Page 1 AIEEE Corner Subjective Questions (Level 1) 1. (a) D v t v t = + 1 1 2 2 = ´ æ è ç ö ø ÷ + ´ æ è ç ö ø ÷ 60 1 80 1 2 km h h km h h =100 km (b) Average speed = = 100km 1.5h 66.67 km/h. 2. (a) Displacement in first two seconds = + ( )( ) ( ) 4 2 1 2 6 2 2 =20 m \ Average velocity = = 20 2 10 m s m/s (b) Displacement in first four seconds = + ( )( ) ( ) 4 4 1 2 6 4 2 =64 m \ Displacement in the time interval t =2 s to t = 4 s = - 64 20 = 44 m \ Average velocity = = 44 2 22 m s m/s. 3. Let the particle takes t time to reach ground. \ - = + - 64 20 1 2 10 2 t t ( ) i.e., 5 20 64 0 2 t t - - = t = 6.1 s If the particle goes h meter above tower before coming down 0 20 2 10 2 = + - ( ) ( ) h Þ h = 20 m (a) Average speed = Total distance moved Time taken = + + ( ) 20 20 60 6.1 =16.4 m/s (b) Average velocity = Net displacement Time taken = -60m 6.1s = -9.8 m/s =9.8 m/s (downwards) 4. Average velocity = S S vt t 2.5v vt vt vT t t T = + + + + 0 0 0 0 2 3 \ 5 3 3 0 0 t T t T + = + 2.5 or T t = 4 0 5. (a) Average acceleration = - + Final velocity initial velocity s ( ) 4 8 = - 0 0 12 =0 m/s 2 (c) a v = = tan max q 4 \ 4 4 = v max (Q a = 4 m/s 2 ) i.e., v max =16 m/s Displacement of particle in 12 seconds = Area under v-t graph = ´ 12 2 v max = ´ 12 16 2 =96 m Average velocity = Displacement m) at Time ( ( ) 96 12 12 s s = +8 m/s 20 | Mechanics-1 h +20 m/s â€“ 60 m v time 12s 8s 4s v max q Page 2 AIEEE Corner Subjective Questions (Level 1) 1. (a) D v t v t = + 1 1 2 2 = ´ æ è ç ö ø ÷ + ´ æ è ç ö ø ÷ 60 1 80 1 2 km h h km h h =100 km (b) Average speed = = 100km 1.5h 66.67 km/h. 2. (a) Displacement in first two seconds = + ( )( ) ( ) 4 2 1 2 6 2 2 =20 m \ Average velocity = = 20 2 10 m s m/s (b) Displacement in first four seconds = + ( )( ) ( ) 4 4 1 2 6 4 2 =64 m \ Displacement in the time interval t =2 s to t = 4 s = - 64 20 = 44 m \ Average velocity = = 44 2 22 m s m/s. 3. Let the particle takes t time to reach ground. \ - = + - 64 20 1 2 10 2 t t ( ) i.e., 5 20 64 0 2 t t - - = t = 6.1 s If the particle goes h meter above tower before coming down 0 20 2 10 2 = + - ( ) ( ) h Þ h = 20 m (a) Average speed = Total distance moved Time taken = + + ( ) 20 20 60 6.1 =16.4 m/s (b) Average velocity = Net displacement Time taken = -60m 6.1s = -9.8 m/s =9.8 m/s (downwards) 4. Average velocity = S S vt t 2.5v vt vt vT t t T = + + + + 0 0 0 0 2 3 \ 5 3 3 0 0 t T t T + = + 2.5 or T t = 4 0 5. (a) Average acceleration = - + Final velocity initial velocity s ( ) 4 8 = - 0 0 12 =0 m/s 2 (c) a v = = tan max q 4 \ 4 4 = v max (Q a = 4 m/s 2 ) i.e., v max =16 m/s Displacement of particle in 12 seconds = Area under v-t graph = ´ 12 2 v max = ´ 12 16 2 =96 m Average velocity = Displacement m) at Time ( ( ) 96 12 12 s s = +8 m/s 20 | Mechanics-1 h +20 m/s â€“ 60 m v time 12s 8s 4s v max q (b)As the particle did not return back distance travelled in 12 s = Displacement at 12 s \ Average speed = 8 m/s. 6. (a) Radius ( ) R of circle = 21 22 m \ Circumference of circle = 2pR = ´ ´ 2 22 7 21 22 m =6 m Speed ( ) v of particle = 1 m/s \ Distance moved by particle in 2 s = 2 m Thus, angle through which the particle moved = ´ = = ° 2 6 2 2 3 120 p p Magnitude of Average velocity = = Magnitude of displacement Time s ( ) 2 = = ° AB R 2 2 60 2 sin =R 3 2 = 21 22 3 2 = 21 3 44 m/s (b) Magnitude of average acceleration = - ½ ½ ½ ½ ½ ½ ½ ½ ® ® v v B A 2s = ° 2 120 2 2 vsin [Q | | | | v v A B v ® ® = = = 1 m/s) = ° vsin 60 = 3 2 m/s 2 7. Position vector at t = 0 s r i j 1 1 2 ® = + ( ) ^ ^ m Position vector at t = 4 s r i j 2 6 4 ® = + ( ) ^ ^ m (a) Displacement from t = 0 s to t = 4 s = - ® ® ( ) r r 2 1 = + - + ( ) ( ) ^ ^ ^ ^ 6 4 1 2 i j i j = + ( ) ^ ^ 5 2 i j m \ Average velocity = + ( ) ^ ^ 5 2 4 i j m s = + ( . . ) ^ ^ 125 05 i j m/s (b) Average acceleration = - Final velocity Initial velocity s 4 = + - + ( ) ( ) ^ ^ ^ ^ 2 10 4 6 4 i j i j = - + 2 4 4 i j ^ ^ = - + ( . ) ^ ^ 05i j m/s 2 (c) We cannot find the average speed as the actual path followed by the particle is not known. Uniform acceleration (a) One dimensional motion 8. If at time t the vertical displacement between A and B is 10 m 1 2 1 2 1 10 2 2 gt gt - - = ( ) or t t 2 2 1 2 - - = ( ) or t t t 2 2 2 1 2 - - + = ( ) or 2 3 t = t =1.5 s 9. The two bodies will meet if Displacement of first after attaining = highest point Displacement of second before attaining highest point Motion in One Dimension | 21 120° O B A(t = 0 s) (t = 2s) v B ® v A ® Page 3 AIEEE Corner Subjective Questions (Level 1) 1. (a) D v t v t = + 1 1 2 2 = ´ æ è ç ö ø ÷ + ´ æ è ç ö ø ÷ 60 1 80 1 2 km h h km h h =100 km (b) Average speed = = 100km 1.5h 66.67 km/h. 2. (a) Displacement in first two seconds = + ( )( ) ( ) 4 2 1 2 6 2 2 =20 m \ Average velocity = = 20 2 10 m s m/s (b) Displacement in first four seconds = + ( )( ) ( ) 4 4 1 2 6 4 2 =64 m \ Displacement in the time interval t =2 s to t = 4 s = - 64 20 = 44 m \ Average velocity = = 44 2 22 m s m/s. 3. Let the particle takes t time to reach ground. \ - = + - 64 20 1 2 10 2 t t ( ) i.e., 5 20 64 0 2 t t - - = t = 6.1 s If the particle goes h meter above tower before coming down 0 20 2 10 2 = + - ( ) ( ) h Þ h = 20 m (a) Average speed = Total distance moved Time taken = + + ( ) 20 20 60 6.1 =16.4 m/s (b) Average velocity = Net displacement Time taken = -60m 6.1s = -9.8 m/s =9.8 m/s (downwards) 4. Average velocity = S S vt t 2.5v vt vt vT t t T = + + + + 0 0 0 0 2 3 \ 5 3 3 0 0 t T t T + = + 2.5 or T t = 4 0 5. (a) Average acceleration = - + Final velocity initial velocity s ( ) 4 8 = - 0 0 12 =0 m/s 2 (c) a v = = tan max q 4 \ 4 4 = v max (Q a = 4 m/s 2 ) i.e., v max =16 m/s Displacement of particle in 12 seconds = Area under v-t graph = ´ 12 2 v max = ´ 12 16 2 =96 m Average velocity = Displacement m) at Time ( ( ) 96 12 12 s s = +8 m/s 20 | Mechanics-1 h +20 m/s â€“ 60 m v time 12s 8s 4s v max q (b)As the particle did not return back distance travelled in 12 s = Displacement at 12 s \ Average speed = 8 m/s. 6. (a) Radius ( ) R of circle = 21 22 m \ Circumference of circle = 2pR = ´ ´ 2 22 7 21 22 m =6 m Speed ( ) v of particle = 1 m/s \ Distance moved by particle in 2 s = 2 m Thus, angle through which the particle moved = ´ = = ° 2 6 2 2 3 120 p p Magnitude of Average velocity = = Magnitude of displacement Time s ( ) 2 = = ° AB R 2 2 60 2 sin =R 3 2 = 21 22 3 2 = 21 3 44 m/s (b) Magnitude of average acceleration = - ½ ½ ½ ½ ½ ½ ½ ½ ® ® v v B A 2s = ° 2 120 2 2 vsin [Q | | | | v v A B v ® ® = = = 1 m/s) = ° vsin 60 = 3 2 m/s 2 7. Position vector at t = 0 s r i j 1 1 2 ® = + ( ) ^ ^ m Position vector at t = 4 s r i j 2 6 4 ® = + ( ) ^ ^ m (a) Displacement from t = 0 s to t = 4 s = - ® ® ( ) r r 2 1 = + - + ( ) ( ) ^ ^ ^ ^ 6 4 1 2 i j i j = + ( ) ^ ^ 5 2 i j m \ Average velocity = + ( ) ^ ^ 5 2 4 i j m s = + ( . . ) ^ ^ 125 05 i j m/s (b) Average acceleration = - Final velocity Initial velocity s 4 = + - + ( ) ( ) ^ ^ ^ ^ 2 10 4 6 4 i j i j = - + 2 4 4 i j ^ ^ = - + ( . ) ^ ^ 05i j m/s 2 (c) We cannot find the average speed as the actual path followed by the particle is not known. Uniform acceleration (a) One dimensional motion 8. If at time t the vertical displacement between A and B is 10 m 1 2 1 2 1 10 2 2 gt gt - - = ( ) or t t 2 2 1 2 - - = ( ) or t t t 2 2 2 1 2 - - + = ( ) or 2 3 t = t =1.5 s 9. The two bodies will meet if Displacement of first after attaining = highest point Displacement of second before attaining highest point Motion in One Dimension | 21 120° O B A(t = 0 s) (t = 2s) v B ® v A ® v t gt v t t g t t 0 2 0 0 0 2 1 2 1 2 - = - - - ( ) ( ) or 0 1 2 2 0 0 0 2 0 = - - - v t g t t t ( ) or gt t gt v t 0 0 2 0 0 1 2 = + or t gt v g = + 1 2 0 0 = + t v g 0 0 2 10. 5 1 2 2 = gt Þ t =1s For A H t gt = + 0 1 2 2 . or H gt = 1 2 2 â€¦(i) For B H g t - = - 25 1 2 1 2 ( ) â€¦(ii) or 1 2 1 2 1 25 2 2 gt gt - - = ( ) [Substituting value of H from Eq. (i)] 1 2 1 25 2 2 g t t [ ( ) ] - - = t t t 2 2 2 1 5 - - + = ( ) 2 1 5 t - = Þ t = 3 s Substituting t = 3 s in Eq. (i) H = ´ ´ = 1 2 10 3 45 2 m 11. s at = 1 2 0 2 Forward motion v at = 0 Backward mo tion - = + - s at t a t ( ) ( ) 0 2 1 2 \ - = - 1 2 1 2 0 2 0 2 at at t at ( ) or - = - t t t t 0 2 0 2 2 or t t t t 2 0 0 2 2 0 - - = t t t t = ± - - - 2 2 41 2 0 0 2 0 2 ( ) ( ) = ± 2 2 2 2 0 0 t t = + t t 0 0 2 (- ive sign being absurd) = + ( ) 1 2 0 t =2141 0 . t From the begining of the motion the point mass will return to the initial position after time 3 141 0 . t . 