Good conceptual problems

Question: If you are riding on a train that speeds past other train moving in the same direction on adjacent train.It appears the other train is moving backward. Why?

Solution:

Your reference frame is that of the train you are riding. If you are traveling with a relatively constant velocity (not over a hill or around a curve or drastically changing speed), then you will interpret your reference frame as being at rest. Since you are moving forward faster than the other train, the other train is moving backwards relative to you. Seeing the other train go past your window from front to rear makes it look like the other train is going backwards. This is similar to passing a semi truck on the interstate out of a passenger window, it looks like the truck is going backwards.

Question:A package is dropped from the plane.Four People sitting in the plane make different statements about the package location when it hits the ground.Air resistance can be ignored

Person A: The package will fall behind the plane
Person B: The package remain below the plane until it hit the ground
Person C:The package move ahead of the plane
Person D: The package location will depend on the speed of the plane

We know that only one person is correct and three are wrong
Who are those three person’s
A) Person A,B,C
b) Person A,C,D
c) Person B,C,D
d) Person A,B,D

Solution:

The package will have same horizontal velocity as of plane.The vertical velocity will be dictated by the force of gravity.
So it will cover the same horizontal distance as plane before hitting the ground.So Person B is correct

Question: Jack and Henry both are good swimmer and can swim with same speed in still water . They setoff across the river at the same time.
Jack moves straight across and Jack is pulled downstream by the current somewhat.Henry heads upstream at an angle so as to arrive at a point directly opposite to starting point .

Who will cross the river first?
A) jack

b) henry

c) Both Jack and Henry will cross the river in same time

Solution:

Both jack and henry need to cover the same “cross river” distance. The swimmer with the greatest speed in the “cross river” direction will be the one that reaches the other side first.The current has no bearing on the problem because the current doesn’t help either of the boats move across the river.
Thus the swimmer heading straight across will reach the other side first. All of his swiming effort has gone into crossing the river. For the upstream swimmer, some of his swimming effort goes into battling the current,and so his “cross river” speed will be only a fraction of his swimming speed.So Jack will come first

Question:

During projection motion,the velocity and acceleration vectors are acting in
different direction at different points.

Which of the following is true

a) v.a =0 at the topmost point

b) v.a < 0 during the upward journey

c) v.a > 0 during the downward journey

d) None of these

Solution:

During Upward journey

v=v_xi+v_yj

g=-gj

v.a < 0

At topmost point

v=v_xi

g=-gj

v.a =0

During downward journey

v=v_xi-v_yj

g=-gj

v.a > 0

Physics 11th Motion in straight Line

Motion in straight Line

1. Introduction

• In our dialy life ,we see lots of things moving around for example car passing through from one place to other ,person riding on a bicycle and many more like this.
• In scientific terms an object is said to be in motion ,if it changes its position with the pasage of time and if it does not change it position with the passage of time then it is said to be at rest
• Both the motion and rest are relative terms for example mobile kept on the table is resting at its position but it is moving in the sense as earth is rotating on its axis.So for a person seeing mobile from earth it is at rest and for person on moon earth seems to change its position with time and so mobile is moving.
• Simplest case of motion is rectilinear motion which is the motion of the object in a straight line
• In our descriotion of object ,we will treat the object as an point object
• Object under consideration can be treated as point object if the size of the object is much smaller than the distance travelled by it in a reasonable time duration for example length of a motor car travelling a distance of 500km can be neglected w.r.t distance travelled by it.
• Here in kinematics ,we study ways to describe the motion without going into the cause of the motion

2.Position and Displacement

(a) Position:

• To locate the position in motion or at rest,we need a frame of refrence.
• Simplest way to choose a frame of refrence is to choose three mutually perpendicular axis labelled as X-,Y- and Z- axis as shown in figure below
  
  
• Such system of labelling position of an object is known as rectangular coordinates system
• If A(x,y,z) be the position of any point in rectangular co-ordinates system it can be labelled as follows
  
  
  
