JEE > DC Pandey Solutions for JEE Physics > DC Pandey Solutions: Laws of Motion- 1

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Page 1 Introductory Exercise 5.1 1. N = Nor mal force on cylinder by plank R = Normal force on cylinder by ground, f = Force of friction by ground by cylinder, w = Weight of cylinder. N f cos q = w N R + = sin q , N R ( ) = Reaction to N, i.e.,normal force on plank by cylinder R¢ = Normal force on plank by ground, w = Weight of plank, f ¢ = frictional force on plank by ground. Resultant of f ¢ and R¢, N R ( ) and w pass through point O. 2. R = Normal force on sphere A by left wall, N = Normal force on sphere A by ground, N ¢ = Normal force on sphere A by sphere B, w A = Weight of sphere A. N R ¢ = cos q N w N A ¢ + = sin q R¢ = Normal force on sphere B by right wall, N R ( ) = Reaction to N i.e. normal force on sphere B by sphere A, w B = Weight of sphere B, R¢, N R ( ) and w B pass through point O, the centre sphere B. 3. N = Normal force on sphere by wall, Laws of Motion 5 N(R) Q R' q O w f' P Ground (Force acting on plank) N q R w A N' Force on sphere A N (R) q R w B O R' B Force of sphere B N O B w C q A w R q f N Force acting on cylinder Page 2 Introductory Exercise 5.1 1. N = Nor mal force on cylinder by plank R = Normal force on cylinder by ground, f = Force of friction by ground by cylinder, w = Weight of cylinder. N f cos q = w N R + = sin q , N R ( ) = Reaction to N, i.e.,normal force on plank by cylinder R¢ = Normal force on plank by ground, w = Weight of plank, f ¢ = frictional force on plank by ground. Resultant of f ¢ and R¢, N R ( ) and w pass through point O. 2. R = Normal force on sphere A by left wall, N = Normal force on sphere A by ground, N ¢ = Normal force on sphere A by sphere B, w A = Weight of sphere A. N R ¢ = cos q N w N A ¢ + = sin q R¢ = Normal force on sphere B by right wall, N R ( ) = Reaction to N i.e. normal force on sphere B by sphere A, w B = Weight of sphere B, R¢, N R ( ) and w B pass through point O, the centre sphere B. 3. N = Normal force on sphere by wall, Laws of Motion 5 N(R) Q R' q O w f' P Ground (Force acting on plank) N q R w A N' Force on sphere A N (R) q R w B O R' B Force of sphere B N O B w C q A w R q f N Force acting on cylinder w = Weight of sphere, T = Tension in string. 4. Component of F 1 ® along x-axis : 4 30 2 3 cos ° = N along y-axis : 4 30 2 sin ° = N Component of F 2 ® along x-axis : 4 120 2 cos ° = - N along y-axis : 4 120 2 3 sin ° = N Component of F 3 ® along x-axis : 6 270 0 cos ° = N along y-axis : 6 270 6 sin ° = - N Component of F 4 ® along x-axis : 4 0 4 cos ° = N along y-axis : 4 0 0 sin ° = N 5. Tak ing mo ment about point A AB l = ( sin ) T l w l 30 2 ° = Þ T w = 6. See figure (answer to question no. 3) sin q = + OA OB BC = + = a a a 1 2 T w cos 30° = or T w 3 2 = or T w = 2 3 7. R f cos cos 30 3 60 ° + = ° i.e., R f 3 2 3 2 + = or R f 3 6 + = …(i) and R f sin sin 30 60 10 ° + ° = i.e., R f 1 2 3 2 10 + = or R f + = 3 20 …(ii) Substituting the value of f from Eq. (i) in Eq. (ii) R R + + = ( ) 3 6 3 20 4 6 3 20 R + = Þ R = - = 20 6 3 4 2.4 N \ f = + ( ) 2.4 3 6 =10.16 N 8.At point B (instantaneous vertical acceleration only) \ mg T ma - ° = sin 45 …(i) At point A (instantaneous horizontal acceleration only) \ T ma cos 45° = …(ii) Combining Eqs. (i) and (ii) mg ma ma - = Þ a g = 2 Laws of Motion 75 30° T sin 30° B O w A T AB = l 30° R f 60° 60° 10 N 3N N A T T 