter-5 5.2 At t = 4 sec slope of AB = 0, acceleration = 0 [ tan 0° = 0] Force = 0 At t = 6 sec, acceleration = slope of BC. In ∆BEC = tan θ = EC BE = 3 15 = 5. Slope of BC = tan (180° – θ) = – tan θ = –5 m/s2 (deceleration) Force = ma = 5 × 10–2 5 = 0.25 N. Opposite to the motion. 7. Let, F  contact force between mA & mB. And, f  force exerted by experimenter. F + mA a –f = 0 mB a –f =0  F = f – mA a ..........(i)  F= mB a .........(ii) From eqn (i) and eqn (ii)  f – mA a = mB a  f = mB a + mA a  f = a (mA + m B).  f = mB F (mB + mA) =          B A m m F 1 [because a = F/mB]  The force exerted by the experimenter is          B A m m F 1 8. r = 1mm = 10–3 ‘m’ = 4mg = 4 × 10–6kg s = 10–3m. v = 0 u = 30 m/s. So, a = 2s v u 2 2  = 3 2 10 30 30     = –4.5 × 105 m/s2 (decelerating) Taking magnitude only deceleration is 4.5 × 105 m/s2 So, force F = 4 × 10–6 × 4.5 × 105 = 1.8 N 9. x = 20 cm = 0.2m, k = 15 N/m, m = 0.3kg. Acceleration a = m F = x kx =   3.0 15 2.0 = 3.0 3  = –10m/s2 (deceleration) So, the acceleration is 10 m/s2 opposite to the direction of motion 10. Let, the block m towards left through displacement x. F1 = k1 x (compressed) F2 = k2 x (expanded) They are in same direction. Resultant F = F1 + F2  F = k1 x + k2 x  F = x(k1 + k2) So, a = acceleration = m F = m x(k k ) 1  2 opposite to the displacement. 11. m = 5 kg of block A. ma = 10 N  a 10/5 = 2 m/s2 . As there is no friction between A & B, when the block A moves, Block B remains at rest in its position. Fig 1 s f m1 m2 F R mBg mBa Fig 3 F R mAg mBa Fig 2 F1 m x K1 F2 K2 10N A B 0.2m Page 2 NETWON'S LAW OF MOTION
Chapter-5 5.3 Initial velocity of A = u = 0. Distance to cover so that B separate out s = 0.2 m. Acceleration a = 2 m/s2  s= ut + 1⁄2 at2  0.2 = 0 + (1⁄2) ×2 × t2  t 2 = 0.2  t = 0.44 sec  t= 0.45 sec. 12. a) at any depth let the ropes make angle θ with the vertical From the free body diagram F cos θ + F cos θ – mg = 0  2F cos θ = mg  F = 2cos mg As the man moves up. θ increases i.e. cos decreases. Thus F increases. b) When the man is at depth h cos  = 2 2 )2/d( h h  Force = 2 2 2 2 d 4h 4h mg h 4 d h mg    13. From the free body diagram  R + 0.5 × 2 – w = 0  R = w – 0.5 × 2 = 0.5 (10 – 2) = 4N. So, the force exerted by the block A on the block B, is 4N. 14. a) The tension in the string is found out for the different conditions from the free body diagram as shown below. T – (W + 0.06 × 1.2) = 0  T = 0.05 × 9.8 + 0.05 × 1.2 = 0.55 N. b)  T + 0.05 × 1.2 – 0.05 × 9.8 = 0  T = 0.05 × 9.8 – 0.05 × 1.2 = 0.43 N. c) When the elevator makes uniform motion T – W = 0  T = W = 0.05 × 9.8 = 0.49 N d) T + 0.05 × 1.2 – W = 0  T = W – 0.05 × 1.2 = 0.43 N. e) T – (W + 0.05 × 1.2) = 0  T = W + 0.05 × 1.2 = 0.55 N s 10N R w ma F  d Fig-1  d/2 F  Fig-2  mg F F  h d/2 A mg 2 m/s2 B A 0.5×2 R W=mg=0.5×10 Fig-1 2m/s2 W 0.05×1.2 T Fig-2 W 0.05×1.2 T Fig-4 –a Fig-3 1.2m/s2 W T Fig-6 a=0 Fig-5 Uniform velocity Fig-7 a=1.2m/s2 W 0.05×1.2 T Fig-8 W 0.05×1.2 T Fig-10 –a Fig-9 1.2m/s2 Page 3 NETWON'S LAW OF MOTION
Chapter-5 5.4 f) When the elevator goes down with uniform velocity acceleration = 0 T – W = 0  T = W = 0.05 × 9.8 = 0.49 N. 15. When the elevator is accelerating upwards, maximum weight will be recorded. R – (W + ma ) = 0  R = W + ma = m(g + a) max.wt. When decelerating upwards, maximum weight will be recorded. R + ma – W = 0 R = W – ma = m(g – a) So, m(g + a) = 72 × 9.9 ...(1) m(g – a) = 60 × 9.9 ...(2) Now, mg + ma = 72 × 9.9  mg – ma = 60 × 9.9  2mg = 1306.8  m = 2 9.9 1306 8.  = 66 Kg So, the true weight of the man is 66 kg. Again, to find the acceleration, mg + ma = 72 × 9.9  a = 9.0 11 9.9 66 72 9.9 66 9.9      m/s2 . 16. Let the acceleration of the 3 kg mass relative to the elevator is ‘a’ in the downward direction. As, shown in the free body diagram T – 1.5 g – 1.5(g/10) – 1.5 a = 0 from figure (1) and, T – 3g – 3(g/10) + 3a = 0 from figure (2)  T = 1.5 g + 1.5(g/10) + 1.5a ... (i) And T = 3g + 3(g/10) – 3a ... (ii) Equation (i) × 2  3g + 3(g/10) + 3a = 2T Equation (ii) × 1  3g + 3(g/10) – 3a = T Subtracting the above two equations we get, T = 6a Subtracting T = 6a in equation (ii) 6a = 3g + 3(g/10) – 3a.  9a = 10 33g  a = 32.34 10 )8.9( 33  a = 3.59  T = 6a = 6 × 3.59 = 21.55 T1 = 2T = 2 × 21.55 = 43.1 N cut is T1 shown in spring. Mass = 8.9 43 1. g wt  = 4.39 = 4.4 kg 17. Given, m = 2 kg, k = 100 N/m From the free body diagram, kl – 2g = 0  kl = 2g  l = 100 19 6. 100 2 8.9 k 2g    = 0.196 = 0.2 m Suppose further elongation when 1 kg block is added be x, Then k(1 + x) = 3g  kx = 3g – 2g = g = 9.8 N  x = 100 8.9 = 0.098 = 0.1 m W T Fig-11 Fig-12 Uniform velocity m W R a ma W R a ma –a Fig-1 1.5g T 1.5(g/10) 1.5a Fig-2 3g T 3(g/10) 3a kl 2g Page 4 NETWON'S LAW OF MOTION