MSBSHSE Class 9 Science Chapter 1 Laws of Motion Solutions are a very crucial study material from the viewpoint of your MSBSHSE Class 9 Science examination. The detailed answers with step by step solutions to the questions provided here will help the students to understand the basic concepts of the chapter.
Laws of motion are a significant concept of Science and it is discussed in detail in this chapter. Students can learn more about the subject by referring to these MSBSHSE Class 9 Solutions of Science Chapter 1 Laws of Motion. These solutions prepared after proper research help students to get a thorough conceptual understanding. The content is so well-structured that it becomes easier for students to learn and comprehend. The content is also as per the updated MSBSHSE Syllabus for Class 9. The solutions provided here are also designed to meet the multiple criteria to ace the exams. Hence, solving these questions with the highly relevant answers provide help to the students to master the subject for the academic year and set the foundation for higher classes.
Maharashtra Board Class 9 Science Chapter 1- BYJU’S Important Questions & Answers
1. Match the first column with correct entries in the second and third columns and make the table again:
| S. No. | COLUMN 1 | COLUMN 2 | COLUMN 3 |
| 1 | Negative Acceleration | The velocity of the object remains constant | A car that was initially at rest reaches a velocity of 50 km/hr in 10 seconds |
| 2 | Positive Acceleration | The velocity of the object reduces | A vehicle is moving with a velocity of 25 m/s |
| 3 | Zero Acceleration | The velocity of the object increases | A vehicle moving with the velocity of 10 m/s, stops after 5 seconds |
Answer:
| S. No. | COLUMN 1 | COLUMN 2 | COLUMN 3 |
| 1 | Negative Acceleration | The velocity of the object reduces | A vehicle moving with the velocity of 10 m/s, stops after 5 seconds |
| 2 | Positive Acceleration | The velocity of the object increases | A car that was initially at rest reaches a velocity of 50 km/hr in 10 seconds |
| 3 | Zero Acceleration | The velocity of the object remains constant | A vehicle is moving with a velocity of 25 m/s |
2. Clarify the differences between the given terms:
A. Distance and displacement
B. Uniform and non-uniform motion.
Answer: A. The actual length of the path travelled by a moving object while going from one point to another is called distance, while displacement is the minimum distance from the starting to the finishing points. Distance is a scalar quantity, while displacement is a vector quantity. Meanwhile, the value of distances is always positive, while that of displacement can be negative, positive or zero.
B. An object is said to be moving with non-uniform speed if an object covers unequal distances in equal time intervals. For example, the motion of a vehicle being driven through heavy traffic. Alternatively, an object covering equal distances in equal time intervals are said to be moving with uniform speed. The time distance graph obtained for uniform speed is a straight line, while the graph of non-uniform speed can take any shape on the basis of how acceleration alters with time.
3. Complete the following table:
| u (m/s) | a (m/s2) | t (sec) | v = u + at (m/s) |
| 2 | 4 | 3 | __ |
| ___ | 5 | 2 | 20 |
| u (m/s) | a (m/s2) | t (sec) | s = ut + ½ at2 (m) |
| 5 | 12 | 3 | __ |
| 7 | ___ | 4 | 92 |
| u (m/s) | a (m/s2) | t (sec) | v2= u2 + 2as (m/s)2 |
| 4 | 3 | ___ | 8 |
| __ | 5 | 8.4 | 10 |
Answer:
First, u=2, a=4 and t=3
If v=u+at
v=2+(4 × 3)
v= 2+ 12 =14
Second,
u=?, a=5 and t=2
If v=u+at = 20
v= u + at
20= u+ (5×2)