45° B mg Page 3 Introductory Exercise 5.1 1. N = Nor mal force on cylinder by plank R = Normal force on cylinder by ground, f = Force of friction by ground by cylinder, w = Weight of cylinder. N f cos q = w N R + = sin q , N R ( ) = Reaction to N, i.e.,normal force on plank by cylinder R¢ = Normal force on plank by ground, w = Weight of plank, f ¢ = frictional force on plank by ground. Resultant of f ¢ and R¢, N R ( ) and w pass through point O. 2. R = Normal force on sphere A by left wall, N = Normal force on sphere A by ground, N ¢ = Normal force on sphere A by sphere B, w A = Weight of sphere A. N R ¢ = cos q N w N A ¢ + = sin q R¢ = Normal force on sphere B by right wall, N R ( ) = Reaction to N i.e. normal force on sphere B by sphere A, w B = Weight of sphere B, R¢, N R ( ) and w B pass through point O, the centre sphere B. 3. N = Normal force on sphere by wall, Laws of Motion 5 N(R) Q R' q O w f' P Ground (Force acting on plank) N q R w A N' Force on sphere A N (R) q R w B O R' B Force of sphere B N O B w C q A w R q f N Force acting on cylinder w = Weight of sphere, T = Tension in string. 4. Component of F 1 ® along x-axis : 4 30 2 3 cos ° = N along y-axis : 4 30 2 sin ° = N Component of F 2 ® along x-axis : 4 120 2 cos ° = - N along y-axis : 4 120 2 3 sin ° = N Component of F 3 ® along x-axis : 6 270 0 cos ° = N along y-axis : 6 270 6 sin ° = - N Component of F 4 ® along x-axis : 4 0 4 cos ° = N along y-axis : 4 0 0 sin ° = N 5. Tak ing mo ment about point A AB l = ( sin ) T l w l 30 2 ° = Þ T w = 6. See figure (answer to question no. 3) sin q = + OA OB BC = + = a a a 1 2 T w cos 30° = or T w 3 2 = or T w = 2 3 7. R f cos cos 30 3 60 ° + = ° i.e., R f 3 2 3 2 + = or R f 3 6 + = …(i) and R f sin sin 30 60 10 ° + ° = i.e., R f 1 2 3 2 10 + = or R f + = 3 20 …(ii) Substituting the value of f from Eq. (i) in Eq. (ii) R R + + = ( ) 3 6 3 20 4 6 3 20 R + = Þ R = - = 20 6 3 4 2.4 N \ f = + ( ) 2.4 3 6 =10.16 N 8.At point B (instantaneous vertical acceleration only) \ mg T ma - ° = sin 45 …(i) At point A (instantaneous horizontal acceleration only) \ T ma cos 45° = …(ii) Combining Eqs. (i) and (ii) mg ma ma - = Þ a g = 2 Laws of Motion 75 30° T sin 30° B O w A T AB = l 30° R f 60° 60° 10 N 3N N A T T 45° B mg Introductory Exercise 5.2 1. Acceleration of system a = + + - + + ( ) ( ) 120 50 1 4 2 =10 m/s 2 Let normal force between 1 kg block and 4 kg block = F 1 \ Net force on 1 kg block = - 120 N \ a F = - 120 1 1 or 10 120 1 = - F i.e., F 1 110 = N Net force on 2 kg block = ´ 2 a = ´ 2 10 =20 N 2. As, 4 30 2 30 g g sin sin ° > ° The normal force between the two blocks will be zero. 3. N R mg ( ) = 4 \ N mg = 4 As lift is moving downward with acceleration a, the pseudo force on A will be ma acting in the upward direction. For the block to be at rest w.r.t. lift. N ma mg + = or mg ma mg 4 + = Þ a g = 3 4 4. Angle made by the string with the normal to the ceiling = = ° q 30 As the train is moving with constant velocity no pseudo force will act on the plumb-bob. Tension in spring = mg = ´ 1 10 = 10 N 5. Pseudo force ( ) = ma on plumb-bob will be as shown in figure T mg ma cos cos ( ) f = + ° - 90 q i.e., T mg ma cos sin f = + q …(i) and T ma sin cos f = q Squaring and adding Eqs. (i) and (ii), T m g m a m ag 2 2 2 2 2 2 2 2 = + + sin sin q q + m a 2 2 2 cos q …(iii) T m g m a m ag 2 2 2 2 2 2 = + + ( Qq = ° 30 ) = + + × m g m g m g g 2 2 2 2 2 4 2 (Qa g = 2 ) = 7 4 2 