Motion in Two Dimension

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Motion In Two Dimensions

In two-dimension motion a particle moves in a plane. e.g. a particle going in a circle, a cricket ball thrown in by a fielder (on a windless day). In the first case the particle can go round with a constant speed (uniform circular motion) or it could move with a non-constant speed (Non uniform circular motion). We will study both. In the second case the cricket ball in a projectile i.e. it has been projected (thrown) and it moves under the influence of gravity. This motion is projectile motion. Before we study this specialized case of two-dimension motion, let us first understand the general features of two-dimension motion.

(i) Position and Displacements: As we had done in one-dimensional motion, the first task for us is to

know where the particle is i.e. its position. The position here is a vector intends from a reference point (usually the origin of the coordinate system) to the object.

In the unit vector notation it can be written, rx i y j 

Where x i, y j are the vector components of rand the coefficients x and y are its scalar components.

The coefficients x and y give the objects location along the axes and relative to the origin. Figure shows particle P whose position vector at an instant is r3i6j

y x 0 3 6 6 j 3 i

As the particle moves its position vector (P.V.) changes such that the vector always extends to the object from origin.

Assuming the particle has a P.V. r1 

at time t1and P.V. r2 

at time t2, its displacement is r r2 r1

 

  

During time interval r.

(i) Velocity and Average velocity : If a particle moves through a displacement



r



in a time intervalt,

then the average velocity is

r V t      x i y j t        x y i j t t          V V ix V jy      

Here V i_x and V j_y are x and y components of average velocity. The instantaneous velocity ,

t 0 r r lim t t d V d          But r=x i y jÙ+ Ù dx dy v i j dt dt  Ù Ù \ = + or v V i V j _{x y} Ù Ù = + , t t x y dx dy V V d d   ,

then components of V along x and y respectively.

Let us understand this using a particle, which is moving in 2D.

Y X 1 r  2 r  r  _{Path of P} P tangent of P(t1)

In following figure particle is at r1 at t1 and at r2 at t2 (= t1+ t2). The vector

r

 is the particle displacement in time t . Then as it is clear from (1) average velocity V points in same direction as r

Now if we reduce t we see that (1) vector r2  moves towards r1  Y X Path of P P V V_y V_x

(2) r becomes smaller and its direction approaches the tangent

(3) In the limit t  0, r, the instantaneous velocity at t1i.e. average velocity becomes instantaneous velocity V, which has the direction along tangent line.

Projectile Motion (Time of Flight, Horizontal Range and Trajectory)

Let P(x, y) represent the position of a projectile after a time t from the time of projection.

At the point O, horizontal component of velocity = u cosand

vertical component of velocity = u sin

Acceleration due to gravity acts vertically downwards. Hence vertical component of velocity changes. While horizontal component of velocity remains constant (u cos) during the complete motion.

u ucos  R usi n  O

Time of flight: It is the time taken by the projectile from the instant it is realized till it strikes a position on

the same horizontal plane as the point of projection.

To calculate time of flight, we can find the time when vertical displacement is zero i.e. 0 = u sin t – _gt2 2 1  (u sin  gt 2 1 _{) t = 0  t = 0 or t =} 2 usinθ g

Thus , Time of flight T =2 usinθ

g

Range: During this time horizontal component of velocity has taken the particle through a distance x

horizontally

Where,x = u cos  T = u cos  2 usinθ

g =

2

u sin2θ g

This distance, we call as range on horizontal plane ,

2 u sin2θ R g  Rmax= g u2 for  = 45

For a given velocity same range can be obtained for an angle  and angle (90 – ) i.e.

R =





2 u sin2 90° - θ g =





2 u sin 180° - 2θ g = 2 u sin2θ g

To calculate maximum height, the vertical component of velocity becomes zero, when the particle is at the highest point from the ground. At that time particle has only horizontal component of velocity i.e. = u cos

0 = (u sin)2_{– 2gH} max Hmax= 2 2 u sin θ 2g

The motion of projectile can be analysed through vector notation also for e.g. velocity of projectile (v) at any time t can be written as





ˆ ˆ

v  u cos i  u singt j

Hence, the magnitude of velocity (speed) v at any time is given as ,

v=



ucos

 

2 usingt



2 = _u2_{g t}2 2_{2u gtsin (The path of a projectile is called its trajectory)}

y = u sin t – 1 2

2gt . . . (1) (vertical displacement at any time t)

x = u cos t . . . (2) (Horizontal displacement at any time t)

