Phy 1 (12)

8/18/2019 Phy 1 (12)

1/50

http://www.rpmauryascienceblog.com/

KINEMATICS PROBLEM AND THEIR SOLUTION

Q.1. A projectile is fired from the top of a tower 40 meter high with an initial speed of 50 m/s at an

unknown angle. Find its speed when it hits the ground.

Q.2. An aeroplane is flying in the horizontal direction with a velocity 540 km/hr at a height of2000 m. When it is vertically above the point 'A' on the ground, a body is dropped from it. Thebody strikes the ground at point B. Calculate the distance AB.

Q.3. The co-ordinates of a moving particle at any time t are given by x = ct2 and y = bt2. Find initialspeed of the particle.

Q.4. A ball is thrown at a speed of 50 m/s at an angle of 600 with the horizontal. Find

(a) the maximum height reached.

(b) the range of ball. (Take g = 10 m/s2)

Q.5. Two cars are moving in the same direction with the same speed 30 km/hr. They are separated by adistance of 5 km. What is the speed of a car moving in the opposite direction if it meets these twocars at an interval of 4 minutes?

Q.6. A stone is projected with a speed of 40 m/s at anangle of 300 with the horizontal from a tower ofheight 100 m above ground. Find(a) the maximum height attained by the stone.(b) the horizontal distance from the tower where it

hits the ground.

100 m

30040 m/s

Q.7. A projectile is fired from the top of a tower 40 meter high with an initial speed of 50 m/s at anunknown angle. Find its speed when it hits the ground.

Q.8. An aeroplane is flying in a horizontal direction with a velocity 600 km/hr at a height of 1960 m.When it is vertically above the point 'A' on the ground, a body is dropped from it. The body strikesthe ground at point B. Calculate the distance AB.

Q.9. Two cars are moving in the same direction with the same speed 30 km/hr. They are separated by adistance of 5 km. What is the speed of a car moving in the opposite direction if it meets these twocars at an internal of 4 minutes?

Q.10. A man standing on a road has to hold his umbrella at 300 with the vertical to keep the rain away. Hethrows the umbrella and starts running at 10 km/hr. He finds raindrops are hitting his head vertically.Find the speed of raindrops with respect to(a) the road (b) the moving man.

http://www.rpmauryascienceblog.com/http://www.rpmauryascienceblog.com/

8/18/2019 Phy 1 (12)

2/50


Q.11. Find the relation between the acceleration of rod A andwedge B in the arrangement shown in the figure. Assume allthe surfaces to be smooth.

B

A

Q.12. The position of a particle at time t = 0 is P = (-1, 2, -1). It starts moving with an initial velocityˆ ˆu 3i 4 j

and with uniform acceleration ˆ ˆ4i 4 j . Find the final position and the magnitude of

displacement after 4 sec.

Q.13. A particle is projected with velocity u and angle with the horizontal. Find the time after which thevelocity will be perpendicular to the initial velocity.

Q.14. A particle moves in x-y plane with constant acceleration ‘a’ directed along the negative y-axis. Theequation of motion of the particle has the form y = x - x2, where and are positive constants.Find the velocity of the particle at the origin.

Q.15. From the velocity time graph shown in figure.Find the distance travelled by the particle duringthe first 40 sec. Also find the average velocityduring this period. 20 40

5 m/s

-5 m/s

sec.

Q.16. A train travels from one station to another at a speed of 40 km/hr and returns to the first station atthe speed of 60 km/hr. Calculate the average speed and average velocity of the train.

Q.17. A body travels 200 cm in the first two seconds and 220 cm in the next four seconds. What will be

the velocity at the end of seventh second from start ?

Q.18. A man standing on a hill top projects a stonehorizontally with speed v0 as shown in figure. Takingthe co-ordinate system as given in the figure find the co-ordinates of the point where the stone will hit the hillsurface.

x

y

(0, 0)

v0

Q.19. A particle of mass 3 kg moves under a force of î4 + 8 ĵ + 10 k̂ . Newton. Calculate theacceleration (as vector) to which the particle is subjected to. If the particle starts from rest and wasat origin initially, what are its new coordinates after 3 seconds?.

Q.20. In a car race, car A takes a time t sec less than car B at the finish and passes the finishing pointwith speed v m/s more than the car B. Assuming that both the cars starts from rest and travel with

constant acceleration a1 and a2 respectively. Show that v=( 21aa ) t

Q.21. On a cricket field the batsman is at the origin of co-ordinates and a fielder stands in position (46 i

+

28 j

) m. The batsman hits the ball so that it rolls along the ground with constant velocity


8/18/2019 Phy 1 (12)

3/50


(7.5 j10i

) m/s. The fielder can run with a speed of 5 m/s. If he starts to run immediately the ball

is hit, what is the shortest time in which he could intercept the ball ?

Q.22. A particle moves in the x - y plane with velocity vx = 8t - 2 and vy = 2. If it passes through the pointx = 14 and y = 4 at t = 2s, Find the equation of the path.

Q.23. (a) A ball rolls off the edge of a horizontal table top 4m high. If it strikes the floor at a point 5m

horizontally away from the edge of the table, what was its speed at the instant it left the table.

Q.24. A farmer has to go 500 m due north, 400 m due east and 200 m due south to reach his field. If hetakes 20 minutes to reach the field,(a) what distance he has to walk to reach the field ?(b) what is his displacement from his house to the field ?(c) what is the average speed of farmer during the walk ?(d) what is the average velocity of farmer during the walk ?

Q.25. Two cars are moving in the same direction with the same speed 30 km/hr. They are separated by a

distance of 5 km. What is the speed of a car moving in the opposite direction if it meets these twocars at an internal of 4 minutes?

Q.26. The equation of motion of a particle moving along a straight line is given as x = ½ vt where x, v, thave usual meaning. Draw its approximate acceleration time graph.

Q.27. A river 400 m wide is flowing at a rate of 4 m/s. A boat is sailing at a velocity of 20 m/s with respect

to the still water in a direction making an angle 370 with the direction of river flow.

(a) Find time taken by the boat to reach the opposite bank.(b) How far from the starting point does the boat reach on the opposite bank.

Q.28. An object projected with the same speed at two different angles covers the same horizontal range

R. If the two times of flight be t1 and t2, prove that R =2

1gt1t2.

Q.29. Velocity-time graph of a particle moving in a

straight line is shown in figure. Plot thecorresponding displacement time graph of a

particle if at t = 0 displacement s = 0.

t (sec.)

v ( m / s )

2 4 6 80

10

20

A B

C

D

Q.30. A particle is projected with velocity v and at angle from the horizontal. Find the instantaneouspower delivered by the gravity at the highest point.

Q.31. Two particles move in a straight line towards each other with initial velocities v1 and v2 and with

constant accelerations a1 and a2 directed against the corresponding velocities at the initial instant.

What must be the initial maximum separation Smax between the two particles for which they meet

during the motion?


8/18/2019 Phy 1 (12)

4/50


Q.32. If an object travels one half of its total path in the last second of its fall from the rest then find

(a) the time and

(b) the height of its fall.

Q.33. A truck moving with constant acceleration covers the distance between two points 180 m apart in 6

seconds. Its speed as it passes the second point is 45 m/s. Find(a) its acceleration

(b) its speed when it was at the first point.

Q.34. A body undergoing uniformly accelerated motion starts moving along +x-axis with a velocity of 5

m/s and after 5 seconds its velocity becomes 20 m/s in the same direction. What is the velocity of

the body 10 seconds after the start of the motion ?

Q.35. What is the speed with which a stone is projected vertically upwards from the ground if it attains a

maximum height of 20 m?

Q.36. A stone is projected from the ground with a velocity of 20 2 m/s at an angle of 450 with thehorizontal? What is the maximum height from the ground attained by the stone?

Q.37. The velocity of a car changes from 72 km/hr due east to 72 km/hr due north in 10 seconds. What is

the average acceleration of the car over this duration of time?

Q.38. A train moving along a straight road with a speed of 108 km/hr is brought to stop to next station

within 120 sec. after applying the brakes for the next station. What is the magnitude of the

retardation of the train.

Q.39. A car moving along a long straight road with a speed of 10 m/s is brought to rest within

10 seconds after applying the brakes. What is the magnitude of the retardation of the car ?

Q.40. A box is sliding on a smooth frictionless surface as shown in

the figure. A particle is projected at any unknown angle w.r.t.box. At the same time another particle of mass m is releasedfrom the ceiling of the block. Find out the relative accelerationof the two particles.

m2m

u

Q.41. A stone is projected from a balloon which is ascending with a velocity 2 m/s. The velocity of the

stone w.r.t. balloon is 2 m/s at an angle of 450. Find out the velocity of stone with respect to

ground.

Q.42. A body is projected with velocity 5 3 m/s at an angle of 600 with the horizontal. Find the angle

between the initial velocity vector and the velocity vector at a height of 2.5 m.


8/18/2019 Phy 1 (12)

5/50


Q.43. The velocity – time graph of a particle movingalong X – axis is shown. Find the displacementand the distance travelled in 10 s.

0 5 10

v 2(m/s)

t(s)

-1

Q.44. A particle of mass m is projected horizontally from certain height with a velocity v0. Find kineticenergy of the particle after t seconds, assuming it is in the air.

Q.45. A particle moves in x-y plane such that x = kt and y = mt2. Where k and m are constants and t isthe time. Find the velocity and the equation of trajectory of the particle.

