Motion And Time 7 & 8

Question 1

Acceleration due to gravity

Answer 1

9.8 m/s

Question 2

Mechanics

Answer 2

Branch of physics which deals with objects in rest or motion

Question 3

The branch - mechanics is divided into 3

Answer 3

Statics
Kinematics
Dynamics

Question 4

Statics subdivision of mechanics -

Answer 4

Deals with the study of obj under rest

Question 5

Kinematics

Answer 5

Deals with the study of obj under without considering the cause of motion
E.g. equations of motion

Question 6

Dynamics

Answer 6

Deals with the study of obj under motion with considering the cause of motion
E.g. Newton’s laws of motion

Question 7

Hooke’s law -

Answer 7

Force exerted by the spring
= spring constant(k) x maximum displacement of the object (x)

F = -(kx)

Question 8

Elastic potential energy of a spring -

Answer 8

1/2 (k) (x) sq

K - spring constant
x - displacement of the object

Question 9

SHM full form

Answer 9

Simple Harmonic motion

Question 10

Kinetic energy of a spring

Answer 10

1/2 (m) (v) sq

Question 11

SHM ?

Answer 11

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.

Question 12

Plumb line ?

Answer 12

Masons use plumb lines to ensure Oda building is vertical.

Question 13

1 st equation of motion ?

Answer 13

v = u + at

Question 14

2nd equation of motion

Answer 14

s = ut + (1/2 .a. t sq)

Question 15

3rd equation of motion

Answer 15

v sq - u sq = 2as

Question 16

Distance travelled in n th second

Answer 16

u + a/2 (2n - 1)

Question 17

2nd equation of motion (position - time relationship)

Answer 17

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.

Question 18

3rd equation of motion (position - velocity relationship)

Answer 18

2as = v² - u²

Here u can find the distance even if u don’t know the time interval as u know the final velocity.

Question 19

Trajectory

Answer 19

Path followed by a projectile

Question 20

Projectile Definition

Answer 20

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.

Question 21

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

Answer 21

Both will land at the same time

Question 22

If a body is moving with uniform acceleration, then , the distance it covers in regular intervals of time is in the ratio of

Answer 22

1:4:9

Question 23

Actually the acceleration due to gravity decreases with rise in height from the Earth’s surface

Answer 23

T. So if u throw a ball from an aeroplane flying very high above , it may not move in uniform acceleration.

Question 24

When u throw a body upwards , at its highest point (v=0) , the acceleration is

Answer 24

0

Question 25

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)

Answer 25

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

Question 26

If you throw a ball upwards with velocity u , then the total distance travelled would be

Answer 26

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

Question 27

During constant acceleration the position velocity formula for denoting the distance covered =

Answer 27

``` 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.

Question 28

How do you derive the distance velocity formula: 2as = v² - u² using a graph

Answer 28

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.

Question 29

How do u derive 2as = v² - u² without using graph ?

Answer 29

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

Question 30

Magnitude of velocity in a projectile motion: | Given height and initial velocity

Answer 30

|v| = √(u²- 2gh)

Question 31

Time taken by a projectile to complete its whole motion

Answer 31

T = 2usin θ / g

Question 32

Change in momentum when the projectile completes its whole journey

Answer 32

Change in momentum (mg) = 2 x m x u sin θ

Question 33

Range of a projectile?

Answer 33

Maximum horizontal distance travelled by it

Question 34

Range of a projectile =

Answer 34

R = u²sin 2θ / g

Question 35

2 x sin θ x cos θ =

Answer 35

Sin 2θ

Question 36

Height at any point P

Answer 36

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.

Question 37

The graph of a trajectory can be expressed as a function of x

Answer 37

y = xtanθ - (gx²/ 2u²cos² θ)

Question 38

The graph of a trajectory can be expressed as a function of x (in terms of range)

Answer 38

y = x tanθ (1 - x/r)

Question 39

Angle of projection

Answer 39

Angle the given projectile make with the x axis when projected.

Question 40

Maximum height of a projectile =

Answer 40

H(max) = u²sin²θ/2g

Question 41

Change in momentum when the projectile completes its half journey

Answer 41

Change in momentum = m x usinθ

Question 42

Path of a projectile when seen from another projectile is a ....... line

Answer 42

Straight

Question 43

The function of the trajectory of a projectile has a .... shape

Answer 43

Parabola

Question 44

At which angle is range of a projectile maximum

Answer 44

45°

Question 45

The max height of the projectile when the range is max. (45°)

Answer 45

u²/4g

Question 46

After completing ground to ground projection , the angle of projection is changed by

Answer 46

2 θ (angle of projection = θ)

Question 47

Read (useful for questions)

Answer 47

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.

Question 48

Velocity at any time ‘t’ :

Answer 48

V = (ucosθ) i^ + (usinθ - gt) j^

Question 49

Velocity at any time ‘t’ (horizontal projection)

Answer 49

|V| = √(u)² + (gt)²)

Question 50

Time of flight (horizontal projection)

Answer 50

T = √(2h/g)

Question 51

Displacement vector of a projectile

Answer 51

Disp = |r| = √[(ut)² + (1/2gt²)²]

Question 52

Equation of trajectory in horizontal projection

Answer 52

y = 1/2 g (x/u)²

Question 53

Average velocity = y = 1/2 g (x/u)²

Answer 53

Avg velocity = Displacement/time Disp = |r| = √[(ut)² + (1/2gt²)²] Time = √(2h/g)

Question 54

To find the angle of a point in the trajectory

Answer 54

Tan α = gt/u

Question 55

Quadratic formula

Answer 55

[−b ± √(b² − 4ac)] / 2a

Question 56

If the vertical component of the initial velocity of 2 projectiles are same then :

Answer 56

Both the time taken to land and the height of the 2 projectiles would be same.

Question 57

Height at any point p =

Answer 57

H = u sinθ x t₁ - 1/2 gt₁ ²

Question 58

The time taken by a projectile to make a 90° change in its direction

Answer 58

u/g sin theta