• Point O is the point of intersection of these mutaully perpendicular axis and is known as refrence point or origin of frame of refrence
• To measure a time ,we can also attach a clock with this frame of refrence
• If any or all co-ordinates of the object under consideration changes with time in this frame of refrence then the object is said to be in a motion w.r.t the frame of the refrence otherwise it is at rest
• For describing motion in one dimension we need one set of co-ordinates axis i.e only one of X,Y and Z axis
• Similary for two and three dimensions we need two or three set of axis respectively
• Motion of an object along a straight line is an example of motion in one dimension
• For such a motion,any one axis say X-axis may be choosen so as to co-incide with the path along which object is moving
• Position of the object can be measured w.r.t origin O shown in the figure
  
  
  
• Position to the right of the origin has positive values and those to the left of origin O has negative values.

(b) Distance and displacement:

• In the graph shown below an object is at position P at time t₁ and at position R at time t₂.
  
  
  
• In the time interval from t₁ to t₂ particle has travelled path PQR and length of the path PQR is the distance travelled by the object in the time interval t₁ to t₂
• Now connect the initial position of the object P with its final position R through a straight line and we get the displacement of the object.
• Displacement of the object has both magnitude and direction i.e., displacement is a vector quantity.
• Magnitude of displacement vector is equal to the length of straight line joining initial and final position and its directionpoints from the initial position of object towards its final position.
• In contrast to displacement distance is scalar quantity.

3. Average velocity and speed

• Consider a particle undergoing motion along a straight line i.e. moving along X-axis.
• X co-ordinate describing motion of the particle from origin O varies with time or we can say that X co-ordinate depends on time.
• If at time t=t₁ particleis at point P , at a distance x₁ from origin and at time t=t₂ it is at point Q at a distance x₂ from the origin then displacement during this time is a vector from point P to Q and is
  Δx=x₂-x₁                             (1)
  
  
  
  
  
  
• The average velocity of the prticle is defined as the ratio of the displacement Δx of the particle in the time interval Δt=t₂-t₁. If v_avg represents average velocity then,
   
• Figure 5b represents the co-ordinate time graph of the motion of the particle i.e., it shows how the value of x-coordinate of moving particle changes with the passage of time.
• In figure 5b average velocity of the particle is represented by the slope of chord PQ which is equal to the ratio of the displacement Δx occuring in the particular time interval Δt.
• Like displacement average velocity v_avg also has magnitude as well as direction i.e., average velocity is a vector quantity.
• Average velocity of the particle can be positive as well as negative and its positive and negative value depends on the sign of displacement.
• If displacement of particle is zero its average velocity is also zero.
• Graphs below shows the x-t graphs of particle moving with positive, negative average velocity and the particle at rest.
  
  
  
  
  
  
  
  
• From graph 5c it is clear that for positive average velocity slope of line slants upwards right or we can say that it has positive slope.
• For negative average velocity slope line slants upwards down to the right i.e. it has negative slope.
• For particles at rest slope is zero.
• So far we have learnt that average speed is the rate of motion over displacement of the object.
• Displacement of the object is different from the actual distance travelled by the particle.
• For actual distance travelled by the particle its average speed is defined as the total distance travelled by the particle in the time interval during which the motion takes place.
• Mathematically,
   
• Since distance travelled by an particle does not involve direction so speed of the particle depending on distance travelled does not involve direction and hence is a scalar quantity and is always positive.
• Magnitude of average speed may differ from average velocity because motion in case of average speed involve distance which may be greater than magnitude of displacement for ex.
  
  
  
  here a man starts travelling from origin till point Q and return to point P then in this case displacement of man is
  Displacement from O to Q is OQ=80m
  Displacement from Q to P is =20m-80m=-60m
  total displacement of particle in moving from O to Q and then moving Q to P is = 80m + (-60m) = 20 m
  Now total distance travelled by man is OQ+OP= 80m +60m = 140m
  Hence diring same cource of motion distance travelled is greater then displacement.
• from this we can say that average speed depending on distance is in general greater than magnitude of velocity.