Q.46. A car starting from rest moving on a straightline has acceleration – time graph as shownin the figure. Draw the velocity – time

graph.

-2

-1

1

2

m/s2

1 2 3 4 5t (sec.)

a

Q.47. The acceleration of a particle varies with timeas shown. If velocity at t = 0 is v0 thencalculate velocity as a function of time.

t

a

450

Q.48. A stone is dropped from a tower of height H. If distance covered during last second is half of

total height H. Find the total time taken by stone to reach the ground. Also find the value of H.

Q.49. A particle moves in a circle of radius 20 cm at a speed given by v = 1 + t + t2 m/s where t is time ins. Find (a) the initial tangential and normal acceleration. (b) the angle covered by the radius in first 2s.

Q.50. A particle of mass m is projected at angle with the horizontal. The speed of a particle, when it isat the greatest height is (2/5)1/2 times its speed when it is at half of its greatest height. Determine itsangle of projection.


8/18/2019 Phy 1 (12)

6/50


Q.51. A particle starts from origin at t = 0 along +ve x axis.It’s velocity –time graph is shown in the figure. Draw(i) a, t graph(ii) x, t graph

-4

O

4

v

2 4

t

Q.52. The motion of a particle along a straight line is described by the function, x = 6 + 4t2 - t4 where x is

in meters and t is in seconds. Find the position, velocity and acceleration at t = 2 sec.

Q.53. A boat travels downstream from point A to point B in two hours and upstream in four hours. Find thetime taken by a log of wood to cover the distance from point A to point B.

Q.54. A particle slides down a smooth inclined plane ofelevation fixed in the elevator going up with anacceleration a0 as shown in figure. The base of theincline has a length L. Find the time taken by the particleto reach the bottom.

m a0

L

Q.55. Two particles of masses m1 and m2 in projectile motion have velocities 1v

and 2v

respectively at

time t = 0. They collide at time t0. Their velocities become 1v

and 2v

at time 2t0 while moving in air.Find the value of |)vmvm()vmvm(| 22112211

.

Q.56. A particle moving with uniform acceleration describes distances S1 and S2 meters in successive

intervals of time t1 and t2 seconds. Prove that the acceleration is2 (S2t1 - S1t2) / t1t2 (t1 + t2)

Q.57. Two cars having masses m1 and m2 move in circles of radii r 1 and r 2 respectively. If they complete

the circles in equal time, find the ratio of the their angular speeds2

1

.

Q.58. A body of mass m is projected vertically upwards with a velocity v0. It goes up and comes back to

the same point. For this motion draw displacement-time, velocity-time, acceleration-time andvelocity-displacement graphs.

Q.59. A man standing on a road has to hold his umbrella at 300 with the vertical to keep the rain away.

He throws the umbrella and running at 10 kmph. He finds that rain drops are hitting his headvertically. find the speed of raindrops with respect to(a) the road(b) the moving man.


8/18/2019 Phy 1 (12)

7/50


Q.60. A projectile travelling in a direction at 300 to the horizontal after 2 seconds of its start. It is travellinghorizontally after one more second. Calculate the speed and angle of projection of the projectile.

Q.61. Find the speed of two objects if, when they move towards each other, they get x meter closer everysecond and when they move uniformly in the same direction with their original speeds, they get ymeter closer every two seconds.

Q.62. From the velocity time graph shown in figure. Find the

distance travelled by the particle during the first 40 sec. Also find the average velocity during this period.

20 40

5 m/s

-5 m/s

sec.

Q.63. A particle A is moving along a straight line with velocity 3 m/s and another particle B has a velocity5 m/s at an angle 300 to the path of A. Find the velocity of B relative to A.

Q.64. One end of a massless spring of spring constant 100 N / m and natural length 0.5 m is fixed and theother end is connected to a particle of mass 0.5 kg lying on a frictionless horizontal table. Thespring remains horizontal. If the mass is made to rotate at an angular velocity of 2 rad/s, find the

elongation of the spring.

Q.65. A ball takes t second to fall from a height h1 and 2t second to fall from a height h2 then what is the

ratio of h1/h2.

Q.66. A projectile is projected with a unknown velocity at an unknown angle . If time of flight is4 sec. What will be the maximum height reach by the projectile.

Q.67. A stone is projected from the ground with a velocity of 10 m/s in the vertically upward direction.How long does it remain in the air ?

Q.68. Two particles A and B are moving in a horizontal planeanticlockwise on two different concentric circles with

different constant angular velocities 2 and respectively.(a) Find the relative velocity of B w.r.t. A after time t = /.(Initial position of particles A and B are shown in figure.)(b) Also find the relative position vector of B w.r.t. A.

A BX

Y

r2r

Q.69. The velocity-time graph of a particle is given as shown inthe figure. Find the distance travelled by the object in 7 th second.

1 sec t

v

7 m/s

4 m/s

3m/s

4 secO 7 sec


8/18/2019 Phy 1 (12)

8/50


Q.70. A plate is moving with a constant velocity v0 in ahorizontal plane. A small particle is moving in a

circular (horizontal) path of radius with constantangular velocity. Find velocity of particle withrespect to ground when line OP makes angle with x-axis.

x

y

P

v0 O

Q.71. A particle is projected with a velocity u at an angle with aninclined plane which makes an angle 45 with thehorizontal. Calculate the radius of curvature of the path ofprojectile when velocity of projectile becomes parallel to theplane.

v

Q.72. A man standing on a road has to hold his umbrella at 300 with the vertical to keep the rain away. Hethrows the umbrella and starts running at 10 km/hr. He finds raindrops are hitting his head vertically.

Find the speed of raindrops with respect to(a) the road (b) the moving man.

Q.73. Two bodies are projected from the same point with equal speeds and different angle of projection.If they both strikes at the same point on an inclined plane whose inclination is . If be the angleof projection of the first body with the horizontal show that the ratio of their times of flight is

cos

sin(

Q.74. The velocities of particles P and Q are in directioninclined at an angle and with the line segment

PQ and if the distance between P and Q remainsconstants and given velocity of P is u. Find angularspeed of Q with respect to P.

u v

P Q

Q.75. (a) From an elevated point A, a stone is projected vertically upwards. When the stone reaches adistance h below A, its velocity is double of what it was at a height h above A. Show that the

greatest height attained by the stone is5

3h .

(b) The dependence of the x coordinate of twobodies moving in a straight line (x - axis) isgiven by curves a and b, respectively. Which

curve corresponds to the accelerated motionand which curve to decelerated motion ?Explain .

t

x b

a

Q.76. An open elevator is ascending with zero acceleration . The speed v = 10m/sec. A ball is thrownvertically up by a boy when he is at a height h = 10m from the ground. The velocity of projection is v= 30m/sec with respect to elevator . Find(a) the maximum height attained by the ball(b) the time taken by the ball to meet the elevator again.


8/18/2019 Phy 1 (12)

9/50


(c) time taken by the ball to reach the ground after crossing the elevator

Q.77. A projectile is launched from on inclined plane with an

initial velocity v0 as shown in the figure. Find the time

after which the projectile hits the plane for the first

time.

90

Q.78. A particle is projected up with a speed of 25 m/s from the ground. What is the maximum height

attained by the stone ? What is the distance travelled by the stone during 3rd second?

Q.79. A car moving with constant acceleration covers a distance of 24 m in first 2 seconds and

51 m in the next 3 seconds. What is the velocity of the car after next 5 seconds.

Q.80. A person walks with constant speed of 5 km/hr. He walks for 1hr. due east, then for 2 hrs. due

north, then for 1 hr. again due east and finally for 2 hr. due south west.

(a) What is the displacement of the person ?(b) What is the total distance travelled by him?

Q.81. A car starts moving from rest with an acceleration whose value linearly increases with time fromzero to 6 m/s2 in 6 sec after which it moves with constant velocity. Find the time taken by the car totravel first 72 m from starting point.

Q.82. In the pulley-block system shown, find theaccelerations of A, B, C and the tension in thestring. Assume the friction to be negligible andthe string to be light and inextensible. Themasses of the blocks are m, 2m and 3mrespectively.

C

B

A

300

Q.83. A farmer has to go 500 m due north, 400 m due east and 200 m due south to reach his field. If hetakes 20 minutes to reach the field,

(a) what distance he has to walk to reach the field ?(b) what is his displacement from his house to the field ?(c) what is the average speed of farmer during the walk ?(d) what is the average velocity of farmer during the walk ?

Q.84. A man wants to reach point B on the opposite bank of ariver flowing at a speed u as shown in the figure. Whatminimum speed relative to water should the man have sothat he can reach point B ? In which direction should heswim ?

B

u450

A


8/18/2019 Phy 1 (12)

10/50


Q.85. Two Blocks A and B of masses 10 kg and 6 kg

respectively are connected through pulley as

shown in figure. Find

(a) acceleration of A and B.

(b) friction force acting on block A and B.

(c) tension between A and B. You may assume

the string was initially in just taut position.

10kg

=0.5

A

4kg

6kg

2kg

B

=0.2

[1+2+1=4]

Q.86. Two particles projected vertically upward from

point (0, 0) and (1, 0) with uniform velocity 10 m/s

and v m/s respectively, as shown in the figure. It is

found that they collide after time t in space. Find v

and t.