Using t = cos x u  We get, y = x tan– 2 2 2 2 cos gx

u , Which suggest that trajectory of projectile is parabola

Example 1: The figure shows two positions A and B at the same

height h above the ground. If the maximum height of the projectile is H, then determine the time t elapsed between the positions A and B in terms of H.

y x A B  H t v0

Solution: Let T be the time of flight. We can now write

2 2 8h T t g   since, 0 2 2 2 0 2 2 sinθ 4 sin θ 8 v T g v H T g g    or t 8



H h



g  

Example 2: It is possible to project a body with a given speed in two possible ways so that it has the

same horizontal range R. The product of the times taken by it in the two possible ways is (g is the acceleration due to gravity)

(A) R g (B) 2R g (C) 3R g (D) 4R g

Solution: (B)If a body is projected with a given velocity u at angles  and (90–) to the horizontal,

it will have the same range R given by ,

2_{sin 2} u R g   The corresponding times of flight are ,t₁ 2 sinu

g 

 and t₂ 2 sin 90u





2 cosu

g g  _    





2 2 1 2 2 2

2u 2 sin cos 2u sin 2

t t g g       2R g 

Example 3: A particle is projected horizontally from the top of a cliff of height H with a speed 2gH.

The radius of curvature of the trajectory at the instant of projection will be

(A) H/2 (B) H (C) 2H (D)  Solution: (C)Since, g  v Radial acceleration ar= g  g r v2

0 _{ where r is the radius of curvature.}

 g ( v 2gH) r gH 2   _  r = 2H

Example 4: A stone is thrown at an angle  to the horizontal reaches a maximum height h. The time

(A) (2h sin ) / gq (B) 2 (2h sin ) / gq (C) 2 (2h) / g (D) (2h) / g

Solution: (C) Vertical height reached by the stone ,

2 2 u sin h u sin 2gh 2g     

time of flight , t 2u sin 2 2gh 2 2h

g g g



  

Example 5: The angle of projection of projectile, for which the horizontal range and the maximum

height are equal, is

(A) _{tan ( 3 )}-1 _{(B) tan ( 4 )}-1 _{(C) tan ( 2)}-1 _{(D) tan}-1 1

3

æ ö

ç ÷

è ø

Solution: (B)Horizontal range = height reached by the projectile

2 2 2 u sin 2 u sin g 2g q q Þ = 2cos sin 2 q q Þ = Þ tanq =4 \ q =tan (4)-1

Example 6: A body is thrown horizontally from the top of a tower of 5 m height. It touches the ground at a

distance of 10 m from the foot of the tower. The initial velocity of the body is : (Given g = 10 ms–2₎

(A) 2.5 ms–1 _{(B) 5 ms}–1 _{(C) 10 ms}–1 _{(D) 20 ms}–1

Solution: (C) Range R u 2h 10 u 2(5)

g g

  

 u = 10 ms–1

Example 7: Four bodies P, Q, R and S are projected with equal velocity having angles of projection

150_{, 30}0_{, 45}0_{and 60}0_{with the horizontal respectively. The body having shortest range is}

(A) P (B) Q (C) R (D) S Solution: (D) If  = 150, Range 2 2 0 u u sin30 g 2g   If  = 150_{, Range} u2 _sin600 3u2 g 2g   If  = 150_{, Range} u2 g  If  = 150_{, Range} 3u2 2g 

 Range is minimum when  = 150_.

Example 8: A stone projected with a velocity u at an angle  with the horizontal reaches maximum

height H1 . When it is projected with velocity u at an angle -2

p q

æ ö

ç ÷

è ø with the horizontal reaches maximum height H2. The relation between horizontal range, R of the projectile, H1and H2is (A) R 4 H H= _{1 2} (B) R 4(H - H )= _{1 2} (C) R 4(H= 1+H )2 (D) 2 1 2 2 H R H =

Solution: (A) Height reached by the stone 1st time ,

2 2 1 u sin H 2g  

Height reached by the stone 2nd time ,

2 2 2 u cos H 2g   2 1 2 u sin cos H H 2g   

But range R u (2 sin cos )2 g

 

 4 H H_{1 2}

Time Of Flight & Horizontal Range Of Projectile Motion On Inclined Plane

Projectile Motion on Inclined Plane: Figure shows an inclined plane

at an angle  and a particle at an angle  with the direction of plane with initial velocity u. In such cases we take our reference x- and y-axes in the direction along and perpendicular to the inclined as shown.

