MECHANICS NOTE PHYSICS
Vectors
1. Vector addition :
A. Mechanics
A and B be vector quantities acting at an angle θ then
Magnitude of resulting (R) = A2 + 2AB cos+ B2
Bsin
α be the angle made by resultant with A then Tan α = A + BCos
Case 1 : If θ = 00 then R = A+B i.e. resultant will be maximum if two vectors are in same direction .
Case 2 : If θ = 180o then R = A-B i.e resultant will be minimum if two vectors are in opposite direction .
Case 3 : If θ = 90o then R = A2 + B2 and direction with A
Tan α = B/A
2. Subtraction of Vectors :
A and
B are two vector quantities acting at an angle then
subtraction of B from A then the result magnitude of resultant
( R) = A2 – 2AB cos + B2, α be the angle made by resultant with A then
Bsin
Tan α = A – B cos
Case 1 : If θ = 00 then R = A-B i.e. resultant will be minimum if two vectors are in same direction .
Case 2 : If θ = 180o then R = A+B i.e resultant will be maximum if two vectors are in opposite direction.
Case 3 : If θ = 90o then R = A2 + B2 and direction with A
Tan α =
3. Product of Vectors :
a. Scalar product or dot product : A and B be two vectors quantities then scalar product =
A. B = |A| |B|cos θ
i. If θ = 0o then , A. B = AB i.e. scalar product of two vectors become maximum .
ii. If θ = 90o then scalar A. B = 0 i.e. scalar product of two vectors will be zero .
iii. If θ = 180o then A. B = -AB i.e. scalar product of two vectors will be –ve.
b. Vectors product or cross product : A and B are two vectors quantities acting at an angle
θ then vectors product , A× B = |A| |B| Sinθn where n = unit of vector perpendicular
to plane containing A and B.
Here ( A × B) = -( ⃗⃗ )
4. Resolution of Vector : When a vector ⃗ is resolved in two component along x-axis & y-axis then
component along x-axis, Rx = R cosθ
i component along y-axis, Ry = R Sinθ
Motion in a straight line
1. The equations of motion for uniformly accelerating body are
a. v = u +at b. S = ut+ at2
|
b. v2 = u2+2as d. S th
a
= u + 2
( 2n-1)
These equations of motion under the action of gravity are written by replacing ‘a’ by ‘g’ and
‘s’ by ‘h’
2. Area under the velocity time curve and time axis gives the displacement of the object for given interval of time .
3. A body falling freely under the action of gravity then the distance covered in subsequent
seconds starting from rest will be in the ratio of 1:3:5:7 etc.
4. Velocity of body falling freely under the action of gravity starting from rest will be in the ratio of 1:2:3:4 etc .
5. If a body is thrown with velocity ‘u’ vertically upward then time to reach at maximum height
u
is g and time to return back to ground is
u
g so time of flight is
2u
g ,if air resistance is
neglected.
6. If air resistance is taken in account then time of ascent (ta) is less than time of descent (td)
7. For a particle having zero initial velocity if
S ∝tx where x > 2 then particle acceleration increases with time .
8. For a particle having zero initial velocity if
S ∝tx where x < 0 then particle acceleration decreases with time.
9. If a body cover half of distance with speed v1 and another half distance with velocity v2 then
Average speed = 2v1v2
v1 + v2
10. If a body travel half time with speed v1 and another half time with speed v2 then average v1 + v2
speed = 2
11. If a particle start from rest and accelerate with α for t1 and cover a distanced x1 then retard
with β for time t2 and comes to rest after covering distance x2 then
(t1 + t2)
vmax = 2x1 = 2 x2 =
2(x1 +x2)
+
= +
12. When a body is thrown vertically upward with velocity ‘u’ then maximum height reached is
u2
H = 2g
Motion in Plane
1. Projectile thrown horizontally from top of tower :- When a projectile is thrown
horizontally from top of tower of height ‘h’ with velocity ‘u’ then
a. Eqn of path , y = g 2 x2
2u
b. Time of flight , T =
c. Horizontal range , R =
2h g
2hu2
g
|
|
d. Velocity at any instant v = v 2
+ v 2
2. Projectile fired making an angle θ with horizontal : A projectile is fired at an angle θ with
horizontal from ground then
|
a. Eqn of path y = x tanθ - g x2
2u cos
2u sin
b. Time of flight T = g
u2sin2
c. Maximum height reached , H = 2g
|
|
u2sin2 u
d. Horizontal range, R =
g for maximum horizontal range Rmax = g if θ = 45
|
e. Velocity at any instant, v = v 2
f. H = , R = 4H cotθ = ,
+ vy2
3. The angle of projection for a projectile with given velocity to have same horizontal range are
θ and 900 –θ .
4. The maximum height is equal to n times the range then the projectile is launched at an angle
θ = tan-1 (4n)
5. If θ and 90o –θ are the two angle of projections for a projectile to have same horizontal range
then
H1 2
a. = tanθ b.