Y

45 30

10 m/sv m/s

x(1, 0)(0, 0)

Q.87. Two smooth wedges of equal mass m are placed as

shown in figure. All surfaces are smooth. Find thevelocities of A & B when A hits the ground.

h

A B

Q.88. A particle moves in the x - y plane with velocity ĵqxîpv where î and ĵ are unit vectors in thedirection of x and y-axis, p and q are constants. At the initial moment of time, the particle waslocated at the point x = y = 0. Find the equation of the trajectory of the particle.

Q.89. The velocity of a boat in still water is n times less than the velocity of flow of a river. At what angle

to the stream direction must the boat move so that drift is minimised ? If n = 2, show that the angle = 1200.

Q.90. The velocity of a particle, when it is at the greatest height is (2/5)1/2 times its velocity when it is athalf of its greatest height. Determine its angle of projection.

Q.91. On frictionless horizontal surface, assumed to be the x-y plane, a

small trolley A is moving along a straight line parallel to the y-axis

(see figure) with a constant velocity of (31) m/s. At a particularinstant, when the line OA makes an angle of 45 with the x-axis, a

ball is thrown along the surface from the origin O. Its velocitymakes an angle with the x-axis and it hits the trolley.

(a) The motion of the ball is observed from the frame of the trolley.

Calculate the angle made by the velocity vector of the ball

with the x-axis in this frame.

O

A

y

x

45

(b) Find the speed of the ball with respect to the surface, if = 34

Q.92. A river 400 m wide is flowing at a rate of 2.0 m/s. A boat is sailing at a velocity of 10 m/sec w.r.t thewater in a direction perpendicular to river, find(a) the time taken by the boat to reach the opposite bank.


8/18/2019 Phy 1 (12)

11/50


(b) how far from the point directly opposite to the starling point does the boat reach on oppositebank.

Q.93. Two guns, situated at the top of a hill of height 10 m, fire one shot each with the same speed 53m/s at some interval of time. One gun fires horizontally and the other fires upwards at an angle of600 with the horizontal. The shots collide in air at a point P. find (I) the time interval between thefringes and (ii) the co-ordinates of the point P, take origin of the co-ordinate system at the foot of thewill right below the muzzle and trajectory in x-y plane.

Q.94. A man can row a boat in still water at 3 km/h He can walk at a speed of 5 km/h on the shore. Thewater in the river flows at 2 km/h. If the man rows across the river and walks along the shore toreach the opposite point on the river bank find the direction in which he should row the boat so thathe could reach the opposite shore in the least possible time. The width of the river is 500 m.

Q.95. There are two parallel planes each inclined to the horizontal at an angle . A particle is projectedfrom a point mid-way between the two planes so that it grazes one of the planes and strikes theother at right angle. Find the angle of projection.

Q.96. A body falling freely from a given height H hits an inclined plane in its path at a height h. As a

result of this impact the direction of the velocity of the body becomes horizontal. For what value ofh/H the body will take maximum time to reach the ground ?

Q.97. The velocity time graph of moving object is given in the figure. Draw the acceleration versus timeand displacement versus time graph. Find the distance travelled during the time interval when theacceleration is maximum. Assume that the particle starts from origin.

80

60

40

O

20

8070605040

Time (s)

302010

V

(m/s

Q.98. A ball is thrown from the origin in the

x - y plane with velocity 28.28 m/s at anangle 45 to the x - axis. At the sameinstant a trolley also starts moving with

uniform velocity of 10m/s along thepositive x - axis. Initially, the trolley islocated at 38m from the origin. Determinethe time and position at which the ball hitsthe trolley.

D

C2mA

B

3m

10m/s

v0

45

xO

y

38m

Q.99. A particle is suspended from a fixed point by a inextensible string of length 5m. It is projected from

its lowest position in the horizontal direction in a vertical plane with such velocity that the string

slackens after the particle has reached a height 8m above the lowest position. Find the velocity of

the particle, just before the string slackens. Find also to what height the particle can rise further.


8/18/2019 Phy 1 (12)

12/50


Q.100. A bus is running along a highway at a speed of v1 = 16 m/s. A man is at a distance a = 60 m from

the highway and at a distance b = 400 m from the bus. In what direction should the man run to

reach any point of the highway, before or at the same time as the bus. The man can run at a speedof v2 = 4 m/s.

Q.101. An aircraft flies at 400 km/hr in still air. A wind of200 2 km /hr is blowing from the south. The pilot

to travel from A to a point B north east of A. Find

the direction he must steer and time of his journey

if AB = 1000 km. (Given cos 150 = 0.9659)

N

AE

450

450

vw

C

vaw

va

B

Q.102. A insect moves with constant speed of 10 m/s. At t = 0. It moves for 3 second due to east, next 3

second due to North and finally for 3 2 second due south west.

(a) What is the displacement of the insect ?

(b) What is the total distance travelled by insect ?

Q.103. A stone ‘A’ is dropped from the top of a tower 20 m high simultaneously another stone ‘B’ is thrown

up from the bottom of the tower so that it can reach just on the top of the tower. What is thedistance of the stones from the ground while they pass one another.

Q.104. A car moving with constant acceleration covers a distance of 100 m in 3 sec and then next 100 m in

2 sec. Find the acceleration of the car.

Q.105. A cyclist moves with constant speed 5 m/s along eastward for 2 seconds, then along southward for

2 seconds, then he moves along west for one second and finally along North- west for 2 seconds.

Find

(a) Distance and displacement of cyclist for whole journey.

(b) Average speed and Average velocity for whole journey

(c) Average acceleration of cyclist for whole journey.

Q.106. A car starts from rest and moves with a constant acceleration of 2.0 m/s2 for 30 seconds. The

brakes are then applied and the car comes to rest in another 60 seconds. Find(a) total distance covered by the car.

(b) Maximum speed attained by the car(c) Find shortest distance from initial point to the point when its speed is half of maximum speed.

Q.107. A particle is projected with speed 60 m/sec at an angle 30 0 from the horizontal. Find

(a) Minimum time taken to reach a height of 25 m.(b) Vertical and horizontal component of velocity at the time found in part (a)

(c) The horizontal displacement covered by particle in the time calculated in part (a).

Q.108. How much high above the ground a monkey can throw a mango if he is able to throw the same

mango upto a maximum horizontal distance of 100 m ?


8/18/2019 Phy 1 (12)

13/50


Q.109. A body of mass ‘m’ is dropped freely from aheight ‘h’. Then draw the following graphsfor the given body ?(a) Displacement - Time graph(b) Kinetic energy - Time graph

(c) Total energy - Time graph(Assume displacement to be zero at groundlevel.)

h

m

Q.110. A batsman hits a ball at a height of 1.22 m above the ground so that ball leaves the bat an angle

45o with the horizontal. A 7.31 m high wall is situated at a distance of 97.53 m from the position ofthe batsman. Will the ball clear the wall if its range is 106.68m.Take g = 10 m/s2.

Q.111. A particle is projected from point 0 on theground with velocity u=55 m/s at angle

1tan (0.5). It strikes at a point C on afixed smooth plane AB having inclinationof 370 with horizontal as shown in figure.Calculate(a) Coordinates of point C in reference to

coordinate system as shown in figure(b) Velocity of particle with which it strikes

inclined plane AB.

37

5 5

O

B

C

x

y

A10/3 m

Q.112. A motor car is travelling at 30 m/s on a circular road of radius 500 m. It is increasing in speed at therate of 2m/s2. What is its acceleration?

Q.113. An object A is kept fixed at the point x =3m and y = 1.25 m on a plank P raisedabove the ground. At time t = 0 theplank starts moving along the +x directionwith an acceleration 1.5 m/s2. At thesame instant a stone is projected from theorigin with a velocity u as shown. Astationary person on the ground observesthe stone hitting the object during itsdownward motion at an angle of 450 to thehorizontal. All the motions are in the x-yplane. Find u and the time after which the

stone hits the object.

u

O 3.0 m

A

P

y

1.25m

P

x

Q.114. Two masses ‘m’ and ‘2m’ are connected by a massless

string which passes over a light frictionless pulley as

shown in fig.1. The masses are initially held with equal

lengths of the strings on either side of the pulley. Find thevelocity of the masses at the instant the lighter mass

moves up a distance of 6.54 m. The string is suddenly cutat that instant. Calculate the time taken by each to reachthe ground. (g = 9.81 m/s2)

13.08 m

2mm

ground

Fig. 1


8/18/2019 Phy 1 (12)

14/50


Q.115. A particle moves in a circle of radius 20 cm at a speed given by v = 1 + t + t2 m/s where t is time in

s. Find(a) the initial tangential and normal acceleration.(b) the angle covered by the radius in first 2 s.

Q.116. A car starts moving from rest with an acceleration whose value linearly increases from zero to 6

m/s2 in 6 sec after which it moves with constant velocity. Find the time taken by the car to travel first

72 m from starting point.

Q.117. A particle is projected vertically with velocity v0. Wind isblowing and is providing a constant horizontalacceleration a0. There is a vertical wall at somedistance from point of projection. If particle strikes thevertical wall perpendicularly then calculate(i) Time of flight.

(ii) Velocity with which particle strikes the vertical wall.(iii) Distance x and y.(iv) If collision at vertical wall is perfectly elastic willparticle retrace its path ?(v) Is path of particle parabolic?

