Page 1 5.1 SOLUTIONS TO CONCEPTS CHAPTER – 5 1. m = 2kg S = 10m Let, acceleration = a, Initial velocity u = 0. S= ut + 1/2 at 2 ? 10 = ½ a (2 2 ) ? 10 = 2a ? a = 5 m/s 2 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’ = s 2 u v 2 2 ? = ? ? 4 2 11 . 11 0 2 2 ? ? = 8 43 . 123 ? = –15.42 m/s 2 (deceleration) So, braking force = F = ma = 2000 × 15.42 = 30840 = 3.08 10 4 N (Ans) 3. Initial velocity u = 0 (negligible) v = 5 × 10 6 m/s. s = 1cm = 1 × 10 –2 m. acceleration a = s 2 u v 2 2 ? = ? ? 2 2 6 10 1 2 0 10 5 ? ? ? ? ? = 2 12 10 2 10 25 ? ? ? = 12.5 × 10 14 ms –2 F = ma = 9.1 × 10 –31 × 12.5 × 10 14 = 113.75 × 10 –17 = 1.1 × 10 –15 N. 4. g = 10m/s 2 T – 0.3g = 0 ? T = 0.3g = 0.3 × 10 = 3 N T 1 – (0.2g + T) =0 ? T 1 = 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 = Tan? OD AD = 3 15 = 5 m/s 2 So, at t = 2sec acceleration is 5m/s 2 Force = ma = 5 × 10 –2 × 5 = 0.25N along the motion fig 1 0.2kg 0.3kg 0.2kg fig 3 0.2g T 1 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 C D E ? ? 4 6 5 15 10 Page 2 5.1 SOLUTIONS TO CONCEPTS CHAPTER – 5 1. m = 2kg S = 10m Let, acceleration = a, Initial velocity u = 0. S= ut + 1/2 at 2 ? 10 = ½ a (2 2 ) ? 10 = 2a ? a = 5 m/s 2 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’ = s 2 u v 2 2 ? = ? ? 4 2 11 . 11 0 2 2 ? ? = 8 43 . 123 ? = –15.42 m/s 2 (deceleration) So, braking force = F = ma = 2000 × 15.42 = 30840 = 3.08 10 4 N (Ans) 3. Initial velocity u = 0 (negligible) v = 5 × 10 6 m/s. s = 1cm = 1 × 10 –2 m. acceleration a = s 2 u v 2 2 ? = ? ? 2 2 6 10 1 2 0 10 5 ? ? ? ? ? = 2 12 10 2 10 25 ? ? ? = 12.5 × 10 14 ms –2 F = ma = 9.1 × 10 –31 × 12.5 × 10 14 = 113.75 × 10 –17 = 1.1 × 10 –15 N. 4. g = 10m/s 2 T – 0.3g = 0 ? T = 0.3g = 0.3 × 10 = 3 N T 1 – (0.2g + T) =0 ? T 1 = 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 = Tan? OD AD = 3 15 = 5 m/s 2 So, at t = 2sec acceleration is 5m/s 2 Force = ma = 5 × 10 –2 × 5 = 0.25N along the motion fig 1 0.2kg 0.3kg 0.2kg fig 3 0.2g T 1 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 C D E ? ? 4 6 5 15 10 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 ? = EC BE = 3 15 = 5. Slope of BC = tan (180° – ?) = – tan ? = –5 m/s 2 (deceleration) Force = ma = 5 × 10 –2 5 = 0.25 N. Opposite to the motion. 7. Let, F ? contact force between m A & m B . And, f ? force exerted by experimenter. F + m A a –f = 0 m B a –f =0 ? F = f – m A a ……….(i) ? F= m B a ……...(ii) From eqn (i) and eqn (ii) ? f – m A a = m B a ? f = m B a + m A a ? f = a (m A + m B ). ? f = B m F (m B + m A ) = ? ? ? ? ? ? ? ? ? B A m m 1 F [because a = F/m B ] ? The force exerted by the experimenter is ? ? ? ? ? ? ? ? ? B A m m 1 F 8. r = 1mm = 10 –3 ‘m’ = 4mg = 4 × 10 –6 kg s = 10 –3 m. v = 0 u = 30 m/s. So, a = s 2 u v 2 2 ? = 3 10 2 30 30 ? ? ? ? = –4.5 × 10 5 m/s 2 (decelerating) Taking magnitude only deceleration is 4.5 × 10 5 m/s 2 So, force F = 4 × 10 –6 × 4.5 × 10 5 = 1.8 N 9. x = 20 cm = 0.2m, k = 15 N/m, m = 0.3kg. Acceleration a = m F = x kx ? = ? ? 