# 01 Kinematics Motion in 1D 2D

July 13, 2017 | Author: Sivakumar Sarma | Category: Acceleration, Velocity, Kinematics, Speed, Dynamics (Mechanics)

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Chapter-1 / Kinematics

Chapter 1

Kinematics Kinematics is the branch of PHYSICS dealing with motion without considering the cause producing motion.

Concept of Point Object A body is considered as a point object depending upon the nature of the motion followed by the body. In general, an object is regarded a point-object when it travels large distances in comparison to its own size and dimensions. Also, in planetary motion, the bodies under consideration can be regarded as point objects when distances of separation are very large.

Difference Between Distance and Displacement Distance

Displacement Vector

i.

The length of the actual path followed between the initial and the final points in the motion is called distance.

i.

The change in position vector of a particle going from initial position (say i) to final position (say f) is called Displacement or r Relative Position Vector of the particle. It is denoted by r , r r r such that r  rf  r

ii.

It is a scalar quantity.

ii.

It is a vector quantity

iii.

It depends on the path followed.

iii.

It is independent of the choice of origin.

iv.

Between two points it is not unique.

iv.

It is unique (one and only one) for any kind of motion between two points.

v.

It can not be negative.

v.

It can be positive, negative and even be zero.

vi.

The magnitude of distance is always more or equal to the displacement.

vi.

The magnitude of the displacement is always less than or equal to the distance for particle’s motion between two points.

vii. A body may displacement.

have

finite

distance

travelled

for

zero

Difference Between Velocity & Speed Velocity i.

Speed

The rate of change of particle’s position with

i.

time is called velocity. It is a vector quantity.

The distance travelled by a particle along the actual path per unit time is called speed. It is a scalar quantity.

ii.

r r r r r rf  r  Average Velocity = vav  t t f  t

ii.

Average Speed = vaverage 

iii.

The velocity of a particle at a particular instant

iii.

The speed of a particle at a particular instant of

of

is called r r r r dr v  lim  t0 t dt

iv.

time

instantaneous

velocity

It can be positive or negative depending on its

iv.

time

is

called

v  lim t0

s ds  t dt

s s  t t

instantaneous

speed.

It can not be negative.

direction.

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SP/CCE/Physics/11

Average Velocity It is that uniform velocity with which if the body would have moved it would have covered the same displacement as it does otherwise by moving with variable velocity. Thus total displacement covered vav  total time taken

Average Velocity in Different Cases (i)

Particles covering different displacement in different times As one a particle covers s1 displacement in t1 and s2 in time t2 and so on then average velocity is s  s  s  .... s  s  s  ... vav  1 2 3  1 2 3 s1 s2 s3 t1  t2  t3 ...    ... v1 v2 v3 Special case if s1 = s2 = s. vav 

2 v1 v2 2s  s s v 1  v2 (harmonic mean)  v1 v2

(ii) Bodies moving with different velocity in different intervals of time A body moves with velocity v1 in time vav 

t1 , v2 in time t2 and so on then vav is given by

v1t1  v2 t2  ... t1  t2 ...

Special case if t1 = t2 = t3 = …. tn = t v1  v2  ...  vn Then vav  (Arithmatic mean) n

Points to Remember

The average speed may or may not be equal to the magnitude of the average velocity.

The instantaneous speed is equal to magnitude of the instantaneous velocity.

An object may have varying velocity without having varying speed but opposite is not possible.

If r is the position vector then velocity in vector form may be written as follows r r dr ˆ  dx  dy  dz v  i    ˆj    kˆ   vx iˆ  vy ˆj  vz kˆ dt  dt   dt  dt

Acceleration The rate of change of velocity with time is called its acceleration. r r v a  (a) Average acceleration = average t r r r v dv  (b) Instantaneous acceleration ains  lim Also, instantaneous acceleration can be expressed as t0 t dt r r r dv d 2 r r ains tan tan eous   2 where r  xiˆ  yjˆ  zkˆ Therefore, dt dt r r 2 2 r d 2 r dv ˆ  d 2 x ˆj  d y  kˆ d z  a iˆ  a ˆj  a kˆ a 2  i  x y z   2 2 2 dt dt  dt   dt   dt

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Chapter-1 / Kinematics

(c)

If the velocity and acceleration of a body are not in the same direction then the equation of motions should be used in the vector form.

