Maharashtra State Board Class 9 Science Solutions Chapter 1 Laws of Motion
Class 9 Science Chapter 1 Laws of Motion Textbook Questions and Answers
1. Match the first column with appropriate entries in the second and third columns and remake the table.
2. Clarify the differences
A. Distance and displacement
Answer:
Distance | Displacement |
(i) Distance is the length of the actual path travelled by an object. | (i) Displacement is the minimum distance between the starting and finishing points. |
B. Uniform and non-uniform motion.
Answer:
Uniform motion | Non-uniform motion |
(i) If an object covers equal distances in equal intervals of time it is said to be in uniform motion. | (i) If an object moves unequal distances in equal intervals of time, its motion is said to be nonuniform. |
3. Complete the following table.
4. Complete the sentences and explain them.
a. The minimum distance between the start and finish points of the motion of an object is called the ……….. of the object.
b. Deceleration is ………………………. acceleration
c. When an object is in uniform circular motion, its ………………………. changes at every point.
d. During collision ………………………. remains constant.
e. The working of a rocket depends on Newton’s ………………………. law of motion.
5. Give scientific reasons.
a. When an object falls freely to the ground, its acceleration is uniform.
Answer:
b. Even though the magnitudes of action force and reaction force are equal and their directions are opposite, their effects do not get cancelled.
Answer:
c. It is easier to stop a tennis ball as compared to a cricket ball, when both are traveling with the same velocity.
Answer:
d. The velocity of an object at rest is considered to be uniform.
Answer:
6. Take 5 examples from your surroundings and give an explanation based on Newton’s laws of motion.
7. Solve the following examples.
a) An object moves 18 m in the first 3 s, 22 m in the next 3 s and 14 m in the last 3 s. What is its average speed? (Ans: 6 m/s)
Answer:
Given:
Total distance (d) = 18 + 22 + 14 = 54 m
Total time taken (t) = 3 + 3 + 3 = 9 sec
To find:
Average speed = ?
The object moves with an average speed of 6 m/s.
b) An object of mass 16 kg is moving with an acceleration of 3 m/s2. Calculate the applied force. If the same force is applied on an object of mass 24 kg, how much will be the acceleration? (Ans: 48 N, 2 m/s2)
Answer:
The force acting on the 1 body is 48 N and the acceleration of the 2” body is 2 m/s2
c) A bullet having a mass of 10 g and moving with a speed of 1.5 m/s, penetrates a thick wooden plank of mass 90 g. The plank was initially at rest. The bullet gets embedded in the plank and both move together. Determine their velocity. (Ans: 0.15 m/s)
Answer:
The plank embedded with the bullet moves with a velocity of 0.15 m/s.
d) A person swims 100 m in the first 40 s, 80 m in the next 40 s and 45 m in the last 20 s. What is the average speed? (Ans: 2.25 m/s2)
Answer:
Given:
Total distance (d) = 100 + 80 + 45 = 225 m
Total time taken (t) = 40 + 40 + 20 = 100 sec
To find:
Average speed =?
The person swims with an average speed of 2.25 m/s.
Class 9 Science Chapter 1 Laws of Motion Intext Questions and Answers
(i) Who will take less time to reach the school and why?
Answer:
Prashant will take less time as the path followed by him is the shortest.
(a) Every morning, Swaralee walks round the edge of a circular field having a radius of 100 m. As shown in figure (a), if she starts from the point A and takes one round, how much distance has she walked and what is her displacement?
Answer:
Radius (r) = 100 m
Distance covered = Circumference of the circle
= 2 nr
= 2 x 3.14 x 100
= 628 m
Displacement = 0 m (Shortest distance between initial and final position is zero)
(b) If a car, starting from point P, goes to point Q (see figure 1.9) and then returns to point P, how much distance has it travelled and what is its displacement?
Answer:
Distance covered = PQ + QP
= 360 + 360
= 720 m
Displacement = 0 m (The shortest distance between initial and final position is zero)
Class 9 Science Chapter 1 Laws of Motion Additional Important Questions and Answers
(A) Choose and write the correct option:
Laws Of Motion Class 9 Questions And Answers Maharashtra Board Question 1.
The displacement that occurs in unit time is called ……………...
(a) displacement
(b) distance
(c) velocity
(d) acceleration
Answer:
(c) velocity
Laws Of Motion Class 9 Maharashtra Board Exercise Answers Question 2
The unit of velocity in the SI system is ……………...
(a) cm/s
(b) rn/s2
(c) um/s2
(d) rn/s
Answer:
(d) m/s
Laws Of Motion Class 9 Maharashtra Board Question 3.
v2= u2+ 2as is the relation between and ……………...
(a) speed and velocity
(b) distance and acceleration
(c) displacement and velocity
(d) speed and distance
Answer:
(c) displacement and velocity
Class 9 Science Notes Chapter 1 Laws Of Motion Question 4.
…………….. is the relation between displacement and time.
(a) v = u + at
(b) v2= u2+ 2as
(c) s = ut + 1/2 at2
(d) v = u + 2as
Answer:
(c) s = ut + 1/2 at2
Class 9 Science Chapter 1 Laws Of Motion Question Answer Question 5.
The force necessary to cause an acceleration of 1 m/s2in an object of mass 1 kg is called ……………...
(a) 1 dyne
(b) 1 m/s
(c) 1 Newton
(d) 1 cm/s
Answer:
(c) 1 Newton.
9th Science Chapter 1 Laws Of Motion Question 6.
Even if the displacement of an object is zero, the actual distance traversed by it ……………...
(a) may not be zero.