3 . 0 2 . 0 15 ? = 3 . 0 3 ? = –10m/s 2 (deceleration) So, the acceleration is 10 m/s 2 opposite to the direction of motion 10. Let, the block m towards left through displacement x. F 1 = k 1 x (compressed) F 2 = k 2 x (expanded) They are in same direction. Resultant F = F 1 + F 2 ? F = k 1 x + k 2 x ? F = x(k 1 + k 2 ) So, a = acceleration = m F = m ) k x(k 2 1 ? opposite to the displacement. 11. m = 5 kg of block A. ma = 10 N ? a 10/5 = 2 m/s 2 . As there is no friction between A & B, when the block A moves, Block B remains at rest in its position. Fig 1 s m 1 m 2 f F R ? m Bg m Ba Fig 3 F R m Ag m Ba Fig 2 F 1 m x K 1 F 2 K 2 10N A B 0.2m Page 3 5.1 SOLUTIONS TO CONCEPTS CHAPTER – 5 1. m = 2kg S = 10m Let, acceleration = a, Initial velocity u = 0. S= ut + 1/2 at 2 ? 10 = ½ a (2 2 ) ? 10 = 2a ? a = 5 m/s 2 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’ = s 2 u v 2 2 ? = ? ? 4 2 11 . 11 0 2 2 ? ? = 8 43 . 123 ? = –15.42 m/s 2 (deceleration) So, braking force = F = ma = 2000 × 15.42 = 30840 = 3.08 10 4 N (Ans) 3. Initial velocity u = 0 (negligible) v = 5 × 10 6 m/s. s = 1cm = 1 × 10 –2 m. acceleration a = s 2 u v 2 2 ? = ? ? 2 2 6 10 1 2 0 10 5 ? ? ? ? ? = 2 12 10 2 10 25 ? ? ? = 12.5 × 10 14 ms –2 F = ma = 9.1 × 10 –31 × 12.5 × 10 14 = 113.75 × 10 –17 = 1.1 × 10 –15 N. 4. g = 10m/s 2 T – 0.3g = 0 ? T = 0.3g = 0.3 × 10 = 3 N T 1 – (0.2g + T) =0 ? T 1 = 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 = Tan? OD AD = 3 15 = 5 m/s 2 So, at t = 2sec acceleration is 5m/s 2 Force = ma = 5 × 10 –2 × 5 = 0.25N along the motion fig 1 0.2kg 0.3kg 0.2kg fig 3 0.2g T 1 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 C D E ? ? 4 6 5 15 10 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 ? = EC BE = 3 15 = 5. Slope of BC = tan (180° – ?) = – tan ? = –5 m/s 2 (deceleration) Force = ma = 5 × 10 –2 5 = 0.25 N. Opposite to the motion. 7. Let, F ? contact force between m A & m B . And, f ? force exerted by experimenter. F + m A a –f = 0 m B a –f =0 ? F = f – m A a ……….(i) ? F= m B a ……...(ii) From eqn (i) and eqn (ii) ? f – m A a = m B a ? f = m B a + m A a ? f = a (m A + m B ). ? f = B m F (m B + m A ) = ? ? ? ? ? ? ? ? ? B A m m 1 F [because a = F/m B ] ? The force exerted by the experimenter is ? ? ? ? ? ? ? ? ? B A m m 1 F 8. r = 1mm = 10 –3 ‘m’ = 4mg = 4 × 10 –6 kg s = 10 –3 m. v = 0 u = 30 m/s. So, a = s 2 u v 2 2 ? = 3 10 2 30 30 ? ? ? ? = –4.5 × 10 5 m/s 2 (decelerating) Taking magnitude only deceleration is 4.5 × 10 5 m/s 2 So, force F = 4 × 10 –6 × 4.5 × 10 5 = 1.8 N 9. x = 20 cm = 0.2m, k = 15 N/m, m = 0.3kg. Acceleration a = m F = x kx ? = ? ? 