`

y

x

v0

Q.118. A particle is projected up a large inclined planefrom a point O on it as shown in the figure. Theprojection velocity has a magnitude of 5.5m/s andits direction makes an angle 37° with the inclinedplane. The inclination of the plane is also 37°. Theinclined plane starts moving towards left with an

acceleration a0 = 5 m/s2 at the moment theparticle is projected. The particle strikes theinclined plane at a point P. Find the time taken bythe particle to move from O to P. Also find themagnitude of displacement along the inclinedplane as it moves from O to P. (Take sin 37 0 =3/5)

P

O370

370u = 5.5 m/s

a0 = 5 m/s2

Q.119. Two inclined planes of inclinations 300 and 600

respectively meet at 900 as shown in figure. A particle isprojected from point P on the first inclined plane with a

velocity u = 10 3 m/s in a direction perpendicular to theinclined plane and it is observed to hit the other inclined

plane at 900.Find (a) the height of point P from ground

(b) the length of PQ.

6030

u

B

Q

O

A

h

P


8/18/2019 Phy 1 (12)

15/50


Q.120. A sleeve ‘A’ can slide freely along the smooth rod bent in the

shape of a half circle of radius R as shown in figure. The system

is set in rotation with a constant angular velocity about thevertical axis OO. Find the angle corresponding to the steadyposition of the sleeve.

R

O’

A

O

Q.121. A large heavy box is sliding without friction down a

smooth inclined plane of inclination . From a pointP on the bottom of the box a particle is projected

inside the box, with speed u (relative to box) atangle with the bottom of the box.

P

Q

u

(a) Find the distance along the bottom of the box between the point of projection P and the point Q

where the particle lands. The particle does not hit any other surface of the box.

(b) If horizontal displacement of the particle with respect to ground is zero. Find the speed of thebox at the moment when particle was projected.

Q.122. A stone is dropped from the top of a tower 20 m high. Simultaneously another stone is thrown up

from the bottom of the tower so that it can reach the top of the tower. What are the speeds of the

stones while they pass one another ?

Q.123. Two rockets are fired vertically from launching pads which are side by side. The first rocket moves

vertically upwards with an acceleration of 6g and second with an acceleration of 8g. If the secondrocket is fired 1 sec. after the first, find how long after its launching the second rocket overtakes the

first.

Q.124. A particle is moving along a vertical circle ofradius R = 20 m with a constant speed v = 31.4m/sec. as shown. Line ABC is horizontal andpasses through the centre of the circle. A shellis fired from point A at the instant when particle

is at C. If distance AB is 20 3 m and shell

collides with the particle at B, calculate

20 m

CB

A

u

(a) Smallest possible value of the angle of projection.(b) Corresponding velocity u of projection

Q.125. A projectile is fired with velocity v0 at an angle with the horizontal on a horizontal plane, Find(a) The average velocity of projectile in half of time of flight.(b) The time in which the velocity of projectile becomes perpendicular to its initial velocity.(c) The radius of curvature of projectile at the instant when it is at its maximum height.

Q.126. A stone is projected from a point of ground in such a direction so as to hit a bird on the top of atelegraph post of height h and then attain the maximum height 2h above the ground. If at theinstant of projection the bird were to fly away horizontally with a uniform speed, find the ratiobetween the horizontal velocities of the bird and the stone, if the stone still hits the bird whiledescending.


8/18/2019 Phy 1 (12)

16/50


Kinematics Solution

Q.1. Initial K.E. = 22 50.m

2

1u.m

2

1

Final K.E. = 2mv2

1

Work done by gravity = +mgh = mg. 40

From W-E principle

mg . 40 = kf - kI =2

1m(v2 - 502)

v = 57.4 m/s.

Q.2. For plane,

Horizontal velocity = 540 18

5 m/s; vertical velocity = 0

Time of flight t =2h 2 2000

g 10

= 20 sec.

Horizontal displacement = 20 540 18

5 = 3000 m

Q.3. zero

Q.4. (i) Maximum height H =2u sin2

2g

H =

22(50) 3

m2 10 2

H = 93.75 m

(ii) Range R =2 2u sin2 (50) sin120

g 10

R = 216.5 m

Q.5. 30v

5

60

4

c

vc = 45 km/hr.

Q.6. (a) Maximum height above the tower, using v2 = u2 + 2as in vertical direction.(u sin 300)2 = 2gh As u = 40 m/s, = 300 40 40 1

4

= 2 10 h h =

160020m

80

height above ground = 100 + 20 = 120m.

(b) Range, time of flight = t, H = u sin t - 21

gt2

, H = - 100 m,


8/18/2019 Phy 1 (12)

17/50


- 100 = (40 1

)t2

- 21

10 t2

- 100 = 20 t - 5t2 t2 – 4t – 20 = 0, t = 6.9 sec.R = ucos t, R distance from tower

R = 40 3 6.92

= 238.9 m.

Q.7. Initial K.E. = 22 50.m2

1u.m

2

1

Final K.E. = 2mv2

1

Work done by gravity = +mgh = mg. 40From w~E principle

mg . 40 = kf - kI =2

1m(v2 - 502)

v = 57.4 m/s.

Q.8. For plane,



Time of flight t =10

19602

g

h2 = 142sec


5

= 3299.83 m 3300 m.

Q.9. 30v

5

60

4

c

vc = 45 km/hr.

Q.10. (a) vmg = vrg sin 300

vrg = 20 km/hr.

(b) vrm = vrg cos 30

vrm = 10 3 km/hr

300

vm, g

vr, mvrg

Q.11. y

tanx

a A = aB tan .

Q.12. Initial position vector of particle

ˆ ˆ ˆi 2 j k Final position of the particle after 4 seconds

2

f i

1S S ut at

2


8/18/2019 Phy 1 (12)

18/50


1ˆ ˆ ˆ ˆ ˆ ˆ ˆi 2 j k 3i 4j 4 4i 3j 162

final position = ˆ ˆ ˆ21i 42j k , Displacement = - 20 î + 40 ̂j .

Magnitude of displacement = 2 2

20 40 20 5 m

Q.13. ĵ)gtsinu(îcosuv

ĵsinuîcosuu

0v.u = u2 cos2 + u2 sin2 - (u sin)gt

t =sing

u

Q.14. Comparing this equation y = x - x2 with equation of projectile.

y = x tan -22

2

cosu2

ax,

we get tan = & = 22 secu2a

or, u2 = )tan1(2

a 2

u = )1(2

a 2

Q.15. Distance = 100 mVelocity = zero

Q.16. Average speed =604060402

vvvv2

v

s

v

s s221

21

21

= 48 km/hr Average velocity = 0

Q.17. 2 = u(2) +2

1a(2)2 2 = 2u + 2a

1 = u + a . . . . (i)

2.2 + 2 = u(6) +2

1a(6)2

4.2 = 6u + 18 a 2.1 = 3u + 9a 0.7 = u + 3a . . . . (ii)Solving (i) and (ii) we geta = -0.15 m/s2. , u = 1.15 m/s.v = 1.15 - 0.15 7 = 0.01 m/s.

Q.18. x = v0 cos t

y = - 2gt2

1

|y| = |x| tan


8/18/2019 Phy 1 (12)

19/50


y = -2

1 g

220

2

cosv

x

2

1g

220

2

cosv

x = x tan

x = 0 or x =g

tancosv2 220

x =g

2sinv20

y =

220

2

240

cosvg

2sinvg

2

1

= -

2

220

cosg2

2sinv

Q.19. TakingF

as the net force

k̂3

10 ĵ

3

8î

3

4a

m/s2

2ta2

1tus

= 9k̂3

10 ĵ

3

8î

3

4

2

10

= k̂15 ĵ12î6 sx = 6m; sy = 12m sz = 15m

Q.20. Let time taken by the car and their final velocities are t1, t2 and v1, v2 respectively.Given t1 = t2 - t and v1 = v2 + v

s1 =1

2

1

21 1

22 2 2

2a t s a t = s (say)

a t a t1 12

2 22 = 2s

also v1 = a1t1, v2 = a2t2 v1t1 = a1t1

2 = 2s and v2t2 = a2t22 = 2s

t1 =2

1

s

vand t2 =

2

2

s

v

so t2 - t1 = 21 1

2 1

s

v v

t

2 1 2

1 2

sv v

v vt

21 2

sv

v vt

v =

v v

st

v v

st1 2 1

222

22 2


8/18/2019 Phy 1 (12)

20/50


=v v

t tt a a t1 2

1 21 2

Q.21. The ball's position is at time t,

(7.5)t. î + (10)t ĵ

Suppose the fielder runs from his position with constant velocity 5 [( sin ĵcosî ] m/s. relative to

the wicket. At interception of the ball by the fielder the position must coincide so equating thecomponents we get7.5 t = 46 + 5t cos . . . (I)and 10t = 28 + 5t sin . . . (ii)

These give 1t5

28t10

t5

46t5.722

which simplifies to t = 4 sec, or

21

116sec.

Hence for earliest interception t = 4s.

Q.22. vx = 8 t - 2x = 8 t2/2 - 2 t + c1 at t = 2x = 14 so c1 = 2so x = 4t2 - 2t + 2 . . . (i)vy = 2so y = 2 t + c2 at t = 2, y = 4 so c2 = 0y = 2t . . . (ii)eliminatating t form (I) and (ii)x = y2 - y+ 2

Q.23. (a) s = ut +2

1at2 for vertical motion

4 = 0 +2

1(10) t2

4m

vtable

5m

t2 =5

4

t =54 5 = v t

v = 5/t =5/4

5 =

2

55

4

55 m/sec.

