CLASS 11 LAWS OF MOTION
LAWS OF MOTION
✤ LINEAR MOMENTUM:-
It is defined as the product of mass (cm) & velocity (V).
P = mv
Units - Kgms⁻¹
Dim - MLT⁻¹
Direction of momentum is same as that of motion.
NOTE :-
① If m = constant
then pαv
P₁/P2 = V₁/V2
② if r = constant
the pαm
P₁/P2 = m₁/m2
✤ Relation b/w K-E & Momentum :>
P = mv
K = ½ mv²
= m²v²/2m
K = (mv)²/2m
K = P²/2m
✤ LAWS OF MOTION:-
1st LAW :-
It states that in the absence of any external force, any body does not change its state of rest on state as motion by itself.
[This also called law of inertia].
2nd LAW:-
It states that rate of change of momentum is always equal to the external force applied on the body
F=P₂-P₁/ t F=dp/dt
✤ CONSERVATION OF MOMENTUM:-
It states that in the absence of external force. Total momentum always remain constant.
We know,
F = dp/dt
=> If F = 0
=> dp/dt = 0
=> P = Constant
(Before) (After)
(m₁) + (m₂) (m1) + (m2)
U₁ U₂ v1 v2
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
NOTE :-
Initial momentum Final momentum
P₁ = mucosθ P2 = -mucosθ
Change in momentum of the ball = P₂ - P₁
= - mucosθ - mucosθ
= - 2mucosθ
momentum transferred to the heat :- +2 mucosθ
✤ IMPULSE :- When a large force is applied on a body for short duration of time then the product of force & time is called impulse.
Impulse = Fxt
SI unit : Ns
Impulse is area enclosed by a rectangle
✤ Impulse momentum theorem:- It states that impulse is always equals to the change in momentum of the body.
i.e Impulse = P₂-P₁
I = mv - mu
= m(v-u)
✤ APPARENT WEIGHT:- It's the weight of a body in motion.
Case I: let a body of mass 'm' is placed in an elevator then in upward motion with acceleration a
F = R - mg
= ma = R - mg
R = mg + ma
R = m(g+a)
Case II: In downward motion, If elevator is going down with acceleration 'a'
So, F = mg - R
ma = mg - R
R = mg - ma
R = m(g-a)
case III: In freefall
a = g
R = m(g-g)
R = 0
case IV: If a > g
in this case
R < 0
Body will be stuck in b/w the ceiling of elevator.
We Know P = mv F = d (mv)/dt
= m (dv/dt)
F = ma
❈ SOME SPECIAL CASES:-
① If f = Constant
ma = constant
a ∝ 1/ mass
② If m = Const
F ∝ a
V = Constant
✤ UNIT OF FORCE :-
→ SI unit is kg m/s²
→ SI unit Newton in MKS system
→ CGS → dyne (gmc m/s²)
→ Gravitational unit : kg f
1 kg f = 9.8 Newton
✤ 3rd LAW :- It states that every action has equal & opposite reaction.
① Prove that second law is the real Eq of motion.
Sol: Case1 => Prog 1st law
We know by 2nd law
F = ma
F=m.dv/dt
If F = 0
m dv/dt = 0 ∴ [Which is the first law]
Here m≠0
Proof of 3rd law:
We known
F₁ = dp₁/dt
dp₁ = f₁·dt -①
dp₂ = f₂·dt
By conservation of Momentum
dp₁+dp₂ = 0
f₁·dt + f₂·dt = 0
dt (F₁ + F₂) = 0
F₁ + F₂ = 0
F₁ = -F₂
Which is 3rd law….
② Is it easier pull man push
IN PUSH
Let force f is applied on the roller then
Apparent Weight = W'=mg+Fsinθ
It shows that weight of the roller increases.
IN PULL
If force is applied is PULL the apparent
Weight ⇒ W' = mg - Fsinθ
i.e height become less. So it is easier pull than push.
③ How does horse pull the cart?
If horse moves forward, H > T
Net force: H-T=ma - ①
Cart moves forward, T > f
Net force T-f=ma - ②
Adding ① & ② H-f= (M+m) a
a=H-f/M+m
∴ H > F
Hence the cart moves forward….
✤ CONNECTING MOTION:-
For upward motion
F = T-m₁g
m₁a = T-m₁g -①
For Downward motion
F = m₂g - T
m₂a = m₂g - T -②
Adding ① & ②
m₁a+m₂a = m₂g - m₁g
a(m₁+m₂) = g (m₂-m₁)
a = g [m₂-m₁/m₂+m₁]
Now from ①
T = m₁a+m₁g
= m₁(a+g)
= m₁ [m₂-m₁/m₂+m₁]g + g
= m₁g [m₂-m₁+1/m₂+m₁]
= m₁g [m₂-m₁+m₂+m₁/m₂+m₁]
T = 2m₁m₂g/m₁+m₂
Acceleration force = m₂g
a = Force/Mass
a = m₂g/m₁+m₂
❈ LAMI'S THEOREM
In equilibrium state
F1/Sin α = F2 /sin ß = F3 / Siny
2)
① aco in the system = Force / m1 + m2 + m3
② T1 = (m2+mg)a
③ T3 = m3a
3) Let acc in the System = a
T1= (m1+m2)(g+a)
T2 = m1(g+a)
❈ FRICTIONAL FORCE
FRICTION
It is a self adjusting force which comes in account when external force act on the body.
