Kerala Syllabus Class 9 Physics Chapter 3 Laws of Motion Notes Solutions

The comprehensive approach in Kerala Syllabus 9th Standard Physics Textbook Solutions Chapter 3 Laws of Motion Notes Questions and Answers English Medium ensures conceptual clarity.

Std 9 Physics Chapter 3 Notes Solutions Laws of Motion

SCERT Class 9 Physics Chapter 3 Notes Solutions Kerala Syllabus Laws of Motion Questions and Answers

Class 9 Physics Chapter 3 Let Us Assess Answers Laws of Motion

Question 1.
A body of mass 5 kg travelling with a velocity 144 km/h comes to rest in 4 s. Calculate its
a) initial momentum
b) final momentum
c) change in momentum
d) rate of change of momentum
Answer:
Mass, m = 5 kg
Initial velocity,
u = 144 km/h
= \(\frac{144 \times 5}{18}\) m/s (1km/h = \(\frac{5}{18}\) m/s) = 40 m/s

Final velocity, v = 0 (since it comes to rest)
Time, t = 4 s

a) Initial momentum, P_initial = m × u = 0 = 5 × 40 = 200 kg m/s
b) Final momentum, P_final = m × v = 0
Since the body comes to rest, the final momentum is 0.

c) Change in momentum, Δ p = P_final – P_initial = 0 – 200 = -200 kg m/s

d) Rate of change of momentum = \(\frac{\Delta p}{\Delta t}\) = \(\frac{-200}{4}\) = -50 kg m/s²

Question 2.
A hockey ball of mass 200 g hits a hockey stick with a speed 20 m/s and returns with the same speed through the same path. What is the change in momentum?
Answer:
Mass of the hockey ball. m = 200 g = 0.2 kg
Initial speed of the hockey ball (u) = 20 m/s
Final speed of the hockey ball (v) = 20 m/s (since it returns with the same speed)
Initial momentum, P_initial = m × u = 0.2 × 20 = 4 kg m/s
Final momentum, P_final = m × v = 0.2 × 20 = 4 kg m/s
Change in momentum, Δ p = P_final – P_initial = 4 – 4 = 0

Question 3.
What is the rate of change of momentum of a loaded truck of mass 10,000 kg, if its velocity changes from 15 m/s to 12 m/s in 4 s?
Answer:

Question 4.
Which of the following does not belong to the group?
Answer:
Speed. (Speed is a scalar quantity)

Question 5.
A cup is covered with a cardboard and a coin is kept on the cardboard.
a) What happens to the coin when the cardboard is struck off suddenly?
b) Why does it happen?
Answer:
a) The coin falls in the cup.
b) When the cardboard is struck off suddenly, it is thrown off because of the force applied on it. But the coin doesn’t get force. It continues in its state of rest due to its inertia of rest.

Question 6.
To clean a carpet, we hold it up and hit it with a stick. The dust falls off. Give reason.
Answer:
When a carpet is hung and hit with a stick, only the carpet gets an external unbalanced force. The dust particles do not get an external unbalanced force. The dust particles fall off due to its inertia of rest.

Question 7.
When a horse pulls the cart, the horse cart moves forward. The cart in turn pulls the horse with an equal and opposite force. But the horse and the cart go ahead. Explain.
Answer:
The horse’s feet exert force on the floor. At this time, an opposite force exerted on the legs by the floor helps the cart to move forward.

Question 8.
The velocity-time graph of a body of mass 250 g moving on a surface is given. Calculate the force of friction exerted by the surface.

Answer:
Mass, m = 250 g = 0.25 kg
From the graph,
Initial velocity, u=0
Final velocity, v = 6 m/s
Time, t = 12 s
Force, F = m × a
= m × \(\frac{v-u}{t}\)
= 0.25 × \(\frac{6-0}{12}\)
= 0.125 N

Question 9.
A body of mass 500 g moves with a velocity of 40 m/s. On applying a force for 4 s. the velocity changes to 80 m/s. Calculate the force applied.
Answer:
Mass, m = 500 g = 0.5 kg
Initial velocity, u = 40 m/s
Final velocity, v = 80 m/s
Time, t = 4s
Force, F = m × a
= m × \(\frac{v-u}{t}\)
= 0.5 × \(\frac{80-40}{4}\)
= 5 N