Unlike to the simple projectile motion, here the x-component of the velocity of the projectile will also be retarded by a gsin. Now y-component of the velocity is retarded by g cos instead of g. As shown here g is resolved in two directions.

A x g v_x v_y v y cos u    0 t=0 sin u  u t=t sin g  _cos g 

Here as y-direction component is retarded by gcos, to find the time of flight and maximum height, we can use equations T= 2 usinθ

g and R =

2

u sin2θ

g , replacing g by gcos,

Time of flight on inclined plane projectile is, 2 sin cos u T g   

Maximum height of the projectile with respect to inclined plane is ,

2_sin2 2 cos u H g    For evaluation of range on inclined plane we cannot use equation ,

2

u sin2θ R

g

 ,

just by replacing g by g cos, as here we also have Acceleration in x-axis ax= -g sin

Now we again find the distance travelled by the particle along x-direction. In this duration, time of flight is

R = _{sin .} 1 _{sin .} 2

2

u T  g T

On substituting the value of time of flight T, we get ,

2 2 2

2

sin2 2 sin sin

cos cos u u R g g       

Students are advised not to apply the above expression of range on inclined plane, as a standard result, it should be processed and evaluated according to the numerical problem. We’ve derived the above results for the projectile thrown up an inclined plane. If projectile is thrown down an inclined plane, the acceleration along the plane gsin will increase the velocity of the particle along the plane, thus in the expression for range we should use +ve sign as

2 2 2

2

sin2 2 sin sin

cos cos u u R g g       

To find the maximum range on inclined plane, One can use maxima-minima as dR

d . The range on inclined

plane has a maximum value given as ,





2 1 sin u R g   

In above equation +ve sign is used for projectile up the plane and –ve sign is used for projectile down the plane.

Projectile from a moving frame:

Consider a boy on a trolley who throws a ball with speed ‘u’ at an angle  w.r.t. trolley which moves with speed v. (a) Ball is projected in the direction of motion of trolley. Horizontal component of ball velocity = u cos + v

Initial vertical component of ball’s velocity = u sin . (b) Ball is projected opposite to direction of motion of trolley.

Horizontal component of ball’s velocity = u cos  – v Initial vertical component of ball’s velocity = usin . (c) Ball projected from a plate form moving upward

Initial vertical component of ball’s velocity = u sin  + v (d) Ball projected from a platform moving downward

Horizontal component of ball’s velocity = u cos  Initial vertical component of ball’s velocity = usin  – 

Example 9: A ball is dropped from a height h above a point on an inclined plane, with angle of inclination .

The ball makes an elastic collision with the surface and rebounds off the plane. Determine the distance from the point of first impact to the point where ball hits the plane second time.

Solution: Take the point of first impact as the origin. Direction along the plane will be the x-axis

and the direction perpendicular to the plane will be the y-axis.

After rebound, the horizontal component of velocity u sin will be accelerated by g sin and the vertical component of the velocity u cos will be retarded by g cos.

Here time of flight from first impact to the second impact is given as,

2 2 cos 2 cos y y u u u T a g g     

In this duration the distance travelled by the horizontal component is R = 2 2 2 1 2 4 sin sin . sin ( ) 2 u u u R u g g g g       8 sinh  (since u= 2gh )

Example 10: A projectile is thrown with a speed u, at an angle  to an

inclined plane of inclination . Find the angle  at which the projectile is thrown such that it strikes the inclined plane normally. x cos u   u sin g  cos g



h Y 0 usin  1   g

Solution: Here the time of flight of particle is , 2 sin

cos u T g   

Thus from speed equation in x-direction, we have 0 = ucos - g sin 2 sin

cos u g        

or cot = tan or  = cot-1_(tan)

X u Y  

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Angular Velocity And Angular Acceleration

The angular velocity (or angular frequency) of a body in a uniform circular motion is the angle swept out by the radius vector per second. If the radius vector sweeps out an angle  (measured in radians) in a time interval t, the angular velocity is given by

t 0 d lim t dt         _{ }  

where  is expressed in rad s–1_{. Linear velocity (), angular velocity () and the radius (r) of the circular} motion are related as



= r



Radian measure of an angle:If the arc of a circle equals the radius, then the angle subtended by that arc

Angular Acceleration:The angular acceleration  is given by , d d t

  