4. Instantaneous velocity and speed

• Velocity of particle at any instant of time or at any point of its path is called instantaneous velocity.
• Again consider the graph 5b and imagine second point Q being taken more and more closer to point P then calculate the average velocity over such short displacement and time interval.
• Instantaneous velocity can be defined as limiting value of average velocity when second point comes closer and closer to the first point.
• Limiting value of Δx/ Δt as Δt aproaches zero is written as dx/dt, and is known as instantaneous velocity.Thus instantaneous velocity is
  
  
• As point Q aproaches point P in figure 5a in this limit slope of the chord PQ becomes equal to the slope of tangent to the curve at point P.
• Thus we can say that instantaneous velocity at any point of a coordinate time graph is equal to the slope of the tangent to the graph at that point.
• Instantaneous speed or speed is the magnitude of the instantaneous velocity unlike the case of average velocity and average speed where average speed over an finite interval of time may be greate than or equal to average velocity.
• Unit of average velocity , average speed, instantaneous velocity and instantaneous speed is ms_-1 in SI system of units.
• Some other units of velocity are ft.s_-1 , cm.s_-1 .

5. Acceleration

• Acceleration is the rate of change of velocity with time.
• For describing average acceleration we first consider the motion of an object along X-axis.
• Suppose at time t₁ object is at point P moving with velocity v₁ and at time t₂ it is at point Q and has velocity v₂. Now average acceleration of object in moving from P to Q is
  
  
  
  
  which is the change in velocity of object with the passage of time.
• Instantaneous acceleration can be defined in the same way as instantaneous velocity
  
  
• The instantaneous acceleration at any instant is the slope of v-t graph at that instant.
  
  
• In figure 7 instantaneous acceleration at point P is equal to the slope of tangent at this point P.
• Since velocity of a moving object has both magnitude and direction likewise acceleration depending on velocity has both magnitude and direction and hence acceleration is a vector quantity.
• Acceleration can also be positive, negative and zero.
• SI unit of acceleration is ms^-2 

6. Motion with constant acceleration

• Motion with constant acceleration or uniformly accelerated motion is that in which velocity changes at the same rate troughout motion.
• When the acceleration of the moving object is constant its average acceleration and instantaneous acceleration are equal. Thus from eq. 5 we have
  
• Let v₀ be the velocity at initial time t=0 and v be the velocity of object at some other instant of time say at t₂=t then above eq. 7 becomes
  
  or , v=v₀+at                         (8)
• Graphically this relation is represented in figure 8 given below.
  
• Thus from the graph it can be seen clearly that velocity v at time t is equal to the velocity v₀ at time t=0 plus the change in velocity (at).
• In the same way average velocity can be written as
  
  where x₀ is the position of object at time t=0 and v_avg is the averag velocity between time t=0 to time t.The above equation then gives
  x=x₀+v_avgt                    (9)
  but for the interval t=0 to t the average velocity is
  
  Now from eq. 8 we find
  v_avg = v₀ + ½(at)                    (11)
  putting this in eq. 9 we find
  x = x₀ + v₀t + ½(at²)
  or,
  x - x₀ = v₀t + ½(at²)                (12)
  this is the position time relation for object having uniformly accelerated motion.
• From eq. 12 it is clear that an object at any time t has quadratic dependance on time, when it moves with constant acceleration along a straight line and x-t graph for such motion will be parabolic in natureas shown below.
  
  
• Equation 8 and 12 are basic equations for constant acceleration and these two equations can be combined to get yet another relation for x , v and a eleminating t so, from 8
  
  putting this value of t in equation 12 and solving it we finally get,
  v² = (v₀)² + 2a ( x - x₀ )                     (13)
• Thus from equation 13 we see that it is velocity displacement relation between velocities of object moving with constant acceleration at time t and t=0 and their corresponding positions at these intervals of time.
• This relation 13 is helpful when we do not know time t.
• Likewise we can also eliminate the acceleration between equation 8 and 12. Thus from equation 8
  
  
  putting this value of a in equation 12 and solving it we finally get,
  ( x - x₀ ) = ½ ( v₀ + v ) t               (14)
• Same way we can also eliminate v₀ using equation 8 and 12. Now from equation 8
  v₀ = v - atputting this value of v₀ in equation 12 and solving it we finally get,
  ( x - x₀ ) = vt + ½ ( at² )                (15)
  thus equation 15 does not involve initial velocity v₀
• Thus these basic equations 8 and 12 , and derived equations 13, 14, and 15 can be used to solve constant acceleration problems.