Question 10.
A person of mass 50 kg runs with a velocity of 8 m/s and makes a long jump. Another person of mass 60 kg makes the jump with a velocity 7 m/s. Compare their momenta. 50 kg
Answer:
Mass of first person, m, = 50 kg
Velocity of first person, v₁ = 8 m/s
Mass of second person, m₂ = 60 kg
Velocity of second person, v₂ = 7 m/s
Momentum of first person, p₁ = m₁ × v₁
=50 × 8
= 400 kg m/s
Momentum of second person, p₂ = m₂ × v₂
60 × 7 = 420 kg m/s
Comparing the two momenta, we can find that the second person has more momentum than the first person.

Question 11.
Calculate the force required to stop a vehicle of mass 14,000 kg by applying a retardation of 1.8 m/s².
Answer:
Mass, m = 14000 kg
Retardation, a = -1.8 m/s²
Force, F = m × a
= 14000 × (- 1.8)
= -25200 N

Question 12.
A force is applied on a body of mass 20 kg for 2s and its velocity changes from 10 m/s to 50 m/s. The same force is applied on another body of mass 10 kg moving with a velocity of 20 m/s for 2s in the direction of its motion. Calculate the final velocity.
Answer:
Mass of first body, m₁ = 20 kg
Initial velocity of first body, u₁ = 10 m/s
Final velocity of first body, v₁ = 50 m/s
Time, t₁ = 2s
Force on first body, F₁ = m × a = m × \(\frac{v_1-u_1}{t}\)
= 20 × \(\frac{50-10}{2}\) = 400 N
Given that force is same, ie., F₁ = F₂ = 400N
Mass of second body, m₂ = 10 kg
Initial velocity of second body, u₂ = 20 m/s
Time, t₂ = 2s
Force = m × \(\frac{v_2-u_2}{t}\)
400 = 10 × \(\frac{\mathrm{v}_2-20}{2}\)
v₂ – 20 = \(\frac{400 \times 2}{100}\)
v₂ = 80 + 20 = 100 m/s

Question 13.
A bullet of mass 20 g hits a wooden block with a velocity of 100 m/s and comes to rest after penetrating 4 cm.
a) What is the acceleration of the bullet?
b) What is the retardation of the bullet?
c) Calculate the force exerted by the bullet on the plank.
Answer:
Mass of bullet, m = 20g = 0.02 kg
Initial velocity, u = 100 m/s
Final velocity, v = 0
Distance penetrated by bullet, s = 4 cm = 0.04 m

a) From equation of motion, v² – u² = 2as
Acceleration, a = \(\frac{v^2-u^2}{2 s}\) = \(\frac{0-100^2}{2 \times 0.04}\) = – 125000 m/s²

b) Retardation of the bullet is 125000 m/s²
c) Force exerted by the bullet,
F = m × a
= 0.02 × (125000)
= – 2500 N

Question 14.
A graph showing the application of force on a body of mass 10 kg is given. The magnitude of force changes as indicated in the graph. (Frictional force need not be considered.)

a) What is the acceleration of the body when it is at 3 m?
b) Which are the instances when the body has uniform velocity?
c) Which are the instances when the body has uniform acceleration?
d) Which is the instance when the body has retardation?
Answer:
Mass of the body, m = 10 kg
a) Force at 3m, F = 8 N
Accelertion, a = \(\frac{F}{m}\)
= \(\frac{8}{10}\) = 0.8 N

b) Uniform velocity is at OA, DE and HI. (Here, force acting is zero. So, acceleration is zero)

c) Uniform acceleration is at BC and FG

d) Retardation is at CD and EF.

Question 15.
Which is the graph showing zero resultant force?

Answer:
Graph A

Question 16.
The figure shows forces applied on an object at rest. What is its acceleration? What is its displacement in 2 s?

Answer:
Mass of the object, m = 10 kg
Resultant force on the object, F = 16 – 7 = 9N
Acceleration, a = \(\frac{F}{m}\) = \(\frac{9}{10}\) = 0.9 m/s²
Initial velocity, u = 0 (since body is at rest)
Time, t = 2s
From equation of motion, S = ut + \(\frac{1}{2}\) at²
=0 × 2 + \(\frac{1}{2}\) × 0.9 × (2)²
= 1.8 m

Question 17.
Observe the figure.