(d)

Velocity and acceleration of a body need not to be zero simultaneously.

(e) A body in equilibrium has zero acceleration only. All other quantities need not be zero. (f) If a body travels with a uniform acceleration a 1 for a time interval t 1 and with uniform acceleration a2 for a time interval t2, then the average acceleration aavg 

a1t1  a2 t2 . t1  t2

(g) For a body moving with uniform acceleration, the average velocity = (u + v)/2, where u is the initial velocity and  is the final velocity.

(h) The distance travelled by the body in successive seconds is in the ratio 1 : 3 : 5 : 7...etc. (i) When the body is starting from rest, the distances travelled by the body in uniform accelerated motion in straight line in the first second, first two seconds, first three seconds,...etc. are in the ratio of 1 : 4 : 9 : 16 : 25...etc.

Important Graphs describing the motion of a body 1. Uniform Motion:

v

x s lo p e = v

a re a = d is p la c e m e n t

x (0 ) t

t

2. Non uniform motion with constant positive acceleration x

v v (0 )

x (0 )

s lo p e = a c c e le ra tio n a r e a = d is p la c e m e n t t

t

a area = change in velocity

t

3. Non uniform motion with constant negative acceleration v v (0 )

x

t

t

a t

4. Graphs for a ball bouncing on the floor without loss of energy

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SP/CCE/Physics/11

v

x

speed

t

t

t

Points to Remember

For applying the above equations greater care has to be taken about the direction of the vector quantities involved.

2

2

For a particle having zero initial velocity if v  t, s  t and v  s then acceleration of particle must be constant i.e. particle is moving rectilinearly with uniform acceleration.

a

For a particle having zero initial velocity if s  t , where a > 2, then particle’s acceleration increases with time.

a

for a particle having zero initial velocity if s  t , where a < 0, then particle’s acceleration decreases with time.

A body can have non zero acceleration without having varying velocity but can not have non-zero

acceleration without having varying speed. r The average slope of a curve in v(t) graph gives average acceleration. r The instantaneous slope of a curve at a point in v(t) graph gives the instantaneous acceleration. r The area under a curve in v(t) graph gives the displacement of the body.

No line in x – t graph can be perpendicular to time axis because it will represent infinite velocity.

Distance time graph cannot be below time axis.

Distance time graph cannot have negative slope because this will mean the decrease of distance covered

• •

with increase of time. •

Distance-time, displacement-time and acceleration-time graphs cannot be closed curves.

Distance time graph of uniformly accelerated motion is a parabola.

Question Solving Map

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Chapter-1 / Kinematics

Check out the form of Motion that is being talked about in the problem

Step I

Step II

A Body is moving with If yes constant velocity or zero acceleration

s = ut

If No check step II A Body is moving with If yes

constant acceleration

Use any one of the following relations

If No check step III

where u is the initial velocity, v is the final velocity, a is the acceleration of the body, t is the time taken by the body, s is the distance covered byn the body and s th is the distance covered by the body in the nth second of motion.

v = u + at

Step III

1 2 s = ut +2 at

1 1 2 2 v – u = 2a .ss = 2 (u + v)t sn= u + 2 a(2n – 1)

A Body is moving with variable acceleration

Acceleration varies as a Acceleration varies as a Acceleration varies as a function of distance function of velocity function of time a = f(v) a = f(t) a = f(x) x t v 2 2 dv v = u + 2 f(x)dx v = f(t)dt t= f(v) x0 t0 u OR

v

x = x0 +

vdv u f(v)

Vertical Motion Under Gravity •

For a body thrown downward with initial velocity u from a height h, the equation of motion are v = u + gt, h  ut 

1 2 gt , v  u 2  2 gh 2

For a body launched up with initial velocity u, the equations of motion before the particle attains maximum height are 1 2 2 v = u – gt, h  ut  gt , v  u  2 gh 2

If a body starts from rest or falls freely or is dropped then, u = 0.

For a body dropped from a height h, the equation of motion are v = gt,

h

1 2 gt , v  2 gh 2

Initial velocity in case of dropping is zero. •

If the body is thrown upwards then it will rise until its vertical velocity becomes zero. Maximum height 2

attained is h = u /2g. •

If a packet is dropped at a height ‘h’, from an aeroplane or balloon ascending with a velocity u, then, the time taken by the packet to reach the ground is given by h   ut 

1 2 gt . 2

If air resistance is negligible the time of rise is equal to time of fall, each is equal to t = u/g. -5-

SP/CCE/Physics/11

The body returns to the starting point with the same velocity with which it was thrown.

