Board of Intermediate and Secondary Education Rawalpindi Class 9th Physics Notes Chapter 2 Kinematics Online read Pdf Download questions, Problems, and Mcqs.
Q.2) Explain translatory motion and give examples of various types of translatory motion.
Answer:
Translatory motion:
In translational motion, a body moves along a line without any rotation. The line may be straight or curved.
Examples:
A man running, a bus moving, a dog walking, a ship sailing, and dropping an object from height.
Types of Translatory motion:
There are three types of translational motion: linear motion, circular motion, and random motion.
Linear motion:
Linear motion is a motion along a straight line.
Examples: Motion of motors on a straight road, march past of soldiers in a parad and falling of apples from a tree.
Circular motion:
The motion of an object in a circular path.
Examples: Motion of moon around earth in a circular orbit. If you tie a stone with string and start moving it around yourself, the stone is said to be in a circular motion.
Random motion:
Random Motion is the motion of an object in a disordered or irregular manner/path.
Examples: Motion of dust particles in air and motion of molecules of gas and liquids.
Q.3) differentiate between the following:
i) Rest and motion.
ii) Circular motion and rotatory motion.
iii) Distance and displacement
iv) Speed and velocity
v) Linear and random motion.
vi) Scalars and vectors
Answer:
i) DIFFERENCE BETWEEN REST AND MOTION
REST
A body is said to be at rest if it does not change its position with respect to its surroundings.
MOTION
A body is said to be in motion if it changes its position with respect to its surroundings.
The state of rest and motion both are relative. For example, a passenger sitting in a moving bus is at rest with respect to the other passengers, because the passenger is not changing position with respect to other passengers and objects in the bus. But the passenger is in motion with respect to the people outside the bus.
ii) DIFFERENCE BETWEEN CIRCULAR MOTION AND ROTATORY MOTION
CIRCULAR MOTION
Circular motion means a body is moving in an orbit and always has a starting point to which it will eventually return.
Examples: Motion of moon around earth in a circular orbit. If you tie a stone with string and start moving it around yourself, the stone is said to be in a circular motion.
ROTATORY MOTION
Rotational motion means the body is turning around itself/its axis.
Examples: Spin top and spinning of earth.
iii) DIFFERENCE BETWEEN DISTANCE AND DISPLACEMENT
DISTANCE
Distance is the length of a path between two points. It is a scalar quantity
DISPLACEMENT
Displacement is a vector quantity and it can be defined as the shortest distance between the initial point and final point of an object. It must be the shortest interval connecting the initial and final points that is a straight line.
iv) DIFFERENCE BETWEEN SPEED AND VELOCITY
SPEED
Speed is defined as “the rate of change of distance with respect to time”
Speed is a scalar quantity that refers to “how fast an object is moving.”
VELOCITY
Velocity is defined as “the rate of change of displacement with respect to time”
Velocity is a vector quantity that refers to “the rate at which an object changes its position.”
v) DIFFERENCE BETWEEN LINEAR MOTION AND RANDOM MOTION
LINEAR MOTION
Linear Motion is the motion of an object along a straight line.
Examples: Objects falling vertically downward and airplanes flying straight in air.
RANDOM MOTION
Random Motion is the motion of an object along a disordered or irregular line.
Examples: Motion of dust particles in air and motion of molecules of gas and liquids.
vi) DIFFERENCE BETWEEN SCALAR AND VECTOR QUANTITIES
SCALAR QUANTITIES
Scalars are quantities that are fully described by a magnitude (numerical value) alone.
Examples: Distance, speed, mass, time and temperature are some scalar physical quantities.
VECTOR QUANTITIES
Vectors are quantities that are fully described by both a magnitude and a direction.
Examples: Force, displacement, velocity, and torque are some vectors physical quantities
Q.4) Define the terms speed, velocity, and acceleration.
Speed is defined as “the rate of change of distance with respect to time”
Speed is a scalar quantity that refers to “how fast an object is moving.” Its S.I unit is ms-1
Velocity:
Velocity is defined as “the rate of change of displacement with respect to time”
It is a vector quantity that refers to “the rate at which an object changes its position”. Its S.I unit is ms-1.
Acceleration:
Acceleration is defined as the rate of change of velocity.
It is a vector quantity and its S.I unit is ms-2.
Q.5) Can a body moving at a constant speed have acceleration ?
Answer:
Yes, when a body is moving with a constant speed, the body can have acceleration if its direction changes because acceleration is a vector quantity.
For example, if the body is moving along a circle with constant speed, it will have accelerated due to change of direction at every instant.
Q.6) How do riders in a Ferris wheel possess translatory motion but not rotatory motion ?
Answer:
Riders in the ferris wheel possess translatory motion because their motion is in a circle without rotation.
Q.7) Sketch a distance-time graph for a body starting from rest. How will you determine the speed of a body from this graph ?
Answer:
Q.8) What would be the shape of a speed – time graph of a body moving with variable speed ?
Answer:
When an object does not cover equal distance in equal interval of time, then its speed is not constant and known as variable speed. In this case, the distance time graph is not a straight line.
Q.9) Which of the following can be obtained from the speed – time graph of a body?
i) Initial speed.
ii) Final speed.
iii) Distance covered in time t.
iv) Acceleration of motion.