A and B are two objects of masses 6 kg and 4 kg respectively. They are placed touching each other on a frictionless surface. Calculate the force exerted by object B on object A, when a force of 15 N is applied on them.
Answer:
Mass of A, m_A = 6 kg
Mass of B, m_B = 4 kg
Force, F = 15 N
Acceleration, a = \(\frac{F}{m_A+m_B}\)
= \(\frac{15}{6+4}\) = 1.5 m/s²
Force exerted by object B on object A,
F_BA = m_A × a
= 6 × 1.5
= 9 N

Class 9 Physics Chapter 3 Extended Activities Answers Laws of Motion

Question 1.
Prepare and present a seminar paper on how overload and overspeed of vehicles affect road safety.
Answer:
HINTS
Title: The Effects of Speeding and Overloaded Vehicles on Traffic Safety

Introduction:

• Begin by providing a brief definition of overload and overspeed in relation to automobiles.
• Stress how crucial road safety is to everyone’s health and safety.

Overloaded Automobiles:

• Describe the hazards connected to overloading
• Give situations or instances to highlight the risks associated with overloading.
• Discuss the significance of adhering to weight restrictions imposed by authorities.

Overspeed Vehicles:

• Talk about the dangers of exceeding the speed limit
• Emphasize how crucial it is to follow speed limits for everyone’s protection.

Effect on Traffic Safety:

• Provide evidence or statistics to back up your claims, if any are available.
• Stress the value of safe driving practices in order to avert collisions and save lives.

Conclusion:

• Recap the key ideas covered in the seminar paper.
• End with a request that everyone make road safety their top priority.

Question 2.
Write a report on how the concept of impulse can be used to explain the working of shock absorber in vehicles and discs in the spinal cord. Present it in the Science Club.
Answer:
HINTS
Title: Explaining Shock Absorbers in Vehicles and Discs in the Spinal Cord Using Impulse

Introduction:

• Define impulse as the change in momentum of an object.
• Explain how vehicle shock absorbers and spinal cord discs use impulse principles to perform properly.

Shock absorbers for vehicles:

• Define shock absorbers and their function in vehicles.
• Relate this process to the concept of impulse.

Discs in the Spinal Cord:

• Discuss the spinal cord’s role in signal transmission between the brain and the body.
• Relate spinal disc function to impulse

Conclusion:

• Summarize major points from the report.
• Emphasize the role of shock absorbers in automobiles and spinal discs in ensuring smooth and safe mobility.
• Emphasize the importance of impulse in both mechanisms, demonstrating how the concept can be utilized in a variety of real-world situations.

Question 3.
Present a seminar on some real life situations in which the concepts related to force are utilised.
Answer:
HINTS
Title: Practical Uses of Forces

Introduction:

• Begin by briefly defining force as a push or pull applied to an item and discussing its significance in day- to-day activities.
• Stress that practically everything we do, from walking to utilizing technology, involves forces.

Forces in Motion:

• Talk about situations in daily life where forces are at work to cause motion
• To demonstrate these points and describe how various forces, such as gravity and friction, affect motion, use demonstrations or movies.

Technological Forces:

• Describe the forces used to run different technology devices and how the engineers designed them to utilise a variety of forces.

Forces in Nature:

• Talk about the ways where natural occurrences are influenced by forces including tension, air resistance, and gravity.

Conclusion:

• Recap the key ideas covered in the seminar.
• Summarise more on how forces are everywhere and vital to life.

Laws of Motion Class 9 Notes Questions and Answers Kerala Syllabus

Question 1.
The rope moves only in one direction even though both the teams are applying force. Why?
Answer:
The Force F1 is greater than the force F2.

Question 2.
Is the force applied by both teams the same?
Answer:
No

Question 3.
Which team applied more force?
Answer:
The team on the left side applied more force.

Question 4.
Wasn’t it the excess force that caused the motion?
Answer:
Yes, it was the excess force that caused the motion.

Question 5.
What will be the resultant force if 100 N force is applied on an object in the east direction and 150 N force in the west direction?
Answer:
The resultant force = 100 N+(-150 N)=-50 N. The force applied in the east direction is positive. The resultant force is 50 N towards the west.

Question 6.
Complete the table by analysing the following figures.