Linear acceleration (a), angular acceleration (



) and radius (r) of the circular motion are related as a = r



Angular acceleration is expressed in rad s–2_.

r r r



Centripetal Acceleration

If a body moves in a circle at a constant speed, it is said to be in uniform circular motion. In such a motion, the magnitude of the velocity (i.e. speed) is constant but the direction of the velocity vector is continually changing. Thus the velocity is changing with time. Hence the motion of the body is accelerated. The acceleration is directed towards the center of the circle and is called centripetal acceleration. The magnitude of the centripetal acceleration is given by ac=  r P Q O 1  2  r

where  is the angular velocity (or angular frequency) and  is the speed along the circle. Since  = r, we have, 2 2 c a r r      

where r is the radius of the circular path. The angular frequency is related to time period T and frequency v

as, 2 2 v

T 

   

Therefore, centripetal acceleration is also given by (since   2 r / T) , 2 2 2

c 2 4 r a 4 r v T      

UNIFORM CIRCULAR MOTION

(i) Velocity remains constant in magnitude but varies in direction (ii) The acceleration is always normal to the velocity vector.

(iii) The acceleration is always directed towards the centre of the circular path.

C a c a c a c v v v

Comparison of uniform circular motion with straight line motion and projectile motion:

v

g

(a) upward Motion θ = 180 θ θ θ (b) Projectile Motion a v θ = 180

(c) Uniform Circular Motion

NON-UNIFORM CIRCULAR MOTION

(i) The velocity changes both in magnitude as well as in direction (ii) The velocity vector is always tangential to the path

(iii) The acceleration vector is not perpendicular to the velocity vector (iv) The acceleration vector has two components

(a) Tangential accelerationatchanges the magnitude of velocity vector, i.e. a = _t dv

dt

(b) Normal acceleration or centripetal acceleration ac changes the direction of the velocity vector, i.e. 2 c v a = r

(v) The total acceleration is the vector sum of the tangential and centripetal acceleration

2 2

t c

a a a

Figure shows an object of mass m whirled with a constant speed  in a vertical circle of centre O with a string of length R. When the object is at top A of the circle, let us say that the tension (force) in the string is T1. Since the weight mg acts vertically downwards towards the centre O, we have,

Force towards centre, F = T1+ mg = 2 m R  _{or T} 1= 2 m R  _{– mg ….(i)} mg O mg A B C T3 T1 T₂ mg

At the point B, where OB is horizontal, the weight mg has no component along OB. Thus, if the tension in the string is T2at B, we have

Force towards centre, F = T2=

2

m R

 _….(ii)

At C, the lowest point of motion, the weight mg acts in the opposite direction to the tension T3 in the string. Thus at C we have,

Force towards centre, F = T3– mg =

2 m R  or T3= 2 m R  _{+ mg ….(iii)}

From (i), (ii) and (iii), we see that the maximum tension occurs at lowest point C of the motion. Here the tension T3must be greater than mg by

2 m

R

 _{to keep the object in a circular path. The minimum tension is given by (i)} when the object is at the highest point A of the motion. Here part of the required centripetal force is provided by the weight and the rest by T1.

In order to keep a body of mass m in a circular path, the centripetal force, at the highest point A, must at least be equal to the weight of the body. Thus

2 A m R  = mg or 2_A = Rg

gives the minimum speed the body must have at the highest point so that it can complete the circle. Then the minimum speed _C the body must have at the lowest point C is given by 2 2

C A 2 2

     Rg where we have used 2 = 2 + 2gh, with H = 2R. Thus

2 C

 = Rg + 4 Rg = 5 Rg

or  _C 5Rg

The tension at this point is given by ,

2 C 3 T m g R    _  _   = m (5g + g) = 6 mg

Figure shows a bucket with water whirled in a vertical circle without water spilling out. When the bucket is vertically above the point of support (as shown), the weight mg of water is less than the required centripetal force

2 m

R

 _{towards the centre and} so the water stays in the bucket without falling out. The rest of the centripetal force is balanced by the reaction of the base of the bucket. If the bucket is whirled slowly so that mg >

2

m R



part of the weight mg provides the necessary force

2 m R  and 2 m R  mg Base of the bucket O

the rest of the weight causes some water to accelerate downwards. This much water will then leave the bucket.