Answer:
In the above speed-time graph, vi and vf show the initial and final speed of the body. Total distance covered by the body is equal to the area under the curve and slope of the line AB gives the acceleration of the body.
Q.10) How can vector quantities be represented graphically ?
Answer:
Graphically, a vector quantity is represented by a straight line with an arrow head on its one end. The length of the straight line represents the magnitude of the vector and the arrowhead shows the direction of the vector.
Q.11) Why vector quantities cannot be added and subtracted like scalar quantities ?
Answer:
Scalars have only magnitude and can be added or subtracted arithmetically. Vector quantities have both direction and magnitude and cannot be processed like scalars. Vectors can be added and subtracted graphically using the ‘head to tail rule’. The resultant line vector gives both magnitude and direction. They can also be processed through vector analysis to find their magnitude and direction.
Q.12) How are vector quantities important to us in our daily life ?
Answer:
Vector quantities are important in our daily life as they describe physical processes in the real world and without understanding them, we cannot perform many tasks efficiently. In the picture above, the knowledge of vectors can guide us to what would be the optimum angle to pull the cart. If the angle with the x-axis is 60/70 degrees, most of the effort would be wasted because maximum force would be acting along the y-axis. Most of the applied force would be along the x-axis if the angle is small. That’s why we have to kneel down to pull a really heavy load to reduce the angle of applied force with the x-axis.
We can make use of the properties of vectors while opening and closing heavy doors, lids, lifting or lowering weights, etc
Q.13) Derive equations of motion for uniformly accelerated rectilinear motion.
Answer:
FIRST EQUATION OF MOTION
Suppose a body is moving with an initial velocity vi and its velocity becomes vf after time t. Clearly,the change in velocity is vf – vi, so acceleration is defined as
SECOND EQUATION OF MOTION
Suppose a body is moving with the initial velocity vi and after a certain time t its velocity becomes vf then the total distance covered by a body in time t is given by
From the first equation of motion v f = vi + at , putting the values of v f in the above equation.
THIRD EQUATION OF MOTION
Suppose a body is moving with initial velocity vi, and after a certain time t its velocity becomes vf, then the distance covered by it is given by
From first equation of motion find the value of t and substituting it in eq (i)
By using formula (a+b)(a-b)=a2-b2
Q.14) Sketch a velocity – time graph for the motion of the body. From the graph explaining each step, calculate the total distance covered by the body.
Answer:
The distance covered by an object can also be determined by using its velocity-time graph.
a) If the object moves at constant velocity v for time t. The distance covered by the object is v × t. This distance can also be found by calculating the area under the velocity-time graph. This area is shaded and equal to v × t.
b) If the velocity of the object increases uniformly from 0 to v in time t. The magnitude of its average velocity is given by
Lesson 3 – Problem
PROBLEMS
Q.1) A train moves with a uniform velocity of 36 km h-1for 10 s. Find the distance traveled by it.
Answer:
Q.2) A train starts from rest. It moves through 1 km in 100 s with uniform acceleration. What will be its speed at the end of the 100s?
Answer:
Q.3) A car has a velocity of 10 ms-1. It accelerates at 0.2 ms-2 for half a minute. Find the distance traveled during this time and the final velocity of the car.
Answer:
Q.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. How long will it take to return to the ground ?
Answer:
Initial velocity; vi = 30 ms-1
Time to reach maximum height; t = 3 s
Acceleration due to gravity; a = g = -10 ms-2
Final velocity; vf = 0 ms-1
i) Maximum height attained by the ball; h =
ii) Time taken to return to ground; t =?
ii)
Total time = Time to reach maximum height + time to return to the ground
t = 3+3
t = 6 s
Q.5) A car moves with uniform velocity of 40 ms-1 for 5 s. It comes to rest in the next 10 s with uniform deceleration. Find
i) Deceleration
ii) total distance traveled by the car.
Answer:
Q.6) A train starts from rest with an acceleration of 0.5 ms-2. Find its speed in km h-1 when it has moved through 100 m.
Answer:
Initial velocity; vi = 0 ms-1
Acceleration; a = 0.5 ms-2
Distance; S = 100 m
Final velocity; vf =?
Using 3rd equation of motion
2as = vf2 – vi2
2 × 0.5 × 100 = vf2 – (0)2
100 = vf2
⇒ vf2 = 100
Taking under-root on both sides, we get
vf = 10 ms-1
Required speed is in kmh-1 unit, so we have to convert ms-1 to kmh-1
Final speed in km h-1
vf = 36 km h-1
Q.7) A train starting from rest, accelerates uniformly and attains a velocity of 48 km h -1 in 2 minutes. It travels at this speed for 5 minutes. Finally, it moves with uniform retardation and is stopped after 3 minutes to find the total distance traveled by train.
Answer:
Q.8) A cricket ball is hit vertically upwards and returns to the ground 6s later. Calculate
i) the initial velocity of the ball.
ii) maximum height reached by the ball
Answer:
Q.9) When brakes are applied, the Speed of a train decreases from 96 km h-1 to 48 km h-1 in 800 m. How much further will the train move before coming to rest? (Assuming the retardation to be constant).
Answer:
Q.10) In the above problem, find the time taken by the train to stop after the application of brakes.
Answer:
