CLASS 11 MOTION IN PLANE
MOTION IN PLANE
Direction
(OP) tail
Head
Pronounced as OP Vector
(1) OP = xi + yj + zk
(2) Magnitude of Vector:-
|OP| = √x² + y² + z² For eg OP = 2i + j + 2k
|OP| = √(2² + 0² + (2)² = √4 + 1 + 4 = √9 = 3 Units
✤ SCALAR COMPONENT
AB = (x₂-x₁)i + (y₂-y₁)j + (z₂-z₁)k
✤ TYPES OF VECTOR :
(1) Unit Vector :- A vector which modulus is one.
Example: A = 1/√3 i + 1/√3 j + 1/√3 k
|A| = √(1/√3)² + (1/√3)² + (1/√3)²
= √3/3
= √1
= 1
Let A is any vector then Unit vector along A :
A = A / |A|
Example:- Find Unit Vector along A = i + 2j + 2k
|A| = √(1)² + (2)² + (2)² = √9 = 3
∴ A = i + 2j + 2k/3 [ By formula A = A / |A| ]
A = 1/3 i + 2/3 j + 2/3 k
(2) Zero Vector: A vector in which initial & final point are same!
(tail & head Coincide each other)
AA = 0
(3) Negative of a Vector : If direction of a vector become reverse [ Change by 180° ] but magnitude remain same.
(4) Collinear Vector:- Those vector which lies in same line
Let a & b any two vectors . If a = λb then a & b are collinear .
For example:
a = 2i + 4j + 6k
b = i + 2j + 3k
So a = λb
(5) Co-Planer Vector :- All Vector should lies in same plane Plane . All resultant forces are zero
(6) Co-initial Vector :- Those vectors which have same initial point
ALGEBRA OF VECTOR:-
A = a₁i + a₂j + a₃k
B = b₁i + b₂j + b₃k
A+B = (a₁+b₁)i + (a₂+b₂)j + (a₃+b₃)k
Example: Unit Vector : A+B/|A+B| = c
A = 2i - 3j + 4k
B = i + 6j - 2k
A+B = 3i + 3j + 2k
Same in Subtraction Rule
SCALAR MULTIPLY :- A = a₁i + a₂j + a₃k λA = λa₁i + λa₂j + λa₃k
Example: A = 2i + 3j + 4k 3A = 6i + 9j + 12k
✤ TRIANGLE LAW :-
It states that when two sides of a Δ are represented by the vectors in some sense of direction then their resultant by 3rd side but opposite in direction.
a+b=c
NOTE :-
① a+b=-c
② a+c=-b
③ b+c=-a
✤ PARALLELOGRAM LAW :-
when adjacent sides of a llgm are represented by 2 vectors in same direction then their resultant is 3rd side but in opposite direction.
c=a+b
a=b-a
✤ REGULAR HEXAGON :-
AD = 2BC
BE = 2CD
FC = 2AB
✤ ANALYTICAL ANALYSIS OF VECTOR ADDITION :-
let two vectors a & b are acting θ angle with each other.
Let R is the resultant
Now |a| = a, |b| = b, |R| = R
In ΔABC,
sinθ = BC/AB
AB sinθ = BC
In ΔOBC, by PGT
OB² = OC² + BC²
R² = (a+bcosθ)²+(b sinθ)²
R² = a² + b²cos²θ + 2abcosθ + b² sin²θ
R² = a² + b² + 2ab cosθ
R = √a² + b² + 2ab cosθ
let R makes α with a then
tanα = BC/OC
tanα = b sinθ/a+b cosθ
✤ RIVER
Case(1) : For shortest distance
let speed of boat in still
water = v₁ & speed of water = v₂
So v' = √(v₁²-v₂²)2
Time to cross the river
t= d/ √(v₁²-v₂²)2
:. Angle from the Vertical
tanθ = v₂/ √(v₁²-v₂²)2
Case(2) : For Shortest time :-
Resultant Velocity = √(v₁²+(v₂)²
Time to cross the river
t= d/ √(v₁²+(v₂)²
Distance from straight Path = v₂.x.t
& tanθ = |v2/v1|
✤ RELATIVE VELOCITY OF RAIN W.R.T MOVING MAN :-
let velocity of rain = Vr
Velocity of man = Vm-Vm
The man hold an umbrella act θ from vertical to protect him from rain
In ΔOAB. By P.C.T
OB² = OA² + AB²
Vrm² = Vr² + Vm²
Vrm = √Vr² + Vm²
In ΔOAB
tanθ = AB/ OA
tanθ = |Vm/Vr|
✤ SCALAR PRODUCT OF TWO VECTORS :
1. a.b = |a||b|cosθ
2. cosθ = a.b/|a||b|
3. If θ=90°, a⊥b
a.b = 0
4. If θ=0°, a||b
a.b = |a||b|
∴ a.b = b.a
5. a.a = |a|²
6. let a = a₁i+a₂j+a₃k
b = b₁i+b₂j+b₃k
a.b = a₁b₁+a₂b₂+a₃b₃
7. Work done by a force F in displacement d
W=F.d
8. Projection of a along b
= a.b/|b|
9. Component of a along b
[a.b/|b|].B
✤ VECTOR PRODUCT OF TWO VECTORS :-
→ Let a & b are any two vectors then a×b = |a||b|Sinθ.n
→ It is vector which is always ⊥ to both a & b
b×a = |a||b|Sinθ.n
In general a×b ≠ b×a
a×b = -(b×a)
|a×b| = |a||b|Sinθ
If a || to b
θ=0°
Sinθ = 0
a×b = 0
If θ=90°
Sinθ = 1
|a×b| = |a|×|b|
* a×a = 0
* a = a₁î + a₂j + a₃k
b = b₁î + b₂j + b₃k
a×b =[■(i&j&k@a1&a2&a3@b1&b2&b3)] = î(a₂b₃-b₂a₃) - j(a₁b₃-b₂a₁) + k(a₁b₂-a₂b₁)
✤ Projectile: when a body goes in air, under the action of gravity only then the body is called projectile.