20= u +10
Hence, u =20-10 = 10
| u (m/s) | a (m/s2) | t (sec) | v = u + at (m/s) |
| 2 m/s | 4 m/s2 | 3 sec | 14 m/s |
| 10 m/s | 5 m/s2 | 2 sec | 20 m/s |
According to the second table, values given are u=5, a=12, t=3
If S=ut + ½ at2
S= (5×3) + ½ × 12× (3)2
S= 15 + ½ × 12 × 9 = 15 + ½ × 108
S = 15 + 54 = 69m
| u (m/s) | a (m/s2) | t (sec) | s = ut + ½ at2 (m) |
| 5 m/s | 12 m/s2 | 3 sec | 69 m |
| 7 m/s | 8 m/s2 | 4 sec | 92 m |
| u (m/s) | a (m/s2) | t (sec) | v2= u2 + 2as (m/s)2 |
| 4 m/s | 3 m/s2 | 8 sec | 8 (m/s)2 |
| 4 m/s | 5 m/s2 | 8.4 sec | 10 (m/s)2 |
4. When an object falls freely to the ground, its acceleration is uniform. Give scientific reason.
Answer: When an object falls freely its acceleration remains uniform since in a free fall the acceleration does not depend on the mass of the body. It is under the effect of a constant force of gravity and there are no other forces acting on it. So, based on Newton’s second law of motion, it is seen that this constant force of gravity helps to accelerate the freely falling object uniformly. This uniform acceleration is known as acceleration due to gravity that acts towards the centre and is denoted by g. Uniform acceleration is generated as a result of gravitational force and here the air resistance is considered negligible. The object also stays near enough to the earth’s surface so that the gravitational field is constant.
5. Even though the magnitude of action force and reaction force are equal and their directions are opposite, their effects do not get cancelled. State the reason.
Answer: When two bodies interact, both the action and reaction forces are taken into consideration. However, despite their magnitudes being equal and their direction opposite, their effects are not cancelled as these action and reaction forces do not act on the same body.
6. It is easier to stop a tennis ball as compared to a cricket ball, when both are travelling with the same velocity. Why?
Answer: The momentum of a body is given as the product of its mass and velocity. Hence, p=mv. Also, the rate of change of momentum of a body is equal to the force applied, see, F=[ m(v-u)] /t. Both the balls have the same velocity and finally stop, when a force directly proportional to the mass of the ball is applied. That is, F∝m. Therefore, since the mass of a cricket ball is more than the tennis ball, its momentum will be greater than the tennis ball. For this reason, a greater force will be required to stop the cricket ball as compared to that needed to stop a tennis ball.
7. The velocity of an object at rest is considered to be uniform. Give a scientific explanation.
Answer: When the speed and direction of an object remains constant, then the velocity of the object is said to be uniform. The object at rest with a uniform speed of zero at all times does not even change the direction. Therefore, the velocity of an object at rest is uniform.
8. 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?
Answer: If the distance travelled and time is given the average speed is calculated using the formula
Average speed= total distance travelled/ total time travelled.
Hence, Average speed = (18 + 22 + 14) / (3+3+3)
Average speed = 54 / 9
Average speed= 6m/s.
9. 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?
Answer: Given that M=16kg and a1=3m/s2 , find the applied force or F.
F is given as M × a1
Hence, F = 16 × 3
Applied force (F) =48N
Also, if F = 48N and m =24 kg , then find the acceleration (a2).
Here, F= m × a2
48 = 24 × a2
So, a2 = 48/24 = 2m/s2
10. 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.
Answer: Mass of the bullet (M) = 10 g
Initial speed of the bullet (u1) = 1.5 m/s
Mass of the plank(m) = 90 g
Initial speed of the plank (u2)= 0 m/s
Based on the law of conservation of momentum.
See the formula Mu1+ mu2= Mv1+ mv2
Because the bullet gets embedded in the plank and both move with same speed v1=v2= v. Hence, the equation Mu1+ mu2= Mv1+ mv2 is rewritten.
Mu1+ mu2= (M+ m) v
10 × 1.5 + 90 × 0= (10 +90) v
Hence, 15 = 100v
-> v= 0.15m/s
11. An athlete is running on a circular track. He runs a distance of 400 m in 25 s before returning to his original position. What is his average speed and velocity?