Figure	Force F1	Force F2	Resultant Force(N)
1
2	+150	-150	0
3
4
5
6

Answer:

Figure	Force F1	Force F2	Resultant Force(N)
1	+150	0	+150 (To the right)
2	+150	-150	0
3	+150	-120	+30 (To the right)
4	0	-300	-300 (To the left)
5	0	-200	-200 (To the left)
6	+200	-200	0

Question 7.
In which of these situations is the resultant force zero?
Answer:
In fig. 2 and fig. 6

Question 8.
Which are the situations where the resultant force is not zero?
Answer:
In fig. 1, fig. 3, fig. 4, fig. 5

Question 9.
Which are the situations where there is no motion?
Answer:
In fig. 2 and fig. 6

Question 10.
In the tug of war shown in the figure below, is the resultant force experienced on the rope balanced or unbalanced?

Answer:
Unbalanced

Activity
To study if all forces cause motion

Fix a pulley each at the both ends of a wooden plank of length about 1.2 m and breadth 10 cm. Keep this plank on a table. Place a toy car in the middle of the plank as shown in the above figure. Hang pans of equal mass on strings attached to the two ends of the toy car. Place 200 g weight each in both the pans.

Question 11.
Does the toy car move?
Answer:
No

Question 12.
Are these forces balanced or unbalanced?
Answer:
Balanced

Question 13.
Add 50 g more to any one of the pans. What do you observe?
Answer:
The toy car moves in the direction in which the extra 50 g is added.

Question 14.
Are the forces balanced or unbalanced in this case?
Answer:
Unbalanced

Question 15.
When the toy car is moving, if 50 g more is added to the pan in the direction in which the car moves, what change can be observed in the motion of the car?
Answer:
Its speed increases.

Question 16.
When the car is moving, a mass of 200 g more is added to the pan attached to the string in the opposite direction of the motion of the car. What is the change observed?
Answer:
It moves in the opposite direction.

Question 17.
Are the forces balanced or unbalanced in the above situation?
Answer:
Unbalanced

Question 18.
What do you understand from these activities?
Answer:
The body moves in the direction of the resultant force.

Question 19.
Does the body move in the direction of resultant force? (moves / does not move)
Answer:
The body moves in the direction of the resultant force.

Question 20.
When does the speed of the car increase?
(when the magnitude of the resultant force increases/decreases)
Answer:
When the magnitude of resultant force increases.

Question 21.
Was the force that moved the car applied from inside the car or from outside?
Answer:
The force that moved the car was applied from outside.

Question 22.
In which situation does the direction of motion change?
Answer:
When the direction of the resultant force changes.
Note: In all the above cases, the force was given externally. Hence, all of them are external forces. An external force can be balanced or unbalanced.

Observe the figure

Question 23.
Can a vehicle move if pushed from inside?
Answer:
No

Question 24.
Isn’t it an internal force?
Answer:
Yes

Question 25.
All internal forces are………………
(balanced/unbalanced)
Answer:
balanced

Question 26.
Complete the chart and redraw it in the science diary

Answer:

Activity
Galileo’s Observations

A wiring channel is used for doing this experiment. The end C of the wiring channel is gradually lowered to horizontal level as shown in the figures.

Question 27.
What do you observe, if in each case, a marble is rolled from the end A in the wiring channel?
Answer:
In figures (a), (b) and (c), the marble that is rolled from A moves towards point C. It does not reach C. In fig (d), the marble moves beyond point C.

Question 28.
Does the distance travelled by the marble increase or decrease in each situation?
Answer:
increases

Question 29.
When did the marble travel the longest distance?
Answer:
fig.(d). The marble covers more distances as the slope decreases.
When the marble falls down it has a tendency to reach the original height. This tendency makes the marble move longer distances as the slope decreases.

Question 30.
Why did the marble come to rest after traversing some distance?
Answer:
The marble come to rest after traversing some distance due to friction.

Question 31.
What would have happened if the force of friction was absent?
Answer:
It will continue in its state of uniform motion.

Question 32.
What would have happened if no external force was applied to the marble?
Answer:
It would have continued in its state of uniform motion.

Question 33.
Write your inference from the above observation.
Answer:
If an unbalanced external force does not act on a body that is in motion, it will continue in its state of uniform motion.

