Page 1 42 | Mechanics-1 10. x gt 1 2 1 2 = x x g t 1 2 2 1 2 2 + = ( ) \ x g t gt 2 2 2 1 2 4 1 2 = × - x gt 2 2 3 2 = x x gt 2 1 2 - = Þ t x x g = - 2 1 Option (a) is correct. 11. y x l 2 2 2 + = \ 2 2 0 y dy dt x dx dt × + × = or dy dt x y dx dt = - = - ° x x dx dt tan30 = -2 3 m/s Option (c) is correct. 12. Maximum separation (x max ) between the police and the thief will be at time t (shown in figure) t v a - = = - - 2 90 5 1 2 kmh ms = ´ 90 5 1000 3600 s i.e., t = 7 s x max [ ] = ´ - ´ ´ 90 7 1 2 5 90 kmh s s kmh –1 –1 = ´ ´ 90 1000 3600 9 2 m =112.5 m Option (a) is correct. 13. If meeting time is t 2 3 30 v u sin sin q = ° i.e., sinq = 3 4 or q = - sin 1 3 4 Option (c) is correct. 14. Distance = + æ è ç ö ø ÷ 1 2 2 ab a b t \ 0 1 2 1 4 1 4 2 = ´ + æ è ç ö ø ÷ t Þ t = 500 s =22.36 s Option (a) is correct. 15. Let BC t = ¢ \ 24 4 t¢ = Þ t¢ =6 s \ OB =50 s Let OA t = \ AB t = - 56 1032 1 2 56 56 24 = + - ´ [ ( )] t Þ t =20 s \ maximum acceleration = 24 20 m / s s =1.2 m/s 2 Option (b) is correct. 16. From DOAB cos AOB OA OB AB OA OB = + - × × 2 2 2 2 v time (t) Thief Police 108 km/h 90 km/h 2 s 7 s t O 24 C B A v (m/s) t (s) 60° O B (3 – 0.2t) m Q – 0.2t 0.2t P (4 – 0.2t) m Page 2 42 | Mechanics-1 10. x gt 1 2 1 2 = x x g t 1 2 2 1 2 2 + = ( ) \ x g t gt 2 2 2 1 2 4 1 2 = × - x gt 2 2 3 2 = x x gt 2 1 2 - = Þ t x x g = - 2 1 Option (a) is correct. 11. y x l 2 2 2 + = \ 2 2 0 y dy dt x dx dt × + × = or dy dt x y dx dt = - = - ° x x dx dt tan30 = -2 3 m/s Option (c) is correct. 12. Maximum separation (x max ) between the police and the thief will be at time t (shown in figure) t v a - = = - - 2 90 5 1 2 kmh ms = ´ 90 5 1000 3600 s i.e., t = 7 s x max [ ] = ´ - ´ ´ 90 7 1 2 5 90 kmh s s kmh –1 –1 = ´ ´ 90 1000 3600 9 2 m =112.5 m Option (a) is correct. 13. If meeting time is t 2 3 30 v u sin sin q = ° i.e., sinq = 3 4 or q = - sin 1 3 4 Option (c) is correct. 14. Distance = + æ è ç ö ø ÷ 1 2 2 ab a b t \ 0 1 2 1 4 1 4 2 = ´ + æ è ç ö ø ÷ t Þ t = 500 s =22.36 s Option (a) is correct. 15. Let BC t = ¢ \ 24 4 t¢ = Þ t¢ =6 s \ OB =50 s Let OA t = \ AB t = - 56 1032 1 2 56 56 24 = + - ´ [ ( )] t Þ t =20 s \ maximum acceleration = 24 20 m / s s =1.2 m/s 2 Option (b) is correct. 16. From DOAB cos AOB OA OB AB OA OB = + - × × 2 2 2 2 v time (t) Thief Police 108 km/h 90 km/h 2 s 7 s t O 24 C B A v (m/s) t (s) 60° O B (3 – 0.2t) m Q – 0.2t 0.2t P (4 – 0.2t) m \ OA OB AB OA OB 2 2 2 + - = × i.e., AB OA OB OA OB 2 2 2 = + - × = - + - ( ) ( ) 4 3 2 2 0.2 0.2 t t + - - ( )( ) 4 3 0.2 0.2 t t or AB t t t 2 2 2 16 9 = + - + + 0.04 1.6 0.04 - + - + 1.2 1.4 0.04 t t t 12 2 or AB t t 2 2 37 = - + 0.12 4.2 For AB to be minimum 0.24 4.2 t - = 0 or t = 17.5 s \ ( ) ( ) ( ) min AB 2 2 37 = - ´ + 0.12 17.5 4.2 17.5 = - 36.75 73.5 +37 ( ) min AB 2 =0.75 m 2 = 7500 2 cm =50 3 cm Option (d) is correct. 