(b) will be zero
(c) will be constant
(d) will be infinity
Answer:
(a) may not be zero
(B) 1. Find the odd man out:
(B) 2. Find out the correlation
(B) 3. Distinguish between:
Positive acceleration | Negative acceleration |
(i) When the velocity of a body increases, acceleration is said to be positive acceleration. | (i) When the velocity of a body decreases, acceleration is said to be negative acceleration. |
Scalar quantity | Vector quantity |
(i) Scalar quantities are physical quantities having magnitude only. | (i) Vector quantities are physical quantities having both magnitude and direction. |
Balanced force | Unbalanced force |
(i) Balanced force keeps the body at rest. | (i) Balanced force keeps the body at rest. |
(B) 4. State whether the following statements are true or false:
(B) 5. Name the following:
(B) 6. Answer the following in one sentence:
Give formula:
Give scientific reasons:
Solve the following numerical:
The initial velocity of the kangaroo must be 7 m/s.
The acceleration is 3 m/s2and distance travelled is 37.5 m.
The momentum of cannon is 125 kg m/s
Answer the following in short:
Complete the flow chart:
(1) Types of force and their effects
Answer:
(2) Newton’s laws
Answer:
Distinguish between:
Speed | Velocity |
(i) Speed is the distance covered by a body in unit time. | (i) The displacement that occurs in unit time is called velocity. |
Balanced force | Unbalanced force |
(i) Two equal forces applied on a body in the opposite direction. | (i) Two unequal forces applied on a body. |
Give examples:
Answer the following questions:
Observe the figure and answer the questions:
(a) Measure the distance between points A and B in different ways as shown in figure (I).
Answer:
Distances measured may be of different lengths depending on the path taken.
(b) Now measure the distance along the dotted line. Which distance is correct according to you and why?
Answer:
Dotted line shows the shortest way of reaching from A to B.
(c) Observe the following figures. If you increase the number of sides of the polygon and make it infinite, how many times will you have to change the direction? What will be the shape of the path?
Answer:
If we increase the number of sides of the polygon and make it infinite, then we will have to change the direction an infinite number of times. The shape of the path thus obtained will be a circle.
Observe the figure and answer the questions
Numerical:
Total time taken (t)
= 4 hours
= 4 x 3600 (v lhr = 3600 sec)
= 14400 s
To find:
Average speed = ?
The person travels with average speed of 5 m/s
Write laws and explain write implications:
Suppose an object of mass ‘m’ has an initial velocity ‘u. When a force ‘F’ is applied in direction of its velocity for time ‘t’, its velocity becomes ‘y’. Then, the total initial momentum of the body = ‘mu’. Its final momentum after time t = ‘mv’.
So, the rate of change of momentum
Hence by Newtons second la of motion, 4he rate of change of momentum is proportional to the applied force.
∴ ma ∝ F
∴ F ∝ ma
∴ F ∝ kma (k = Constant of proportionaLity and value is 1).
∴ F = ma
Let mass of object A and B be m1and m1respectively
Let their initial velocity be u1and u2Let their final velocity be v1and v2
We know,
P = mv
Let their initial momentum be m1u1and m2u2
Let their final momentum be m1v1and m2v2
Total initial momentum = (m1u1+ m2u2)
Total final momentum = (m1v1+ m2v2)
If F2is the force that acts on object B,
i.e. The magnitude of total of total final momentum = the magnitude of total initial momentum
Complete the paragraph:
The force necessary to cause a change in the momentum of an object depends upon the rate of change of momentum. Every action force has an equal and opposite reaction force which acts simultaneously. As the mass of the gun is much higher than the mass of the bullet, the velocity of the gun is much smaller than the velocity of the bullet. The magnitude of the momentum of the bullet and that of the gun are equal and their directions are opposite. Thus, the total momentum is constant. Total momentum is also constant during the launch of a rocket.
Answer the following in detail:
The speed of a body is the distance travelled in unit time. The units of speed in CGS system is cm/s and in SI system is m/s.
There are two types of speed :
The speed of the body at any instant is called instantaneous speed. Average speed is the ratio of total distance covered to total time taken.
The velocity of a body is the distance travelled by a body in a particular direction in unit time. Thus, rate of change of displacement is called velocity.
v = s/t
where: s = displacement; t = time taken; v = velocity
(MKS unit: m/s CGS unit: cm/s)
There are two types of velocities :
(ii) Types of acceleration:.
(a) Uniform acceleration : If the change in velocity is equal in equal intervals of time, the acceleration is uniform acceleration.
(b) Non-uniform acceleration : If the change in velocity is unequal in equal intervals of time, the acceleration is a non-uniform acceleration.
(iii) Kinds of acceleration:
Positive acceleration : When the velocity of an object goes on increasing, it is said to have Positive acceleration.
Negative acceleration : When the velocity of an object goes on decreasing, it is said to have negative acceleration or retardation or deceleration.
Zero acceleration : If the velocity of the object does not change with time, it has zero acceleration.
Suppose an object of mass’m’ has an initial velocity ‘u’. When a force ‘F’ is applied in the direction of its velocity for time’t’, its velocity becomes ‘v’. Then, the total initial momentum of the body = ‘mu’. Its final momentum after time t = ‘mv’.
So, the rate of change of momentum
Hence by Newton’s second law of motion, the rate of change of momentum is proportional to the applied force.
∴ ma ∝ F
∴ F ∝ ma
∴ F = kma (k = Constant of proportionality and value is 1).
∴ F = ma
(ii) We know,
P = mv
Let their initial momentum be m1u1and m2u2
Let their final momentum be m1v1and m2v2
(iii) Total initial momentum = (m1u1+ m2u2)
Total final momentum = (m1v1+ m2v2)
(iv) If F2is the force that acts on object B,
i.e. The magnitude of total final momentum = the magnitude of total initial momentum.
Question 6
Maharashtra Board Solution
Obtain the equations of motion by graphical method: |
(a) Equation for velocity-time relation.