Question 34.
What is the importance of Newton’s first law of motion?
Answer:
The first law of motion helps us to define the physical quantities like inertia and force

Question 35.
Passengers standing in a bus tend to fall backwards when the bus at rest moves forward suddenly. Why?
Answer:
Before the bus moved forward, the passengers and the bus were stationary. When the bus moves forward suddenly, the passengers tend to fall backwards because of the tendency to continue in the state of rest. This tendency is the inertia of rest.

Question 36.
Why do the passengers standing in a bus tend to fall forward when the moving bus stops suddenly?
Answer:
The passengers and the bus were moving. When the bus stops suddenly, the passengers tend to fall. forward because of the tendency to continue in the state of motion. This tendency is the inertia of motion.

Activity

Place a paper on a table. Keep a closed flat bottomed bottle filled with water over the paper. Quickly pull the paper horizontally.

Question 37.
What happened to the bottle?
Answer:
The bottle continues in the state of rest.

Question 38.
Name the inertia possessed by the bottle.
Answer:
The bottle possesses inertia of rest.

Activity

Place a glass filled with water on a desk. Slowly move it forward and gradually increase its speed. Stop it suddenly.

Question 39.
What do you observe? Name the inertia possessed by the water.
Answer:
The water in the cup splashes forward. The water possesses inertia of motion.

Activity

Stack some carrom coins, one above the other, as shown in the figure. Place a plastic cup filled with water above it. Using a long plastic scale, quickly strike out the carrom coins one by one from the bottom.

Question 40.
Does the cup possess inertia? Which type?
Answer:
Yes, the cup possesses inertia. The cup possesses inertia of rest.

Question 41.
Write down the following statements related to inertia in the table appropriately. Expand the table by including more examples for inertia of rest and inertia of motion.

• On shaking the branch of a mango tree, the mangoes get detached and fall down.
• A participant in long jump competition, runs some distance and then jumps.
• Travelling in a car. without wearing seat belt is dangerous.

Inertia of rest	Inertia of motion
When a bus moves forward suddenly, the standing passengers tend to fall backward.	A ball set rolling on a horizontal floor keeps moving forward.

Answer:

Inertia of rest

Inertia of motion

When a bus moves forward suddenly, the standing passengers tend to fall backwards. On shaking the branch of a mango tree, the mangoes get detached and fall down. Put a coin on a cardboard placed on the top of a glass. Strike off the cardboard quickly. The coin will fall into the cup due to its inertia of rest. When a carpet is hung and tapped with a stick, the dust particles fall down due to its inertia of rest.

A participant in long jump competition, runs some distance and then jumps. Travelling in a car without wearing seat belt is dangerous. A ball set rolling on a horizontal floor keeps moving forward If we try to jump out of a moving bus, we will fall forward due to inertia of motion. To avoid this, we have to step forward a little further. Sitting on a moving train, if we throw an object upwards, it falls back into our hands.

Activity

Place a paper on a table. Take two identical flat bottomed bottles. Fill one of them with sand. Place the bottles vertically on the paper. Quickly pull the paper horizontally.

Question 42.
Which bottle does not topple over?
Answer:
The bottle filled with sand does not topple over.

Question 43.
Which bottle has a higher mass?
Answer:
The bottle filled with sand has a higher mass.

Question 44.
Which bottle possesses more inertia?
Answer:
The bottle filled with sand possesses more inertia.

Question 45.
Based on the above observations, what is the relation between mass and inertia?
Answer:
As mass increases, inertia increases.

Question 46.
Which one possesses greater inertia – an empty barrel or a tar filled barrel? Give reason.
Answer:
The tar-filled barrel possesses greater inertia. Since the tar-filled barrel has more mass due to. the added weight of the tar, it will have greater inertia compared to the empty barrel.

Question 47.
Why people run in a zig-zag manner to escape from an elephant attack?
Answer:
As the mass of an elephant is greater, its inertia is also greater. So elephant cannot turn and run easily. Because we have relatively less mass, we can turn and run more easily than an elephant.