17. Velocity of ball w.r.t. elevator = 15 m/s Acceleration of ball w.r.t. elevator = - - + ( ) ( ) 10 5 = - 15 m/s 2 Final displacement of ball w.r.t. elevator =-2 m \ - = + - 2 15 1 2 15 2 t t ( ) i.e., 15 30 4 0 2 t t - - = \ t = + + ´ ´ ´ 30 900 4 4 15 2 15 =2.13 s Option (a) is correct. 18. Velocity of ball w.r.t. ground ( ) v BE = + ( ) 15 10 m/s =25 m/s Now v u as 2 2 2 = + \ 0 25 2 10 2 2 = + - ( ) ( )H or H =31.25 m Maximum height by ball as measured from ground = + + 31.25 2 50 = 83.25 m Option (c) is correct. 19. Displacement of ball w.r.t. ground during its flight = = H 31.25 m Option (d) is correct. 20. Dis place ment of ball w.r.t. floor of el e va tor at time t s t t = + - + 15 1 2 15 2 2 ( ) s will be maximum, when ds dt = 0 i.e., 15 15 0 - = t i.e., t = 1s \ s max = ´ + - + ( ) ( )( ) 15 1 1 2 10 1 2 2 = 12 m Option (a) is correct. 21. Let the particles meet at time t i.e., displacement of the particles are equal and which is possible when Area of D ACB = Area of D A B C ¢ ¢ [Area OBCA O ¢ q being common] or Area of D APC = Area A P C ¢ ¢ 1 2 1 2 AP PC A P P C ´ = ¢ ¢ ´ ¢ or AP PC A P P C ´ = ¢ ¢ ´ ¢ or ( tan ) ( tan ) PC PC P C P C q q = ¢ ¢ or PC P C = ¢ ¢ Motion in One Dimension | 43 2 m 10 m/s 15 m/s = v BE H 2 5 m/s + Max. height attained by ball 50 m v B B' P' u B u A C q q P A B v time (s) Q R 4 O v A A' Page 3 42 | Mechanics-1 10. x gt 1 2 1 2 = x x g t 1 2 2 1 2 2 + = ( ) \ x g t gt 2 2 2 1 2 4 1 2 = × - x gt 2 2 3 2 = x x gt 2 1 2 - = Þ t x x g = - 2 1 Option (a) is correct. 11. y x l 2 2 2 + = \ 2 2 0 y dy dt x dx dt × + × = or dy dt x y dx dt = - = - ° x x dx dt tan30 = -2 3 m/s Option (c) is correct. 12. Maximum separation (x max ) between the police and the thief will be at time t (shown in figure) t v a - = = - - 2 90 5 1 2 kmh ms = ´ 90 5 1000 3600 s i.e., t = 7 s x max [ ] = ´ - ´ ´ 90 7 1 2 5 90 kmh s s kmh –1 –1 = ´ ´ 90 1000 3600 9 2 m =112.5 m Option (a) is correct. 13. If meeting time is t 2 3 30 v u sin sin q = ° i.e., sinq = 3 4 or q = - sin 1 3 4 Option (c) is correct. 14. Distance = + æ è ç ö ø ÷ 1 2 2 ab a b t \ 0 1 2 1 4 1 4 2 = ´ + æ è ç ö ø ÷ t Þ t = 500 s =22.36 s Option (a) is correct. 15. Let BC t = ¢ \ 24 4 t¢ = Þ t¢ =6 s \ OB =50 s Let OA t = \ AB t = - 56 1032 1 2 56 56 24 = + - ´ [ ( )] t Þ t =20 s \ maximum acceleration = 24 20 m / s s =1.2 m/s 2 Option (b) is correct. 16. From DOAB cos AOB OA OB AB OA OB = + - × × 2 2 2 2 v time (t) Thief Police 108 km/h 90 km/h 2 s 7 s t O 24 C B A v (m/s) t (s) 60° O B (3 – 0.2t) m Q – 0.2t 0.2t P (4 – 0.2t) m \ OA OB AB OA OB 2 2 2 + - = × i.e., AB OA OB OA OB 2 2 2 = + - × = - + - ( ) ( ) 4 3 2 2 0.2 0.2 t t + - - ( )( ) 4 3 0.2 0.2 t t or AB t t t 2 2 2 16 9 = + - + + 0.04 1.6 0.04 - + - + 1.2 1.4 0.04 t t t 12 2 or AB t t 2 2 37 = - + 0.12 4.2 For AB to be minimum 0.24 4.2 t - = 0 or t = 17.5 s \ ( ) ( ) ( ) min AB 2 2 37 = - ´ + 0.12 17.5 4.2 17.5 = - 36.75 73.5 +37 ( ) min AB 2 =0.75 m 2 = 7500 2 cm =50 3 cm Option (d) is correct. 