CHAPTER 2 CLASS 9TH ENGLISH MEDIUM NOTES 2020
E-CLASS NCRT NEET PHYSICS IT NOTES CHAPTER 2
physics chapter 2 class 9 free pdf download
PHYSICS CHAPTER CLASS 9TH SOLVED EXERCISE PDF DOWNLOAD
(PREPARED BY )
- FAREED FIDA IQ LEARNING SCHOOL 18 HAZARI
- Muhammad.RASHEED THE EDUCATOR RACHNA CAMPUS 18 HAZARI BHAKKAR ROAD
 |
Muhammad Rasheed |
CH#2 KINEMATICS
Different CHOICE
Pick the right answer from the accompanying decisions:
I. A body has translatory movement on the off chance that it moves along a
straight line
circle
line without revolution
bent way
ii. The movement of a body about a pivot is called:
round movement
rotatory movement
vibratory movement
irregular movement
iii. Which of coming up next is a vector amount?
speed
separation
removal
power
iv. If a thing is moving with a consistent speed, by then, its partition time outline will be a straight
along with separation hub
corresponding to time-hub
slanted to time-hub
v.A straight line comparing to time-turn on a partition time graph tells that the article is:
very still
moving with variable speed
moving
vi. The speed-time chart of a vehicle is appeared in the figure, which of the accompanying explanation is valid?
the vehicle has an increased speed of 1.5 ms-2
the vehicle has a consistent speed of 7.5 ms-1
separation went by the vehicle is 75 m
the normal speed of the vehicle is 15 ms-1
vii. The speed-time chart of a vehicle is showed up in the figure, which of the going with declaration is substantial?
diagram a
diagram b
diagram c
diagram d
viii. By partitioning the uprooting of a moving body with time, we acquire:
speed
speeding up
speed
deceleration
ix. A ball is tossed vertically upward. Its speed at the most noteworthy point is:
- 10 ms-1
zero
10 ms-2
none of these
x. An adjustment in position is called:
speed
speed
removal
separation
xi. A train is moving at a speed of 36 kmh-1. Its speed communicated in ms-1 is:
10 ms-1
20 ms-1
25 ms-1
30 ms-1
4.Scalars and Vectors
a. The physical quantities which can be expressed completely by a magnitude only are called scalars e.g Mass, speed and pressure.
b.The physical quantities that can be expressed completely by magnitude and direction are called a vector. e.g force and weight.
Q#3: Define the terms speed velocity and acceleration?
Ans:
Speed
Distance covered in one second is called speed V= s/t. it is a scalar quantity.
Velocity:
Displacement covered in one second is called velocity V= d/t .it is a vector quantity
Acceleration:
Rate of change of velocity is called acceleration a= it is a vector quantity.
Q#4:Can a body moving at constant speed has acceleration?
Ans: Yes, a body moving in a circular track with a constant speed have acceleration called “Centripetal Acceleration” it is ac= V2/ r
Q#5: How does rider in a ferries wheel possess translatory motion but not rotatory motion?
Ans: Since rider in a Ferris wheel moves in a circular track without rotation so it posses translatory motion but not rotatory motion.
Q#6; Sketch a distance-time graph for a body starting from rest. give the speed of the body from the graph
Ans: This graph gives velocity d/t at each point. As graph A.
Q#7: What would be the shape of a speed-time graph of a
Body moving with variable speed?
Ans: If the body is moving with variable speed then speed time graph A
The graph will not be straight line it can be shown in figure.
Q#8: Which of the following can be obtained from Speed-time graph of a body?
Ans: the following things can be obtained from Speed-time graph of a body . 1. Initial speed
2. Final Speed
3. Distance covered in time (t)
4. Acceleration of motion.
We can find initial, final speeds acceleration & distance covered by a body by the velocity-time graph. graph for variable speed
Q# 10:How can vector quantities be represented graphically?
Ans: Graphically a vector quantity is represented by a straight line with an arrowhead at its one end. The length of the straight line represents magnitude and arrowhead represents its direction vector A is given as
Q#11: Why vector quantities cannot be added and subtracted like scalar quantities?
Ans: As Scalar can be added and subtracted by simple rules of mathematics while vectors are added and subtracted by a rule called head to tail rule. So vector quantities cannot be added and subtracted like scalar quantities
Q#12 How are vector quantities important in our daily life?
Ans: Force, weight, torque, displacement, velocity and acceleration are all vector quantities.
In our daily life, we deal with different kinds of forces velocities, acceleration and torques etc. For example, due to the force of gravity, we can walk due to torque we can open and close a door.
Q#13: Sketch a velocity-time graph for the motion of a body? From graph explain each step. Calculate the total distance covered?
Ans: Graph shows velocity is increasing uniformly the shaded area under the curved line describes the total distance covered.
Total distance covered = Area of triangle OAB.
2.1: A train moves with a uniform velocity of 36km/h for 10s.find distance?