7. Free fall acceleration

• Freely falling motion of any body under the effect of gravity is an example of uniformly accelerated motion.
• Kinematic equation of motion under gravity can be obtained by replacing acceleration 'a' in equations of motion by acceleration due to gravity 'g'.
• Value of g is equal to 9.8 m.s^-2.
• Thus kinematic equations of motion under gravity are
  v = v₀ + gt                          (16a)
  x = v₀t + ½ ( gt² )          (16b)
  v² = (v₀)² + 2gx           (16c)
• The value of g is taken positive when the body falls vertically downeards and negative when the body is projected up against gravity.

8. Relative velocity

• Consider two objects A and B moving with uniform velocities v_Aand v_B along two straight and parallel tracks.
• Let x_OAand x_OB be their distances from origin at time t=0 and x_Aand x_B be their distances from origin at time t.
• For object A
  x_A = x_OA + v_At               (18)
  and for object B
  x_B = x_OB + v_Bt                (19)
  subtracting equation 18 from 19
  x_B - x_A = ( x_OB - x_OA ) + ( v_B - v_A) t                          (20)
• Above equation 20 tells that as seen from object A , object B seems to have velocity ( v_B - v_A) .
• Thus ( v_B - v_A ) is the velocity of object B relative to object A. Thus,
  v_BA = ( v_B - v_A )                (21)
• Similarly velocity of object A relative to object B is
  v_AB = ( v_A - v_B )                (22)
• If v_B = v_A then from equation 20
  x_B - x_A = ( x_OB - x_OA )
  i.e., two objects A and B stays apart at constant distance.
• v_A > v_B then (v_B - v_A ) would be negative and the distance between two objects will go on decreasing by an amount ( v_A- v_B ) after each unit of time. After some time they will meet and then object A will overtake object B.
• If v_A and v_B have opposite signs then magnitude of v_BA or v_AB would be greater then the magnitude of velocity of A or that of B and objects seems to move very fast

Solved Examples Part 1

Question 1.

 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 v varies with the height h as









Solution(1):
. Before hitting the ground ,the velocity v is given by v²=2gd Which is a quadratic equation and hence parabolic path. Downward direction means negative velocity , After collison ,the velocity becomes positive and velocity decreases Further v₁²=2g(d/2)=gd Therefore v=v₁√2 As the direction is reversed and speed is decreased, Hence (a) is the answer 

Question 2. The displacement -time graph of a moving particle is shown below.The instantanous velocity of the particle is negative at the point



a. C
b. D
c. E
d. B

Solution(2):
. The instantanous velocity is given by the slope of the displacement time graph.
Since slope is negative at point B,instantanous velocity is negative at B
Hence (d) is correct

Question 3.

The velocity -time graph of a moving particle is shown below.Total displacement of the particle during the time interval when there is nonzero acceleration and retardation is 



a. 60m
b. 40m
c. 50m
d. 30m

Solution(3):
. Acceleration is given by the slope of the velocity time graph
Non zero acceleration happened in the time interval 20 to 40 sec as the slope of the graph is non zero in that time interval

Now displacement in that interval is given by the area enclosed the v-t curve during that interval

So area=(1/2)20*3 +20*1=50m
Hence (c) is correct 
Question 4. Figure below shows the displacement -time graph of two particles.Mark the correct statement about their relative velocity


a. It first increases and then decreases
b. It is a non zero constant
c. it is zero
d. none of the above
Solution(4):
. The instantanous velocity is given by the slope of the displacement time graph.
As slope of both the particle displacement time graph is constant.That means there individual velocities are constant.
So relative velocity is also constants
so it is a non zero constant as they have different velocity
Hence (b) is correct 

Four position -time graph are shown below.
a.

b.

c.

d.


Question 5 What all graph shows motion with positive velocities
a. a and c only
b. all the four
c. b and D only
d. b only

Solution(5):
. The instantanous velocity is given by the slope of the displacement time graph.
Since slope is positive in graph a and c.Positive velocity is there in a and c curve
Hence (a) and (c) are correct 

Question 6.