17. Velocity of ball w.r.t. elevator = 15 m/s Acceleration of ball w.r.t. elevator = - - + ( ) ( ) 10 5 = - 15 m/s 2 Final displacement of ball w.r.t. elevator =-2 m \ - = + - 2 15 1 2 15 2 t t ( ) i.e., 15 30 4 0 2 t t - - = \ t = + + ´ ´ ´ 30 900 4 4 15 2 15 =2.13 s Option (a) is correct. 18. Velocity of ball w.r.t. ground ( ) v BE = + ( ) 15 10 m/s =25 m/s Now v u as 2 2 2 = + \ 0 25 2 10 2 2 = + - ( ) ( )H or H =31.25 m Maximum height by ball as measured from ground = + + 31.25 2 50 = 83.25 m Option (c) is correct. 19. Displacement of ball w.r.t. ground during its flight = = H 31.25 m Option (d) is correct. 20. Dis place ment of ball w.r.t. floor of el e va tor at time t s t t = + - + 15 1 2 15 2 2 ( ) s will be maximum, when ds dt = 0 i.e., 15 15 0 - = t i.e., t = 1s \ s max = ´ + - + ( ) ( )( ) 15 1 1 2 10 1 2 2 = 12 m Option (a) is correct. 21. Let the particles meet at time t i.e., displacement of the particles are equal and which is possible when Area of D ACB = Area of D A B C ¢ ¢ [Area OBCA O ¢ q being common] or Area of D APC = Area A P C ¢ ¢ 1 2 1 2 AP PC A P P C ´ = ¢ ¢ ´ ¢ or AP PC A P P C ´ = ¢ ¢ ´ ¢ or ( tan ) ( tan ) PC PC P C P C q q = ¢ ¢ or PC P C = ¢ ¢ Motion in One Dimension | 43 2 m 10 m/s 15 m/s = v BE H 2 5 m/s + Max. height attained by ball 50 m v B B' P' u B u A C q q P A B v time (s) Q R 4 O v A A' \ OR PC P C = + ¢ ¢ =2PC = 8 s. Option (c) is correct. 22. Area of D A B O ¢ ¢ = Area of D ABO \ 1 2 4 1 2 4 ( ) ( ) v v u u B A A B - × = - × i.e., v v u u B A A B - = - = - 5 15 = - 10 Þ v v A B - = - 10 1 ms Option (b) is correct. 23. u A = - 6 1 ms , u B = - 12 1 ms and at t = 4 s common velocity = - 8 1 ms To find velocity of A at t = 10 s v u A A - - = - 8 10 4 8 4 or v A - = - 8 6 6 8 4 \ v A = - 5 1 ms Option (d) is correct. More than One Cor rect Op tions 1. Q a v = -a 1 2 / …(i) i.e., dv dt v = -a 1 2 / or v dv dt - ò ò = - 1 2 / a 2 1 2 1 v t C / = - + a Now, at t = 0, v v = 0 \ C v 1 0 1 2 2 = / Þ 2 2 1 2 0 1 2 v t v / / = - + a i.e., the particle will stop at t v = 2 0 1 2 / a Option (a) is correct. Option (b) is incorrect. From Eq. (i), v dv dt v = -a 1 2 / or v dv dx 1 2 / ò ò = -a or 2 3 3 2 2 v C / = - + a As at x = 0, v v = 0 , C v 2 0 3 2 2 3 = / \ 2 3 2 3 3 2 0 3 2 v x v / / = - + a i.e., when the particle stops x v = 2 3 0 3 2 a / Option (d) is correct. Option (c) is incorrect. 2. a t = - 0.5 (m/s 2 ) dv dt t = - 2 or dv tdt ò ò = - 1 2 v t C = - + 2 1 4 At t =0, v=16 m/s \ v t = - + 2 4 16 …(i) From above relation v is zero at t = 8 s Options (a) is correct. From relation (i), ds dt t = - + 2 4 16 \ ds t dt ò ò = - + æ è ç ö ø ÷ 2 4 16 i.e., s t t C = - + + 3 2 12 16 s t t = - + 3 12 16 [As at t = 0, s = 0] At t = 4 s s = 58.67 m Option (b) is correct. The particle returns back at t = 8 s From relation (ii) s 8 3 8 12 16 8 = - + ´ = ( ) 85.33 m s 10 3 10 12 16 10 = - + ´ = ( ) 76.66 m Distance travelled in 10 s = + - s s s 8 8 10 ( ) = ´ - ( ) 2 85.33 76.66 =94 m Option (c) is correct. Velocity of particle at t = 10 s 44 | Mechanics-1 Page 4 42 | Mechanics-1 10. x gt 1 2 1 2 = x x g t 1 2 2 1 2 2 + = ( ) \ x g t gt 2 2 2 1 2 4 1 2 = × - x gt 2 2 3 2 = x x gt 2 1 2 - = Þ t x x g = - 2 1 Option (a) is correct. 