Question 48.
From the following, write the situations where an unbalanced force is experienced.
a) Brakes are applied on a car moving with a velocity of 20 m/s.
b) A book is supported by the hand.
c) An artificial satellite travels with uniform speed.
Answer:
(a) and (c)

Question 49.
A force of 200 N is applied on a body in one direction and another force of 250 N in the opposite direction.
a) Calculate the resultant force.
b) If it moves, what will be the direction of motion?
Answer:
a) F₁ = + 200 N
F₂ = -250 N
Resultant force = +200 N + -250 N = -50 N
b) It moves in the direction of 250 N force.

Activity
Observe the given figure.

Question 50.
Pull back the balls in a Newton’s cradle and release them in the following order. Write down the observations.
Answer:
First ball alone – When the first ball alone hits the next ball, the momentum thus transferred by it reaches the last ball through the other balls and results in the last ball being tossed off.

First two balls – When the first two balls together hit the next ball, they form a moving system and the momentum of this system is transferred to the last two balls. Hence, the last two balls are tossed off.

First three balls – When the first three balls together hit the next ball, they form a moving system and the momentum of this system is transferred to the last three balls. Hence, the last three balls are tossed off.

First four balls – When the first four balls together hit the next ball, they form a moving system and the momentum of this system is transferred to the last four balls. Hence, the last four balls are tossed off.

Factors influencing momentum

Activity
In the Newton’s cradle, pull back one ball alone to a small distance and release it. The last ball moves out only a little. When the same ball is pulled back further and released, it hits with a greater velocity. Now we can see that the last ball covers a greater distance on moving out. It is due to the increase in the velocity of the first ball.

Question 51.
Here, which factor influenced the momentum of the ball?
Answer:
Velocity of the ball.
We have seen that when a system of two balls together hit the others in Newton’s cradle, the last two balls are tossed out. This is due to the increase in the mass of the hitting system.

Question 52.
In this case, what decided the momentum of the balls?
Answer:
Mass of the ball.
Momentum is a vector quantity.
The direction of momentum is the same as that of its velocity.
Unit of momentum = Unit of mass × Unit of velocity Kg m/s

Question 53.
Calculate the momentum of a body of mass 200 kg moving with a velocity 16 m/s. 16 m/s
Answer:
m = 200 kg
v = 16 m/s
p = m v = 200 kg × 16 m/s =3200 kg m/s

Question 54.
The momentum of a body moving with a velocity 20 m/s is 200 kg m/s. What is its mass?
Answer:
p = 200 kg m/s
v = 20m/s
p=mv
m = p/v
=200/20
= 10 kg

Question 55.
Calculate the momentum of a bullet of mass 60 g moving with a velocity 200 m/s. What is its momentum when it is at rest?
Answer:
m = 60 g = \(\frac{60 \mathrm{~g}}{1000}\) = 0.06 kg
v = 200 m/s
p = mv = 0.06 kg × 200 m/s = 12 kg m/s
p = 0 (at rest)

Question 56.
A body of mass 20 kg is at rest. When a force is applied on it for 5 s, its velocity changes to 30 m/s. Find the change in momentum of the body.
Answer:
Initial momentum = mu = 20 kg × 0 = 0
Final momentum = mv =20 kg × 30 = 600 kg m/s
Change in momentum = mv – mu = 600 kg m/s – 0 = 600 kg m/s

Question 57.
What will be its change in momentum in unit time or its rate of change of momentum?
Answer:
Rate of change of momentum = change of momentum/time
= 600 / 5 = 120 kg m/s²

Question 58.
A body of mass 100 kg starts from rest and acquires a velocity of 30 m/s in the fourth second. If so,
a) what is its initial momentum?
b) what is its final momentum?
c) what is the change in momentum?
d) what is the rate of change of momentum?
Answer:
m = 100 kg, u = 0,v = 30 m/s, t = 4s
a) Initial momentum = mu = 100 kg × 0 = 0

b) Final momentum = mv = 1oo kg × 30 = 3000 kg m/s

c) Change in momentum = mv – mu = 3000 kg m/s – 0
= 3000 kg m/s

d) Rate of change of momentum = (mv – mu)/t = 3000/4 = 750 kg m/s²

Question 59.
A body of mass 20 kg is at rest. The velocity at various instances when a force of varying magnitude is applied on it for a time interval of 5 s is given. Calculate the initial momentum, final momentum and the rate of change of momentum of the body in each case. Complete the table and find the relation between the rate of change of momentum and the force applied on them.