2 m g or T mg = 7 2 =5 7 N Dividing Eq. (i) by Eq. (ii), tan cos sin f = + ma mg ma q q 76 | Mechanics-1 mg N (R) = N ma N a A q = 30° v (constant) mg T q q = 30° f a 90° – q ma T mg Page 4 Introductory Exercise 5.1 1. N = Nor mal force on cylinder by plank R = Normal force on cylinder by ground, f = Force of friction by ground by cylinder, w = Weight of cylinder. N f cos q = w N R + = sin q , N R ( ) = Reaction to N, i.e.,normal force on plank by cylinder R¢ = Normal force on plank by ground, w = Weight of plank, f ¢ = frictional force on plank by ground. Resultant of f ¢ and R¢, N R ( ) and w pass through point O. 2. R = Normal force on sphere A by left wall, N = Normal force on sphere A by ground, N ¢ = Normal force on sphere A by sphere B, w A = Weight of sphere A. N R ¢ = cos q N w N A ¢ + = sin q R¢ = Normal force on sphere B by right wall, N R ( ) = Reaction to N i.e. normal force on sphere B by sphere A, w B = Weight of sphere B, R¢, N R ( ) and w B pass through point O, the centre sphere B. 3. N = Normal force on sphere by wall, Laws of Motion 5 N(R) Q R' q O w f' P Ground (Force acting on plank) N q R w A N' Force on sphere A N (R) q R w B O R' B Force of sphere B N O B w C q A w R q f N Force acting on cylinder w = Weight of sphere, T = Tension in string. 4. Component of F 1 ® along x-axis : 4 30 2 3 cos ° = N along y-axis : 4 30 2 sin ° = N Component of F 2 ® along x-axis : 4 120 2 cos ° = - N along y-axis : 4 120 2 3 sin ° = N Component of F 3 ® along x-axis : 6 270 0 cos ° = N along y-axis : 6 270 6 sin ° = - N Component of F 4 ® along x-axis : 4 0 4 cos ° = N along y-axis : 4 0 0 sin ° = N 5. Tak ing mo ment about point A AB l = ( sin ) T l w l 30 2 ° = Þ T w = 6. See figure (answer to question no. 3) sin q = + OA OB BC = + = a a a 1 2 T w cos 30° = or T w 3 2 = or T w = 2 3 7. R f cos cos 30 3 60 ° + = ° i.e., R f 3 2 3 2 + = or R f 3 6 + = …(i) and R f sin sin 30 60 10 ° + ° = i.e., R f 1 2 3 2 10 + = or R f + = 3 20 …(ii) Substituting the value of f from Eq. (i) in Eq. (ii) R R + + = ( ) 3 6 3 20 4 6 3 20 R + = Þ R = - = 20 6 3 4 2.4 N \ f = + ( ) 2.4 3 6 =10.16 N 8.At point B (instantaneous vertical acceleration only) \ mg T ma - ° = sin 45 …(i) At point A (instantaneous horizontal acceleration only) \ T ma cos 45° = …(ii) Combining Eqs. (i) and (ii) mg ma ma - = Þ a g = 2 Laws of Motion 75 30° T sin 30° B O w A T AB = l 30° R f 60° 60° 10 N 3N N A T T 45° B mg Introductory Exercise 5.2 1. Acceleration of system a = + + - + + ( ) ( ) 120 50 1 4 2 =10 m/s 2 Let normal force between 1 kg block and 4 kg block = F 1 \ Net force on 1 kg block = - 120 N \ a F = - 120 1 1 or 10 120 1 = - F i.e., F 1 110 = N Net force on 2 kg block = ´ 2 a = ´ 2 10 =20 N 2. As, 4 30 2 30 g g sin sin ° > ° The normal force between the two blocks will be zero. 3. N R mg ( ) = 4 \ N mg = 4 As lift is moving downward with acceleration a, the pseudo force on A will be ma acting in the upward direction. For the block to be at rest w.r.t. lift. N ma mg + = or mg ma mg 4 + = Þ a g = 3 4 4. Angle made by the string with the normal to the ceiling = = ° q 30 As the train is moving with constant velocity no pseudo force will act on the plumb-bob. Tension in spring = mg = ´ 1 10 = 10 N 5. Pseudo force ( ) = ma on plumb-bob will be as shown in figure T mg ma cos cos ( ) f = + ° - 90 q i.e., T mg ma cos sin f = + q …(i) and T ma sin cos f = q Squaring and adding Eqs. (i) and (ii), T m