Thursday, May 21, 2009

IIT JEE 2011 Physics Study Diary - Ch.3 Rest and Motion - Day 4

Day 4 - Study Plan

3.7 Motion in a plane
Ex. 3.8
3.8 Projectile motion
Ex. 3.9
WOE 11,12, 14

Points to Note

Motion in a plane

Motion in plane is described by x coordinate and y coordinate, if we choose X-Y plane. You can imagine time t is on the third axis.

The position of the particle or the body can be described by x and y coordinates.

r = xi = yj

Displacement during time period t to t+Δt can be represented by Δr

Δr = Δxi = Δyj

Then Δr/Δt = (Δx/Δt)i = (Δy/Δt)j

Taking the limits as Δt tends to zero

v = dr/dt = (dx/dt)i+(dy/dt)j ... (3.15)

Hence x component of velocity is dx/dt

The x-coordinate, the x component of velocity, and the x component of acceleration are related by equations of straight line motion along X axis.
Similarly y components.


Projectile motion is an important example of motion in a plane.

What is a projectile? When a particle is thrown obliquely near the earth's surface, it is called a projectile. It moves along a curved path. If we assume the particle is close to the earth and negligible air resistance to the motion of the particle, the acceleration of the particle will be constant. We solve projectile problems with the assumption that acceleration is constant.

Vertical motion of the projectile is the motion along Y axis and horizontal motion is motion along X axis.

Terms used in describing projectile motion

Point of projection
Angle of projection
Horizontal range
Time of flight
Maximum height reached

The motion of projectile can be discussed separately for the horizontal and vertical parts.

The origin is taken as the point of projection.
The instant the particle is projected is taken as t = 0.
X-Y plane is the plane of motion.
The horizontal line OX is taken as the X axis.
Vertical line OY is the Y axis.
Vertically upward direction is taken as positive direction of Y

Initial velocity of the particle = u
Angle between the velocity and horizontal axis = θ

ux – x-component of velocity = u cos θ
ax – x component of acceleration = 0

uy – y component of velocity = u sin θ
ay = y component of acceleration = -g

Horizontal motion of the projectile – Equations of motion

ux = u cos θ
ax = 0
vx = ux +axt = ux = u cos θ (as ax = 0)
Hence x component of the velocity remains constant.
Displacement in horizontal direction = x = uxt+1/2ax t²
As ax = 0, x = ux t = ut cos θ

Vertical motion – Equations of motion
uy = u sin θ
ay = -g
vy = uy – gt
Displacement in y direction = y = uyt – ½ gt²
vy² = uy² - 2gy

22. Time of flight of the projectile = (2u sin θ)/g

23. OB = (u²sin 2θ)/g

24. t = (u sin θ)/g
At t vertical component of velocity is zero.

25. Maximum height reached by the projectile = (u² sin²θ)/2g

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