R1
H2 = tan θ
2R
b. R2 = 1 d. T1 . T2 = g
c. R = 4 H1H2 f. H1+H2 = u2/2g
6. If two bodies are thrown in opposite direction from top of tower with velocity u1 and u2 then
u1u2
a. Time after which velocity become perpendicular to each other is t = g
b. Time after which position vectors becomes perpendicular to each other is t = √
Circular motion
1. When a body is moving along the circumstance of circular path with uniform speed then velocity and acceleration has equal magnitude but they are changing due to change in direction.
3. ac= rω2 =
= ω.v = 4π2f2r =
42
T2 r
f = frequency , T = Time period
→ →
∴ ar =
× v
4. When body is moving with variable speed then acceleration is a = ac2 + a t2 at = tangential acceleration
5. A body is moving in a vertical circle with speed v by making an angle θ with vertical then
mv2
T = r + mg cos
a. If θ = 0o then Tmax =
mv2
r + mg , at lowest point of vertical circle .
|
mv2
b. If θ = 180o then Tmin =
2
-mg , at highest point of vertical circle
c. If θ = 90o then T = mv , at horizontal position . r
6. Minimum velocity of body in vertical circle.
a. The minimum velocity at lowest point, v = 5gr
b. The minimum velocity at highest point,
v = gr
c. The minimum velocity at horizontal point, v = 3gr
7. Motion in a conical pendulum : When a body is revolved in a circular path of radius r by
string of length ‘l’ by making an angle θ with vertical then
v2
a. tanθ = rg
b. Time period (T) = 2π
l cos
g
8. Maximum safe velocity of a cyclist for turning without inclination
v = rg , this is also velocity to avoid skidding while turning the vehicle .
9. Banking of roads : The angle banking of road is θ in which outer edge is raised at an angle θ
then
a. v = rg tan
b. tanθ = h/b, b = width of road
Laws of motion and fraction
1. Momentum (P) = mv
dp dp
dv dm
2. F = dt = dt (mv) = m dt + v dt
dv
a. If mass is constant then F = m dt = ma
4. Conservation of linear momentum : If no force act on the system of colliding body then
⃗
= = 0 or p = constant or mv = constant
5. Apparent weight in a lift : When a person is in a lift then a. For uniform motion or at rest
R = mg
b. For lift accelerating upward
R = m (g+a)
c. For lift accelerating downward
R = m (g-a)
d. For lift in free fall, a = g so
R = 0
6. When a body of mass ‘m’ is tied with a mass less string then tension on string is
a. for uniform motion or rest T = mg
b. For accelerating upward T = m(g+a)
c. For accelerating downward T = m (g-a)
7. When a chain or rope is held horizontally and pulled : A chain of length L, mass M is pulled by force F at one end then tension at x is T = ( )F
dm
8. Rocket Propulsion when
dt is rate of ejection of fuel with velocity v then
dm
a. Force on rocket in absence of gravity F =
dt v
b. Acceleration of rocket in absence of gravity a = F/M =
dm
c. Force on rocket in presence of gravity F =
dt v-Mg
d. acceleration of rocket in presence of gravity a = F/M = – g
9. When a jet of liquid of density ρ is escaping from pipe of cross sectional area ‘A’ with velocity v then Force (F) = ρAv2
Power (P) = Fv= ρAv3
10. When a block is pulled by a force F & just start to slide then
Limiting force of friction (F) = μR
For body moving with uniform velocity
μk = F/R
For body rolling with uniform speed
μr = F/R
11. Angle of friction (θf): angle between resultant of normal reaction and limiting force of friction with normal reaction .
∴tanθf = μ
12. Angle of repose (α ) Angle of inclined plane with horizontal at which body placed on it just
start to slide down the plane
∴ tanα = μ
13. When a block of mass m is pulled by force F inclined at an angle θ with horizontal then
Minimum force (F) = cos+ sin
14. Minimum horizontal force to be applied on block of mass m held against a vertical wall so as to prevent it from sliding down the wall is mgcotθ or .
15. A uniform chain L is hanging with ‘x’ down the edge then , μ =
x L –x
a. Fraction hanging x/L =
+ 1
16. When a block takes n time to slide down a rough inclined plane of inclination θ as it takes to
1
slide down a perfectly smooth inclined plane then coefficient of friction , μ = tanθ 1 –n2
17. When bottom of rough inclined plane with velocity v having inclination θ andblock reach
with velocity nv at bottom of smooth inclined plane of same inclination then coefficient of
1
friction μ = tanθ 1 –n2
18. When a block slide a distance x along rough inclined of inclination θ and slide nx along
1
smooth inclined plane of same inclination in same time then μ = tanθ 1 –n
Work, Energy, Power & Collision
→ →
1. Work done by force (W) = F . s = FS cosθ
2. Work done by variable force
W = ∫
w
3. Power (p) = t =
→ →
change in energy
time If a body is moving under the action of constant force
then P = F . v
4. Change in PE of body at a height ‘h’ close to surface of earth is
∆PE = mgh
|
1 p2
5. Kinetic energy (KE) = 2 mv
6. Work energy theorem ,
= 2m
Work done = change in energy
7. A light and heavy body have equal KE then
i.e. m1> m2 so P1>P2 Hence heavy body has greater linear momentum than lighter
one .
8. Coefficient of restitution ( e) =
v2–v1
u1–u2
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