g m a m ag 2 2 2 2 2 2 2 2 = + + sin sin q q + m a 2 2 2 cos q …(iii) T m g m a m ag 2 2 2 2 2 2 = + + ( Qq = ° 30 ) = + + × m g m g m g g 2 2 2 2 2 4 2 (Qa g = 2 ) = 7 4 2 2 m g or T mg = 7 2 =5 7 N Dividing Eq. (i) by Eq. (ii), tan cos sin f = + ma mg ma q q 76 | Mechanics-1 mg N (R) = N ma N a A q = 30° v (constant) mg T q q = 30° f a 90° – q ma T mg = + a g a cos sin q q = + cos sin q q 2 = ° + ° cos sin 30 2 30 = 3 5 i.e., f = é ë ê ù û ú - tan 1 3 5 6. a = + + 8 2 1 1 =2 m/s 2 Net force on 1 kg mass = - 8 2 T \ 8 1 2 2 - = ´ T Þ T 2 6 = N Net force on 1 kg block = T 1 \ T a 1 2 = = ´ = 2 2 4 N Introductory Exercise 5.3 1. F g g = ° = 2 30 sin For the system to remain at rest T g 2 2 = …(i) T F T 2 1 + = …(ii) or T g T 2 1 + = …[ii (a)] T mg 1 = …(iii) Substituting the values of T 1 and T 2 from Eqs. (iii) and (i) in Eq. [ii(a)] 2g g mg + = i.e., m =3 kg 2. As net downward force on the system is zero, the system will be in equilibrium \ T g 1 4 = and T g 2 1 = \ T T 1 2 4 = 3. 2 2 g T a - = Laws of Motion 77 2kg 1kg F = 8 N A B 2kg 1kg 8 N T 1 T 1 T 2 T 2 B A T 1 T 1 mg T 2 T 2 2 g T 1 T 2 T 2 F T 1 T 1 T 1 T 2 T 2 T 1 4g 1g 3g T T T T 1g 2g a a Page 5 Introductory Exercise 5.1 1. N = Nor mal force on cylinder by plank R = Normal force on cylinder by ground, f = Force of friction by ground by cylinder, w = Weight of cylinder. N f cos q = w N R + = sin q , N R ( ) = Reaction to N, i.e.,normal force on plank by cylinder R¢ = Normal force on plank by ground, w = Weight of plank, f ¢ = frictional force on plank by ground. Resultant of f ¢ and R¢, N R ( ) and w pass through point O. 2. R = Normal force on sphere A by left wall, N = Normal force on sphere A by ground, N ¢ = Normal force on sphere A by sphere B, w A = Weight of sphere A. N R ¢ = cos q N w N A ¢ + = sin q R¢ = Normal force on sphere B by right wall, N R ( ) = Reaction to N i.e. normal force on sphere B by sphere A, w B = Weight of sphere B, R¢, N R ( ) and w B pass through point O, the centre sphere B. 3. N = Normal force on sphere by wall, Laws of Motion 5 N(R) Q R' q O w f' P Ground (Force acting on plank) N q R w A N' Force on sphere A N (R) q R w B O R' B Force of sphere B N O B w C q A w R q f N Force acting on cylinder w = Weight of sphere, T = Tension in string. 4. Component of F 1 ® along x-axis : 4 30 2 3 cos ° = N along y-axis : 4 30 2 sin ° = N Component of F 2 ® along x-axis : 4 120 2 cos ° = - N along y-axis : 4 120 2 3 sin ° = N Component of F 3 ® along x-axis : 6 270 0 cos ° = N along y-axis : 6 270 6 sin ° = - N Component of F 4 ® along x-axis : 4 0 4 cos ° = N along y-axis : 4 0 0 sin ° = N 5. Tak ing mo ment about point A AB l = ( sin ) T l w l 30 2 ° = Þ T w = 6. See figure (answer to question no. 3) sin q = + OA OB BC = + = a a a 1 2 T w cos 30° = or T w 3 2 = or T w = 2 3 7. R f cos cos 30 3 60 ° + = ° i.e., R f 3 2 3 2 + = or R f 3 6 + = …(i) and R f sin sin 30 60 10 ° + ° = i.e., R f 1 2 3 2 10 + = or R f + = 3 20 …(ii) Substituting the value of f from Eq. (i) in Eq. (ii) R R + + = ( ) 3 6 3 20 4 6 3 20 R + = Þ R = - = 20 6 3 4 2.4 N \ f = + ( ) 2.4 3 6 =10.16 N 8.At point B (instantaneous vertical acceleration only) \ mg T ma - ° = sin 45 …(i) At point A (instantaneous horizontal acceleration only) \ T ma cos 45° = …(ii) Combining Eqs. (i) and (ii) mg ma ma - = Þ a g = 2 Laws of Motion 75 30° T sin 30° B O w A T AB = l 30° R f 60° 60° 10 N 3N N A T T 45° B mg Introductory