12. 5 2 1015 2 2 = + - u ( ) Þ u 2 325 = (a) For H 0 2 10 2 2 = + - u H ( ) i.e., 20 325 H = or H =1625 . m (b) For t 0 325 10 = + - ( )t \ t = 325 10 =18 . s 13. (a) 15 2 60 2 2 = + ´ u a â€¦(i) and 15 6 = + ´ u a â€¦(ii) Substituting the value of 6a from Eq. (ii) in Eq. (i) 225 20 15 2 = + - u u ( ) 22 | Mechanics-1 T O W E R H A 5m 25 m u = 0 B Time interval t second for A (tâ€“1) second for B A B 1 s H 15m v = 0 5 m/s u x 60 m 6.0 s At rest a u a 15 m/s Page 4 AIEEE Corner Subjective Questions (Level 1) 1. (a) D v t v t = + 1 1 2 2 = ´ æ è ç ö ø ÷ + ´ æ è ç ö ø ÷ 60 1 80 1 2 km h h km h h =100 km (b) Average speed = = 100km 1.5h 66.67 km/h. 2. (a) Displacement in first two seconds = + ( )( ) ( ) 4 2 1 2 6 2 2 =20 m \ Average velocity = = 20 2 10 m s m/s (b) Displacement in first four seconds = + ( )( ) ( ) 4 4 1 2 6 4 2 =64 m \ Displacement in the time interval t =2 s to t = 4 s = - 64 20 = 44 m \ Average velocity = = 44 2 22 m s m/s. 3. Let the particle takes t time to reach ground. \ - = + - 64 20 1 2 10 2 t t ( ) i.e., 5 20 64 0 2 t t - - = t = 6.1 s If the particle goes h meter above tower before coming down 0 20 2 10 2 = + - ( ) ( ) h Þ h = 20 m (a) Average speed = Total distance moved Time taken = + + ( ) 20 20 60 6.1 =16.4 m/s (b) Average velocity = Net displacement Time taken = -60m 6.1s = -9.8 m/s =9.8 m/s (downwards) 4. Average velocity = S S vt t 2.5v vt vt vT t t T = + + + + 0 0 0 0 2 3 \ 5 3 3 0 0 t T t T + = + 2.5 or T t = 4 0 5. (a) Average acceleration = - + Final velocity initial velocity s ( ) 4 8 = - 0 0 12 =0 m/s 2 (c) a v = = tan max q 4 \ 4 4 = v max (Q a = 4 m/s 2 ) i.e., v max =16 m/s Displacement of particle in 12 seconds = Area under v-t graph = ´ 12 2 v max = ´ 12 16 2 =96 m Average velocity = Displacement m) at Time ( ( ) 96 12 12 s s = +8 m/s 20 | Mechanics-1 h +20 m/s â€“ 60 m v time 12s 8s 4s v max q (b)As the particle did not return back distance travelled in 12 s = Displacement at 12 s \ Average speed = 8 m/s. 6. (a) Radius ( ) R of circle = 21 22 m \ Circumference of circle = 2pR = ´ ´ 2 22 7 21 22 m =6 m Speed ( ) v of particle = 1 m/s \ Distance moved by particle in 2 s = 2 m Thus, angle through which the particle moved = ´ = = ° 2 6 2 2 3 120 p p Magnitude of Average velocity = = Magnitude of displacement Time s ( ) 2 = = ° AB R 2 2 60 2 sin =R 3 2 = 21 22 3 2 = 21 3 44 m/s (b) Magnitude of average acceleration = - ½ ½ ½ ½ ½ ½ ½ ½ ® ® v v B A 2s = ° 2 120 2 2 vsin [Q | | | | v v A B v ® ® = = = 1 m/s) = ° vsin 60 = 3 2 m/s 2 7. Position vector