KINEMATICS PREVIOUS EAMCET BITS ENGINEERING PAPER

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1 KINEMATICS PREVIOUS EAMCET BITS ENGINEERING PAPER. A body is projected vertically upwards at time t = 0 and is seen at a heiht at time t and t seconds durin its fliht. The maximum heiht attained is [ = acceleration one to ravity] [EAMCET 009E ] ( t ) ( ) ( ) ( ) t t+ t t+ t t t ) ) 3) ) Ans: When a body is projected vertically upwards it occupies the same position while oin up and comin down after time of t and t = ut t t ut + = 0 Sum of roots = max ( + t ) u t t+ t = u = [ + t ] u t. A body thrown vertically up to reach its maximum heiht in t seconds. The total time from the time of projection to reach a point at half of its maximum heiht while returnin (in seconds) is [EAMCET 00 E] ) ) + t 3) 3t Ans: Time taken to reach the maximum heiht = time of ascent = t h t h t as h t = = h t h/ t t = t Time taken to reach half distance of its maximum heiht Total time t+ t = + t 3. A body is projected from the earth at anle 30 with the horizontal with some velocity. If its rane is 0m, the maximum heiht reached by it is [ in metres] [EAMCET 006 E] ) 5 3 ) 5 3 3) 0 3 ) 0 3 Ans : 37 = t ) t

2 Kinematics u sin θ tan θ R u sin θcosθ tan30 5 = = m 0 3. A body is projected vertically upwards at time t = 0 and is seen at a heiht at time t and t seconds durin its fliht. The maximum heiht attained is [ = acceleration due to ravity] [EAMCET 005 E] ( t ) ( ) ( ) ( ) t t+ t t+ t t t ) ) 3) ) Ans: 3 When a body is projected vertically upwards it occupies the same position while oin up and comin down after time of t and t = ut t t ut + = 0 Sum of roots = max ( + t ) u t t+ t = u = [ + t ] u t 5. The horizontal and vertical displacement x and y of a projectile at a ive time t are iven by x = 6t metres and y = t 5t metres. The rane of the projectile in metres is [EAMCET 00 E] ) 9.6 ) 0.6 3) 9. ) 3. Ans: Comparin x = 6t with x = ( u cos θ) t, u cos θ= 6 ( ) ucos [ θ][ usinθ ] 6 Rane = 9.6m 0 y = t 5t with y = u sin θ t t,u sin θ= 6. The equations of motion of a projectile are iven by x = 36t metre and y = 96t- 9.t metre. The anle of projection is [EAMCET 003 E] ) sin 5 Ans: ) sin 5 3 3) sin 3 Comparin x = 36t with x = ( u cos θ) t, u cos θ= 36..() y 96t 9.t y = t.9t ) sin 3 3

3 y= usinθ t t Comparin the equation with ( ) usinθ= () usinθ = tan θ= u cosθ Kinematics (or) sin θ= (or) θ= sin The horizontal and vertical displacement of a projectile at time t are x = 36t and y = t-.9t respectively. Initial velocity of the projectile in m/s is [EAMCET 00 E] ) 5 ) 30 3) 5 ) 6 0 Ans: comparin the equation x = 36t with x = ( ucosθ ) t u cos θ= 36 () y = t.9t with y = ( u sin θ) t t usinθ=..() Squarin () and () and addin u = 60 ms. An object is projected with a velocity of 0 m/s makin an anle of 5 with horizontal. The equation for the trajectory is h = Ax Bx where h is heiht, x is horizontal distance, A and B are constants. The ratio A : B is ( = 0 m/s ) [EAMCET 00 E] ) : 5 ) 5 : 3) : 0 ) 0 : Ans: x h = Ax Bx compared with h = xtanθ u cos θ tan θ= A, B = u cos θ A tan θ A u sin θcos θ 0 = B = B = u cos θ 9. Four bodies P, Q, R and S are projected with equal velocities havin anles of projection 5, 30, 5 and 60 with the horizontal respectively. The body havin shortest rane is [EAMCET 000 E] ) P ) Q 3) R ) S Ans: As Rane = u sin θ R sin θ As θ increases sinθ also increases Shortest rane is for P. 0. A body is thrown horizontally from the top of a tower of 5m heiht. It touches the round at a distance of 0 m from the foot of the tower. The initial velocity of the body is : ( = 0 ms ) [EAMCET 000 E ] ).5 ms ) 5 ms 3) 0 ms ) 0 ms Ans: 3