It is for the same reason that the pilot of an aircraft who is not tied to his seat can loop a vertical circle in air without falling out at the top of the loop. He must have a certain minimum velocity at the bottom of the loop in order to clear the loop without any mishap. This exercise is known as looping the loop.

Example 11: A particle is moving in a circular path of radius 10 cm. Its linear speed is given by

t 4

Solution: Radial acceleration 2 2 R 6.4 cm/s R v a   Tangential acceleration _{4 /} 2 t dv a cm s dt   4 tan 6.4 t R a a     ; 8 5 tan1  

Example 12: A stone is tied up to the end of a string and whirled in a vertical circle. The tension

of the string is T1when the stone is at highest point and T2 when it is at the lowest point; then

(A) T1= T2 (B) T1= -T2 (C) T1> T2 (D) T1< T2

Solution: (D) Tension at the highest point T1= mg

Tension at the lowest point T2= 6 mg  T1< T2

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When two objects moves in the same straight line, we compare their motion in terms of their relative velocity. If two objects A and B are moving in a straight line with velocities VA and VB respectively, the relative velocity of object A with respect to object B is given by , V_AB = V - V_{A B}

where VB is called reference object velocity

It follows that the relative velocity of object B with respect to object A will be , V = V - V_{BA B A}

where VA is called reference object velocity

Key points regarding relative motion while calculating relative velocity:

(i) Relative velocity of a particle = velocity of a particle - velocity of reference object

(ii) If the velocity of a particle be VA and the velocity of a reference object be VB then the relative velocity of the particle V_AB = V - V_{A B}

(iii) Relative velocity of a particle while moving in the same direction. Relative velocity = V - V_{A B}

(iv) Relative velocity of a particle while moving in the opposite direction. Relative velocity = V_AV_B

Example 13: Two boats A and B move in perpendicular direction to a boy and anchored at some point O on

a river. They travel with velocity 1.2v, where v is the stream velocity. Boat A moves along the river, whereas boat B moves perpendicular to it. After traversing an equal distance from O the two boats return. Find the ratio of the time taken by the two boats.

Solution: Let l = distance covered by the boat A along the river as well as by the boat B across the river.

Relative velocity of boat A with respect to stream when the boat goes along the river = 1.2 v –v Relative velocity of boat A with respect to stream when the

boat goes against the stream = 1.2 v + v

 Time taken by the boat A to cover the whole journey is 

v V 1.2 v O A V B 1.2 1.2     A l l t v v v v









2 2 2 1.2 1.2 2.4 5.45 0.44 1.2        l v v v v vl l v v v v . . . (i)

Boat B is moving from O perpendicular to the direction of flow of stream. Its velocity must be at an angle



to the direction of the stream velocity so that the resultant velocity is directed perpendicular to the flow of stream

 Resultant speed of boat is given by ,





2 2

1.2 0.44 0.66

   

v v v v v

 Time taken by the boat B to cover the whole journey is , 2 2

0.66 0.33   B l l l t v v v . . . (ii)

From (1) and (2), we have A 5.45 0.33 1.80 B

t l v

Example 14: A boat which has a speed of 5 Km/hr in still water crosses a river of width 1 km along the shortest possible path in 15 minutes. The velocity of the river water in km/hr is :

(A) 1 (B) 3 (C) 4 (D) 14 Solution: (B) t d v  2 2 1 1 4 _{u v} Þ = - 2 1 1 16 25 v Þ = -v 3 km / h Þ = U2_-v2 v u

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LEVEL - I

1. If for a particle position x  t2then :

(A) Velocity is const (B) acceleration is const

(C) Acceleration is variable (D) None

2. A bomb is dropped from an aircraft travelling horizontally at 150 m/s at a height of 490 m. The

horizontal distance travelled by the bomb before it hits the ground is, in metres

(A) 1000 (B) 1200 (C) 1500 (D) 1800

3. The directions of velocity and acceleration of a projectile at the highest point on the trajectory are

(A) Parallel to each other (B) antiparallel to each other

(C) perpendicular to each other (D) no specific relationship exists between them

4. Two vectors of same magnitude have a resultant equal to either, then the angle between the vectors will be

(A) 200 _{(B) 135}0 _{(C) 120}0 _{(D) 45}0

5. A person throws a bottle into a dustbin at the same height as he is 2 m away at an angle of 45o .

The velocity of the throw is

(A) g (B) g (C) 2 g (D) 2g

6. A stone tied to string is rotated in a vertical circle. The minimum speed with which the string has to be rotated

(A) decreases with increasing mass of the stone (B) is independent of the mass of the stone