When a body is dropped freely from the top of the tower and another body is projected horizontally from the same point, both will reach the ground at the same time.

Motion in a Plane Any type of planar motion can be resolved into two mutually perpendicular independent motions resolved along x and y axis (since x and y components do not have any dependence in each other). Hence all equation of motion can be applied as given below along x and y axis separately.

x-component of equations of motion

y-components of equations of motion

Vx = ux + axt

Vy = uy + ayt

X = uxt +

1 2 axt 2

y = uyt +

1 2 ayt 2

v2y  u 2y  2a y y

v 2x  u x2  2a x x  u x  v x  t  2 

 u y  v y t 2  

x 

y

Projectile Motion If a particle is given an initial velocity from a point on the surface of the earth at any angle with the horizontal, the particle is said to be under projectile motion. Projectile motion is a two dimensional motion under a constant one dimensional acceleration. Projectile motion can be taken as the combination of two and one dimensional motion such that along x-axis the velocity remains constant whereas along y-axis it is accelerated.

A-Oblique Projection of body with velocity u at an angle  with horizontal y

Assumptions:

g

(i)

g = constant

(ii)

no air resistance

vy

In this case ux = u cos  and uy = u sin .

u y= u s in  O

Time of flight T

2u sin  T u sin  and tascent  tdescent   g 2 g

uy2

u2 sin 2  or H = 2g 2g

Range(R) R

-6-

g

or T 

Maximum height reached above the ground H

2uy

2 ux uy g

or R 

u P

2 u2 sin 2 (u cos )(u sin ) = g g

v

v y= 0 u x= u c o s 

y

H

u x= u c o s 

x

Chapter-1 / Kinematics

Range is maximum at  = 45° and Rmax  •

u2 g

Equation of trajectory y  x tan  

gx 2 , 2u 2 cos 2 

which is a parabola. x, y are chosen with point of projection as origin. •

Equation of trajectory of an oblique projectile in terms of range (R) is x  y = x tan  1    R  

where R =

u2 sin 2 g

at any instant if v is the velocity of projectile making angle  with the horizontal, then vx = vcos = ucos and vy = vsin = usin – gt Range is same for complimentary angles

R  R90 

   

u 2 sin (2) g

R

If H and H90– are corresponding maximum heights for angles  and 90 – , then R  R90  4 H  H 90

For complimentary angles  and 90 – , if T and T90 –  are the times of flight and R is the range, then T  T90 

2 R g

2 R90

g

2R g –1

Maximum height is equal to range when the projectile is launched at an angle  = tan (4) = 76° with the horizontal.

If K is the kinetic energy at the point of launch then kinetic energy at the highest point is K' 

1 1 mux2  mu 2 cos 2   2 2 

2

K = K cos 

If  and  on the angle of elevations of a point from initial and final positions then angle of projection  is given by tan  = tan  + tan  •

The trajectory of a projectile as viewed from another projectile is a straight line of constant slope.

During projectile motion the momentum change along x-axis is zero whereas along y-axis is mg t

B-Horizontal projection of body from a certain height with velocity ‘u’ Assumptions:

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SP/CCE/Physics/11

(i)

g = constant

(ii)

no air resistance

Motion is very close to surface of earth. So

O

g

u x= u

x y

acceleration due to gravity is taken constant. •

In this the initial velocity is given along the horizontal i.e. ux = u

P (x, y)

x

 v y= g t

y

v x = u =x u v

Horizontal motion is a non-accelerated motion with uniform velocity i.e. ax = 0, vx = ux = u and x = uxt

vertical motion is accelerated motion under influence of gravity. Therefore, v y = gt

Time taken by a horizontal projectile to strike the ground is T 

Horizontal distance at which it strikes the ground is range R  uT  u

Equation of trajectory is y  

1 2 2 2 Velocity at any instant of time is v  u  g t , inclined at an angle   tan 

2h g 2h g

g 2 x which is equation for a parabola. 2 2  u 

 gt  , with the horizontal.  u

C-Projection of Body on Inclined Plane •

an angle with the horizontal on an inclined plane making an angle  with the horizontal. Then ( – ) is the angle with which the projectile is launched with respect to the inclined plane. Resolution gives ux = u cos ( – )

along +x-axis

uy = u sin ( – )

along +y-axis

ax = – g sin 

along x-axis

ay = – g cos 

along y-axis

Time of flight 2u sin      T= g cos   , 2 2u 2u   g cos  g sin 

T is maximum when     

   hence Tmax 2

Range of the projectile along the incline plane 2u 2 sin (  )cos  R g cos 2 

-8-

u

Consider a projectile to be launched with an initial velocity u making y

x

Chapter-1 / Kinematics

R is maximum when  

u2    , hence Rmax  g(1  sin ) 4 2

Maximum height above the incline plane At the maximum height, we have uy = 0. So, H 

u 2 sin 2 (  ) 2 g cos 

Points to Remember

H equals R, when 4cos = sin( – ) cos

If tan  = 2 tan  the projectile hits the inclined plane horizontaly at the final point.

If tan  = cot  + 2 tan  then the projectile hits the inclined plane normally.

If a body is dropped on an inclined plane then its range along the inclined

h

plane is 8h sin  (where h is the height of fall) R 

Circular Motion Circular motion is another example of motion in a plane and it takes place when a body moves at a fixed r r r distance from a given point. If v is the linear speed, r is the radius and  is the angular velocity then v    r . During circular motion the direction of angular velocity is given by Right Hand Thumb Rule.

During circular motion the body generally experienced the following type of acceleration. 

Centripetal acceleration (ac) This acceleration changes the direction of velocity only without changing its magnitude. v ac   r 2 r 2

v

r r r in vector form ac   v

Tangential acceleration (at) This acceleration changes the magnitude of velocity without changing its direction. r r r r d| v| at  also at    r where  is angular acceleration dt

Points to Remember

If a body has only centripetal acceleration it must be in uniform circular motion.

If a body has both centripetal as well as tangential acceleration it must be in non uniform circular motion. a net

ac 

anet  at 2  acp 2

at

and tan  = ac/ at

For circular motion we have : -9-

SP/CCE/Physics/11

r r r v r r (ii) r antiparallel to ac r r (iii) ac  v r r (iv) ac  at r r r r (v) r , ac , at and v lie in the same plane. (i)

When a body rotates with uniform angular velocity, its different particles have centripetal acceleration directly proportional to the radius ( ac  r ).

Centripetal force cannot change the kinetic energy of the body.

Centrifugal Force

   

It exists only in circular moving frame.

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A pseudo force, that is equal and opposite to the centripetal force is called centrifugal force. The centrifugal force appears to act on the agency which exerts the centripetal force. The centrifugal force cannot balance the centripetal force because they act on the different bodies.

Chapter-1 / Kinematics

Exercise - I Multiple choice questions with only ONE option correct 1.

A ball is thrown vertically upwards. The quantity which stays constant is (a)

2.

displacement

(b)

acceleration

(c)

speed

(d) velocity

A stone is released from an elevator going up with an acceleration a. The acceleration of the stone after the release is (a)

3.

a upward

(b)

(g – a) upward

(c)

(g – a) downward

(d) g downward

A 150 m long train is moving to north at a speed of 10 m/s. A pigeon flying towards south with a speed of 5 m/s crosses the train. The time taken by the pigeon to cross the train is (a)

4.

30 s

(b)

15 s

(c)

8s

(d) 10 s

Water drops fall at regular intervals from a roof. At an instant when a drop is about to leave the roof, the separations between 3 successive drops below the roof are in the ratio (a)

5.

1:2:3

(b)

1 : 4: 9

(c)

1:3:5

(d) 1 : 5 : 13

A train travels on a straight track passing station A at 20 m/s. If accelerates uniformly at 2 m/s2 and reaches station B 100 m away from A. At B the velocity of the train in m/s is approximately (a)

6.

10

(b)

20

(c)

28

(d) 56

A body throws a ball in air in such a manner that when the ball is at its maximum height he throws another ball. If the balls are thrown after the time difference of 1s then what will be the height attained by them (a)

7.