at t = 0 s r i j 1 1 2 ® = + ( ) ^ ^ m Position vector at t = 4 s r i j 2 6 4 ® = + ( ) ^ ^ m (a) Displacement from t = 0 s to t = 4 s = - ® ® ( ) r r 2 1 = + - + ( ) ( ) ^ ^ ^ ^ 6 4 1 2 i j i j = + ( ) ^ ^ 5 2 i j m \ Average velocity = + ( ) ^ ^ 5 2 4 i j m s = + ( . . ) ^ ^ 125 05 i j m/s (b) Average acceleration = - Final velocity Initial velocity s 4 = + - + ( ) ( ) ^ ^ ^ ^ 2 10 4 6 4 i j i j = - + 2 4 4 i j ^ ^ = - + ( . ) ^ ^ 05i j m/s 2 (c) We cannot find the average speed as the actual path followed by the particle is not known. Uniform acceleration (a) One dimensional motion 8. If at time t the vertical displacement between A and B is 10 m 1 2 1 2 1 10 2 2 gt gt - - = ( ) or t t 2 2 1 2 - - = ( ) or t t t 2 2 2 1 2 - - + = ( ) or 2 3 t = t =1.5 s 9. The two bodies will meet if Displacement of first after attaining = highest point Displacement of second before attaining highest point Motion in One Dimension | 21 120° O B A(t = 0 s) (t = 2s) v B ® v A ® v t gt v t t g t t 0 2 0 0 0 2 1 2 1 2 - = - - - ( ) ( ) or 0 1 2 2 0 0 0 2 0 = - - - v t g t t t ( ) or gt t gt v t 0 0 2 0 0 1 2 = + or t gt v g = + 1 2 0 0 = + t v g 0 0 2 10. 5 1 2 2 = gt Þ t =1s For A H t gt = + 0 1 2 2 . or H gt = 1 2 2 â€¦(i) For B H g t - = - 25 1 2 1 2 ( ) â€¦(ii) or 1 2 1 2 1 25 2 2 gt gt - - = ( ) [Substituting value of H from Eq. (i)] 1 2 1 25 2 2 g t t [ ( ) ] - - = t t t 2 2 2 1 5 - - + = ( ) 2 1 5 t - = Þ t = 3 s Substituting t = 3 s in Eq. (i) H = ´ ´ = 1 2 10 3 45 2 m 11. s at = 1 2 0 2 Forward motion v at = 0 Backward mo tion - = + - s at t a t ( ) ( ) 0 2 1 2 \ - = - 1 2 1 2 0 2 0 2 at at t at ( ) or - = - t t t t 0 2 0 2 2 or t t t t 2 0 0 2 2 0 - - = t t t t = ± - - - 2 2 41 2 0 0 2 0 2 ( ) ( ) = ± 2 2 2 2 0 0 t t = + t t 0 0 2 (- ive sign being absurd) = + ( ) 1 2 0 t =2141 0 . t From the begining of the motion the point mass will return to the initial position after time 3 141 0 . t . 12. 5 2 1015 2 2 = + - u ( ) Þ u 2 325 = (a) For H 0 2 10 2 2 = + - u H ( ) i.e., 20 325 H = or H =1625 . m (b) For t 0 325 10 = + - ( )t \ t = 325 10 =18 . s 13. (a) 15 2 60 2 2 = + ´ u a â€¦(i) and 15 6 = + ´ u a â€¦(ii) Substituting the value of 6a from Eq. (ii) in Eq. (i) 225 20 15 2 = + - u u ( ) 22 | Mechanics-1 T O W E R H A 5m 25 m u = 0 B Time interval t second for A (tâ€“1) second for B A B 1 s H 15m v = 0 5 m/s u x 60 m 6.0 s At rest a u a 15 m/s i.e., u u 2 20 75 0 - + = ( )( ) u u - - = 15 5 0 \ u = 5 m/s (15 m/s being not possible) (b) Using Eq. (ii) a = 5 3 m / s 2 (c) u ax 2 2 0 2 = + i.e., x u a = = ´ 2 2 2 5 2 5 3 ( ) =75 . m (d) s at = 1 2 2 = ´ ´ 1 2 5 3 2 t = 5 6 2 t â€¦(iii) t( ) s 0 1 3 6 9 12 v (m/s) 0 5/6 7.5 30 67.5 120 Differentiating Eq. (iii) w.r.t. time t v t = 5 3 t( ) s 0 3 6 9 12 