Q.24. (a) Distance = 500 + 400 + 200 = 1100m

(b) Displacement = ĵ200î400 j500

= î400 j300

Magnitude of displacement = 22 300400 = 500m

y

x

N

E


8/18/2019 Phy 1 (12)

21/50


(c) Average speed =timeTotal

distanceTotal =

6020

1100

=

12

11m/s

(d) Average velocity =Time

placementisD =

6020

500

=

12

5m/s

Q.25. 30v

5604

c

vc = 45 km/hr.

Q.26. x =2

t.

dt

dx

x = kt2 k some constant

k2dt

xd2

2

.t

a

Q.27. (a) Resultant velocity of the boat is

v = (vR + vB cos 370 ) i + vB sin 37

0 j

4i + 20 5

4i + 20

5

3 j

v =20 i + 12 j m/s

time taken by boat to cross the river

=directionyinvelocity

directionyintravelledcetandis

t =12

400 =

3

100sec.

370

vB

vR400 m

Y

X

(b) Displacement along x = v t

= 20 3

100 =

3

2000 m

distance from starting point

3

109400

3

2000)400(

2

2

m.

Q.28. For same range, angles are and 900 -

R =g

2sinu2

t1 =g

sinu2 , t2 =

g

cosu2

g

90sin(u2

t1t2 =g

R2

g

cossinu42

2

R = 21tgt2

1.


8/18/2019 Phy 1 (12)

22/50

8/18/2019 Phy 1 (12)

23/50


u = gh2 = 20102 = 20 m/s

Q.36. uy = u sin = 20 2 sin 45-0 = 20 m/s

maximum height =

102

2020

g2

sinu 22

= 20 m.

Q.37. t

vva 12av

, s/m20v2

ĵ

s/m20v1

î

Thus ĵ2î210

î20 ĵ20aav

hence a = 2 2 m/s due north west.

Q.38. vi = 108 km/hr =6060

1000108

= 30 m/sec.

t = 120 sec.

vf = 0vf = vi + at

or, 0 = 30 m/s + 100 sec,

or, a = - 2s/m4

1

.sec120

sec/m30

or a = -(0.25) m/sec2 magnitude of retardation = 0.25 m/s2

Q.39. u = 10 m/s, v = 0, t = 10 s

a = 2s/m110

10

t

uv

|a| = 1m/s2

Q.40. Acceleration of particle of mass m w.r.to ground = g Acceleration of particle mass (2m) w.r.to ground = gThus ar = g – g = 0

Q.41. 0 0SBˆ ˆv v cos 45 i v sin45 j

1 1

ˆ ˆ2 i 2 j2 2 ˆ ˆ( i j)m / s

BGˆv 2jm / s

Thus S,G SB BGv v v

= 2 ˆ ˆ ˆ j ( i j) = ( ˆ ˆ(i 3 j)

v = 2 21 3 10 m / s and tan =3

1

= tan-1(3).

3 ̂j

î


8/18/2019 Phy 1 (12)

24/50


Q.42. For vertical component of motionvy

2 = uy2 + 2gh

v2sin21 = u2 sin21 2gh

v2sin22 = 2)35( sin260 2 10 (2.5)

v sin 2 = 5/2

Initially velocity vector u =

ĵ2

15î

2

35

1

O

2

2

v2

v1

2.5m

Velocities vector at height 2.5 m v1 =

ĵ

2

5î

2

35 and v2 =

ĵ

2

5î

2

35

Now the angle between initial velocity and velocities at height 2.5 m is

cos1 =|v||u|

v.u

1

1 =

4

25

4

75

4

225

4

75

4

75

4

75

=

32

3

2

10

2

300

4

752

cos1 = 3030cos23

1

cos2 =

900

4

100

4

300

4

75

4

75

2

Q.43. Displacement = 1 10 1 10 15

2 5 1 5 12 3 2 3 6

m

Distance travelled =

1 10 1 10 55

2 5 1 5 12 3 2 3 6

m

For plane,



Time of flight t =2 20002h

g 10

= 20 sec.


5=

10000

3m

Q.44. vx (horizontal) = v0 vy (vertically downward) = gt

v2 = 2y2x vv =

220 )gt(v

K.E. of particle after time ‘t’ =2

1mv2 =

2

1m( ))gt(v 220

Q.45. vx =dx

kdt

; vy =dy

mtdt

ˆ ˆv ki 2mtj

t =x

k & y = mt2 = m

2 2

2

x mx

k k


8/18/2019 Phy 1 (12)

25/50


Equation of trajectory, y =2

2

mx

k

Q.46.

1

2

4

5

3

1 2 3 4 5t (sec.)

v (m/s)

Q.47.

t

0

v

v

tdtdv

0

v = v0 + t2/2

Q.48. 21

H gt2

2H 1

g t 12 2

t

2t 1

tH

t 2t 2 2 1 t 2

2

t sec2 1 =3.4 sec

H = 21

gt 57.8m2

Q.49. Tangent acceleration at =dt

dv = 2t +1

normal acceleration an =R

v 2 (at)t=0 = 1 m/s

2

(an)t=0 = 2.01

R

v2

0

= 5 m/s2

(b) v = Rdt

d

R d = (1+t+t2 )dt R

'

0

2

0

2 dttt1d

= 33.3 rad

Q.50. Let u and are projection speed and angle of projection respectively.vx = u cos and vy = u sin


8/18/2019 Phy 1 (12)

26/50


At greatest height h =g2

sinu 22

vertical velocity at half the greatest height

v2 = 2/gh2v2y

v2 = u2 sin2 -2

sinug2

singu2222

v =2

sinu

Resultant velocity at half the greatest height

vR =

2222

22 cosu2

sinu)cosu(v

u cos =5

2

2222

cosu2

sinu

tan = 3 , = 600

Q.51. (i) Velocity is decreasingso, a = -4/2 = -2

-2

O

a

4t

(ii)

4

t

x

Q.52. At, t = 2 sec, x = 6 + 16 - 16 = 6 m

velocity, v =dt

dx = 8t - 4t3

at t = 2 sec, v = 16 - 32 = -16 m/sec.

Acceleration,dt

dv = 8 - 12 t2

at t = 2 sec. Acceleration = 8 - 12 (2)2 = -40 m/s2

Q.53. 8 hrs.


8/18/2019 Phy 1 (12)

27/50


Q.54.

2/1

0 cossin)ag(

L2

Q.55. 2 (m1 + m2)gt0

Q.56. Let u be the initial velocity. Then S1 = ut1 +2

1f 21t where f is the acceleration . In the next t2

seconds the point moves to S2. It moves through distances S1 + S2 in (t1 + t2) seconds

S1+S2 = u(t1 + t2 )+2

1 f (t2 + t1)

2

=(ut1 +2

1f 21t ) + ut2 +

2

1ft2 (t2 + 2t1)

S2 = ut2 +2

1ft2 (t2 + 2t1)

Hence S2t1 - S1t2 =

2

1ft1t2 (t1 + t2 )

or f =)tt(tt

)tStS(2

2121

2112

Q.57. 1

Q.58. displacement time velocity displacement

s

t

v

d velocity time acceleration time

v

t

a

t

Q.59. Velocity of rain w.r.t. road is r v

andvelocity of rain w.r.t. moving man is rmv

but

rmv

= mr vv

=-vr sin 30 î - vr cos 30 ĵ - 10 î

= (- vr sin 30 – 10 ) î - vr cos 30 ĵ

300

vm

vrvrm

But - vr sin 30 – 10 = 0 vr sin 30 = -10

vr =30sin

10 = - 20 m/s. But vr is not negative î10vm


8/18/2019 Phy 1 (12)

28/50


and ]30cos20[vrm

= 20 cos 30 ĵ

= 10 3 ĵ .

Q.60. After 2 seconds, vertical component of velocity, vy = v

0 sin - g(2) = (v

0 sin - 2g)

Horizontal component of velocity = v0 cos

Hence,

cosv

g2sinv

0

0 = tan 300 =3

1 . . . (i)

Again, velocity becomes horizontal when projectile reaches maximum height. Hence, time to reachmaximum height = 3 seconds.

Therefore,g

sinv0 = 3 v0 sin = 3g . . . (ii)

From (i) and (ii), v0 cos = 3 g, v0 sin = 3gHence, squaring, and adding, v0 = 2g3 = 20 3 m/sec.Dividing, tan = 3 = tan 600 = 600

Q.61. Let v A and vB are the speeds of the objectsWhile moving towards each other

v A + vB = x (m/s) . . . (i)While moving in same direction (Assuming v A > vB )

v A - vB = y/2 (m/s) . . . (ii)Solving (i) and (ii),

v A = )yx2(4

1

vB = )yx2(4

1 .

Q.62. Distance = 100 mVelocity = zero

Q.63. |vB - v A| =2 A

2B vv - 2v AvB cos 30

0

= 52 + 32 - 2 5 3 (3/2)= 8.02

using sine rule030sin

832.2

sin

3

= 320

300

vB

-v A

5 m/s

3m/s

vB - v A

Q.64. kx = m2 (0 + x)

x 1 cm.

Q.65. h1 =2gt

2

1, h2 =

222 gt2)t(g

2

1

4

1

gt2

gt)2/1(

h

h2

2

2

1

Q.66. Time of flight T =g

sinv2 0 … (i)


8/18/2019 Phy 1 (12)

29/50


Maximum height H =g2

sinv 220 … (ii)

eliminating v0 sin

H =g8gT

g22

Tg22

f

2

=8

gT

2

At T = 4 sec.,

H =8

4410

8

gT2 = 20 m

Q.67. u = 10 m/s, t = ? y = 0 thus

0 = 0 + 10t - 2gt2

1 10t =

2

1 10 t2 t = 2 seconds.