Answer:

Question 60.
A body of mass 12 kg is moving with an acceleration of 4 m/s2. Calculate the force applied.
Answer:
m = 12 kg a = 4 m/s² F = ?
F = ma = 12 × 4 = 48 N

Question 61.
A force of 40 N is applied on a body of mass 20 kg. Calculate the acceleration produced.
Answer:
m = 20 kg F = 40 N a = ?
a = F/m = 40 N / 20 kg =2 m/s²

Question 62.
A vehicle of mass 1000 kg is travelling with a velocity of 90 km/h. The vehicle comes to rest when brakes are applied for 5 s. Calculate the force applied.
Answer:
Initial velocity u = 90 km/h = 90 × 5/18 m/s = 25 m/s
Final velocity v=0
Mass m = 1000 kg
F = ma
= m (v-u)/t
= 1000 (0-25)/5 = -5000 N

Question 63.
There is negative sign for this force. Why?
Answer:
Force is a vector quantity. The negative sign indicates that the force applied in a direction opposite to the motion of the vehicle.

Question 64.
The velocity of a body of mass 10 kg changes from 6 m/s to 18 m/s in 4 s.
a) What is the rate of change of momentum?
b) What is the force applied?
c) Calculate the acceleration of the body.
d) What will be its velocity if this force is applied for 6 s?
Answer:
mass = 10 kg
Initial velocity u = 6m/s
Final velocity v = 18 m/s

a) Rate of change of momentum
= m (v – u)/t
=10 (18 – 6)/4 = 30 N

b)Force, F = Rate of change of momentum = 30 N

c) Acceleration, a = F/m = 30 N/ 10 kg = 3 m/s²

d)Final velocity v=u+at
= 6 m/s +3 m/s² × 6 s = 24 m/s

Question 65.
A shot of mass 7 kg rolled on level ground, with a velocity 2 m/s came to rest in 5 s.
a) Which force brought it to rest?
b) Calculate the magnitude of this force.
Answer:
m = 7 kg, u = 2 m/s, t = 5 s, v = 0
a) Frictional Force
b) F = ma
= m(v – u)/t
7(0 – 2)/5 N
= -2.8 N

Question 66.
Can you find out the peculiarities of forces applied in following situations?
hitting the ball with cricket bat
kicking the ball while playing football
Nailing the wall
Answer:
Large forces are applied here for a short interval of time. Such forces are impulsive forces. Impulsive force

Question 67.
A ball of mass 200 g is moving with a velocity of 30 m/s. A person catches the ball.
a) If the time taken to bring the ball to rest is as follows, what will be the force felt on the arm in each case?
i) 0.3 s
ii) 0.2 s
iii) 0.1 s
b) The magnitude of the force is negative. This indicates that the force is applied in the opposite direction of the object’s motion.
c) As the time taken to bring the object to rest decreases, the force felt on the hand increases.
Answer:
a) m = 200g = 200/1000 = 0.2 kg
u = 30 m/s, v = 0
(i) F m (v – u) / t
= 0.2 (0 – 30) / 0.3
=0.2 × – 30/0.3
=- 20N

(ii)F = m(v – u)/t
= 0.2 (0 – 30) / 0.2
=0.2 × – 30/0.2
= – 30N

(iii)F =m(v – u)/t
0.2 (0 – 30) / 0.1
=0.2 × – 30/0.1
= -60N

b) The magnitude of the force is negative. This indicates that the force is applied in the opposite direction of the object’s motion.
c)As the time taken to bring the object to rest., decreases, the force felt on the hand increases.

Question 68.
Based on the conclusions formulated, find the reason for the following statements.

a) Cricket players move their hands backwards along with the ball while catching a fast moving ball.
b) In the game of football, while the goalkeeper catches the ball coming into the goal post, he moves his hands backwards along with the ball.
c) A foam bed is placed in a pole vault pit.
d) Sponge or hay is kept between glass vessels while packing.
Answer:
a) Cricket players move their hands backwards along with the ball while catching a fast-moving ball to increase the time taken to bring the moving ball to rest. This helps to decrease the impact or force felt on the hand.

b) In the game of football, while the goalkeeper catches the ball coming into the goal post, he moves his hands backwards along with the ball, increasing the time taken to bring the moving ball to rest. This helps to decrease the impact or force felt on the hand.

c) A foam bed is placed ¡n a pole vault pit.