Answer: Given here is Total distance travelled (d) = 400 m
Total displacement is denoted as 0, as he returns to his original position.
Total time taken (t) = 25 seconds.
Hence, calculate the Average speed and Average velocity = ?
Now, Average speed = Total distance / Total time
Hence, Average speed = 400 / 25 = 16 m/s.
Alternatively, Average velocity = Total displacement / time taken
Therefore, Average velocity = d/t = 0/ 25 = 0m/s.
12. A kangaroo can jump 2.5 vertically. What must be the initial velocity of the kangaroo?
Answer: a is given as 9.8 m/s2
When, s = 2 5 m, v = 0, then u = ?
Take the equation v2= u2 + 2as
Replacing the values in the equation, you get
(0)2 = u2 + 2 × (-9 8) (2 .5)
Here, the negative sign used before the acceleration denotes the direction opposite to that of velocity.
0 = u2 – 49
u2 = 49
u = 7 m/s
13. What is Newton’s second law of motion?
Answer: ‘The rate of change of momentum is proportional to the applied force and the change of momentum occurs in the direction of the force.’
14.Given that an object of mass m has an initial velocity u and when a force F is applied in the direction of its velocity for time t, its velocity becomes v. Then how to calculate the rate of change of momentum?
Answer: Suppose, the initial momentum of the object = mu,
Its final momentum after time t = mv
Hence, Rate of change of momentum= Change in momentum / time
Rate of change of momentum = (mv-mu) / t
= m (v-u) / t
You know that (v-u) / t = a
So, m (v-u) / t = ma
15. According to Newton’s second law of motion, what is K in the equation F = k ma?
Answer: According to Newton’s second law of motion, the K from F = k ma is the constant of proportionality and its value is 1.
16. What is a dyne?
Answer: In the CGS system the unit of force is a dyne. The force necessary to cause an
acceleration of 1 cm/s2 in an object of mass 1 gm is called 1 dyne. 1 dyne = 1 g × 1 cm/s2.
17. What idea is expressed in Newton’s third law of motion?
Answer: ‘Every action force has an equal and opposite reaction force which acts simultaneously,’ is the third law of motion by Newton. In nature force cannot act alone as it is a reciprocal action between two objects and is always applied in pairs. When force is applied on one object by another object, the former object also reciprocates and simultaneously applies force on the latter object. The forces between two objects are equal and opposite and this idea is expressed in Newton’s third law of motion. Force applied on the first object is called the action force, while the force applied to the second object is called the reaction force.
18. The mass of a cannon is 500 kg and it recoils with a speed of 0.25 m/s. What is the momentum of the cannon?
Answer: The mass of the cannon (m) is denoted as 500 kg and has recoil speed = 0.25 m/s, So, calculate Momentum = ?
Given that equation Momentum = m × v
Replace values to the equation and then momentum= 500 x 0.25 = 125 kg m/s
19. 2 balls have masses of 50 gm and 100 gm and they are moving along the same line in the same direction with velocities of 3 m/s and 1.5 m/s, respectively. They collide with each other and after the collision, the first ball moves with a velocity of 2.5 m/s. Calculate the velocity of the other ball after collision.
Answer: The mass of first ball ( m1) is 50 g = 0 05 kg, while
mass of the second ball(m2)= 100 g = 0 1 kg. Here, the
Initial velocity of the first ball (u1) = 3 m/s and
Initial velocity of the second ball (u2) = 1.5 m/s
Given here are also Final velocity of the first ball(v1) = 2 5 m/s, then find the final velocity of the second ball (v2) = ?
According to the law of conservation of momentum, total initial momentum = Total final momentum.
Hence, m1u1 + m2u2 = m1v1 + m2v2. .Replace the values to the equation
m1u1 + m2u2 = m1v1 + m2v2
(0.05 × 3) + (0.1 × 1.5) = (0.05 × 2.5) + (0.1 × v2)
(0.15)+(0.15) = 0.125 + 0.1v2
Hence, 0.3 = 0.125 + 0.1v2
0.1v2= 0.3-0.125
Therefore, v2= 0.175/ 0.1 = 1.75m/s
20. Explain Motion.
Answer: Motion is a relative concept. If the position of an object is altering with respect to its surroundings, then the object is in motion. Otherwise, it is in rest.