v (m/s) 0 5 10 15 20 14. Jour ney A to P v xt max = + 0 1 â€¦(i) and v xs max 2 1 2 = â€¦(ii) Þ t v x 1 = max Jour ney P to B 0 2 = + - v y t max ( ) â€¦(iii) and v ys max 2 2 2 = â€¦(iv) Þ t v y 2 = max \ v x v y t t max max + = + = 1 2 4 or v x y max 1 1 4 + é ë ê ù û ú = â€¦(v) From Eq. (ii) and Eq. (iv) s s v x v y 1 2 2 2 2 2 + = + max max or 4 2 1 1 2 = + é ë ê ù û ú v x y max â€¦(vi) Dividing Eq. (vi) by Eq. (v) v max = 2 Substituting the value of v max in Eq. (v) 1 1 2 x y + = (Proved) 15. Let acceleration of the particle be a using v u at = + 0 6 = + u a \ a u = - 6 (a) At t = 10 s, s = - 2 m - = ´ + ´ 2 10 1 2 10 2 u a or - = - + 2 6 10 50 ( ) a a or - = - 10 2 a or a = 0.2 m/s 2 Motion in One Dimension | 23 20 15 10 5 3 6 9 12 v (m/s) t (s) Starts a = + x t 1 t 2 a = â€“y Stops 4 km B s 1 s 2 4 min B A 120 90 60 30 15 3 6 9 12 t (s) s (m) 2m 4 0 t = 10 s t = 6 s t = 0 s v = 0 +ive x-axis Page 5 AIEEE Corner Subjective Questions (Level 1) 1. (a) D v t v t = + 1 1 2 2 = ´ æ è ç ö ø ÷ + ´ æ è ç ö ø ÷ 60 1 80 1 2 km h h km h h =100 km (b) Average speed = = 100km 1.5h 66.67 km/h. 2. (a) Displacement in first two seconds = + ( )( ) ( ) 4 2 1 2 6 2 2 =20 m \ Average velocity = = 20 2 10 m s m/s (b) Displacement in first four seconds = + ( )( ) ( ) 4 4 1 2 6 4 2 =64 m \ Displacement in the time interval t =2 s to t = 4 s = - 64 20 = 44 m \ Average velocity = = 44 2 22 m s m/s. 3. Let the particle takes t time to reach ground. \ - = + - 64 20 1 2 10 2 t t ( ) i.e., 5 20 64 0 2 t t - - = t = 6.1 s If the particle goes h meter above tower before coming down 0 20 2 10 2 = + - ( ) ( ) h Þ h = 20 m (a) Average speed = Total distance moved Time taken = + + ( ) 20 20 60 6.1 =16.4 m/s (b) Average velocity = Net displacement Time taken = -60m 6.1s = -9.8 m/s =9.8 m/s (downwards) 4. Average velocity = S S vt t 2.5v vt vt vT t t T = + + + + 0 0 0 0 2 3 \ 5 3 3 0 0 t T t T + = + 2.5 or T t = 4 0 5. (a) Average acceleration = - + Final velocity initial velocity s ( ) 4 8 = - 0 0 12 =0 m/s 2 (c) a v = = tan max q 4 \ 4 4 = v max (Q a = 4 m/s 2 ) i.e., v max =16 m/s Displacement of particle in 12 seconds = Area under v-t graph = ´ 12 2 v max = ´ 12 16 2 =96 m Average velocity = Displacement m) at Time ( ( ) 96 12 12 s s = +8 m/s 20 | Mechanics-1 h +20 m/s â€“ 60 m v time 12s 8s 4s v max q (b)As the particle did not return back distance travelled in 12 s = Displacement at 12 s \ Average speed = 8 m/s. 6. (a) Radius ( ) R of circle = 21 22 m \ Circumference of circle = 2pR = ´ ´ 2 22 7 21 22 m =6 m Speed ( ) v of particle = 1 m/s \ Distance moved by particle in 2 s = 2 m Thus, angle through which the particle moved = ´ = = ° 2 6 2 2 3 120 p p Magnitude of Average velocity = = Magnitude of displacement Time