Ans:
Data
V = 36km/h
V = 36 × 1000 m/60×60s
V = 10 m/s
Time t = 10s
As we know that
S = v × t
S = 10m/s× 10s
S = 100m
2.2: Train starts from rest.Moves through 1km in 100s with uniform acceleration what will be its speed at end of 100s?
solution:
Data:
Vi = 0
Distance S = 1km
S = 1000m
Time t = 100s
speed =?
As
S = Vit + ½ at2
1000 = 0 + ½ a (100)2
1000 = 5000 a
a = 1000/5000
a = 0.2 m/s2
Vf =Vi + at
Vf=0 + 0.2 × 100
= 20 m/s
2.3:A car has velocity 10m/s for half minutes.Find distance during
this time and final velocity of car?acceleration is 0.2ms-2 ?solution:
data:
Vi = 10 ms-1
Time t = 30 sec
a =0.2ms-2
distance S=?
as
Distance S= vi t + at2
S = 10(30) + (0.2) (30)2
S = 300 + 90
S = 390 meters
Vf = vi + at
Vf = 10 +0.2(30)
Vf = 10m/s+6 ms-1
Vf = 16m/s
2.4: a tennis ball is hit vertically upward with a velocity of 30 ms-1, .it takes 3 s to reach the highest point. Calculate the maximum height reached by the ball. give the time for this to reach a height
Solution:
Data:
Vi = 30 ms-1
Time t = 3 s
a =g = -10ms-2
Distance S=h =?
as
Distance S= vi t + at2
h = 30(3) +( -10) (3)2
h = 90 - 45
h = 45 m
Time for maximum height to surface of the earth
t2 =?
a = g
vi = 0
Vf=30ms-1
As t2 =
t2 =
t2 = 3 s
Total time to reach the surface of earth = 3s + 3s
= 6 s
2.5: IF the car is moving with a uniform velocity of 40meter per second for 5seconds. comes to rest in next
10s with uniform deceleration find?
1:deceleration 2:distance travelled
solution:
velocity v = 40 ms-1
t = 5 s
comes to rest in 10 seconds
t = 10 s
vf = 0
a= ?as a =
a =
a = -4 ms-2
distance travelled in the first 5 seconds
S1 = vi×t
S1 = 40×5 = 200m
Distance during deceleration -4 ms-2 for 10 seconds is
2a S2 =vf2 – vi2
2 (- 4)S2 = 0 – (40)2
-8 S2 = 0 - 1600
S2= 200 m
total Acceleration” it is ac= V2/ r
Q#5: How does a rider in a ferries wheel possess translatory motion but not rotatory motion?
Ans: Since rider in a Ferris wheel moves in a circular track without rotation so it posses translatory motion but not rotatory motion.
Q#6; Sketch a distance-time graph for a body starting from rest. calculate the speed of the body from the graph
Ans: This graph gives velocity d/t at each point. As graph A.
Q#7: What would be the shape of a speed-time graph of a
Body moving with variable speed?
Ans: if the body is moving with variable speed then speed time graph A
The graph will not be straight line it can be shown in the figure.
Q#8: Which of the following can be obtained from Speed-time graph of a body?
Ans:
the following things can be obtained from Speed-time graph of a body.
1. Initial speed
2. Final Speed
3. Distance covered in time (t)
4. Acceleration of motion.
We can find initial, final speeds acceleration & distance covered by a body by the velocity-time graph.
Q# 10: how can vector quantities be represented graphically?
Ans: Graphically a vector quantity is represented by a straight line with an arrowhead at its one end. The length of the straight line represents magnitude and arrowhead represents its direction vector A is given as
Q#11: Why vector quantities cannot be added and subtracted like scalar quantities?
Ans: As Scalar can be added and subtracted by simple rules of mathematics while vectors are added and subtracted by a rule called head to tail rule. So vector quantities cannot be added and subtracted like scalar quantities
Q#12 How are vector quantities important in our daily life?
Ans: Force, weight, torque, displacement, velocity and acceleration are all vector quantities.
In our daily life, we deal with different kinds of forces velocities, acceleration and torques etc. For example, due to the force of gravity, we can walk due to torque we can open and close a door.
Q#13: Sketch a velocity-time graph for the motion of a body? From graph explain each step. Calculate the total distance covered?
Ans:
The graph shows velocity is increasing uniformly shaded area under the curved line describes the total distance covered.
Total distance covered = Area of triangle OAB.
2.1: A train moves with a uniform velocity of 36km/h for 10s.find distance?
Ans:
Data
V = 36km/h
V = 36 × 1000 m/60×60s
V = 10 m/s
Time t = 10s
As we know that
S = v × t
S = 10m/s× 10s
S = 100m
2.2: Train starts from rest.Moves through 1km in 100s with uniform acceleration what will be its speed at end of 100s?
solution:
Data:
Vi = 0
Distance S = 1km
S = 1000m
Time t = 100s
speed =?