20 m

(b)

10 m

(c)

5m

(d) 2.5 m

From the top of a tower, a stone is thrown up and reaches the ground in time t1. A second stone is thrown down with the same speed and reaches the ground in time t2. A third stone is released from rest and reaches the ground in time t3. (a)

8.

t3 =

1 (t1 + t2) 2

(b)

t3 =

t1t 2

(c)

1 1 1   t 3 t 2 t1

(d)

t 32  t12  t 22

A balloon is ascending at the rate of 10 m/s. When it is at a height of 75 m from the ground, a packet is dropped from it. The packet will reach the ground after time of (a)

9.

10.

4s

(b)

3s

(c)

5s

(d) 7.5 s

slope of a velocity time graph

The acceleration of a moving object is equal to the (a)

slope of a displacement time graph

(b)

(c)

area below a displacementtime graph

(d) area below a velocity time graph

The variation of velocity with time of a particle moving along a straight line is given by following graph. The distance travelled by particle in 4 second will be (a)

44 m

(b)

32 m

(c)

60 m

(d) 64 m

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SP/CCE/Physics/11

11.

A train can accelerate at 1 m/s2 and retard at 3 m/s2. The shortest time in which it can travel between two stations 1200 m apart is (a)

12.

40 s

(b)

20 s

(c)

40

2 s

(d) 40

3 s

A car is moving along a straight road with a uniform acceleration. It passes through two points P and Q separated by a distance with velocity 30 km/hr and 40 km/hr respectively. The velocity of the car midway between P and Q is (a)

13.

33.3 km/hr

(b)

20

3 km/hr

(c)

25

2 km/hr

(d) 35 km/hr

A particle is falling freely under gravity. In first t second it covers s1 and the next t seconds it covers s2, then t is given by (a)

14.

s2  s1 g

(b)

s2  s1 g

(c)

s2s1 g

(d)

s22  s12 g

A parachutist falls freely from an aeroplane for 10s before the parachute opens out. Then he descends with a net retardation of 2.5 m/s2. If he bails out of the plane at a height of 2495 m, his velocity on reaching the ground will be (a)

15.

5 m/s

(b)

10 m/s

(c)

15 m/s

(d) 20 m/s

For a particle moving along a straight line, the displacement x depends on time t as x = t3 + t2 + t + . The ratio of its initial acceleration to its initial velocity depends (a)

16.

only on  and 

(b)

only on  and 

(c)

only on  and 

(d) only on 

Two balls of same masses are shot upward one after another at an interval of 2 second along the same vertical line with same initial velocity of 40 m/s. The height at which they collide (a)

17.

40 m

(b)

75 m

(c)

80 m

(d) 125 m

A particle is projected vertically upwards and it reaches the maximum height H in time T. The height of the particle at any time t will be (a)

18.

1 g t  T  2 2

(b)

 1 H -   g(t - T)2  2

(c)

 1 2   g(t - T)  2

(d)

H - g(t - T)

A bird flies for 4 s with a velocity of |t - 2| m/s in a straight line, where t is the time in second. It covers a distance of (a)

19.

2m

(b)

4m

(c)

6m

(d) 8 m

An aluminum ball X and an iron ball Y of the same volume are thrown horizontally with the same velocity from the top of a building. Neglecting air resistance, X reaches the ground (a)

before Y and at the same distance from the building

(b)

at the same time as Y and at a near distance from the building

(c)

at the same time as Y and at a same distance from the building

(d) at the same time as Y and at a farther distance from the building 20.

Three particles A, B and C are thrown from the top of a tower with the same speed. A is thrown straight up, B is thrown straight down and C is thrown horizontally. They hit the ground with speeds vA, vB and vC respectively. (a)

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vA = vB = vC

(b)

vB > vC > vA

(c)

vA = vB > vC

(d) vA > vB = vC

Chapter-1 / Kinematics

21.

The equation of projectile is y = (a)

22.

1 m/s

(b)

3 x

1 2 gx . The velocity of projection is 2

2 m/s

(c)

3 m/s

(d) 4 m/s

An aeroplane is flying in a horizontal direction with a velocity of 600 km/h and 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 strikes the ground at point B. The distance AB will be (a)

23.

10 km 3

(b)

10 km

(c)

1 km 3

(d)

5 km 3

Two particles are projected simultaneously in the same vertical plane from the same point, with different speeds u1 and u2, making angles 1 and 2 respectively with the horizontal, such that u1 cos1 = u2 cos2. The path followed by one, as seen by the other (as long as both are in flight), is (a)

a horizontal straight line

(b)

a vertical straight line

(c)

a parabola

(d) a straight line making an angle | with the horizontal. 24.