 What all graph shows motion with negative velocities
a. a and b only
b. all the four
c. a and c only
d. c only

Solution(6):
. The instantanous velocity is given by the slope of the displacement time graph.
Since slope is negative in graph b and d.Negative velocity is there in b and d curve
Hence (b) and (d) are correct 
Question 7.which of the folliwing graph correctly represents velocity-time relationships for a particle released from rest to fall under gravity
a.

b.

c.

d.


Solution(7):
. 
The velocity will increase with time so b is the correct answer
Hence (b) is correct 
Question 8.The v-x graph of a particle moving along a straight line is shown below.Which of the below graph shows a-x graph



a.

b.

c.

d.


Solution(8):
. The equation for the given graph is
v=-(v₀/x₀)x + v₀ ---(1)
Differentiating both sides we get
dv/dx=-v₀/x₀ ---(2) 
Now
a=v(dv/dx)
or
a=(-v₀/x₀)[-(v₀/x₀)x + v₀ ]
or
a=mx+c
where m=v₀²/x₀)²
and c=-v₀²/x₀)²

So that means slope is positive and intercept is negative
So (d) is correct

Question 9.

A truck accelerates from rest at the constant rate a for some time after which it decelerates at a constant rate of b to come to the rest.If the total time elapsed is t ,then find out the maximum velocity attains by the truck
a. (ab/a+b)t
b.(a+b/ab)t
c. (a²+b²/ab)t
d.(a²-b²/ab)t

Solution(9):
. Let t₁ and t₂ be the the time for acceleration and deccleration.
Let v be the maximum velocity attained

Then
v=at₁ or t₁=v/a
v=bt₂ or t₂=v/b

Now t=t₁ + t₂
or t=v/a + v/b

or v=abt/(a+b)
Hence (a) is correct 

Question 10.Displacement(y) of the particle is given by
y=2t+t²-2t³
The velocity of the particle when acceleration is zero is given by
a. 5/2
b. 9/4
c. 13/6
d. 17/8

Solution(10):
. given

y=2t+t²-2t³

Velocity is given 
v=dy/dt=2+2t-6t²

a=dv/dt=2-12t

Now acceleratio is zero
2-12t=0
t=1/6

Putting this value velocity equation
v=2+2/6-6(1/6)²
=2+1/3-1/6
=13/6
Hence (c) is correct 


Question 11.Mark out the correct statement
a. Instantaneous Velocity vector is always in the direction of the motion
b. Instantaneous acceleration vector is always in the direction of the motion
c. Acceleration of the moving particle can change its direction without any change in direction of velocity
d. None of the above

Solution(11):
. Take the case of uniform circular motion,Instantanous Velocity vector and acceleration vector at any point is tangent and radial to the circle.So it is not along the direction of the circle
Take the case the moving car in one direction.If the car accelerated,acceleration is along the direction of velocity.if car driver put a brake then it deacclerates without any change in the direction of the velocity

Hence (c) is correct 

Matrix match type

Question 12.In a free fall motion from rest,Match column I to column II

column I 
A) Graph between displacement and time
B) Graph between velocity and time
C) Graph between velocity and displacement
D) Graph between KE and displacement

column II
P) Parabola
Q) Straight line
C) Circle
D) No appropiate match given

Solution(12):
. Equation of motion for a free fall from rest
x=(1/2)gt².It is a parabola
v=gt it is a straight line
v²=2gx it is a parabola
KE=(1/2)mv²=mgx..... it is a straight line
So the correct answers are A -> P
B -> Q
C -> P
D -> Q 

Physics 11th Motion in a Plane

Motion in a Plane

1. Introduction

• In previous chapter we have learned about the motion of any particle along a straight line
• Straight line motion or rectilinear motion is motion in one dimension.Now in this chapter ,we will consider both motion in two dimension and three dimension.
• In two dimensional motion path of the particle is constrained to lie in a fixed plane.Example of such motion motion are projectile shot from a gun ,motion of moon around the earth,circular motion and many more.
• To solve problems of motion in a plane,we need to generalize kinematic language of previous chapter to a more general using vector notations in two and three dimensions.