21. What is velocity? How do you calculate it?
Answer: The distance that an object travels in a particular direction is known as its velocity. Here, the unit time is considered as one second, one minute, one hour and so on. If large
units are used, one year can also be used as a unit of time. Velocity is the displacement that occurs in unit time. To calculate velocity, you just have to divide displacement by time. Hence, velocity = displacement / time.
22. When are the values of speed and velocity the same?
Answer: If the motion is along a straight line, then the values of speed and velocity are the same. If not, then they can be different.
23. Velocity depends on speed as well as direction. According to what does the velocity change?
Answer: 1. changing the speed while keeping the direction same
2. changing the direction while keeping the speed same
3. changing the speed as well as the direction
24. What is the speed of revolution of the earth around the sun?
Answer: The speed of revolution of the earth around the sun is about 29770 m/s.
25. What is acceleration? How do you calculate it?
Answer: Acceleration is the rate of change of velocity. Acceleration is the change in velocity divided by time. Given that the initial velocity is u, time is t and final velocity is v, then acceleration (a) = final velocity(v)- initial velocity(u) / time.
Hence , a = (v-u) / t
26. Name the two types of acceleration found in an object in motion.
Answer: An object in motion can have two types of acceleration. They are the uniform acceleration and non-uniform acceleration.
27. Write the first equation of motion describing the relation between velocity and time. Draw the velocity time graph.
Answer: The object with the velocity u starts from the point D in the graph and its velocity keeps increasing with the time t. It reaches the point B given on the graph. Here, the initial velocity (u)= OD, while final velocity (v) = OE. Meanwhile, Time (t) = OE.
Acceleration (a) = change in velocity / time.
I.e a = (final velocity – initial velocity) / time
a=(v-u)/ t = (OC-OD)/OE
Therefore, CD = at ———– (i) OC-OD=CD
Draw a line parallel to the Y axis passing through B, which also crosses the X axis in E. Draw a line parallel to X-axis passing through D. This will cross the line BE at A.
Now, in the graph BE = AB + AE
Hence, v = CD +OD___, (where AB= CD and AE=OD).
Therefore, v= at + u (from i)
u= u + at
Thus, the first equation of motion is derived.
28. Give the reasons for the following:
1. A static object does not move without the application of a force.
2. The force which is sufficient to lift a book from a table is not sufficient to lift the table.
3. Fruits on a tree fall down when its branches are shaken.
4. An electric fan keeps on rotating for some time even after it is switched off.
Answer: If we take all the above, you will see that all objects have some inertia. We have learnt that inertia is related to the mass of the object. Newton’s first law of motion that describes this property, is also known as the law of inertia. All the instances of inertia are examples for Newton’s first law of motion.
29. ‘An object continues to remain at rest or in a state of uniform motion along a straight line unless an external unbalanced force acts on it.’ Explain.
Answer: When an object is in uniform motion or is at rest along a straight line, then it does not mean that no force is acting on it. The truth is there are a number of forces acting on it, however, they cancel one another so that the net force is zero. Newton’s first law of motion explains the phenomenon of inertia, that is the inability of an object to change its state of motion on its own. It also explains the unbalanced forces that cause a change in the state of an object at rest or in uniform motion.
30. What property of the object was termed as momentum by Newton?
Answer: The effect of one object striking another object depends on both the mass of the former object as well as on its velocity. This means that the effect of the force is dependent on a property related to both mass and velocity of the striking object. This property is termed as momentum by Newton. Momentum is a vector quantity and it is the product of mass and velocity of an object. That is P = m v.
31. What is the unit of momentum?
Answer: The unit of momentum In the SI system is kg m/s, while in the CGS system, it is g cm/s.