A particle moves along the positive branch of the curve y = x2/2 with x governed by x = t2/2 where x and y are measured in metre and t in second. At t = 2s, the velocity of the particle is (a)

25.

ˆ 4ˆ 2i j m/ s

(b)

ˆ 4ˆ 2i j m/ s

(c)

ˆ 2ˆ 4i j m/ s

(d)

ˆ 2ˆ 4i j m/ s

A ball rolls from the top of a stair way with a horizontal velocity u. If the steps are h high and b wide, the ball will hit the edge of the nth step if (a)

26.

n = 2hu/gb2

(b)

n = 2hu 2/gb2

(c)

n = 2hu2/gb

(d) n = hu2 /gb2

A river is flowing from west to east at a speed of 5 m/min. A man on the south bank of the river capable of swimming at 10m/min in still water wants to swim across the river in shortest possible time. In what direction should he swim ?

27.

(a)

due north

(b)

at an angle 30° west of south

(c)

at an angle 30° east of north

(d) at 60° west of north

To a man walking at the rate of 3 km/h the rain appears to fall vertically. When he increases his speed to 6 km/h it appears to meet him at an angle of 45° with vertical from the front. The actual speed of rain is (a)

28.

3 km/hr

(b)

4 km/hr

(c)

3 2 km /h

(d)

2 3 km/hr

A particle starts from the origin of coordinates at time t = 0 and moves in the xy plane with a constant acceleration  in the y–direction. Its equation of motion is y = x2. Its velocity component in the x–direction is (a)

variable

(b)

2 

(c)

 2

(d)

 2

*****

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SP/CCE/Physics/11

Exercise - II More than one Correct 1. A bead is free to slide down a smooth wire tightly stretched between the points p 1 and p2 on a vertical circle of radius R. If the bead starts from rest from p 1, the highest point on the circle and p 2 lies anywhere on the circumference of the circle. Then P

1

R cos q g

(a) time taken by bead to go from p1 to p2 is dependent on position of p2 and equals 2 (b) time taken by bead to go from p1 to p2 is independent on position of p2 and equals 2

R g

(c) acceleration of bead along the wire is g cos  (d) velocity of bead when it arrives at p2 is 2 gR cos q 2. The velocity of a particle moving along a straight line increases according to the linear law v = v0 + kx where k is a constant. Then (a) the acceleration of the particle is k ( v0 + kx) (b) the particle takes a time

æv1 ö 1 ÷ ÷ log e ç ç ÷ ç ÷to attain a velocity v1. k v è 0ø

(c) velocity varies linearly with displacement with slope of velocity displacement curve equal to k (d) data is insufficient to arrive at a conclusion 3. A car starts moving rectilinearly (invitial velocity zero) first with an acceleration of 5 ms –2 then uninformaly and finally decelerating at the same rate till it stops. Total time of journey is 25 second and average velocity during the journey is 72 km h–1. then (a) total distance travelled by the car is 500 m (b) maximum speed attained during the journey is 25 ms–1 (c) Car travels with uniform speed for 15 sec. (d) Car accelerates for 5 sec and decelerates also for 5 sec 4. Two second after projection, a projectile is travelling in a direction inclined at 30° to the horizon. After one more second it is travelling horizontally. Then (a) The velocity of projection is 20 ms–1 (b) The velocity of projection is (c) The angles of projection is 30° with the vertical (d) The angles of projection is 30° with the horizon 5. Two particles is projected from the same point with same speed u at angle of projection  and  strike the horizontal ground at the same point. If h 1 and h2 are maximum heights attained by the projectiles, R be the range for both and t1 and t2 be their time of flights respectively then (a) µ +b =

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p 2

(b)

R = 4 h1 h2

(c)

t1 = tan µ t2

(d)

tan µ =

h1 h2

Chapter-1 / Kinematics

Matrix Match 1.

Trajectory of particle is a projectile motion is given as y = x -

x2 . Here, x and y are is metres. For this 80

projectile motion match the following with g = 10 m/sec2. Column – I (A)

2.