Examples: when a stone is thrown in air - then it is called Projectile.
✤ Horizontal Projectile: when a projectile is thrown from some height above ground.
Let after 't' time P(x,y). So from horizontal motion
S = ut + 1/2 at²
x = ut + 0
t = x/u - (1)
For vertical motion
S = ut + 1/2 gt²
y = 1/2 g(x/u)²
y = 1/2 g x²/ u²
x² = (2u² / g) y
It is represent a parabolic path so path of projectile is parabolic.
✤Time of Flight: It is the time taken by the projectile in air.
S = ut + 1/2 at²
h = 0 + 1/2 gt²
T = √2h/g
✤ Range: It is the horizontal distance covered by projectile.
S = ut + 1/2 gt², For horizontal
R = ut + 0
R = ut
R = u √2h/g
✤ Velocity: It is the velocity of projectile at any instant.
v = √vx² + vy²
& vy = vy + gt
vy = gt
v = √u² + g²t²
tan θ = |vy/vx|
We know v = u+at
Vx = u + 0
✤ VERTICAL PROJECTILE: when a body is thrown from ground into air & it move under the action of gravity only then it is called vertical Projectile.
ux = u cos θ
vy = u sin θ
Let after 't' time P(x,y) is a point.
For horizontal motion For Vertical motion
S = ut + 1/2 at² S = uyt + 1/2 at²
x = uxt + 1 x 0 y = u sin θ.t - 1/2 gt²
x = ux t
x = u cos θ.t y = u sin θ.x - 1/2 g(x/u cos θ)²
t = x/u cos θ - (1)
y = u sin θ.x/u cos θ - 1/2 g(x²/u² cos² θ)
y = x tan θ - x²(g/2u² cos² θ)
It is the path of projectile (trajectory). It is parabolic.
✤ Time of Flight: it is the time taken by the projectile in air.
Let Time as of flight = T
Time of ascent = T/2
We know, vy = Uy + ayt
At max-height
Vy = 0
Uy = u sin θ
ay = -g
t = T/2
θ = u sin θ - gT/2
T = 2u sin θ / g
✤ Height of Projectile :- It is the maximum vertical displacement of projectile to bind height Sy = Uyt+1/2 ayt²
Here Sy=H
Uy = USinθ
ay=-g
t=T/2
H = USinθ.T/2 - 1/2 g (T/2)²
= USinθ (USinθ/g) - 1/2 g (USinθ/g)²
= u²Sin²θ/g - u²Sin²θ/2g
= u²Sin²θ (1-1/2)
H= u²Sin²θ/2g
✤ Range:- It is the horizontal distance covered by Projectile
S = Uxt + 1/2 axt²
R = Ucosθ.T
= Ucosθ (2Sinθ/g)
R = u²(2Sincosθ)/g
R = u²Sin2θ/g
✤ Velocity:-
v = √rx² + ry²
Now, rx = Ux + axt & ry=Uy + ayt
rx = Ucosθ ry = Usinθ - gt
v = √(ucosθ)² + (Usinθ - gt)²
= u²cos²θ + u²Sin²θ + g²t²- 2UgtSinθ
v = √u²+ g²t²- 2UgtSinθ
For Max. Range :-
Sin2θ = 1
Sin2θ = 90°
2θ = 90°
θ = 45°
:. Rmax = u²/g
When angle is given from neutral s
(i) In this Case, T= 2u Sin(90-θ)/g
T = 2u cosθ/g
(ii) Height :-
H = u²Sin²(90-θ)/2g
H = u²Cos²θ/2g
(ii) R = u²Sin2(90-θ)/g
= u²Sin(180-2θ)/g
R = u²Sin2θ/g
✤ Circular Motion :-
(i) Angular Displacement :-
θ = x/r unit is Radian
(ii) Angular Velocity :- It is defined as rate of Change of angular displacement
ω = θ₂ - θ₁/t unit is Rad/second
ω = dθ/dt
(iii) Relation b/w ω & v
We know, θ = s/r
Diff. both sides w.r.t t
dθ/dt = (1/r)dx/dt
Here dθ/dt = ω
dx/dt = v
ω = 1/r x v => v = rω
✤ Angular Acceleration :-
It is defined as the rate of change of angular velocity.
:: α = W₂ - W₁ / t Rad/sec²
⇒ α = dω/dt
Relation blw α & linear acceleration (a) :-
We know
v = rω
diff w.r.t t
dv/dt = r dω/dt
Here dv/dt = a
dω/dt = α
∴ a = rα
Let time period of revolution is T
∴ω = 2π/T
ω = 2π x ῡ
Here ῡ = r.p.s
(Rotation per Second)
✤ Equation of Motion :-
① ω = ω₀ + αt
② θ = ω₀t + 1/2 αt²
③ ω² = ω₀² + 2αθ
✤ Centripetal Force :-
If a body of mass 'm' as revolving round in the circle then ;
* Centripetal Force
F = mv²/r
* Centripetal Acceleration :-
We Know P = mv²/r
F = ma
On Comparing ac = v²/r
We have, ῡ = rω₀
ac = r2ω²/r
ac = rω²
* * * * * * * * * * * * * * *