2.Average velocity

• Consider a particle moving along a curved path in x-y plane shown below in the figue
• Suppose at any time,particle is at the point P and after some time 't' is at point Q where points P and Q represents the position of particle at two different points.
  
  
  
• Position of particle at point P is described by the Position vector r from origin O to P given by
  r=xi+yj
  where x and y are components of r along x and y axis
• As particle moves from P to Q,its displacement would be would be Δr which is equal to the difference in position vectors r and r'.Thus
  Δr = r'-r = (x'i+y'j)-(xi+yj) = (x'-x)i+(y'-y)j = Δxi+Δyj                                          (1)
  where Δx=(x'-x) and Δy=(y'-y)
• If Δt is the time interval during which the particle moves from point P to Q along the curved path then average velocity(v_avg) of particle is the ratio of displacement and corresponding time interval
  
  
  since v_avg=Δr/Δt , the direction of average velocity is same as that of Δr 
• Magnitude of Δr is always the straight line distance from P to Q regardless of any shape of actual path taken by the particle.
• Hence average velocity of particle from point P to Q in time interval Δt would be same for any path taken by the particle.

3.Instantaneous velocity

• We already know that instantaneous velocity is the velocity of the particle at any instant of time or at any point of its path.
• If we bring point Q more and more closer to point P and then calculate average velocity over such a short displacement and time interval then
  
  
  where v is known as the instantaneous velocity of the particle.
• Thus, instantaneous velocity is the limiting value of average velocity as the time interval aproaches zero.
• As the point Q aproaches P, direction of vector Δr changes and aproaches to the direction of the tangent to the path at point P. So instantaneous vector at any point is tangent to the path at that point.
• Figure below shows the direction of instantaneous velocity at point P.
  
  
  
• Thus, direction of instantaneous velocity v at any point is always tangent to the path of particle at that point.
• Like average velocity we can also express instantaneous velocity in component form
  
  
  where v_x and v_y are x and y components of instantaneous velocity.
• Magnitude of instantaneous velocity is
  |v|=√[(v_x)²+(v_y)²]
  and angle θ which velocity vector makes with x-axis is
  tanθ=v_x/v_y
• Expression for instantaneous velocity is
  
  
  Thus, if expression for the co-ordinates x and y are known as function of time then we can use equations derived above to find x and y components of velocity.

4. Average and instantaneous acceleration

• Suppose a particle moves from point P to point Q in x-y plane as shown below in the figure
  
  
  
• Suppose v₁ is the velocity of the particle at point P and v₂ is the velocity of particle at point Q
• Average acceleration is the change in velocity of particle from v₁ to v₂ in time interval Δt as particle moves from point P to Q. Thus average acceleration is
  
  
  Average accelaration is the vector quantity having direction same as that of Δv.
• Again if point Q aproaches point P, then limiting value of average acceleration as time aproaches zero defines instantaneous acceleration or simply the acceleration of particle at that point. Ths, instantaneous acceleration is
  
   
• Figure below shows instantaneous acceleration a at point P.
  
  
  
• Instantaneous acceleration does not have same direction as that of velocity vector instead it must lie on the concave side of the curved surface.
• Thus velocity and acceleration vectors may have any angle between 0 to 180 degree between them.