Angle of projection

Column – II (p)

20 m

(B) Angle of velocity with horizontal after 4s

(q)

80 m

(C) Maximum height

(r)

45°

(D) Horizontal range

(s)

æ1 ö ÷ tan- 1 ç ÷ ç ÷ è2 ø

A balloon rises up with constant net acceleration of 10m/sec 2. After 2 second a particle drops from the balloon. After further 2 second match the following : (Take g = 10 m/sec2) Column – I (A)

Height of the particle from ground

Column – II (p)

zero

(B) Speed of particle

(q)

10 S.I. units

(C) Displacement of particle

(r)

40 S.I. units

(D) Acceleration of particle

(s)

20 S.I. units

Integer Type 1.

A Police Jeep is chasing a culprit going on a motor bike. The motor bike crosses a furning at a speed of 72 km/h. The jeep follows it at a speed of 90 km/h, crossing the turning ten seconds later than the bike. Assuming that they travel at constant speeds, how far (in km) from the turning will the jeep catch up with the bike?

2.

The benches of a gallery in a cricket stadium are 1 m wide and 1 m high. A batsman strikes the ball at a level one metre above the ground and hits a mammoth sixer. The ball starts at 35 m/s at an angle of 53° with the horizontal. The benches are perpendicular to the plane of motion and the first bench is 110 m from the batsman. On which bench will the ball hit?

Assertion and Reason In the following question, a statement of Assertion (A) is given which is followed by a corresponding statement of reason (R). Mark the correct answer out of the following options/codes. (a)

If both (A) and (R) are true and (R) is the correct explanation of (A).

(b)

If both (A) and (R) are true but (R) is not correct explanation of (A).

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SP/CCE/Physics/11

(c)

If (A) is true but (R) is false.

(d) If (A) is false but (R) is true. 1.

A.: A body can have acceleration even if its velocity is zero at a given instant of time. R.:

2.

A body is momentarily at rest when it reverses its direction of motion.

A.: The two bodies of masses M and m (M > m) are allowed to fall from the same height if the air resistance for each be the same then both the bodies will reach the earth simultaneously. R.:

3.

For same air resistance, acceleration of both the bodies will be same.

A.: A particle in motion may have variable velocity but constant speed. R.:

A particle in motion may have non-zero acceleration but constant velocity.

4.

A.: A body having uniform speed in circular path has a constant acceleration. R.: Direction of acceleration is always away from the centre.

5.

A.: A particle in xy-plane is governed by x = a sin t and y = a – cos t, where a and  are constants then the particle will have parabolic motion. R.: A particle under the influence of mutually perpendicular velocities has parabolic motion.

Passage Based Questions Passage-1 A particle is moving along X- axis under a force such that its position-time graph is as shown in figure. 1.

As the particle passes through position (1) (a)

it is moving along negative X-direction with a speed that is increasing with time

(b)

it is moving along positive X-direction with a speed that is decreasing with time

(c)

it is moving along negative X-direction with a speed that is decreasing with time

(d) it is moving along positive X-direction with a speed that is increasing with time 2.

As the particle passes through position (2) (a)

it is moving along positive X-direction with a maximum speed

(b)

it is moving along positive X-direction and its speed is zero at this point

(c)

it is moving along the negative X-direction and its speed is zero at this point

(d) it is moving along negative X-direction with a minimum speed 3.

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As the particle passes through position (3) (a)

it is moving along positive X-direction with a maximum speed

(b)

it is moving along negative X-direction with a speed that is increasing with time

Chapter-1 / Kinematics

(c)

it is moving along positive X-direction with a speed that is decreasing with time

(d) it is moving along negative X-direction with a speed that is decreasing with time 4.

As the particle passes through position (4) (a)

it is instantaneously at rest and will now move along negative X-direction

(b)

it is instantaneously at rest and will now move along positive X-direction

(c)

it is moving along positive X-direction with a maximum speed

(d) it is moving along negative X-direction with a maximum speed 5.

As the particle passes through position (5) (a)

it is instantaneously at rest and will now move along positive X-direction

(b)

it is moving along positive X-direction with a speed that is decreasing with time

(c)

it is moving along negative X-direction with a maximum speed.

(d) it is moving along negative X-direction with a minimum speed

Passage-2 A person standing at the corner of the roof of a building throws a ball vertically upward at an instant t = 0. The ball leaves his hand with an upward speed 20 m/s. The ball rises to a certain height and then moves downward to the ground. The ball hits the ground at t = 5 s. Assuming that (i) the vertically upward direction is the positive Y-direction (ii) the position of the ball at t = 0 is the origin (iii) the ball does not rebound and comes to rest at the same place where it hits earth and (iv) air resistance is negligible, and take g = 10 ms –2 1.