Laws of Friction:
* Direction of frictional force is always opposite to the motion.
* Frictional force is always directly proportional to the normal reaction
F α R
F= μR
Here μ is called coefficient of friction
μ = F/R μ is unit less.
NOTES: It is depend upon the nature of surface (roughness or smoothness)
* Force of friction does not depend upon the area of contact b/w two bodies but it depend upon the molecular attraction b/w two surfaces.
TYPES OF FRICTION:-
STATIC FRICTION:- It is the friction which comes In account upon external force act on the body & body's in rest.
FL > PS > FK LIMITING FRICTION:- It is maximum static friction.
KINETIC FRICTION:- It is the friction when body is in motion.
(1) F=uR
② F=umg
③ we have by 2nd law
F=ma
ma = umg
a=ug OR u = a/ g
④
Net Accelerating Force → f' = f-f
✤ ANGLE OF FRICTION:-
It is the angle made by the resultant with normal reaction
tanθ = F/R
We have F/R = u
tanθ = u
✤ ANGLE OF REPOSE
It is the minimum angle of inclination also that any body can move without any external force.
:. mg cosx = R - (1)
mg sinx = F - (2)
tanx = F/R
Relation b/w θ & α
tanθ = F/R
tanα = F/R
∴tanθ = tanα
θ = α
✤ ACCELERATION IN A BODY IN AN INCLINED PLANE:-
Let a body of mass 'm' is placed on an inclined plane at θ
∴ mg cosθ = R - ①
& f = mg sinθ - F
ma = mg sinθ - μR
ma = mg sinθ - μmg cosθ
IN DOWNWARD MOTION:- a = g (sinθ - μcosθ)
IN UPWARD MOTION:-
In Balancing condition
R = mg cosθ - ①
F = mg sinθ + f
ma = mg sinθ + μR
ma = mg sinθ + μmg cosθ
a = g (sinθ + μcosθ)
✤ WORK DONE IN A BODY :-
In Upward motion :
We know, W = F x S
= m x g x S
W = mgs (sinθ + μcosθ)
In Downward Motion :
We know, N = mgs (sinθ - μcosθ)
✤ SPEED OF CAR ON A CIRCULAR TRACK TO TAKE SAFE TURN :
Let a car of mass 'm' is moving with velocity
v on a circular track of radius 'r' in this case
F = μR Centripetal force in this
F = μmg - (1) cases:-
F = mv²/r - (2)
To Take safe turn
mv²/r ≤ μmg
v² ≤ μrg
v ≤ √μrg
Maximum Velocity to take safe turn : V = √μrg
✤ BANKING ON ROAD :- It is defined as the phenomenon by which Outer edge of a circular track is made at some height as compare to inner edge to provide centripetal force to take safe turn.
__ Friction Acting on
Banking
Let a body of mass m is taking turn on the banked road with velocity in this case.
Rcosθ = mg + Fsinθ & Rsinθ + Fcosθ = mv² / r
Rcosθ - Fsinθ = mg Rsinθ + Rµcosθ = mv² / r
Rcosθ - uRsinθ = mg R sinθ + Rµcosθ = mv² / r
R (cosθ - u sinθ) = mg - ① R(sinθ + µcosθ) = mv² / r - ②
Dividing ② by ①
( Sinθ + u cosθ ) / ( cosθ - u sinθ ) = v² / rg
v² = rg ( Sinθ+ u cosθ )/ cosθ-u sinθ
v = √( rg (Sinθ + u cosθ ) /cosθ - u sinθ ) - Dividing eq by cosθ
v = √ rg ( tanθ + u )/ (1- u tanθ)
❈ let width of road = la
& height = h
OA = √b²-h²
:. tanθ = h / √b²-h²
❈ Bending Of Cyclist
Let a cyclist of mass m is taking turn on a circular track of radius r with velocity V in balancing
Condition
Rcosθ = mg - ①
Rsinθ = mv² / r - ②
Dividing ② by ①
tanθ = v² / rg
✤ MOTION IN A VERTICAL CIRCLE :-
① Velocity of the body at any point at a height h from the lowest point :-
v = √u²-2gh
② Tension in the string at any point ;
T = m / r ( u²-3gh+gr )
③ Tension at lowest point ;
Tʟ = m / r ( u²+gr )
④ Tension at the highest point ;
Tʜ = m / r ( u²-5gr )
⑤ Difference in tensions at the highest & lowest point
Tʟ - Tʜ = 6mg
⑥ Minimum Velocity at the lowest point for looping the vertical loop,
vʟ = √5gr
⑦ Velocity at the highest point for looping the loop
vʜ = √gr
* * * * * * * * * * * * * *