11. y x l 2 2 2 + = \ 2 2 0 y dy dt x dx dt × + × = or dy dt x y dx dt = - = - ° x x dx dt tan30 = -2 3 m/s Option (c) is correct. 12. Maximum separation (x max ) between the police and the thief will be at time t (shown in figure) t v a - = = - - 2 90 5 1 2 kmh ms = ´ 90 5 1000 3600 s i.e., t = 7 s x max [ ] = ´ - ´ ´ 90 7 1 2 5 90 kmh s s kmh –1 –1 = ´ ´ 90 1000 3600 9 2 m =112.5 m Option (a) is correct. 13. If meeting time is t 2 3 30 v u sin sin q = ° i.e., sinq = 3 4 or q = - sin 1 3 4 Option (c) is correct. 14. Distance = + æ è ç ö ø ÷ 1 2 2 ab a b t \ 0 1 2 1 4 1 4 2 = ´ + æ è ç ö ø ÷ t Þ t = 500 s =22.36 s Option (a) is correct. 15. Let BC t = ¢ \ 24 4 t¢ = Þ t¢ =6 s \ OB =50 s Let OA t = \ AB t = - 56 1032 1 2 56 56 24 = + - ´ [ ( )] t Þ t =20 s \ maximum acceleration = 24 20 m / s s =1.2 m/s 2 Option (b) is correct. 16. From DOAB cos AOB OA OB AB OA OB = + - × × 2 2 2 2 v time (t) Thief Police 108 km/h 90 km/h 2 s 7 s t O 24 C B A v (m/s) t (s) 60° O B (3 – 0.2t) m Q – 0.2t 0.2t P (4 – 0.2t) m \ OA OB AB OA OB 2 2 2 + - = × i.e., AB OA OB OA OB 2 2 2 = + - × = - + - ( ) ( ) 4 3 2 2 0.2 0.2 t t + - - ( )( ) 4 3 0.2 0.2 t t or AB t t t 2 2 2 16 9 = + - + + 0.04 1.6 0.04 - + - + 1.2 1.4 0.04 t t t 12 2 or AB t t 2 2 37 = - + 0.12 4.2 For AB to be minimum 0.24 4.2 t - = 0 or t = 17.5 s \ ( ) ( ) ( ) min AB 2 2 37 = - ´ + 0.12 17.5 4.2 17.5 = - 36.75 73.5 +37 ( ) min AB 2 =0.75 m 2 = 7500 2 cm =50 3 cm Option (d) is correct. 17. Velocity of ball w.r.t. elevator = 15 m/s Acceleration of ball w.r.t. elevator = - - + ( ) ( ) 10 5 = - 15 m/s 2 Final displacement of ball w.r.t. elevator =-2 m \ - = + - 2 15 1 2 15 2 t t ( ) i.e., 15 30 4 0 2 t t - - = \ t = + + ´ ´ ´ 30 900 4 4 15 2 15 =2.13 s Option (a) is correct. 18. Velocity of ball w.r.t. ground ( ) v BE = + ( ) 15 10 m/s =25 m/s Now v u as 2 2 2 = + \ 0 25 2 10 2 2 = + - ( ) ( )H or H =31.25 m Maximum height by ball as measured from ground = + + 31.25 2 50 = 83.25 m Option (c) is correct. 19. Displacement of ball w.r.t. ground during its flight = = H 31.25 m Option (d) is correct. 20. Dis place ment of ball w.r.t. floor of el e va tor at time t s t t = + - + 15 1 2 15 2 2 ( ) s will be maximum, when ds dt = 0 i.e., 15 15 0 - = t i.e., t = 1s \ s max = ´ + - + ( ) ( )( ) 15 1 1 2 10 1 2 2 = 12 m Option (a) is correct. 21. Let the particles meet at time t i.e., displacement of the particles are equal and which is possible when Area of D ACB = Area of D A B C ¢ ¢ [Area OBCA O ¢ q being common] or Area of D APC = Area A P C ¢ ¢ 1 2 1 2 AP PC A P P C ´ = ¢ ¢ ´ ¢ or AP PC A P P C ´ = ¢ ¢ ´ ¢ or ( tan ) ( tan ) PC PC P C P C q q = ¢ ¢ or PC P C = ¢ ¢ Motion in One Dimension | 43 2 m 10 m/s 15 m/s = v BE H 2 5 m/s + Max. height attained by ball 50 m v B B' P' u B u A C q q P A B v time (s) Q R 4 O v A A' \ OR PC P C = + ¢ ¢ =2PC = 8 s. Option (c) is correct. 