(C) decreases with increasing in length of the string (D) is independent of the length of the string

7. A gun mounted on the top of a moving truck is aimed in the backward direction at an angle of 300to

the vertical. If the muzzle velocity of the gun is 4 m/s, the value of the speed of the truck that will make the bullet come out vertically is

(A) 1 m/s (B) 3m / s

2 (C) 0.5 m/s (D) 2 m/s

8. A ball is thrown upwards. If air resistance is taken into account the time reaching maximum height

(A) is equal to time for falling (B) is less than time for falling

(C) is greater than time for falling (D) none

9. An object start sliding on a frictionless inclined plane and from the same height another object starts

falling freely:

(A) both will reach with same speed (B) both will reach with same acceleration

(C) both will reach in same time (D) none of above

10. The displacement of a particle is given by : y = a + bt + ct2- dt4

The initial velocity and acceleration are respectively

(A) b, -4d (B) -b, 2c (C) b, 2c (D) 2c, -4d

11. A coin falls faster than a scrap of paper when dropped from the same height because for coin :

(A) gravitational acceleration is more (B) gravitational acceleration is less

(C) air resistance is less (D) None of these

12. A body projected vertically upwards with a velocity u returns to the staring point in 4 second. If g =

10 ms2_{the value of u is (ms}–1₎

(A) 5 (B) 4 (C) 40 (D) 10

13. A body released from a great height falls freely towards earth. Another body is released from the

same height exactly one second later. The separation between the two bodies two second after the release of the second body is

(A) 9.8 m (B) 49 m (C) 24.5 m (D) 19.6 m

14. The slope of the distance - time graph of two bodies are 300and 600. Their velocities are in ratio

(A) 1: 3 (B) 3 : 3 (C) 3 : 1 (D) 1 : 3

15. A body is falling freely under gravity. The ratio of distance covered by it in Ist, IIndand IIIrdsecond respectively is

(A) 1: 3 (B) 3 : 3 (C) 3 : 1 (D) 1 : 3

16. If a particle tied to the end of string is set in circular motion then the tension of the string is

(B) always perpendicular to the velocity of the particle

(C) perpendicular to the velocity of the particle only at one instant (D) parallel to the velocity of the particle only at one instant

17. For an electron circulating around the nucleus, the centripetal force is supplied by

(A) electromagnetic force (B) electrostatic force

(C) gravitational force (D) magnetic force

18. Two cars going round curve with speeds one at 90 km/hr and other at 15 km/h. Each car

experiences same acceleration. "The radii of curves are in the ratio of

(A) 4 : 1 (B) 2 : 1 (C) 16 : 1 (D) 36 : 1

19. The minimum speed for a particle at the lowest point of a vertical circle of radius R, to describe the

circle is 'V'. If the radius of the circle is reduced to one-fourth its value, the corresponding minimum speed will be

(A) V

4 (B)

V

2 (C) 2 V (D) 4 V

20. The speed of a projectile at its maximum height is 3 / 2 times its initial speed. If the range of the

projectile is 'P' times the maximum height attained by it, P (A) 4

3 (B) 2 3 (C) 4 3 (D)

3 4

LEVEL - II

1. What can be the possible velocity displacement (v – s) graph of a particle moving in a straight line

under constant acceleration:

(A) straight line (B) parabola (C) ellipse (D) circle

2. A particle is moving eastward with a speed of 5 m/s. After 10 seconds, the direction changes towards

north, but speed remains same. The average acceleration in this time is (A) zero (B) 2 1 m/s2_{towards N-W (C)} 2 1 m/s2_{towards N-E (D)} 2 1 m/s2_{towards S-W}

3. A ballast bag is dropped from a balloon that is 300 m above the ground and rising at 13 m/s. The

time before the bag hits the ground is [take g = 10 m/s2_]:

(A) 10 sec (B) 9.8 sec (C) 9.5 sec. (D) 9.15 sec

4. If a particle takes t second less and acquires a velocity of v ms-2more in falling through the same

distance on two planets where the accelerations due to gravity are 2g and 8g respectively then:

(A) v = 4gt (B) v = 5gt (C) v = 2gt (D) v = l6gt

5. An aeroplane flies along a horizontal circle of circumference 10 km, at a constant speed of 100

km/hr. The change in velocity in one fourth of a revolution is: (A) zero

(B) 141 km/hr at 90 from the original direction (C) 141 km/hr at 135 from the original direction. (D) 200 km/hr at 180 from the original direction.