3 . 0 2 . 0 15 ? = 3 . 0 3 ? = –10m/s 2 (deceleration) So, the acceleration is 10 m/s 2 opposite to the direction of motion 10. Let, the block m towards left through displacement x. F 1 = k 1 x (compressed) F 2 = k 2 x (expanded) They are in same direction. Resultant F = F 1 + F 2 ? F = k 1 x + k 2 x ? F = x(k 1 + k 2 ) So, a = acceleration = m F = m ) k x(k 2 1 ? opposite to the displacement. 11. m = 5 kg of block A. ma = 10 N ? a 10/5 = 2 m/s 2 . As there is no friction between A & B, when the block A moves, Block B remains at rest in its position. Fig 1 s m 1 m 2 f F R ? m Bg m Ba Fig 3 F R m Ag m Ba Fig 2 F 1 m x K 1 F 2 K 2 10N A B 0.2m 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/s 2 ? s= ut + ½ at 2 ? 0.2 = 0 + (½) ×2 × t 2 ? t 2 = 0.2 ? t = 0.44 sec ? t= 0.45 sec. 12. a) at any depth let the ropes make angle ? with the vertical From the free body diagram F cos ? + F cos ? – mg = 0 ? 2F cos ? = mg ? F = ? cos 2 mg As the man moves up. ? increases i.e. cos ??decreases. Thus F increases. b) When the man is at depth h cos ? = 2 2 h ) 2 / d ( h ? Force = 2 2 2 2 h 4 d h 4 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/s 2 B A 0.5×2 R W=mg=0.5×10 Fig-1 2m/s 2 W 0.05×1.2 T Fig-2 W 0.05×1.2 T Fig-4 –a Fig-3 1.2m/s 2 W T Fig-6 a=0 Fig-5 Uniform velocity Fig-7 a=1.2m/s 2 W 0.05×1.2 T Fig-8 W 0.05×1.2 T Fig-10 –a Fig-9 1.2m/s 2 Page 4 5.1 SOLUTIONS TO CONCEPTS CHAPTER – 5 1. m = 2kg S = 10m Let, acceleration = a, Initial velocity u = 0. S= ut + 1/2 at 2 ? 10 = ½ a (2 2 ) ? 10 = 2a ? a = 5 m/s 2 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’ = s 2 u v 2 2 ? = ? ? 4 2 11 . 11 0 2 2 ? ? = 8 43 . 123 ? = –15.42 m/s 2 (deceleration) So, braking force = F = ma = 2000 × 15.42 = 30840 = 3.08 10 4 N (Ans) 3. Initial velocity u = 0 (negligible) v = 5 × 10 6 m/s. s = 1cm = 1 × 10 –2 m. acceleration a = s 2 u v 2 2 ? = ? ? 2 2 6 10 1 2 0 10 5 ? ? ? ? ? = 2 12 10 2 10 25 ? ? ? = 12.5 × 10 14 ms –2 F = ma = 9.1 × 10 –31 × 12.5 × 10 14 = 113.75 × 10 –17 = 1.1 × 10 –15 N. 4. g = 10m/s 2 T – 0.3g = 0 ? T = 0.3g = 0.3 × 10 = 3 N T 1 – (0.2g + T) =0 ? T 1 = 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 = Tan? OD AD = 3 15 = 5 m/s 2 So, at t = 2sec acceleration is 5m/s 2 Force = ma = 5 × 10 –2 × 5 = 0.25N along the motion fig 1 0.2kg 0.3kg 0.2kg fig 3 0.2g T 1 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 C D E ? ? 4 6 5 15 10 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 ? = EC BE = 3 15 = 5. Slope of BC = tan (180° – ?) = – tan ? = –5 m/s 2 (deceleration) Force = ma = 5 × 10 –2 5 = 0.25 N. Opposite to the motion. 