22. Area of D A B O ¢ ¢ = Area of D ABO \ 1 2 4 1 2 4 ( ) ( ) v v u u B A A B - × = - × i.e., v v u u B A A B - = - = - 5 15 = - 10 Þ v v A B - = - 10 1 ms Option (b) is correct. 23. u A = - 6 1 ms , u B = - 12 1 ms and at t = 4 s common velocity = - 8 1 ms To find velocity of A at t = 10 s v u A A - - = - 8 10 4 8 4 or v A - = - 8 6 6 8 4 \ v A = - 5 1 ms Option (d) is correct. More than One Cor rect Op tions 1. Q a v = -a 1 2 / …(i) i.e., dv dt v = -a 1 2 / or v dv dt - ò ò = - 1 2 / a 2 1 2 1 v t C / = - + a Now, at t = 0, v v = 0 \ C v 1 0 1 2 2 = / Þ 2 2 1 2 0 1 2 v t v / / = - + a i.e., the particle will stop at t v = 2 0 1 2 / a Option (a) is correct. Option (b) is incorrect. From Eq. (i), v dv dt v = -a 1 2 / or v dv dx 1 2 / ò ò = -a or 2 3 3 2 2 v C / = - + a As at x = 0, v v = 0 , C v 2 0 3 2 2 3 = / \ 2 3 2 3 3 2 0 3 2 v x v / / = - + a i.e., when the particle stops x v = 2 3 0 3 2 a / Option (d) is correct. Option (c) is incorrect. 2. a t = - 0.5 (m/s 2 ) dv dt t = - 2 or dv tdt ò ò = - 1 2 v t C = - + 2 1 4 At t =0, v=16 m/s \ v t = - + 2 4 16 …(i) From above relation v is zero at t = 8 s Options (a) is correct. From relation (i), ds dt t = - + 2 4 16 \ ds t dt ò ò = - + æ è ç ö ø ÷ 2 4 16 i.e., s t t C = - + + 3 2 12 16 s t t = - + 3 12 16 [As at t = 0, s = 0] At t = 4 s s = 58.67 m Option (b) is correct. The particle returns back at t = 8 s From relation (ii) s 8 3 8 12 16 8 = - + ´ = ( ) 85.33 m s 10 3 10 12 16 10 = - + ´ = ( ) 76.66 m Distance travelled in 10 s = + - s s s 8 8 10 ( ) = ´ - ( ) 2 85.33 76.66 =94 m Option (c) is correct. Velocity of particle at t = 10 s 44 | Mechanics-1 v 10 2 10 4 16 = - + ( ) = - + 25 16 = -9 m/s \ Speed of particle at t = 10 s is 9 m/s. Option (d) is correct. 3. | | v ® is scalar. \ Option (a) is incorrect. d dt v a ® ® = (by definition). \ Option (b) is correct. v 2 is scalar. \ Option (c) is incorrect. v v ® ® | | is v ^ (unit vector) \ d dt v a a ^ ^ = ¹ ® \ Option (d) is incorrect. 4. v j A ® - = -20 1 ^ kmh v i j B ® = ° + ° ( cos ) ( sin ) ^ ^ 40 37 40 37 kmh -1 = + - ( ) ^ ^ 32 24 1 i j kmh v v v AB A B ® = - ® ® = - - + ( ) ( ) ^ ^ ^ 20 32 24 j i j = - - 32i j ^ ^ 44 Option (a) is correct. Option (c) is incorrect. At any time S i j j A t ® = + + - 3 4 20 ^ ^ ^ ( ) S i j B t ® = + ( ) ^ ^ 32 24 Position of A relative to B : S S S AB A B ® ® ® = - = + - - + ( ) ( ) ^ ^ ^ ^ ^ 3 4 20 32 24 i j j i j t t t = - + - ( ) ( ) ^ ^ 3 32 4 44 t t i j Option (b) is correct. Option (d) is incorrect. 