Q.68. A = 2

= 2, v A = 2r ̂

B =

= , vB = 2r(- ̂j )

(a) BA B Aˆ ˆv v v 2 r( j) 2 r( j)

= -4r ̂ j

(b) ABBA r r r

= 2r ( )î - (r) î = - 3r î .

A BX

Y

r2r

Q.69. 2dv 7 4

1m/sdt 4 1

t 7

as u (2n 1)

2

13 (2 7 1) 9.5m

2 .

Q.70. Velocity of particle w.r.t. ground

= v0 )î(sinî + cos ĵ

= (v0 - sin ) î + cos ĵ

P

v0 O

Q.71. Along y,usin gcos t 0

u

t tang

Along x,

xv ucos gsin .t u

ucos gsin . tang

=ucos usin ucos2

cos cos

X

g sin g cos

Y v


8/18/2019 Phy 1 (12)

30/50


2

2 2x

3n

v u cos 2r

a gcos

Q.72. (a) vrm = vrg cos 30

vrm = 10 3 km/hr(b) vmg = vrg sin 300

vmg = 20 km/hr.

300

vm, g

vr, mvrg

Q.73. Let be the angle of projection of the second body

R =

sin)2[sin(cosg

u2

2

since range of both the bodies is same. Thereforesin (2 - ) = sin (2 - )

or 2 - = - (2 - ) =

2

- ( - )

u

Now T =

cosg

sin(u2

T =

cosg

sin(u2

cos

sin(

T

T

Q.74. Let v be the velocity of Q

if is constant

u cos = v cos v =

cos

cosu

=

sinusinv

=

)sincos(tanu

Q.75. (a) If B & D be the points h above and h below A, then in the stone’s downward motion

vD = 2vB v v g hD B

2 2 2 2

4 42 2v v ghB B

v ghB2 4

3 h

h

C

B

A

D If C be the highest point attained by the stone then,

v v ghC B2 2 2 ' [when h = BC]

v gh h gh ghB2 2

4

6

2

3 ' '


8/18/2019 Phy 1 (12)

31/50


Greatest height = h + h =5

3h

(b) The acceleration of a body moving along straight line is the second derivative of its positioncoordinate. For a concave surface , the second derivative is +ive while it is negative for the convexsurface. Hence ‘a’ corresponds to accelerated while ‘b’ corresponds to decelerated motion.

Q.76. (a) Velocity of ball relative to elevator = 30 m/sVelcoity of ball relative to ground = 10 + 30 = 40 m/s

Maxmum height attained by ball =

102

4010

2

= 90m

(b) When they meet again their displacement is same

10t = 40t -1

2 10 t2

t = 6 sec

(c) 70 = 40t +1

2 10 t2

5t2 + 40t - 70 = 0

Q.77. Let the projectile hit the plane after time t.

The horizontal displacement x = (v0 sin) t

The vertical displacement y = (v0 cos) t -2

1gt2

y = -(tan) x for the plane

t =cosg

v2 0

Y

X

V0

O

Q.78. Maximum height h =102

625g2

u2

= 31.25 m

time after which v = 0, t =10

25 = 2.5 second.

Distance travelled in 3rd second = |y1| + |y2|where,y1 = y(2.5) - y(2)y2 = y(3) - y(2.5)

y(2) = 25 2 - m3020502102

1 2

y(3) = 25 3 - m30457531021 2

y(2.5) = 31.25 m

thus distance during 3rds = 1.25 + 1.25 = 2.5 m

Q.79. Let initial speed of the car is u and acceleration be a. then

2u +2

1 a 22 = 24

2u + 2a = 244 + a = 12 …(i)


8/18/2019 Phy 1 (12)

32/50


and 5u +2

1 a 52 = 75

10 u + 25 a = 150 …(ii)from (i) and (ii) a = 2 m/s2 s and u = 10 m/safter 10 s = 10 + 2 10 = 30 m/s

Q.80. Unit vector due east be î and due west be ĵ . then

net displacement = 5 )] ĵ(45sin25)î(45cos25[î5 ĵ10î

= (10 – 5) ĵ)510(î

= 5 ĵ5î

(a) thus net displacement = 5 2 due north east

(b) distance = 5 + 10 + 5 2 + 5 = 20 + 5 2 m

Q.81. Since acceleration varies linearly soa t m/s2 by given condition

a = kt 6 6

0 0

da k dt k = 1

then, tdt

dv v =

2

t2 m/s

then,2

t

dt

ds 2 or, s =

6

t3

at the end of t = 6 sec. Acceleration becomes zero.Distance moved by car at t = 6 sec is

S1 =

6

666 = 36 m

Speed of the car =2

66 = 18 m/s

Remaining distance = 72 – 36 = 36 m.

so time taken to cover this distance = t2 = .sec2.sec18

36

Total time = 6+2 = 8 sec.


8/18/2019 Phy 1 (12)

33/50


Q.82. The FBD of A,B,C are shownT = ma1 . . . (1)2mg – 2T = 2ma2 . . . (2)T + 3 mg sin 300 = (3m)a3 . . . (3)constraint relation :a1+ a3 = 2a2 . . . (4)

solving the equations

a1 =9

g20

, a2 =11

g20

, a3 =13

g20

, T =9

mg20

N1

T

a1

mg

2mg

2T

a2

N3

3mg

T

a3

Q.83. (a) Distance = 500 + 400 + 200 = 1100m

(b) Displacement = ĵ200î400 j500

= î400 j300

Magnitude of displacement = 22 300400 = 500m

(c) Average speed =timeTotal

distanceTotal =

6020

1100

=

12

11m/s

y

x

N

E

(d) Average velocity =Time

placementisD =

6020

500

=

12

5m/s

= tan-1300

400

= 370 due North of East.

Q.84. vx = u – v sin vy = v cos

tan 450 =x

y

v

v = 1

vy = vx u – v sin = v cos

v = cossin

u

450

v

u

=)45sin(2

u0

clearly minimum value of v =2

u

for = 450 .

Q.85. f max = 1000.5+600.2 = 62 NFx = 40 – 20 = 20 N

f max > Fx

So acceleration of Block = zero.

10kg 6kg 4020

f B f A

f B = 12 N

40 – T – 12 = 0

T = 28 N

f A + 20 = T

f A = 8 N


8/18/2019 Phy 1 (12)

34/50


Q.86. 1x 10 cos30 t

2x v cos 45 t

21

1y 10 sin30 t gt2

22

1y v sin45 t gt

2

For collision

1 2y y

1 v10

2 2

v 5 2 m/s

1 2x x 1

10cos30 t 5 2 cos 45 t 1

t 5 3 5 1

and t =

1

5 3 1sec

Q.87. Writing constraint relationy A = yB tan differentiate w.r.t. t we getv A = vB tan … (i)

using COE

y A

yB

mgh = 2 A2

B mv2

1)v(m

2

1 … (ii)

putting value v A in equation (ii) and solving we get

vB = gh2 cos , v A = gh2 sin

Q.88. ĵqxîpv It is clear that p is the x-component of velocity and qx is the y-component

pdt

dx or dx = p.dt

or t

0

x

0dtpdx

or x = pt … (1)

anddt

dy = qx or dy = qx.dt

or dy = q.p.t dt x = p.t

y

0

t

0t.q.pdy dt

y =2

qt.p 2 … (2)


8/18/2019 Phy 1 (12)

35/50


Combining (1) and (2)

2y = pq2 2

2

x x q

pp

.

Q.89. Given vb =nvR

bv

= (-vb sin ) ĵ)cosv(î b Resultant velocity of boat

= Rb vv

= (vR - vb sin ) î + (vb cos ) ĵ

Bvb

A

vR

If w = width of the river, time for crossing is

T =cosv

W

b

Drift during time T is (vR - vb sin ) T

Drift x = vb(n - sin )cosv

w

b

= w(n sec - tan )

For x to be minimum,d

dx = 0 lead to = sin-1 (1/ n)

Direction of boat w.r.t. stream is900 + = 900 + sin-1 (1/n)For n = 1/2, the required angle = 900 + 300 = 1200

Q.90. Let u and are projection velocity and angle of projection respectively.vx = u cos and vy = u sin

At greatest height h = g2

sinu 22

vertical velocity at half the greatest height

v2 = 2/gh2v2y

v2 = u2 sin2 -2

sinu

g2

singu 2222

v =2

sinu

Resultant velocity at half the greatest height

vR =

22

2222

cosu2

sinu

)cosu(v

given, u cos =5

2

2222

cosu2

sinu

tan = 3 , = 600


8/18/2019 Phy 1 (12)

36/50


Q.91. (a) From the diagram BTv

makes an angle of 45 withthe x-axis.

(b) Using sine rule

s/m2v

15sin

v

135sin

vB

TB

`

450 600

O

A

B

vT

vB

vBT

Q.92. Let VB,R = velocity of boat w.r.t river = 10 m/secVR,G = velocity of river w.r.t ground = 2 m/secVB,G = velocity of boat w.r.t ground.From the figure,

Time t =R,BVriver of width

(a) t = 4010

400 sec

400 m

d

VR

VB,R

VB,G

(b) Let the drift be d, then from geometry R,B

R

V

V tan =

river of width

d

R,B

R

V

V width of river =

10

2 400 = 80 m

Alternatively d = (VR,G) (t) = 80 m.