Maximum displacement of the ball from the initial position is (a)

2.

(b)

45 ˆj m

(c) 15 ˆj m

(d)

25 ˆj m

(c)  3 ˆj m/s

(d)

 9 ˆj m/s

Average velocity of the ball from t = 0 to t = 5 s is (a)

3.

5 ˆj m

1ˆj m/s

(b)

5 ˆj m/s

Position-time graph for the given motion of the ball is

(a)

(b)

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SP/CCE/Physics/11

(c)

4.

5.

(d)

Velocity of the ball will vary with time as

(a)

(b)

(c)

(d)

Acceleration of the ball will vary with time as

(a)

(b)

(c)

(d)

Passage-3 A and B are two fixed spots in a river in which water has a steady speed uw. A person who can swim with a speed v relative to water swims from A to B and back to A along the shortest path. If the water is still, the person will take a time 30 minute in swimming from A to B and back to A along the shortest path. But we know that water is actually not still. The person also knows the technique of flowing with water without making his own effort. Using this technique, he takes 20 minute in moving from A to B. As shown, X and Y are two places directly opposite to each other on opposite banks. Assuming that width of the river is the same as the distance between A and B, answer these questions. (assume v > uw)

1.

Time taken by the person to swim from A to B (in running water) if the person makes his own efforts also, will be nearly (a)

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15.2 min

(b)

16.5 min

(c) 8.5

(d)

12 min

Chapter-1 / Kinematics

2.

Time taken by the person to swim from B to A (in running water is) (a)

3.

60 min

(c) 30 min

(d)

12 min

60 min 12

(b)

60 min 5

(c)

60 min 2

(d)

60 min 7

Shortest time in which the person could swim from one bank to the other (in running water) will be (a)

5.

45 min

Time taken by the person to swim from X to Y (in running water) along the shortest path is (a)

4.

(b)

12 min

(b)

15 min

(c) 8 min

(d)

25 min

While swimming as required in above question, the person swims an actual distance 1.2 km. Width of the river is (a)

1040 m

(b)

960 m

(c) 875 m

(d)

750 m

Passage-4 A ball of mass 200 g is thrown at an angle 45° above the horizontal at t = 0. It hits a wall at a horizontal distance 15 m at a point 5 m above the point of projection. Take g = 10 m/s2, ignore air resistance and answer these questions. 1.

Magnitude of initial velocity of the ball is (a)

2.

8 m/s

(b)

21 m/s

(c) 12 m/s

(d)

15 m/s

If a second wall of height slightly less than 5 m is constructed in front of the first wall such that the ball just misses to hit both the walls, then horizontal distance of the second wall from the point of projection will be (a)

7.5 m

(b)

12 m

(c) 10.5 m

(d)

6.5 m

*****

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SP/CCE/Physics/11

Answers Exercise - I Only One Option is correct 1. 6. 11. 16. 21. 26.

(b) (c) (c) (b) (b) (a)

2. 7. 12. 17. 22. 27.

(d) (b) (c) (b) (a) (c)

3. 8. 13. 18. 23. 28.

(d) (c) (b) (b) (b) (d)

4. 9. 14. 19. 24.

(c) (b) (a) (c) (b)

5. 10. 15. 20. 25.

(c) (a) (b) (c) (b)

Exercise - II More than one correct 1. (b, c, d) 2.

(a, b, c)

3.

(a, b, c, d)

4.

(b, c)

5.

(a, b, c, d)

(d)

3.

(c)

4.

(c)

5.

(d)

2.

(a)

3.

(c)

4.

(a)

5.

(c)

2.

(b)

3.

(a)

4.

(a)

5.

(c)

2.

(a)

3.

(d)

4.

(b)

5.

(b)

2.

(a)

Matrix Match 1. (A – r), (B – r), (C – p), (D – q) 2.

(A - r), (B – p), (C – s), (D – q)

Integer Type 1. 1 km

2.

Assertion and Reason 1. (a) 2. Passage-1 1. (d) Passage-2 1. (d) Passage-3 1. (c) Passage-4 1. (d)

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6