5. Case I. v a t = 1 1 v a t = 2 2 s a t a t 1 1 1 2 2 2 2 1 2 1 2 = + Case II. v a t ¢ =2 1 1 =2v 2 2 3 v a t = 2 2 2 2 3 ( ) a t a t = Þ t t 3 2 2 = s a t at 2 1 1 2 23 2 1 2 2 1 2 = + ( ) = + a t a t 1 1 2 2 2 2 1 2 2 ( ) = + a t a t 1 1 2 2 2 2 2 2 2 1 1 2 2 2 2 s a t a t = + \ s s 1 1 2 > …(i) 4 2 2 1 1 1 2 2 2 2 s a t a t = + \ s s 2 1 4 < …(ii) Combining Eqs. (i) and (ii), 2 4 1 2 1 s s s < < Option (d) is correct. In Case I. v s t t av = + 1 1 2 i.e., v a t a t t t 1 1 1 2 2 2 2 1 2 1 2 1 2 = + + Motion in One Dimension | 45 East X (km) 20 km/h A (3, 4) North Y (km) South B 37° 40 km/h – a 2 a 1 t 1 t 2 v = (Let) S 1 Rest Average velocity = v 1 Rest – a 2 2a 1 t 1 t 3 S 2 Rest Average velocity = v 2 v' = 2v Page 5 42 | Mechanics-1 10. x gt 1 2 1 2 = x x g t 1 2 2 1 2 2 + = ( ) \ x g t gt 2 2 2 1 2 4 1 2 = × - x gt 2 2 3 2 = x x gt 2 1 2 - = Þ t x x g = - 2 1 Option (a) is correct. 11. y x l 2 2 2 + = \ 2 2 0 y dy dt x dx dt × + × = or dy dt x y dx dt = - = - ° x x dx dt tan30 = -2 3 m/s Option (c) is correct. 12. Maximum separation (x max ) between the police and the thief will be at time t (shown in figure) t v a - = = - - 2 90 5 1 2 kmh ms = ´ 90 5 1000 3600 s i.e., t = 7 s x max [ ] = ´ - ´ ´ 90 7 1 2 5 90 kmh s s kmh –1 –1 = ´ ´ 90 1000 3600 9 2 m =112.5 m Option (a) is correct. 13. If meeting time is t 2 3 30 v u sin sin q = ° i.e., sinq = 3 4 or q = - sin 1 3 4 Option (c) is correct. 14. Distance = + æ è ç ö ø ÷ 1 2 2 ab a b t \ 0 1 2 1 4 1 4 2 = ´ + æ è ç ö ø ÷ t Þ t = 500 s =22.36 s Option (a) is correct. 15. Let BC t = ¢ \ 24 4 t¢ = Þ t¢ =6 s \ OB =50 s Let OA t = \ AB t = - 56 1032 1 2 56 56 24 = + - ´ [ ( )] t Þ t =20 s \ maximum acceleration = 24 20 m / s s =1.2 m/s 2 Option (b) is correct. 16. From DOAB cos AOB OA OB AB OA OB = + - × × 2 2 2 2 v time (t) Thief Police 108 km/h 90 km/h 2 s 7 s t O 24 C B A v (m/s) t (s) 60° O B (3 – 0.2t) m Q – 0.2t 0.2t P (4 – 0.2t) m \ OA OB AB OA OB 2 2 2 + - = × i.e., AB OA OB OA OB 2 2 2 = + - × = - + - ( ) ( ) 4 3 2 2 0.2 0.2 t t + - - ( )( ) 4 3 0.2 0.2 t t or AB t t t 2 2 2 16 9 = + - + + 0.04 1.6 0.04 - + - + 1.2 1.4 0.04 t t t 12 2 or AB t t 2 2 37 = - + 0.12 4.2 For AB to be minimum 0.24 4.2 t - = 0 or t = 17.5 s \ ( ) ( ) ( ) min AB 2 2 37 = - ´ + 0.12 17.5 4.2 17.5 = - 36.75 73.5 +37 ( ) min AB 2 =0.75 m 2 = 7500 2 cm =50 3 cm Option (d) is correct. 17. Velocity of ball w.r.t. elevator = 15 m/s Acceleration of ball w.r.t. elevator = - - + ( ) ( ) 10 5 = - 15 m/s 2 Final displacement of ball w.r.t. elevator =-2 m \ - = + - 2 15 1 2 15 2 t t ( ) i.e., 15 30 4 0 2 t t - - = \ t = + + ´ ´ ´ 30 900 4 4 15 2 15 =2.13 s Option (a) is correct. 18. Velocity of ball w.r.t. ground ( ) v BE = + ( ) 15 10 m/s =25 m/s Now v u as 2 2 2 = + \ 0 25 2 10 2 2 = + - ( ) ( )H or H =31.25 m Maximum height by ball as measured from ground = + + 31.25 2 50 = 83.25 m Option (c) is correct. 19. Displacement of ball w.r.t. ground during its flight = = H 31.25 m Option (d) is correct. 20. Dis place ment of ball w.r.t. floor of el e va tor at time t s t t = + - + 15 1 2 15 2 2 ( ) s will be maximum, when ds dt = 0 i.e., 15 15 0 - = t i.e., t = 1s \ s max = ´ + - + ( ) ( )( ) 15 1 1 2 10 1 2 2 = 12 m Option (a) is correct. 