As
S = Vit + ½ at2
1000 = 0 + ½ a (100)2
1000 = 5000 a
a = 1000/5000
a = 0.2 m/s2
Vf =Vi + at
Vf=0 + 0.2 × 100
= 20 m/s
2.3: A car has velocity 10m/s for half minutes. Find distance during this time and final velocity of car? acceleration is 0.2ms-2?solution:data:
Vi = 10 ms-1
Time t = 30 sec
a =0.2ms-2
distance S=?
as
Distance S= vi t + at2
S = 10(30) + (0.2) (30)2
S = 300 + 90
S = 390 meters
Vf = vi + at
Vf = 10 +0.2(30)
Vf = 10m/s+6 ms-1
Vf = 16m/s
2.4: a tennis ball is hit vertically upward with a velocity of 30 ms-1 .it takes 3 s to reach the highest point. Calculate maximum height reached by the ball. calculate the time required for its maximum height?
Solution:
Data:
Vi = 30 ms-1
Time t = 3 s
a =g = -10ms-2
Distance S=h =?
as
Distance S= vi t + at2
h = 30(3) +( -10) (3)2
h = 90 - 45
h = 45 m
Time for maximum height to surface of the earth
t2 =?
a = g
vi = 0
Vf=30ms-1
As t2 =
t2 =
t2 = 3 s
Total time to reach the surface of earth = 3s + 3s
= 6 s
2.5: consider when a car moves with a uniform velocity of 40meter/s for 5second. comes to rest in next10s with uniform deceleration find?1:deceleration 2:distance travelled
solution:
velocity v = 40 ms-1
t = 5 s
comes to rest in 10 seconds
t = 10 s
vf = 0
a= ?as a =
a =
a = -4 ms-2
distance travelled in the first 5 seconds
S1 = vi×t
S1 = 40×5 = 200m
Distance during deceleration -4 ms-2 for 10 seconds is
2a S2 =vf2 – vi2
2 (- 4)S2 = 0 – (40)2
-8 S2 = 0 - 1600
S2= 200 m
Total distance covered S = S1 + S2
S = 200m + 200m
S = 400m
2.6: Train starts from rest within an acceleration of 0.5ms-2.Find its speed in
km/h when it has moved through a distance of 100m?
solution:
vi= 0
a = 0.5ms-2
S = 100m
Vf =?
2aS = vf2- vi2
2(0.5) 100 =vf2-0
100 = vf2
10ms-1= vf
Vf= 10×60×60 / 1000
Vf= 36 kmh-1
2.7: A train starting from rest acceleration uniformly and attains a velocity of 48km/h in 2 minutes. It travels at this speed at 5 minutes. Finally moves with uniform retardation and stopped after 3 minutes. Find the total distance covered?solution:
vi =0
t = 2min = 120s
Vf = 48kmh-1= 48×1000m/60×60s
Vf= 13.33ms-1
a =?
a = Vf – Vi / t
a = 13.33ms-1-0 / 120
a =0.111ms-2
S1 =vit +1 at2/2
S1 = 0 + 0.5 (0.111) (120)2
S1 = 800m
t2 =5×60=300m
S2 = vf ×t
S2 = 13.33× 300s
S2 = 4000m
T3 = 3min = 180s
Vf= 0
Retardation a=?
a = 0 -13.33 / 180
a = - 0.074ms-2
2aS= vf2 – vi2
2(-0.074)S3= 0(- 13.33)2
-0.148S3 = -177.69
S3 = 1200m
S =S1 + S2+ S3
S = 800m + 4000m +1200m
S = 6000m
2.8: A cricket ball hit vertically up word returns to ground 6s later calculate
1:Max height
2:Initial velocity
solution:
Max. Height h =?
Time to reach max.height t = 3 s
a =-g=- 10ms-2
vf=0
vi =?
Vf= vi + at
Vi =vf - gt
Vi = 0 – (-10)3
Vi= 30 ms-1
Maximum height
h = vit - gt2
h = 30(3) - (10) 32
h = 90 -45
h= 45 m
2.9: when brakes are applied, the speed of train decreases from 96 kmh-1to 48 kmh-1 in 800m.how much distance is covered before coming to rest?
Data:
Vi= 96 kmh-1
Vi=
Vi = 26.67ms-1
Vf = 48 kmh-1
Vf =
Vf = 13.33ms-1
S = 800m
2as = vf2-vi2
2a(800) = (13.33)2- (26.67)2
1600 a = 177.69 – 711.29
a =
a = -0.33ms-2
(b) if Vi = 13.33ms-1
Vf =0
S =?
Since 2aS = Vf2- Vi2
2(-0.33) S = - 177.69
-0.66 S = - 177.69
S = -177.69/ -0.66
S = 268m
2.10: in the above problem, find the time taken by the train to stop after the applications of brakes.
solution: Data;
After application of breaks
Initial speed Vi = 26.67ms-1
The final speed is Vf = 0
t =?
As a =
t =
t = 80 s
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