d) Sponge or hay is kept between glass vessels while packing.
Answer:
a) Cricket players move their hands backwards along with the ball while catching a fist-moving ball to increase the time taken to bring the moving balito rest. This helps to decrease the impact or force felt on the hand.

b)In the game of football, while the goalkeeper catches the ball coming into the goal post, he moves his hands backwards along with the ball, increasing the time taken to bring the moving ball to rest. This helps to decrease the impact or force felt on the hand.

c) A foam bed is placed in a pole vault pit to increase the time taken to bring the athlete to rest. This helps to decrease the impact or force felt on the body.

d) Sponge or hay is kept between glass vessels while packing to increase the colliding time between the glass containers and thus reduce the impact or force.
Answer:
F = ma
a = F/m
when F = 0 then
a = 0
That means there will be no acceleration if no force is applied. That is, an object in motion with no acceleration will continue in its motion in a straight line or an object at rest will continue in its state of rest. This is the first law of motion. That means that the second law of motion is on par with the first law of motion.

Activity
A straw is passed through a smooth plastic thread tied diagonally. Attach an inflated balloon in the straw with cello tape and release the air from the balloon.

Question 70.
What is the direction of airflow from the balloon?
Answer:
The air flows backwards. (In the opposite direction of motion of the balloon).

Question 71.
What is the direction of motion of the balloon?
Answer:
The balloon moves forward. (In the opposite direction of motion of the air flow).

Activity

A and B are two identical spring balances.

Fix one end of the spring balance B firmly to the grill of a window. Apply force of 40 N on the spring balance B using A.

Question 72.
What is the reading shown by each spring balance?
Answer:
40 N

Question 73.
Are the readings the same?
Answer:
Yes

Question 74.
Are forces in same direction or the opposite direction?
Answer:
The forces are in the opposite direction.

Question 75.
A car will not move if we push it sitting inside. But we can move the front seat by pushing it sitting on the back seat. How is it possible?
Answer:
While sitting on the back seat and pushing the front seat, we are actually outside the front seat. Hence, we are able to exert an unbalanced external force. But when we are pushing the car sitting inside, the same force that is exerted on the car is transferred through our body to the platform of the car. Thus, the forces become balanced. Hence, the car will not move. On pushing the car by standing on the road the car moves as an unbalanced external force is acting on it.

Question 76.
Which happens first – action or reaction?
Answer:
Action and reaction act simultaneously.

Question 77.
While rowing a boat the water is pushed back, but the boat moves forward.
Answer:
When we row a boat, we push the water backwards, and the water pushes the boat forward. The boat moves forward due to the force exerted by water.

Question 78.
When a rocket is launched, gases are produced in its chamber by the combustion of fuels. These gases which are at high pressure move in one direction at high speed. But the rocket is propelled in the opposite direction.
Answer:
The high-temperature and high-pressure gases are produced by burning the fuel inside the rocket’s combustion chamber. These gases are ejected at very high speed through the nozzle of the rocket engine. The exhaust gases will exert a high force on the rocket. This will be in the opposite direction to the outflow of gases. As a result of this force, a forward acceleration is created in the rocket, and it moves forward.

Question 79.
When a person jumps from a boat onto a shore, the boat moves backwards.
Answer:
When a person jumps from a boat onto a shore, action and reaction will be exerted on the legs and boat. As a result of the action and reaction, the person jumps forward, and the boat moves backwards.

Question 80.
In the diagram below, are the forces in both directions equal?

Answer:
Forces occur only in pairs.
F₁₂ is the force exerted by the first body on the second body.
F₂₁, is the force exerted by the second body on the first body. If so, according to Newton’s third law of motion, F₁₂ – F₂₁
Answer the following questions and justify them.

Question 81.
Action and reaction are equal and opposite. If so, will they cancel each other?
Answer:
No, they will not cancel each other because action and reaction are acting on different bodies.

Question 82.
If you are pushing a vehicle standing on ice, will the vehicle move?
Answer:
No, Ice-covered surfaces have less frictional force. So, reaction force will not be obtained. The vehicle will not move.

Question 83.
Based on the third law of motion, establish how an internal force becomes balanced force.
Answer:
When an internal force is applied, both action and reaction are acting on the same body. They will cancel each other. Internal forces are always balanced.