21. Let the particles meet at time t i.e., displacement of the particles are equal and which is possible when Area of D ACB = Area of D A B C ¢ ¢ [Area OBCA O ¢ q being common] or Area of D APC = Area A P C ¢ ¢ 1 2 1 2 AP PC A P P C ´ = ¢ ¢ ´ ¢ or AP PC A P P C ´ = ¢ ¢ ´ ¢ or ( tan ) ( tan ) PC PC P C P C q q = ¢ ¢ or PC P C = ¢ ¢ Motion in One Dimension | 43 2 m 10 m/s 15 m/s = v BE H 2 5 m/s + Max. height attained by ball 50 m v B B' P' u B u A C q q P A B v time (s) Q R 4 O v A A' \ OR PC P C = + ¢ ¢ =2PC = 8 s. Option (c) is correct. 22. Area of D A B O ¢ ¢ = Area of D ABO \ 1 2 4 1 2 4 ( ) ( ) v v u u B A A B - × = - × i.e., v v u u B A A B - = - = - 5 15 = - 10 Þ v v A B - = - 10 1 ms Option (b) is correct. 23. u A = - 6 1 ms , u B = - 12 1 ms and at t = 4 s common velocity = - 8 1 ms To find velocity of A at t = 10 s v u A A - - = - 8 10 4 8 4 or v A - = - 8 6 6 8 4 \ v A = - 5 1 ms Option (d) is correct. More than One Cor rect Op tions 1. Q a v = -a 1 2 / …(i) i.e., dv dt v = -a 1 2 / or v dv dt - ò ò = - 1 2 / a 2 1 2 1 v t C / = - + a Now, at t = 0, v v = 0 \ C v 1 0 1 2 2 = / Þ 2 2 1 2 0 1 2 v t v / / = - + a i.e., the particle will stop at t v = 2 0 1 2 / a Option (a) is correct. Option (b) is incorrect. From Eq. (i), v dv dt v = -a 1 2 / or v dv dx 1 2 / ò ò = -a or 2 3 3 2 2 v C / = - + a As at x = 0, v v = 0 , C v 2 0 3 2 2 3 = / \ 2 3 2 3 3 2 0 3 2 v x v / / = - + a i.e., when the particle stops x v = 2 3 0 3 2 a / Option (d) is correct. Option (c) is incorrect. 2. a t = - 0.5 (m/s 2 ) dv dt t = - 2 or dv tdt ò ò = - 1 2 v t C = - + 2 1 4 At t =0, v=16 m/s \ v t = - + 2 4 16 …(i) From above relation v is zero at t = 8 s Options (a) is correct. From relation (i), ds dt t = - + 2 4 16 \ ds t dt ò ò = - + æ è ç ö ø ÷ 2 4 16 i.e., s t t C = - + + 3 2 12 16 s t t = - + 3 12 16 [As at t = 0, s = 0] At t = 4 s s = 58.67 m Option (b) is correct. The particle returns back at t = 8 s From relation (ii) s 8 3 8 12 16 8 = - + ´ = ( ) 85.33 m s 10 3 10 12 16 10 = - + ´ = ( ) 76.66 m Distance travelled in 10 s = + - s s s 8 8 10 ( ) = ´ - ( ) 2 85.33 76.66 =94 m Option (c) is correct. Velocity of particle at t = 10 s 44 | Mechanics-1 v 10 2 10 4 16 = - + ( ) = - + 25 16 = -9 m/s \ Speed of particle at t = 10 s is 9 m/s. Option (d) is correct. 3. | | v ® is scalar. \ Option (a) is incorrect. d dt v a ® ® = (by definition). \ Option (b) is correct. v 2 is scalar. \ Option (c) is incorrect. v v ® ® | | is v ^ (unit vector) \ d dt v a a ^ ^ = ¹ ® \ Option (d) is incorrect. 4. v j A ® - = -20 1 ^ kmh v i j B ® = ° + ° ( cos ) ( sin ) ^ ^ 40 37 40 37 kmh -1 = + - ( ) ^ ^ 32 24 1 i j kmh v v v AB A B ® = - ® ® = - - + ( ) ( ) ^ ^ ^ 20 32 24 j i j = - - 32i j ^ ^ 44 Option (a) is correct. Option (c) is incorrect. At any time S i j j A t ® = + + - 3 4 20 ^ ^ ^ ( ) S i j B t ® = + ( ) ^ ^ 32 24 Position of A relative to B : S S S AB A B ® ® ® = - = + - - + ( ) ( ) ^ ^ ^ ^ ^ 3 4 20 32 24 i j j i j t t t = - + - ( ) ( ) ^ ^ 3 32 4 44 t t i j Option (b) is correct. Option (d) is incorrect. 