Exercise 5.2 1. Acceleration of system a = + + - + + ( ) ( ) 120 50 1 4 2 =10 m/s 2 Let normal force between 1 kg block and 4 kg block = F 1 \ Net force on 1 kg block = - 120 N \ a F = - 120 1 1 or 10 120 1 = - F i.e., F 1 110 = N Net force on 2 kg block = ´ 2 a = ´ 2 10 =20 N 2. As, 4 30 2 30 g g sin sin ° > ° The normal force between the two blocks will be zero. 3. N R mg ( ) = 4 \ N mg = 4 As lift is moving downward with acceleration a, the pseudo force on A will be ma acting in the upward direction. For the block to be at rest w.r.t. lift. N ma mg + = or mg ma mg 4 + = Þ a g = 3 4 4. Angle made by the string with the normal to the ceiling = = ° q 30 As the train is moving with constant velocity no pseudo force will act on the plumb-bob. Tension in spring = mg = ´ 1 10 = 10 N 5. Pseudo force ( ) = ma on plumb-bob will be as shown in figure T mg ma cos cos ( ) f = + ° - 90 q i.e., T mg ma cos sin f = + q …(i) and T ma sin cos f = q Squaring and adding Eqs. (i) and (ii), T m g m a m ag 2 2 2 2 2 2 2 2 = + + sin sin q q + m a 2 2 2 cos q …(iii) T m g m a m ag 2 2 2 2 2 2 = + + ( Qq = ° 30 ) = + + × m g m g m g g 2 2 2 2 2 4 2 (Qa g = 2 ) = 7 4 2 2 m g or T mg = 7 2 =5 7 N Dividing Eq. (i) by Eq. (ii), tan cos sin f = + ma mg ma q q 76 | Mechanics-1 mg N (R) = N ma N a A q = 30° v (constant) mg T q q = 30° f a 90° – q ma T mg = + a g a cos sin q q = + cos sin q q 2 = ° + ° cos sin 30 2 30 = 3 5 i.e., f = é ë ê ù û ú - tan 1 3 5 6. a = + + 8 2 1 1 =2 m/s 2 Net force on 1 kg mass = - 8 2 T \ 8 1 2 2 - = ´ T Þ T 2 6 = N Net force on 1 kg block = T 1 \ T a 1 2 = = ´ = 2 2 4 N Introductory Exercise 5.3 1. F g g = ° = 2 30 sin For the system to remain at rest T g 2 2 = …(i) T F T 2 1 + = …(ii) or T g T 2 1 + = …[ii (a)] T mg 1 = …(iii) Substituting the values of T 1 and T 2 from Eqs. (iii) and (i) in Eq. [ii(a)] 2g g mg + = i.e., m =3 kg 2. As net downward force on the system is zero, the system will be in equilibrium \ T g 1 4 = and T g 2 1 = \ T T 1 2 4 = 3. 2 2 g T a - = Laws of Motion 77 2kg 1kg F = 8 N A B 2kg 1kg 8 N T 1 T 1 T 2 T 2 B A T 1 T 1 mg T 2 T 2 2 g T 1 T 2 T 2 F T 1 T 1 T 1 T 2 T 2 T 1 4g 1g 3g T T T T 1g 2g a a T g a - = 1 1 Adding above two equations 1 3 g a = \ a g = 3 Velocity of 1kg block 1 section after the system is set in motion v at = + 0 = × g 3 1 = g 3 (upward) On stopping 2 kg, the block of 1kg will go upwards with retardation g. Time ( ) t¢ taken by the 1 kg block to attain zero velocity will be given by the equation. 0 3 = æ è ç ö ø ÷ + - ¢ g g t ( ) Þ t¢ = 1 3 s If the 2 kg block is stopped just for a moment (time being much-much less than 1 3 s), it will also start falling down when the stopping time ends. In t¢ = æ è ç ö ø ÷ 1 3 s time upward displacement of 1 kg block = = = u a g g g 2 2 2 3 2 18 ( / ) Downward displacement of 2 kg block = = × æ è ç ö ø ÷ = 1 2 1 2 3 18 2 2 at g g g As the two are just equal, the string will again become taut after time 1 3 s. 4. F g T a + - = 1 1 …(i) and T g a - = 2 2 …(ii) Adding Eqs. (i) and (ii), F g a - = 1 3 \ a = - 20 10 3 = 10 3 ms -2 Introductory Exercise 5.4 1. 2 2 T a = ´ …(i) and 1 2 g T a - = …(ii) Solving Eqs. (i) and (ii), a g = 3 \ Acceleration of 1 kg block 2 2 3 20 3 a g = = ms -2 Tension in the string T g = = 3 10 3 N 2. Mg T Ma - = …(i) 78 | Mechanics-1 T T T T F 2g a 1g a T T 2 kg 2T a 1g T T T T 2a 2T T T mg T T a M T T aRead More

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