s ( ) 2 = = ° AB R 2 2 60 2 sin =R 3 2 = 21 22 3 2 = 21 3 44 m/s (b) Magnitude of average acceleration = - ½ ½ ½ ½ ½ ½ ½ ½ ® ® v v B A 2s = ° 2 120 2 2 vsin [Q | | | | v v A B v ® ® = = = 1 m/s) = ° vsin 60 = 3 2 m/s 2 7. Position vector at t = 0 s r i j 1 1 2 ® = + ( ) ^ ^ m Position vector at t = 4 s r i j 2 6 4 ® = + ( ) ^ ^ m (a) Displacement from t = 0 s to t = 4 s = - ® ® ( ) r r 2 1 = + - + ( ) ( ) ^ ^ ^ ^ 6 4 1 2 i j i j = + ( ) ^ ^ 5 2 i j m \ Average velocity = + ( ) ^ ^ 5 2 4 i j m s = + ( . . ) ^ ^ 125 05 i j m/s (b) Average acceleration = - Final velocity Initial velocity s 4 = + - + ( ) ( ) ^ ^ ^ ^ 2 10 4 6 4 i j i j = - + 2 4 4 i j ^ ^ = - + ( . ) ^ ^ 05i j m/s 2 (c) We cannot find the average speed as the actual path followed by the particle is not known. Uniform acceleration (a) One dimensional motion 8. If at time t the vertical displacement between A and B is 10 m 1 2 1 2 1 10 2 2 gt gt - - = ( ) or t t 2 2 1 2 - - = ( ) or t t t 2 2 2 1 2 - - + = ( ) or 2 3 t = t =1.5 s 9. The two bodies will meet if Displacement of first after attaining = highest point Displacement of second before attaining highest point Motion in One Dimension | 21 120° O B A(t = 0 s) (t = 2s) v B ® v A ® v t gt v t t g t t 0 2 0 0 0 2 1 2 1 2 - = - - - ( ) ( ) or 0 1 2 2 0 0 0 2 0 = - - - v t g t t t ( ) or gt t gt v t 0 0 2 0 0 1 2 = + or t gt v g = + 1 2 0 0 = + t v g 0 0 2 10. 5 1 2 2 = gt Þ t =1s For A H t gt = + 0 1 2 2 . or H gt = 1 2 2 â€¦(i) For B H g t - = - 25 1 2 1 2 ( ) â€¦(ii) or 1 2 1 2 1 25 2 2 gt gt - - = ( ) [Substituting value of H from Eq. (i)] 1 2 1 25 2 2 g t t [ ( ) ] - - = t t t 2 2 2 1 5 - - + = ( ) 2 1 5 t - = Þ t = 3 s Substituting t = 3 s in Eq. (i) H = ´ ´ = 1 2 10 3 45 2 m 11. s at = 1 2 0 2 Forward motion v at = 0 Backward mo tion - = + - s at t a t ( ) ( ) 0 2 1 2 \ - = - 1 2 1 2 0 2 0 2 at at t at ( ) or - = - t t t t 0 2 0 2 2 or t t t t 2 0 0 2 2 0 - - = t t t t = ± - - - 2 2 41 2 0 0 2 0 2 ( ) ( ) = ± 2 2 2 2 0 0 t t = + t t 0 0 2 (- ive sign being absurd) = + ( ) 1 2 0 t =2141 0 . t From the begining of the motion the point mass will return to the initial position after time 3 141 0 . t . 12. 5 2 1015 2 2 = + - u ( ) Þ u 2 325 = (a) For H 0 2 10 2 2 = + - u H ( ) i.e., 20 325 H = or H =1625 . m (b) For t 0 325 10 = + - ( )t \ t = 325 10 =18 . s 13. (a) 15 2 60 2 2 = + ´ u a â€¦(i) and 15 6 = + ´ u a â€¦(ii) Substituting the value of 6a from Eq. (ii) in Eq. (i) 225 20 15 2 = + - u u ( ) 22 | Mechanics-1 T O W E R H A 5m 25 m u = 0 B Time interval t second for A (tâ€“1) second for B A B 1 s H 15m v = 0 5 m/s u x 60 m 6.0 s At rest a u a 15 m/s i.e., u u 2 20 75 0 - + = ( )( ) u u - - = 15 5 0 \ u = 5 m/s (15 m/s being not possible) (b) Using Eq. (ii) a = 5 3 m / s 2 (c) u ax 2 2 0 2 = + i.e., x u a = = ´ 2 2 2 5 2 5 3 ( ) =75 . m (d) s at = 1 2 2 = ´ ´ 1 2 5 3 2 t = 5 6 2 t â€¦(iii) t( ) s 0 1 3 6 9 12 v (m/s) 0 5/6 7.5 30 67.5 120 Differentiating Eq. (iii) w.r.t. time t v t = 5 3 t( ) s 0 3 6 9 12 v (m/s) 0 5 10 15 20 14. Jour ney A to P v xt max = + 0 1 â€¦(i) and v xs max 2 1 2 = â€¦(ii) Þ t v x 1 = max Jour ney P to B 0 2 = + - v y t max ( ) â€¦(iii) and v ys max 2 2 2 = â€¦(iv) Þ t v y 2 = max \ v x v y t t max max + = + = 1 2 4 or v x y max 1 1 4 + é ë ê ù û ú = â€¦(v) From Eq. (ii) and Eq. (iv) s s v x v y 1 2 2 2 2 2 + = + max max or 4 2 1 1 2 = + é ë ê ù û ú v x y max â€¦(vi) Dividing Eq. (vi) by Eq. (v) v max = 2 Substituting the value of v max in Eq. (v) 1 1 2 x y + = (Proved) 15. Let acceleration of the particle be a using v u at = + 0 6 = + u a \ a u = - 6 (a) At t = 10 s, s = - 2 m - = ´ + ´ 2 10 1 2 10 2 u a or - = - + 2 6 10 50 ( ) a a or - = - 10 2 a or a = 0.2 m/s 2 Motion in One Dimension | 23 20 15 10 5 3 6 9 12 v (m/s) t (s) Starts a = + x t 1 t 2 a = â€“y Stops 4 km B s 1 s 2 4 min B A 120 90 60 30 15 3 6 9 12 t (s) s (m) 2m 4 0 t = 10 s t = 6 s t = 0 s v = 0 +ive x-axis (b) v t u a ( ) at s = = + 10 10 = - + 6 10 a a = 4a =08 . m/s (c) Two or three dimensional motion 16. a F ® ® = = m 10 2 N north kg =5 m/s 2 , north =5j ^ m/s 2 u ® =10 m/s, east =10i ^ m/s using v u a ® ® ® = + t v i j ® = + ´ 10 5 2 ^ ^ ( ) = + 10 10 i j ^ ^ | | v ® =10 2 m/s v ® =10 2, north-east using s u a ® ® ® = + t t 1 2 2 = ´ + ( ) ( ) ^ ^ 10 2 1 2 5 2 2 i j = + ( ) ^ ^ 20 10 i j m | | s ® = + 20 10 2 2 =10 5 m cotq = = 20 10 2 q = - cot 1 2 s ® =10 5 m at cot ( ) -1 2 from east to north. 17. s i j ® = + 0 2 4 ( ) ^ ^ m a i ® = 1 2 ^ m/s 2 (t = 0 s to t = 2 s) t 1 2 = s u 0 ® ® = m/s a j ® = - 2 4 ^ m/s 2 (t = 2 s to t = 4 s) t 2 2 = s (a) Velocity v u a ® ® ® = + 1 1 1 t = + ® 0 i ( ) ^ 2 2 = 4i ^ v v a a ® ® ® ® = + + 2 1 1 2 2 ( )t = + - 4 2 4 2 i i j ^ ^ ^ ( ) or v i j ® = - 2 8 8 ( ) ^ ^ m/s (b) Co-ordinate of particle s s u a ® ® ® ® = + + 1 0 1 1 1 2 1 2 t t = + + + ( ) ( )( ) ( ) ^ ^ ^ 2 4 0 2 1 2 2 2 2 i j i = + + 2 4 4 i j i ^ ^ ^ = + 6 4 i j ^ ^ s s v ® ® ® = + + + 2 1 1 2 1 2 2 2 1 2 t a a t ( ) = + + + - 6 4 4 2 1 2 2 4 2 2 i j i i j ^ ^ ^ ^ ^ ( ) ( ) = + + + - 6 4 8 4 8 i j i i j ^ ^ ^ ^ ^ = - 18 4 i j ^ ^ Co-ordinate of the particle [ , ] 18 4 m m - 18. u i j ® = - ( ) ^ ^ 2 4 m/s, s 0 ® ® = 0 m a i j ® = + ( ) ^ ^ 4 m/s 2 (a) Velocity v u a ® ® ® = + t = - + + ( ) ( ) ^ ^ ^ ^ 2 4 4 2 i j i j = - ( ) ^ ^ 10 2 i j m/s (b) Co-ordinates of the particle s s u a ® ® ® ® = + + 0 1 2 2 t t = + - + + ® 0 i j i j ( ) ( $ ) ^ ^ ^ 2 4 1 2 4 2 2 24 | Mechanics-1 O North East q s ® j ^ i ^Read More

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