7. Let, F ? contact force between m A & m B . And, f ? force exerted by experimenter. F + m A a –f = 0 m B a –f =0 ? F = f – m A a ……….(i) ? F= m B a ……...(ii) From eqn (i) and eqn (ii) ? f – m A a = m B a ? f = m B a + m A a ? f = a (m A + m B ). ? f = B m F (m B + m A ) = ? ? ? ? ? ? ? ? ? B A m m 1 F [because a = F/m B ] ? The force exerted by the experimenter is ? ? ? ? ? ? ? ? ? B A m m 1 F 8. r = 1mm = 10 –3 ‘m’ = 4mg = 4 × 10 –6 kg s = 10 –3 m. v = 0 u = 30 m/s. So, a = s 2 u v 2 2 ? = 3 10 2 30 30 ? ? ? ? = –4.5 × 10 5 m/s 2 (decelerating) Taking magnitude only deceleration is 4.5 × 10 5 m/s 2 So, force F = 4 × 10 –6 × 4.5 × 10 5 = 1.8 N 9. x = 20 cm = 0.2m, k = 15 N/m, m = 0.3kg. Acceleration a = m F = x kx ? = ? ? 3 . 0 2 . 0 15 ? = 3 . 0 3 ? = –10m/s 2 (deceleration) So, the acceleration is 10 m/s 2 opposite to the direction of motion 10. Let, the block m towards left through displacement x. F 1 = k 1 x (compressed) F 2 = k 2 x (expanded) They are in same direction. Resultant F = F 1 + F 2 ? F = k 1 x + k 2 x ? F = x(k 1 + k 2 ) So, a = acceleration = m F = m ) k x(k 2 1 ? opposite to the displacement. 11. m = 5 kg of block A. ma = 10 N ? a 10/5 = 2 m/s 2 . As there is no friction between A & B, when the block A moves, Block B remains at rest in its position. Fig 1 s m 1 m 2 f F R ? m Bg m Ba Fig 3 F R m Ag m Ba Fig 2 F 1 m x K 1 F 2 K 2 10N A B 0.2m 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/s 2 ? s= ut + ½ at 2 ? 0.2 = 0 + (½) ×2 × t 2 ? t 2 = 0.2 ? t = 0.44 sec ? t= 0.45 sec. 12. a) at any depth let the ropes make angle ? with the vertical From the free body diagram F cos ? + F cos ? – mg = 0 ? 2F cos ? = mg ? F = ? cos 2 mg As the man moves up. ? increases i.e. cos ??decreases. Thus F increases. b) When the man is at depth h cos ? = 2 2 h ) 2 / d ( h ? Force = 2 2 2 2 h 4 d h 4 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/s 2 B A 0.5×2 R W=mg=0.5×10 Fig-1 2m/s 2 W 0.05×1.2 T Fig-2 W 0.05×1.2 T Fig-4 –a Fig-3 1.2m/s 2 W T Fig-6 a=0 Fig-5 Uniform velocity Fig-7 a=1.2m/s 2 W 0.05×1.2 T Fig-8 W 0.05×1.2 T Fig-10 –a Fig-9 1.2m/s 2 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 = 9 . 9 2 8 . 1306 ? = 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 9 . 9 66 9 . 9 72 ? ? ? ? ? m/s 2 . 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 g 33 ? a = 34 . 32 10 33 ) 8 . 9 ( ? ?a = 3.59 ? T = 6a = 6 × 3.59 = 21.55 T 1 = 2T = 2 × 21.55 = 43.1 N cut is T 1 shown in spring. Mass = 8 . 9 1 . 43 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 6 . 19 100 8 . 