5. Case I. v a t = 1 1 v a t = 2 2 s a t a t 1 1 1 2 2 2 2 1 2 1 2 = + Case II. v a t ¢ =2 1 1 =2v 2 2 3 v a t = 2 2 2 2 3 ( ) a t a t = Þ t t 3 2 2 = s a t at 2 1 1 2 23 2 1 2 2 1 2 = + ( ) = + a t a t 1 1 2 2 2 2 1 2 2 ( ) = + a t a t 1 1 2 2 2 2 2 2 2 1 1 2 2 2 2 s a t a t = + \ s s 1 1 2 > …(i) 4 2 2 1 1 1 2 2 2 2 s a t a t = + \ s s 2 1 4 < …(ii) Combining Eqs. (i) and (ii), 2 4 1 2 1 s s s < < Option (d) is correct. In Case I. v s t t av = + 1 1 2 i.e., v a t a t t t 1 1 1 2 2 2 2 1 2 1 2 1 2 = + + Motion in One Dimension | 45 East X (km) 20 km/h A (3, 4) North Y (km) South B 37° 40 km/h – a 2 a 1 t 1 t 2 v = (Let) S 1 Rest Average velocity = v 1 Rest – a 2 2a 1 t 1 t 3 S 2 Rest Average velocity = v 2 v' = 2v = + + ( ) ( ) ( ) a t t a t t t t 1 1 1 2 2 2 1 2 2 = + + vt vt t t 1 2 1 2 2( ) = v 2 In Case II. v s t t av = + 2 1 3 i.e., v a t a t t t 2 1 1 2 2 2 2 1 3 2 = + + = + + ( ) ( ) a t t a t t t t 1 1 1 2 2 2 1 2 2 2 = + + vt v t t t 1 2 1 2 2 2 ( ) =v =2 1 v Option (a) is correct. Option (b) is incorrect. 6. If the particle’s initial velocity is + ive and has some constant - ive acceleration the particle will stop somewhere and then return back to have zero displacement at same time t ( ) > 0 . Also if particle’s initial velocity is - ive and has some constant + ive acceleration the particle will stop somewhere and then return back to have zero displacement at same time t ( ) > 0 . \ Options (b) and (c) are correct. and options (a) and (d) are incorrect. 7. F t = a Þ ma t = a or a m t = a …(i) \ Graph between a (acceleration) and time ( ) t will be as curve 1. \ Option (a) is correct. From equation dv dt m t = × a \ dv m tdt ò ò = a i.e., v m t c = × + a 2 2 \ Graph between velocity ( ) v and time ( ) t will be as curve 2. Option (b) is correct. 8. a s = - + 6 30 6 5 30 a s = - + 5 30 × = - + v dv ds s or 5 30 vdv s ds ò ò = - + ( ) or 5 2 2 30 2 2 1 v s s c = - + + or 5 2 2 30 2 2 v S s = - + …(i) [C 1 0 = as at s = 0 particle is at rest] Substituting s = 10 m in Eq. (i), 5 2 10 2 300 2 2 v = - + ( ) i.e., v=10 m/s \ Option (b) is correct. From Eq. (i) v to be maximum - + = s 30 0 s =30 \ 5 2 30 2 30 30 2 2 v max ( ) ( ) = - + ´ or 5 2 450 2 v max = or v max = 180 m/s Option (c) is correct. 9. If particle’s path is (i) straight with backward motion (ii) not straight somewhere. Distance moved will be greater than the modulus of displacement \ | | v av av ® < v Option (a) is correct. If particle returns to its initial position, the value of v av ® will be zero while its average speed (v av ) will not be zero. \ Option (c) is correct. 10. If u = 0, v at = and s at = 1 2 2 \ v-t graph will be as shown in (a). s-t graph will be as shown in (d). 46 | Mechanics-1Read More

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