4 Kinematics u x = y iven x = 0 m, y = 5 m, = 0 ms u = 0ms. A body is thrown vertically upwards with an initial velocity u' reaches maximum heiht in 6 seconds. The ratio of distances travelled by the body in the first second and seventh second is [EAMCET 000 E] ) : ) : 3) : ) : Ans: As time of ascent = 6 s u t = u = t = 6 h = distance travelled by the body in the first Second () () ut t = 6 = h = distance travelled in the seventh second is same as distance travelled in st second in downward motion = / h = : h MEDICAL. A body of mass k is projected from the round with a velocity 0 ms at an anle 30 C with the vertical. If t is the time in seconds at which the body is projected and t is the time in seconds at which it reaches the round, the chane in momentum in k ms durin the time t t is [EAMCET 006 M] ( ) ) 0 ) 0 3 3) 50 3 ) 60 Ans : u = ucosθ ˆi+ usinθ ˆj v= ucosθ ˆ i+ usinθ ˆj Initial velocity ( ) ( ) Final velocity ( ) ( )( ) Chane in velocity = v u = usinθ( ˆj) Manitude of chane in momentum = musinθ as θ = 30 with the vertical. with the horizontal θ = 60 Δ p= 0 3 = A body is projected vertically upwards with a velocity u. It crosses a point in its journey at a heiht h twice, just after and 7 seconds. The value of u in ms is : ( = 0 ms ) [EAMCET 006 M] ) 50 ) 0 3) 30 ) 0 Ans: As the body is projected upwards we can write the equation as 0

5 h = ut t t ut + h = 0 t+ t are the roots of the equation u t+ t = u = ( t+ t) u = 0ms Kinematics. Two balls are projected simultaneously in the same vertical plane from the same point with velocities V and V with anles θ and θ respectively with the horizontal. If V cos θ = V cos θ, the path of one ball as seen from the position of other ball is [EAMCET 006 M] ) Parabola ) orizontal straiht line 3) Vertical straiht line ) Straiht line makin 5 with the vertical Ans:3 It is iven that vcos θ = vcosθ. ence the relative velocity between them is zero. The path of one ball as seen from the position of the other ball is vertical straiht line. 5. The maximum heiht reached by a projectile is metres. The horizontal rane is metres. Velocity of project in ms is ( acceleration due to ravity) [EAMCET 00 M] ) 5 ) Ans: u sin θ tanθ R u sin θcosθ tanθ = tan θ= (or) 3 u 5 u = 5 3 3) 3 sin θ= 5 ) 5 6. Two stones are projected with the same speed but makin different anles with the orizontal. Their horizontal ranes are equal. The anle of projection of one is π/3 and the maximum heiht reached by it is 0 metres. Then the maximum heiht reached by the other in metre is [EAMCET 003 M] ) 336 ) 3) 56 ) 3 Ans: As ranes are equal, the anles of projection are θ and 90 θ One anle is 60 the other anle is 30

6 h u sin 60 sin 60 h u sin 30 sin = h = 3m h Kinematics 7. A projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of 3 ms for 0.5 minutes. If the maximum heiht reached by it is 0 m then the anle of projection is : ( = 0 ms ) [EAMCET 00 M ] ) tan 3 ) tan 3) tan 9 Ans: 3 v = u + at = = 90ms u cos θ= 90ms () u sin θ = 0= usinθ= 0.() 0 dividin () and () usinθ = tan θ= ucosθ 9 3 ) sin 9 θ= tan 9. It is possible to project particle with a iven speed in two possible ways so that it has the same horizontal rane R. The product of the times taken by it in the two possible ways is ( = acceleration due to ravity) [EAMCET 00 M] ) R ) R 3) 3R ) R Ans: For same rane two anles of projection are θ and 90 - θ usin θ usin( 90 θ) u cosθ t =,t u sinθcosθ R tt = = 9. The initial velocity of a particle, u = i + 3j. It is movin with uniform acceleration, a = 0.i + 0.3j. Its velocity after 0 seconds is [EAMCET 00 M] ) 3 units ) units 3) 5 units ) 0 units Ans: v = u+ at = i ˆ+ 3j ˆ + 0.i ˆ+ 0.3j ˆ 0 From ( ) ( ) v = 0 units = i ˆ+ 6j ˆ

7 Kinematics 0. A body of mass m projected vertically upwards with an initial velocity u reaches a maximum heiht h. Another body of mass m is projected alon an inclined plane makin an anle 30 with the horizontal and with speed u. The maximum distance travelled alon the incline is [EAMCET 00 M] ) h ) h 3) h/ ) h/ Ans: u h =.() From the equation v u = as ( ) = ( θ ) 0 u sin.h u h h = h [from ()] sin θ sin 30. A stone projected with a velocity u at an anle θ with the horizontal reaches maximum heiht. π When it is projected with velocity u at an anle θ with the horizontal reaches maximum heiht. The relation between horizontal rane, R of the projectile and is [EAMCET 000 M] ) R = ) R = ( ) 3) R ( ) Ans: u sin θ u sin ( 90 θ) u cos θ =, = + ) u sin θcosθ but R = on solvin R =. For a projectile the ratio of maximum heiht reached to the square of time of fliht is [ = 0 ms ] [EAMCET 000 M] ) 5 : ) 5 : 3) 5 : ) 0 : Ans: u sin θ 0 5 = = T u sin θ 3

8 Kinematics 3. The averae velocity of a body movin with uniform acceleration after travelin a distance of 3.06m is 0.3 ms. If the chane in velocity of the body is 0. ms durin this time its uniform acceleration is [EAMCET 000 M] ) 0.0 m/s ) 0.0 ms 3) 0.03 ms ) 0.0 ms Ans: u+ v u+ v s = t 3.06 =.t 306 = 3.06 = ( 0.3) t t 93 3 v u 0. but a = 0.0ms t 9