5. Motion with constant acceleration

• Motion in two dimension with constant acceleration we we know is the motion in which velocity changes at a constant rate i.e, acceleration remains constant throughout the motion
• We should set up the kinematic equation of motion for particle moving with constant acceleration in two dimensions.
• Equation's for position and velocity vector can be found generalizing the equation for position and velocity derived earliar while studying motion in one dimension
  Thus velocity is given by equation
  v=v₀+at                                          (8)
  where
  v is velocity vector
  v₀ is Intial velocity vector
  a is Instantanous acceleration vector
  Similary position is given by the equation
  r-r₀=v₀t+(1/2)at²                                          (9)
  where r₀ is Intial position vector
  i,e
  r₀=x₀i+y₀j
  and average velocity is given by the equation
  v_av=(1/2)(v+v₀)                                          (10)
• Since we have assumed particle to be moving in x-y plane,the x and y components of equation (8) and (9) are
  v_x=v_x0+a_xt                                          (11a)
  x-x₀=v_0xt+(1/2)a_xt²                                          (11b)
  and
  v_y=v_y0+a_yt                                           (12a)
  y-y₀=v_0yt+(1/2)a_yt²                                           (12b)
• from above equation 11 and 12 ,we can see that for particle moving in (x-y) plane although plane of motion can be treated as two seperate and simultanous 1-D motion with constant acceleration
• Similar result also hold true for motion in a three dimension plane (x-y-z)

6. Projectile Motion

• Projectile motion is a special case of motion in two dimension when acceleration of particle is constant in both magnitude and direction
• An object is referred as projectile when it is given an intial velocity which subsequently follows a path determined by gravitational forces acting on it.For example bullet fired from the rifle,a javalin thrown by the athelite etc
• Path followed by a projectile is called its trajectory
• In this section we will study the motion of projectile near the earth surface neglecting the air resistance.
• Acceleration acting on a projectile is constant which is acceleration due to gravity (g=9.81 m/s) directed along vertically downward direction.
• we shall treat the projectile motion ina cartesian co-ordinates system taking y axis in vertically upwards direction and x axis alomg horizontal directions
• Now x and y components of acceleration of projectile is
  a_x=0 and a_y=-g
  Since acceleration in horizontal direction is zero,this shows that horizontal component of velocity is constant and verical motion is simply a case of motion with constant acceleration.
• Suppose at time t=0 object is at origin of co-ordinate system and velocity components v_0x and v_0y.From above components of acceleration are a_x=0 and a_y=-g.From equation 11 and 12 in the previous section components of position and velocity are
  x=v_0xt                                          (14a)
  v_x=v_0x                                           (14b)
  and v_y=v_0y-gt                                          (15b)
  y=v_0y-(1/2)gt²                                          (15a)
• Figure below shows motion of an object projected with velocity v₀ at an angle θ₀
  
  
  
• In terms of initial velocity v₀ and angle θ₀ components of initial velocity are
  v_0x=v₀cosθ₀                                          (16a)
  v_0y=v₀sinθ₀                                           (16b)
  
• Using these relations in equation 14 and 15 we find
  x=(v₀cosθ₀)t                                          (17a)
  y=(v₀sinθ₀)t-(1/2)gt²                                           (17b)
  v_x=v₀cosθ₀                                           (17c)
  v_y=v₀sinθ₀-gt                                           (17d)
  Above equations describe the position and velocity of projectile as shown in fig 5 at any time t.

(A)Equation of Path of projectile(Trajectory)

• From equation 17a
  t=x/v₀cosθ₀
  now putting this value of t in equation 17b,we find
  y=(tanθ₀ )x-[g/2(v₀cosθ₀)²]x²                                           (18)
  In equation (18),quantities θ₀,g and v₀ are all constants and equation (18) can be compared with the equation
  y=ax-bx²
  where a and b are constants
• This equation y=ax-bx² is the equation of the parabola.From this we conclude that path of the projectile is a parabola as shown in figure 5

B) Time of Maximum height

• At point of maximum height v_y=0.Thus from equation (17d)
  v_y=v₀sinθ₀-gt
  0=v₀sinθ₀-gt -
  or t_m=v₀sinθ₀/g                                          (19)
• Time of flight of projectile which is the total time during which the projectile is in flight can be obtained by putting y=0 because when projectile reaches ground ,verical distance travelled is zero.This from equation (17b)
  
  t_f=2(v₀sinθ₀)/g                                          (20)
  or
  t_f=2t_m
• Maximum height reached by the projectile can be calculated by substituting t=t_m in equation 17b
  y=H_m=(v₀sinθ₀)(v₀sinθ₀/g)-(g/2)(v₀sinθ₀/g)²
  or
  H_m=v₀²sin²θ₀/2g                                          (21)

(C) Horizontal Range of Projectile

• Since acceleration g acting on the projectile is acting vertically ,so it has no component in horizontal direction.
• So, projectile moves in horizontal direction with a constant velocity v₀cosθ₀. So range R is
  R=OA=velocity x time of flight
  
   
• Maximum range is obtained when sin2θ₀=1 or θ₀=45⁰. Thus when θ₀=45⁰ maximum range achieved for a given initial velocity is (v₀)²/g.