Q.93. Let the 1st

and 2nd

shorts take time t and t respectively to collide at point P(x, y).Ist shotx = 53 t . . . (I)

y1 = 10 -2

1gt2 . . . (ii)

For 2nd shotx2 = (53 cos 60

0) t . . . (iii)

y2 = 10 + (52 sin 600)t -

2

1gt2 . . . (iv)

602

n shot

10 mP(x, y)

Is shot

For collision, we have x1 = x2 and y1 = y2.

From (I) and (iii), t = 2t and from (ii) & (iv), t = 1 sec, t = 2 sec.Now (I) Time interval = 2 - 1 = 1 sec.(ii) x1 = 53 t = 53

y1 = 10 -2

1(10) (1)2 = 10 - 5 = 5

Required co-ordinates = (5 3 m, 5m).


8/18/2019 Phy 1 (12)

37/50


Q.94. Let the points towards B and reches at Ct1 : the time taken by the boat to reach C

t1 =cosu

AD CD = (v - u sin )t1

t1 =

cos3

10500 3 hr =

cos6

1

CD = (-3 sin + 2)cos6

1

= - 0.5 tan +cos3

1

B D C

A

v

u

t2 : time taken by the man from C to D

t2 =sv

CD = -

5cos3

1

5

tan5.0

= tan

10

1 +

cos15

1

= -

cos15

1

cos10

sin

=

cos30

)2sin3(

total time t = t1 + t2 =

cos30

sin3

cos6

1 =

cos30

sin37

=30

7sec -

10

1 tan

for minimum t

0d

dt

30

7sec tan -

10

1sec2 = 0

10

1 sec

sectan

3

7 = 0

37 tan - sec = 0

cos3

sin7 = 0 = sin-1 (3/7)

Q.95. Let the angle of projection be and velocitybe u. The velocity parallel andperpendicular to the planes are u cos( -) and u sin ( - ). The component u sin( - ) becomes zero at the first planewhere as the component ucos ( - ) at thesecond plane.

-

(i)(ii)

If the time required in the first case is t1 and the second is t2.

Then 0 = u sin ( - ) – g cos t1 t1 =

cosg

)sin(u . . . (i)

and 0 = u cos ( - ) - g sin t2 t2 =

sing

cos(u . . . . (ii)

As the particle is mid way between the planes

u sin ( - )t1 -2

1g cos 21t = - {(u sin ( - ) t2 -

2

1g cos 22t }


8/18/2019 Phy 1 (12)

38/50


u sin ( - ) (t1 + t2) =2

1g cos )tt( 22

21 . . . (iii)

putting (i) and (ii) in (iii) we obtain

u sin ( - )

sing

cos(u

cosg

sin(u

=

22

22

22

22

sing

(cosu

cosg

(sinucosg

2

1

canceling u2/g from both the sides and rearranging we get

2

222

sin2

cos(cos

cos2

(sin

sin

cos(sin(

cos

(sin = 0

sin2 ( - ) sin2 + 2 sin ( - ) cos ( - ) sin cos - cos2 ( - ) cos2 = 0dividing each side by cos2 ( - ) sin2 tan2 ( - ) + 2 tan ( - ) cot - cot2 = 0

tan ( - ) =

2

cot4cot4cot2 22 = cot 2 cot

since ( - ) is an acute angle tan ( - ) is + ve - = tan1(cot (2 – 1)) = + tan1 {cot (2 – 1)}.

Q.96. Let t1 be the time taken to fall through a distance (H - h)

(H - h) =2

1g 21t t1 =

g

)hH(2

Further, the time taken by the body to fall through a distance h is given by t2 =g

h2

Total time taken by the body = t = t1 + t2

= hhHg

2

For the time to be maximum,dh

dt = 0

2/12/1 h

2

1)hH(

2

1

g

2 = 0

H = 2h

or h/H =2

1.


8/18/2019 Phy 1 (12)

39/50


Q.97.

80

60

40

O

20

8070605040

Time (s)

302010

O

1

8070605040302010

2

O

200

8070605040302010

6

400

900

Time (s)

Time (s)

a

m/s2

(m)

x

4100

D

C

2mA

B3m

10m/sv0

45

xO

y

38m

Q.98. Let t be the instant at which the ball hits rear face AB of the trolley.Then (v0cos45 - u0)t = 38


8/18/2019 Phy 1 (12)

40/50


or t =38

45

38

28 28 45 1038

0 0v us

cos . cos.

At t = 3.8 s, the y - coordinate of the ball is

y = (v0sin45)t -1

2gt2 = 20t - 5t2

or y = 20(3.8) - 5(3.8)2 = 3.8 mSince 3.8m > 2m, therefore, the ball cannot hit the rear face of the trolley.Now, we assume that the ball hits the top face BC of the trolley , and let t be that instant. Then,

y = 2 = 20t - 5 t 2 or t 2 - 4 t + 0.4 = 0

t = 3.9 sLet d be the distance from the point B at which the ball hits the trolley. Then,

d = (v0cos45 - u0) (t - t) = (20 - 10) (3.9 - 3.8) = 1m

Q.99. (a) Equation of motion along radius at point Q.

T + mg sin =

2mv

When the string slackens i.e. T = 0

0 + mg sin =

2mv

vQ =5

3510sing = 30 = 5.48 m/s

(b) Vertical velocity at point Q = vQ cos v2 = u2 + 2gh

0 = 2Qv cos

2 + 2(g)h

h = g2

cosv 22Q = 102

251630

= 25

24

= 0.96 m

mg3m

mg sin

T

90

vQ cos

Q

P

5m

Q.100.

sin

BD

sin

AD

sin = sintv

tv

22

11 sina

b

v

v

t

t

1

2

2

1

since t1 t2 . Then sin 2

1

bv

av

sin 0.6 370 1430

v2

B (Man)

(Bus) A

v1

O D

Q.101.

sin

BC

45sin

AC0

= 300

00

45sin

400

105sin

v v0 = 546.47 km/hr

t =47.546

1000

v

AB

a

= 1.83 hr.


8/18/2019 Phy 1 (12)

41/50


Q.102. (a) displacement = zero

(b) distance = OA + AB + BO

3 + 3 + 3 2

= 6 + 3 2 cm

B

AOW

N

S

E

Q.103. Let t be the time when they pass one another

for stone B, y= vBt +2t)g(

2

1 … (i)

for stone A, H – y =

2

gt2

1

… (ii)from (i) and (ii),H = vBt … (iii)

Stone B can reach just one the top of tower. We can calculatethe velocity of stone B,

ya2vv y2i

2f

uf = 0, for ymax = H = 20 m

vi = vB ay = -g

vB = 20 m/s

H C

A

B

y

H- y

g

O

y

from (iii) t = .sec1s/m20

m20

from equation (i), the required distance (BC) from ground = 20 1 -2

1 10 12 = 15 m

Q.104. If u is initial velocity, a is acceleration

100 = u 3 + 2)3(a2

1 = 3u +

2

a9 …(i)

200 = 5u +2

1a(5)2 = 5u +

2

a25 …(ii)

5u + 2

a18

642

a25

u2

a7

Now, putting value of v in equation (i)

100 = 3 2

a30

2

a9

2

a7

a = 2ms15

100

= 6.67 m/s2


8/18/2019 Phy 1 (12)

42/50


Q.105. (a) In figure shown final displacement

m ĵ5DO

Distance = OA + AB + BC + CD

= (25 + 5 2 ) m OD

CB

A

10 m52

10 m x

y North

East

5 m

(b) Average speed =25

2525

timetotal

cetandistotal

= 5 m/s.

Average velocity =)25(

ĵ5

m/s

(c) For average acceleration =t

vv if

, ĵ5î5v f vi = 5 î

average =)25(

ĵ5 ĵ5î5

=

25

ĵ5î10

m/s2

Q.106. Final velocity at A v A = 2 t1 = 2 30 = 60 m/sec.For AB, Let the retardation be ‘b’

0 = v A + bt2 b = -

t

v A = -60

60 = -1 m/s2

O

A

B

(a) total distance = OA + AB

OB = )bt2

1tv(at

2

1 222 A

21

= ( )606012160606030302

21

= 900 + 3600 – 1800 = 2700 m.

(b) Maximum speed v A = 60 m/s.(c) v2 = 2 a s

s =22

3030

a2

)2/v( 2 A

= 225 m.

Q.107. (a) Analyzing the vertical motion

y = uyt +2t)g(

2

1 , Uy = u sin 30 = 60

2

1 = 30 m/s

25 = 30 t -2

1 10t2 5t2 - 30 t + 25 = 0

solving for t t = 1 sec. & 5 secso minimum time = 1 sec.

(b) vertical velocity vy = uy – gt = 30 – 10 1 = 20 m/s

horizontal velocity vx = u cos 30 = 60 2

3 = 30 3 m/s


8/18/2019 Phy 1 (12)

43/50


(c) Horizontal distance covered = ux t = 602

3 1 = 30 3 m

Range =10

60sin6060

g

2sinu2

= 60 6

2

3 = 180 3 m

Q.108. Max. range, R =g

u2 [ R =

g

2sinu2

100g

u2

Now, max. height for a projectile is H =g2

sinu 22

H will be max. for = 900, Hmax =g2

u2

Hmax =

g

u21

2

= )100(21 = 50 m.