9 2 k g 2 ? ? ? = 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-12 Fig-11 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 5 5.1 SOLUTIONS TO CONCEPTS CHAPTER – 5 1. m = 2kg S = 10m Let, acceleration = a, Initial velocity u = 0. S= ut + 1/2 at 2 ? 10 = ½ a (2 2 ) ? 10 = 2a ? a = 5 m/s 2 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’ = s 2 u v 2 2 ? = ? ? 4 2 11 . 11 0 2 2 ? ? = 8 43 . 123 ? = –15.42 m/s 2 (deceleration) So, braking force = F = ma = 2000 × 15.42 = 30840 = 3.08 10 4 N (Ans) 3. Initial velocity u = 0 (negligible) v = 5 × 10 6 m/s. s = 1cm = 1 × 10 –2 m. acceleration a = s 2 u v 2 2 ? = ? ? 2 2 6 10 1 2 0 10 5 ? ? ? ? ? = 2 12 10 2 10 25 ? ? ? = 12.5 × 10 14 ms –2 F = ma = 9.1 × 10 –31 × 12.5 × 10 14 = 113.75 × 10 –17 = 1.1 × 10 –15 N. 4. g = 10m/s 2 T – 0.3g = 0 ? T = 0.3g = 0.3 × 10 = 3 N T 1 – (0.2g + T) =0 ? T 1 = 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 = Tan? OD AD = 3 15 = 5 m/s 2 So, at t = 2sec acceleration is 5m/s 2 Force = ma = 5 × 10 –2 × 5 = 0.25N along the motion fig 1 0.2kg 0.3kg 0.2kg fig 3 0.2g T 1 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 C D E ? ? 4 6 5 15 10 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 ? = EC BE = 3 15 = 5. Slope of BC = tan (180° – ?) = – tan ? = –5 m/s 2 (deceleration) Force = ma = 5 × 10 –2 5 = 0.25 N. Opposite to the motion. 7. Let, F ? contact force between m A & m B . And, f ? force exerted by experimenter. F + m A a –f = 0 m B a –f =0 ? F = f – m A a ……….(i) ? F= m B a ……...(ii) From eqn (i) and eqn (ii) ? f – m A a = m B a ? f = m B a + m A a ? f = a (m A + m B ). ? f = B m F (m B + m A ) = ? ? ? ? ? ? ? ? ? B A m m 1 F [because a = F/m B ] ? The force exerted by the experimenter is ? ? ? ? ? ? ? ? ? B A m m 1 F 8. r = 1mm = 10 –3 ‘m’ = 4mg = 4 × 10 –6 kg s = 10 –3 m. v = 0 u = 30 m/s. So, a = s 2 u v 2 2 ? = 3 10 2 30 30 ? ? ? ? = –4.5 × 10 5 m/s 2 (decelerating) Taking magnitude only deceleration is 4.5 × 10 5 m/s 2 So, force F = 4 × 10 –6 × 4.5 × 10 5 = 1.8 N 9. x = 20 cm = 0.2m, k = 15 N/m, m = 0.3kg. Acceleration a = m F = x kx ? = ? ? 3 . 0 2 . 0 15 ? = 3 . 0 3 ? = –10m/s 2 (deceleration) So, the acceleration is 10 m/s 2 opposite to the direction of motion 10. Let, the block m towards left through displacement x. F 1 = k 1 x (compressed) F 2 = k 2 x (expanded) They are in same direction. Resultant F = F 1 + F 2 ? F = k 1 x + k 2 x ? F = x(k 1 + k 2 ) So, a = acceleration = m F = m ) k x(k 2 1 ? opposite to the displacement. 11. m = 5 kg of block A. ma = 10 N ? a 10/5 = 2 m/s 2 . As there is no friction between A & B, when the block A moves, Block B remains at rest in its position. Fig 1 s m 1 m 2 f F R ? m Bg m Ba Fig 3 F R m Ag m Ba Fig 2 F 1 m x K 1 F 2 K 2 10N A B 0.2m 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/s 2 ? s= ut + ½ at 2 ? 