Motion And Time 7 & 8 Flashcards
Acceleration due to gravity
9.8 m/s
Mechanics
Branch of physics which deals with objects in rest or motion
The branch - mechanics is divided into 3
Statics
Kinematics
Dynamics
Statics subdivision of mechanics -
Deals with the study of obj under rest
Kinematics
Deals with the study of obj under without considering the cause of motion
E.g. equations of motion
Dynamics
Deals with the study of obj under motion with considering the cause of motion
E.g. Newton’s laws of motion
Hooke’s law -
Force exerted by the spring
= spring constant(k) x maximum displacement of the object (x)
F = -(kx)
Elastic potential energy of a spring -
1/2 (k) (x) sq
K - spring constant
x - displacement of the object
SHM full form
Simple Harmonic motion
Kinetic energy of a spring
1/2 (m) (v) sq
SHM ?
Simple harmonic motion is any motion where a restoring force is applied that is proportional to the displacement and in the opposite direction of that displacement.
Plumb line ?
Masons use plumb lines to ensure Oda building is vertical.
1 st equation of motion ?
v = u + at
2nd equation of motion
s = ut + (1/2 .a. t sq)
3rd equation of motion
v sq - u sq = 2as
Distance travelled in n th second
u + a/2 (2n - 1)
2nd equation of motion (position - time relationship)
S = ut + 1/2 at²
Here u can find the distance if the time is given even if u don’t know the final velocity.
3rd equation of motion (position - velocity relationship)
2as = v² - u²
Here u can find the distance even if u don’t know the time interval as u know the final velocity.
Trajectory
Path followed by a projectile
Projectile Definition
If an object s given an initial velocity in any direction and then allowed travel freely under gravity , then the object is called projectile.
The motion of a projectile is called projectile motion.
If there is no air resistance and you drop a 10 kg ball and a feather from the same height , then which will reach the surface first
Both will land at the same time
If a body is moving with uniform acceleration, then , the distance it covers in regular intervals of time is in the ratio of
1:4:9
Actually the acceleration due to gravity decreases with rise in height from the Earth’s surface
T. So if u throw a ball from an aeroplane flying very high above , it may not move in uniform acceleration.
When u throw a body upwards , at its highest point (v=0) , the acceleration is
0
If you throw a ball upwards with velocity u , what time does it take for it to reach its highest point and then come back(a = 10)
If u = 40m/s , then after every second the velocity gets reduced by 10. So after 4 seconds the velocity becomes 0 (at its highest point).
So time taken to reach highest point = u/a = u/g
Now time taken for the ball to reach its initial velocity is going to be the same. (Think)
So total time taken = 2u/g
If you throw a ball upwards with velocity u , then the total distance travelled would be
So use the formula v² - u² = 2as
When the object reaches its highest point
v = 0
a = -g
s = h (height to which the body travelled)
Thus : -u² = -2gh
u²/2g = h
During constant acceleration the position velocity formula for denoting the distance covered =
D = ut + 1/2 at²
= t (u + 1/2 at)
= t (u + 1/2(v-u))
= t (u + 1/2 v - 1/2 u)
= t ( 1/2 v + 1/2 u)
= t [ 1/2(v + u)]
Distance covered = time taken x average velocity
U should remember this only holds true if the acceleration is constant.
How do you derive the distance velocity formula: 2as = v² - u² using a graph
We know the distance covered = area under a graph. Now if u calculate the area using the formula for a trapezium: 1/2 h (a+b)
U would get this formula.
How do u derive 2as = v² - u² without using graph ?
S = t x avg velocity ( a basic formula derived using 2nd formula of motion)
= (v-u)/a x 1/2 x (v+u)
= [(v-u)(v+u)] / 2a
2as = v² - u²
Magnitude of velocity in a projectile motion:
Given height and initial velocity
|v| = √(u²- 2gh)
Time taken by a projectile to complete its whole motion
T = 2usin θ / g
Change in momentum when the projectile completes its whole journey
Change in momentum (mg) = 2 x m x u sin θ
Range of a projectile?
Maximum horizontal distance travelled by it
Range of a projectile =
R = u²sin 2θ / g
2 x sin θ x cos θ =
Sin 2θ
Height at any point P
H = 1/2 x g x t₁ x t₂
t₁ - time taken to reach point P
t₂ - time taken complete journey from P to end point.
The graph of a trajectory can be expressed as a function of x
y = xtanθ - (gx²/ 2u²cos² θ)
The graph of a trajectory can be expressed as a function of x (in terms of range)
y = x tanθ (1 - x/r)
Angle of projection
Angle the given projectile make with the x axis when projected.
Maximum height of a projectile =
H(max) = u²sin²θ/2g
Change in momentum when the projectile completes its half journey
Change in momentum = m x usinθ
Path of a projectile when seen from another projectile is a ……. line
Straight
The function of the trajectory of a projectile has a …. shape
Parabola
At which angle is range of a projectile maximum
45°
The max height of the projectile when the range is max. (45°)
u²/4g
After completing ground to ground projection , the angle of projection is changed by
2 θ (angle of projection = θ)
Read (useful for questions)
If velocity is made n times the maximum height attained, then the range becomes n² times the initial value and the time ; n times the initial value.
Velocity at any time ‘t’ :
V = (ucosθ) i^ + (usinθ - gt) j^
Velocity at any time ‘t’ (horizontal projection)
|V| = √(u)² + (gt)²)
Time of flight (horizontal projection)
T = √(2h/g)
Displacement vector of a projectile
Disp = |r| = √[(ut)² + (1/2gt²)²]
Equation of trajectory in horizontal projection
y = 1/2 g (x/u)²
Average velocity = y = 1/2 g (x/u)²
Avg velocity = Displacement/time
Disp = |r| = √[(ut)² + (1/2gt²)²]
Time = √(2h/g)
To find the angle of a point in the trajectory
Tan α = gt/u
Quadratic formula
[−b ± √(b² − 4ac)] / 2a
If the vertical component of the initial velocity of 2 projectiles are same then :
Both the time taken to land and the height of the 2 projectiles would be same.
Height at any point p =
H = u sinθ x t₁ - 1/2 gt₁ ²
The time taken by a projectile to make a 90° change in its direction
u/g sin theta