7. Uniform circular motion

• When an object moves in a circular path at a constant speed then motion of the object is called uniform circular motion.
• In our every day life ,we came across many examples of circular motion for example cars going round the circular track and many more .Also earth and other planets revolve around the sun in a roughly circular orbits
• Here in this section we will mainly consider the circular motion with constant speed
• if the speed of motion is constant for a particle moving in a circular motion still the particles accelerates becuase of costantly changing direction of the velocity.
• Here in circular motion ,we use angular velocity in place of velocity we used while studying linear motion

(A) Angular velocity

• Consider an object moving in a circle with uniform velocity v as shown below in the figure
  
  
  
  
  The velocity v at any point of the motion is tangential to the circle at that point.Let the particle moves from point A to point Balong the circumference of the circle .The distance along the circumference from A to B is
  s=Rθ                                          (23)
  Where R is the radius of the circle and θ is the angle moved in radian's
• Magnitude of velocity is
  v=ds/dt=Rdθ/dt                                          (24)
  Since radius of the circle remains constant quantity,
  ω=dθ/dt                                          (25)
  is called the angular velocity defined as the rate of change of angle swept by radius with time.
• Angular velolcity is expressed in radians per second (rads^-1)
• From equation 24 and 25,we find the following
  v=ωR                                          (26)
• Thus for a particle moving ain circular motion ,velocity is directly proportional to radius for a given angular velocity
• For uniform circular motion i.e, for motion with constant angular velocity the motion would be periodic which means particle passes through each point of circle at equal intervals of time
• Time period of motion is given by
  T=2π/ω                                          (27)
  Since 2π radians is the angle θ in one revolution
• If angular velocity ω is constant then integrating equation (25) with in limits θ₀ to θ,we find
  
  
  where θ₀ is the angular position at time t₀ and θ is the angular position at time t .The above equation is similar to rectilinear motion result x-x₀=v(t-t₀)

(B) Angular acceleration

• Angular acceleration is defined as the rate of change of angular velocity moving in circular motion with time.
  Thus
  α=dω/dt=d²θ/dt²                                          (29)
  Unit of angular acceleration is rads^-2
• For motion with constant angular acceleration
  
  
  or
  ω=ω₀+α(t-t₀)                                          (30)
  where ω₀ is the angular velocity at time t₀
  Again since
  ω=dθ/dt
  or dθ=ωdt then from equation 30
  
  If in the begining t₀=0 and θ₀=0 the angular position at any time t is given by
  θ=ωt+(1/2)αt²
  This result is of the form similar to what we find in case of uniformly accelerated motion while studying rectilinear motion

8. Motion in three dimensions

• We have already studied physical quantities like displacement ,velocity,acceleration etc in one and two dimension
• In this topic ,we will generalize our previous knowlegde of motion in 1 and 2 -dimension to three dimension's
• As we have used vectors to represent motion in a plane,we can freely use vectors and its properties in 3-dimension as we have done in case of motion in a plane 
• In three dimensions ,we have three units vectors i ,j and k associated with each co-ordinate axis of cartesian co-ordinates system shown below in the figure
  
  
  
• Consider a particle moving in 3-D space .Let P be its position at any point t.Position vector of this particle at point P would be
  r=xi+yj+zk
  Where x,y and z are co-ordinates of point P
• Similarly velocity and acceleration vectors of particle moving in 3-D space are
  v=v_xi+v_yj+v_zk where v_x=dx/dt,v_y=dy/dt and v_z=dz/dt
  and
  a=v_xi+a_yj+a_zk
  where a_x=dv_x/dt,a_y=dv_y/dt and a_z=dv_z/dt
• All the relations we have derived incase of motion in plane are valid for 3-D motion with one added co-ordinate