Q.109. (a) S = 0 + ½ gt2 S t h

O

S

T

(b) KE = ½ mv

2

KE v2

v= 0 + gt KE t2

mgh

TO

KE

(c) TE = constant alwaysTE

TO

mgh

Q.110. R(range) =g

2sinv20

Rg2sin

Rgv20

as = 45o.

vo = Rg . . . (1)

Equation of trajectory

106.68 m

BA

x

yvo

45o

1.22 m


8/18/2019 Phy 1 (12)

44/50


Y = xtan45o -ocosv

gx

452 220

2

= x -

2

12

2

Rg

gx using (1)

Putting x = 97.53, we get

y = 97.53 -

35.81068.106

53.97102

Hence height of the ball from the ground level is h= 8.35 + 1.22 = 9.577m.as height of the wall is 7.31m so the ball will clear the wall.

Q.111. Let (x,y) be the coordinates of point C.Now x component of point C=OA+AD

0 10 4y10x ycot373 3

(1)

As point C lies on the trajectory of a parabola we have2

2 2

gxy x tan

24 cos

(2)

370

55

O A D

C

y

Given that1 2

tan ,cos2 5

From (1) & (2) we get y=1.25 mHence x=5 mHence the coordinates of point C are (5m,1.25m)(b) Let vy be the vertical component of velocity of the particle just before collision at C. fromV=u+at

y

xv usin g

ucos

y

1 5v 5 5 10 0

5 5 5 cos

Thus at C the particle has only horizontal component of velocity, x2

v ucos 5 5 10m / s5

Q.112. Instantaneous speed of the car = 30 m/s = VRadius = 500 m

dt

dV = 2 m/s2 = at (Tangential acceleration)

Normal or radial acceleration or centripetal acceleration

=

5

9

500

)30(

r

v 22 m/s2 = an

Total acceleration 2/1t2n 2aaa

=

2/12

2

5

9)2(

=

2/1

25

814

=

2/1

25

181

= 2.7 m/s2

Let direction of this acceleration make an angle with the velocity then


8/18/2019 Phy 1 (12)

45/50


tan =10

9

2

5/9

a

a

t

n

= tan –1

10

9

Q.113. Let ĵuîuu yx

At the instant of impact (x)stone = (x0) plank + (x) plankuxt = x0 + (1/2) at

2 uxt = 3 + 0.75 t

2 . . . (i)In y-directionuyt – (1/2) gt

2 = y0 uyt – 5t

2 = 1.25 . . . (ii)also (vy/ux) = tan (-45

0){(uy – gt)/ux} = -1 . . .. (iii)

solving (I), (ii) and (iii), we get

t = 1 s, ux = 3.75 m/s, uy = 6.25 m/s u = 7.28 m/s and 0 = tan1 (5/3).

Q.114.

FBD of m;

mg

T

and T – mg = ma …(1)

FBD of 2m;

2mg

T

and 2mg – T = 2ma …(2)

Adding (1) and (2) mg = 3ma a = g/3Hence velocity of ‘m’ after moving up 6.54 m is;

V2 = 2 (g/3) (6.54) = 2

3

81.9 (6.54) = 2(3.27) (6.54)

V = 6.54 m/s upwardsVelocity of ‘2m’ at that instant = 6.54 m/s downwards.When string is cut ‘m’ falls to the ground from a height of 13.08 + 6.54 = 19.62 m

19.62 = – 6.54t + 1/2 (9.81) t2 where ‘t’ is the time taken to reach the groundt = 2.7 seconds

‘2m’ falls a distance of 13.08 – 6.54 = 6.54 m

t = 0.6 seconds.

Q.115. Tangent acceleration at =dt

dv = 2t +1

normal acceleration an =R

v 2

(at)t=0 = 1 m/s2

(an)t=0 = 2.0

1

R

v2

0 = 5 m/s2


8/18/2019 Phy 1 (12)

46/50


(b) v = Rdt

d

R d = (1+t+t2 )dt R

'

0

2

0

2 dttt1d

= 33.3 rad

Q.116. Since acceleration varies linearly so

a = t m/s2 by given condition

or, tdt

dv

v =2

t2 m/s

or2

t

dt

ds 2 or, s =

6

t 3

at the end of t = 6 sec. Acceleration becomes zero.Distance moved by car at t = 6 sec is

S1 =6

666 = 36 m

Speed of the car =2

66 = 18 m/s

Remaining distance = 72 – 36 = 36 m.

so time taken to cover this distance

= t2 = .sec2.sec18

36

Total time = 6+2 = 8 sec.

Q.117. v0 – gt = 0 t = v0/gv = 0 + a0t = a0v0/g

x =2

1a0t

2 =g2

va 200 ; y = v0t - 2

20

g

vg

2

1 =

g2

v20

yes it will retrace its path, path is not parabolic.

Q.118. Let us take x – y axes as shown in the figure

The magnitude of pseudo force acting on theparticle has a magnitude of ma0 and its direction

will be towards right as shown in the free bodydiagram

P

O37

0

370

u = 5.5 m/sa0 = 5 m/s

2

XY

The components of the acceleration ofthe particle are

ax =m

37sinmg37cosmao

= 2s/m25

310

45

XY

370

370

ma0

mg

F.B.D. of the particle


8/18/2019 Phy 1 (12)

47/50


ay =m

)37sinma37cosmg( oo

o

=

5

35

5

410 -11m/s2

ux = ucos37° = 5.5 54

= 4.4 m/s.

uy = usin37° = 5.5 5

3 = 3.3 m/s

Displacement of the particle along y – axis

y = uy t +2

1ayt

2

y = 3.3t -2

1 11t2

When the particle strikes the plane y = 0

t =y

y

a

u2

=

11

3.32 = 0.6 sec.

OP = x = uxt +2

1axt

2

= 4.4 0.6 -2

1 2 (0.6)2 = 2.28 m.

Q.119. (a) We observe the motion of projectile fixing Y-axis with OP and x-axis with OQ. Hence;

velocity at any instant t along x-axis:

vx =10 3 - (g sin 600)t

vy = 0 – (g cos 600)t

As vx = 0 at the time of hitting;Time of flight = T = 2 sec.Displacement OP during this time

=2

1(g cos 600)t2 =

2

1 10

2

1 4 = 10 m

Hence; h = OP sin 300 = 10 2

1 = 5m

(b) Similarly displacement OQ = (10 3 )(2) -2

1 10

2

3 4= 10 3 m

Hence PQ = 22 OQOP = 20 m.

Q.120. N sin = m2 R sin … (i)and N cos = mg ….(ii)from (i) and (ii)

= cos-1

R

g2

g can not be greater than 2R.

N

O’

A

O

mg

If 2 R > g then there are two equilibrium position.


8/18/2019 Phy 1 (12)

48/50


1 = 0 and 2 = cos-1

R

g2

Q.121. For the part (a) choose the co-ordinate axes parallel to the plane and perpendicular to the inclined

plane and for the part (b) choose the horizontal axis and vertical axis as co-ordinate axes.

(a) îsingab

ĵsinuîcosuv )0t(b/p

ĵcosgîsingaP

ĵcosga b/P

XY

ĵ)tcosg2

1tsinu(îtcosuR 2)t(b/P

At Q, u sin t 2

1 g cos t2 = 0

t =

cosg

sinu2 ; PQ =

cosg

2sinu2.

(b) Let ĵsinvîcosvv G/b

ĵ)sin(uî)cos(uv b/p

ĵ]sinv)sin(u[î]cosvcos(u[v G/p

For x to be 0u cos ( + ) = v cos

v =

cos

)cos(u.

X

Y

Q.122. Let they meet after time t, and let the speed with the stone is thrown up be u. then

h = g2

u2

u = gh2 = 20102 = 20 m/s

Now 20 t - 20gt2

1gt

2

1 22

t = 1s.Hence they pass other after 1sspeed of the stone projected up = 20 – 10 1 = 10 m/sspeed of the stone dropped = 0 + 10 1 = 10 m/s.

Q.123. Let the second rocket overtakes the first t second. after it is projected When they meet their displacement are equal

22 )1t)(g6(21t)g8(21

3

4

6

8

t

1t2

3

2

t

1t

2

3

1t

t


8/18/2019 Phy 1 (12)

49/50


t =32

3

2

31

2/3

t = .,sec

32

3

.,sec

32

3

(not possible)

= 46.6268.0

732.1 sec.

Q.124. Before collision number of revolution completed by the particle is odd multiple of half or

2

1n2

where n is an integer.

Period of revolution of the particle =v

R2 = 4 seconds.

Time of flight of the shell = )4(2

1n2 = (4n - 2) second

Range isg

2sinu2 = 20 3

Hence tan =3

)1n2( 2

For to be smallest; n = 1 ; therefore = 300 hence u = 20 m/sec.

Q.125. (a)

g2

sinv

ĵyîx

time

ntdisplacemev

0av

x = half of range =g2

2sinv2

0

y = Max. height =g2

sinv 220

y

xx î

y ĵ

g2

sinv

ĵg2

sinvî

g2

2sinv

v0

220

20

av

= ĵsin

sinvî

sin

2sinv 200

= 2v0 cos î + v0 sin ĵ

(b) Let the perpendicular velocity be v

v v = 0 v v = (v0 cos î + v0 sin ĵ ). (v0 cos î + (v0 sin - gt) ĵ

0)gtsinv

Phy 1 (12)

Documents

Transcript of Phy 1 (12)