32. What causes a change in momentum of an object?
Answer: Upon application of an unbalanced force on an object, a change in the velocity of the object is caused, also leading to a change in its momentum. The force necessary to cause a change in the momentum of an object is dependant upon the rate of change of momentum.
33. If the same force is applied on different objects, the change in momentum is the same. Explain.
Answer: Take the example of two objects with different masses and which are initially at rest. The initial momentum for both is zero. Now, if a force ‘F’ acts for time ‘t’ on both objects, then the lighter object starts moving faster than the heavier object. However, from the F = m × a formula, it is deduced that the rate of change of momentum i.e. ‘F’ in both objects is the same as well as the total change in their momentum will also be the same. That is ‘Ft’. Hence, it is confirmed that if the same force is applied on different objects, the change in momentum is the same.
34. What is Newton (N) ?
Answer: The unit of force in the SI system is called newton and the force necessary to cause an acceleration of 1 m/s2 in an object of mass 1 kg is called 1 newton. Hence, 1 N = 1 kg × 1 m/s2.
35. Explain the law of conservation of momentum with equation.
Answer: Given here is an object A with mass m1 and initial velocity u1, while the object B has a mass of m2 and initial velocity of u2. As per the formula for momentum, the initial momentum of A and B are m1u1 and m2u2, respectively. Now, if these objects collide the force on A due to B is F1, causing an acceleration in A and its velocity will become v1. So, the momentum of A following the collision will be m1v1. Now, if we take Newton’s third law of motion, A also exerts an equal force in the opposite direction on B, causing a change in the momentum of B. If its velocity after collision is v2. The momentum of B, following the collision is m2v2. Here, if F2 is the force that acts on B then F2 = – F1
That is m2a2 = – m1a1 ( given that F=ma)
Hence, m2 × (v2-u2)/ t = – m1 (v1-u1)/t ( replacing a = (v-u) / t)
m2(v2-u2)= -m1(v1-u1)
m2v2-m2u2=-m1v1+ m1u1
Therefore, m2v2+m1v1=m1u1+m2u2
That is the magnitude of total final momentum = the magnitude of total initial momentum.
Thus, if there is no external force acting on two objects, then their total initial momentum and their total final momentum are equal. This statement is true for any number of objects
36. What is the corollary to Newton’s third law of motion?
Answer: ‘When no external force acts on two interacting objects, their total momentum remains constant. It does not change.’ This is a corollary to Newton’s third law of motion. According to it, the momentum remains the same after the collision and it gets redistributed between the colliding objects. This causes the momentum of one of the objects to decrease while that of the other increases. Thus, this corollary can also be stated as given. ‘When two objects collide, the total momentum before collision is equal to the total momentum after collision.’
37. What is the recoil of a gun?
Answer: Before firing a bullet, both the gun and the bullet are at rest with the total initial momentum as zero. According to the above law that states that the total momentum before collision is equal to the total momentum after collision of two objects when they collide, the total final momentum also has to be zero. Hence, the forward moving bullet causes the gun to move backward after firing. This backward motion of the gun is known as its recoil.
Frequently Asked Questions on Maharashtra State Board Solutions for Class 9 Science Chapter 1 Laws of Motion
Will these Maharashtra State Board Solutions for Class 9 Science Chapter 1 Laws of Motion help to prepare for the exams?
Students are highly recommended practising these solutions after revising the subjects, as they set the basis for the questions that could get asked often in the board exams. They act as the perfect guide for Class 9 students during their board exam preparation because it is created by qualified subject teachers according to the latest class 9 Science Syllabus. Students can discover various solved and unsolved questions and exercises that will aid them to prepare well for board exams.
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How are these Maharashtra State Board Solutions for Class 9 Science Chapter 1 Laws of Motion useful?
Students are encouraged to answer the questions at first, and then they to refer back to the solutions to analyse their performance. This will also help them to rectify the mistakes, so that they can avoid making any during the board exams. Timing the process also helps them to manage time better while answering the exams.
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