6. A body falls freely under gravity. The distance travelled by it in the last second of its journey equals

the distance travelled by it in the first three seconds of its free fall. The total time taken by the body to reach the ground is:

(A) 5s (B) 8s (C) 12s (D) 15s

7. Ratio of minimum kinetic energies of two projectiles of same mass is 4 : 1. The ratio of the maximum

height attained by them is also 4 : 1. The ratio of their ranges would be:

(A) 16 : 1 (B) 4 : 1 (C) 8 : 1 (D) 2 : 1

8. A particle of mass m attached to a string of length I is describing circular motion on a smooth plane

inclined at an angle  with the horizontal. For the particle to reach the highest point its velocity at the lowest point should exceed.

(A) 5gl (B) 5gl(cos +1) (C) 5gl tan  (D) 5glsin 

9. A particle is projected from the ground with an initial speed of v at an angle  with horizontal. The

average velocity of the particle between its point of projection and highest point of trajectory is: (A) v _{1 2cos}2 2   (B) 2 v 1 cos 2   (C) 2 v 1 3 cos 2   (D) v cos 

10. A particle is projected with a certain velocity at an angle  above the horizontal from the foot of an inclined

plane of inclination 30. If the particle strikes the plane normally then  is equal to: (A) _{30 + tan}-1 1 2 3        _{(B) 45 (C) 60 (D) 30 + tan 2 3} -1

 

11. With what minimum speed must a particle be projected from origin so that it is able to pass through

a given point (30m, 40m). Take g  10 m/s2

(A) 15 m/s (B) 24 m/s (C) 40 m/s (D) 50 m/s

12. A ball is dropped vertically from a height h above the ground. It hits the ground and bounces up

vertically to a height h/2. Neglecting subsequent motion and air resistance, its velocity  varies with the height h as:

h  h  h  h  (A) (B) (C) (D)

13. A ball is projected at an angle of 45, so as to cross a wall at a distance from the point of projection. It falls

at a distance b on the other side of the wall. If h is the height of the wall then:

(A) ha 2 (B) hb 2 (C) 2ab a + b  h (D) ab a + b  h

14. A man walking at velocity iˆ2 m/s observes that rain is falling at velocity



2iˆ1jˆ



m/s. The velocity of

man w.r.t. rain will be:

(A) jˆ1 m/s (B)



4iˆ1jˆ



m/s (C) 



2iˆ1jˆ



m/s (D) 1jˆ m/s

15. A string of length L = 1 m is fixed at one end and carries a mass of 100g at the other end. The string

makes 5 / revolutions per second about a vertical axis passing through its second end. What is the angle of inclination of the string with the vertical? Take g = 10 ms–2_:

(A) 30 (B) 45 (C) 60 (D) 75

16. Two bodies fall freely from the same height, but the second body starts falling T seconds after the

first. The time (after which the first body begins to fall) when the distance between the bodies equals L is: (A) 1 gh (B) 1 2gh (C) 1 gt (D) 1 2gt

17. Two projectiles, one fired from the surface of the earth with speed 5 m/s and the other fired from the surface of a planet with initial speed 3 m/s, trace identical trajectories. Neglecting air resistance, the value of acceleration due to gravity on the planet will be if g = 10 m/s2_{on earth}

(A) 5.9m/s2 _{(B) 3.6 ms/}2 _{(C) 16.3 m/s}2 _{(D) 8.5 m/s}2

18. A particle moves along a parabolic path y = 9x2 in such a way that the x component of velocity

remain constant and has a value 1 1

ms 3

 _{. The acceleration of the particle is}

(A) 1ˆ_jms-1 3 (B) ˆ -2 3jms (C) 2ˆ_jms-2 3 (D) ˆ -2 2jms

19. The co-ordinate of the particle in x-y plane are given as , x 2 + 2t + 4t2 and y 4t + 8t2

The motion of the particle is

(A) along a straight line (B) uniformly accelerated

(C) along a parabolic path (D) non-uniformly accelerated

20. A projectile has the same range R for two angles of projections. If T1 and T2 be the time of flight in the two cases, then

(A) T1T2 R (B) T1T2 R2 (C) 1 2 1 T T R  (D) T T_{1 2} 1₂ R 