0.2 = 0 + (½) ×2 × t 2 ? t 2 = 0.2 ? t = 0.44 sec ? t= 0.45 sec. 12. a) at any depth let the ropes make angle ? with the vertical From the free body diagram F cos ? + F cos ? – mg = 0 ? 2F cos ? = mg ? F = ? cos 2 mg As the man moves up. ? increases i.e. cos ??decreases. Thus F increases. b) When the man is at depth h cos ? = 2 2 h ) 2 / d ( h ? Force = 2 2 2 2 h 4 d h 4 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/s 2 B A 0.5×2 R W=mg=0.5×10 Fig-1 2m/s 2 W 0.05×1.2 T Fig-2 W 0.05×1.2 T Fig-4 –a Fig-3 1.2m/s 2 W T Fig-6 a=0 Fig-5 Uniform velocity Fig-7 a=1.2m/s 2 W 0.05×1.2 T Fig-8 W 0.05×1.2 T Fig-10 –a Fig-9 1.2m/s 2 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 = 9 . 9 2 8 . 1306 ? = 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 9 . 9 66 9 . 9 72 ? ? ? ? ? m/s 2 . 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 g 33 ? a = 34 . 32 10 33 ) 8 . 9 ( ? ?a = 3.59 ? T = 6a = 6 × 3.59 = 21.55 T 1 = 2T = 2 × 21.55 = 43.1 N cut is T 1 shown in spring. Mass = 8 . 9 1 . 43 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 6 . 19 100 8 . 9 2 k g 2 ? ? ? = 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-12 Fig-11 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 Chapter-5 5.5 18. a = 2 m/s 2 kl – (2g + 2a) = 0 ?kl = 2g + 2a = 2 × 9.8 + 2 × 2 = 19.6 + 4 ? l = 100 6 . 23 = 0.236 m = 0.24 m When 1 kg body is added total mass (2 + 1)kg = 3kg. elongation be l 1 kl 1 = 3g + 3a = 3 × 9.8 + 6 ? l 1 = 100 4 . 33 = 0.0334 = 0.36 Further elongation = l 1 – l = 0.36 – 0.12 m. 19. Let, the air resistance force is F and Buoyant force is B. Given that F a ? v, where v ? velocity ? F a = kv, where k ? proportionality constant. When the balloon is moving downward, B + kv = mg …(i) ? M = g kv B ? For the balloon to rise with a constant velocity v, (upward) let the mass be m Here, B – (mg + kv) = 0 …(ii) ? B = mg + kv ? m = g kw B ? So, amount of mass that should be removed = M – m. = g kv 2 g kv B kv B g kv B g kv B ? ? ? ? ? ? ? ? = G ) B Mg ( 2 ? = 2{M – (B/g)} 20. When the box is accelerating upward, U – mg – m(g/6) = 0 ? U = mg + mg/6 = m{g + (g/6)} = 7 mg/7 …(i) ? m = 6U/7g. When it is accelerating downward, let the required mass be M. U – Mg + Mg/6 = 0 ? U = 6 Mg 5 6 Mg Mg 6 ? ? ? M = g 5 U 6 Mass to be added = M – m = ? ? ? ? ? ? ? ? ? 7 1 5 1 g U 6 g 7 U 6 g 5 U 6 = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? g U 35 12 35 2 g U 6 = ? ? ? ? ? ? ? ? ? g 1 6 mg 7 35 12 from (i) = 2/5 m. ? The mass to be added is 2m/5. 2a a kl 2g a 2×2 kl 3g 2m/s 2 M Fig-1 v kV mg B Fig-2 v kV mg B Fig-1 g/6T mg/6 mg V Fig-2 g/6T mg/6 mg VRead More

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