How Forces Affect Motion Class 9 NCERT Solutions

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How Forces Affect Motion Class 9 NCERT Solutions provide easy-to-understand, step-by-step answers to all the NCERT textbook questions from Chapter 10: How Forces Affect Motion of the Class 9 Science textbook. These solutions are prepared according to the latest NCERT and CBSE syllabus and help students understand important concepts such as force, balanced and unbalanced forces, Newton’s laws of motion, inertia, momentum, and friction.

How Forces Affect Motion Class 9 NCERT Solutions

1. Why does a canoe move forward when the canoeist pushes water backwards with their paddle, and why does it move faster when they push harder?

Answer: When the canoeist pushes water backwards with the paddle, then the water pushes the canoe forward. This happens because of Newton’s 3rd Law, which states that “every action has an equal and opposite reaction”. If the canoeist pushes the water harder, then the force on the water will be big. A bigger force on water means a bigger reaction force on the canoe, and because of this, the canoe moves faster.

2. Suppose the same canoeist uses the same paddle force in two different canoes, one empty and one carrying another passenger. In which case will the canoe move faster?

Answer: If canoe is empty or full with passengers, in both cases acceleration depends on force ÷ mass which is Newto’s 2nd law.

\[ a = \frac{F}{m} \]

• In an empty canoe, the acceleration is bigger, so the canoe moves fast.
• In a canoe with a passenger, the acceleration is smaller, so the canoe moves slower.

3. A weightlifter lifts a barbell (Fig. 6.8). List two forces that are acting on the barbell. Are these forces balanced if the weightlifter keeps the barbell steady?

Answer: The two forces that are acting on the barbell are

• Gravitational force. The gravitational force pulls the barbell downward with a force equal to its weight.
• Upward force: The weightlifter’s hands apply an upward force to hold the barbell.

Yes, the weightlifter keeps the barbell steady, meaning not moving up or down; here, upward force equals to the downward gravitational force. It is known as a balanced force.

4. Two players R and S are participating in an arm-wrestling match (Fig. 6.9). At the instant, when the arms tilt to the front direction (out of the page towards you), are the forces exerted by the players balanced? If not, which player exerted the larger force?

Answer: In arm wrestling, the two players push against each other with equal force than the arm stays in the middle. This is called a balanced force, but if one pushes stronger, then the arm moves. Then the forces are unbalanced.

5. The force of friction disappears in the world? How will the motion of objects be impacted?

Answer: If the friction disappears, then the object will not stop on its own and will keep moving once started. Without friction people cannot walk or drive because there is no grip. Motion will become uncontrolled, and everything keeps sliding until blocked.

6. An object is moving with a constant velocity. Is there a net force acting upon it?

Answer: Newton’s 1st law says that “an object keeps moving with constant velocity unless a net force acts on it”. So, if an object is moving with constant velocity, there is no net force acting on it, and then the net force will be 0.

7. Suppose, no net force is acting on an object. Which of the following situations are possible?
(i) Object remains at rest if at rest.
(ii) Object keeps moving with a constant velocity if already moving.
(iii) Object is moving with a constant acceleration.

Answer: The situations which are possible are –

• (i) Object remains at rest if at rest: Yes, it is possible because, if the object was already at rest, it will continue to stay at rest.
• (ii) Object keeps moving with a constant velocity if already moving.: Yes, it is possible, if the object is already moving then it will keep moving at the same speed and in the same direction.
• (iii) Object is moving with a constant acceleration: This is not possible, because acceleration requires a net force.

8. In the real world, it is difficult to find a situation where no forces are acting on an object. But by applying additional forces, a condition can be achieved where the net force on the object is zero. Explain with the help of an example.

Answer: In the real world, it is very hard to find an object with no forces at all acting on it. But sometimes, by applying some additional forces, we can make the net force = 0. For example, A book kept on a table table where downword gravitational force is applied. The table applies an equal upward force. These two forces are equal and opposite, so they cancel each other, and the book remains at rest.

9. How much does a force of 1 N feel? If you hold a 100 g mass in your palm, the upward force your palm applies on the mass is around 1 N.

Answer: If you hold a 100 g object like small apple, chocolate bar or biscuit packet in your hand, you palm pushes upward force with 1 N to stop from falling. So, the feel of 1 N is just like holding a light object in your hand.

10. A toy car of mass 100 g is moving with a constant velocity of 0.5 m s–1. What is the net force acting on the toy car?

Answer:

• Mass of toy car = 100 g = 0.1 kg
• Velocity = 0/5 m/s (constant)

If velocity is constant, the car is not accelerating.

Newton’s 2nd Law: F = m . a

• 𝐹 = force
• 𝑚 = mass
• 𝑎 = acceleration

Here, acceleration is a = 0

• F = 0.1 x 0
• = 0 N

The net force acting on the toy car is 0 N.

11. Two children of different masses are sitting on identical swings. To impart identical initial acceleration, for which child would you require to apply a larger force? Explain why.

Answer: To impart the same initial acceleration, you must apply a larger force on the child with greater mass. This is because force depends on mass ( 𝐹 = 𝑚 ⋅ 𝑎 ).

12. How are glass items packed for transportation using a bubble wrap or hay protected from damage?

Answer: The glass item are packed with bubble wrap or hay because these materials act as shock absorbers. They helps to reduce vibration and prevent direct impact, so the glass does not break during transportation.

13. Why does a fireperson sometimes struggle when holding the pipe issuing water?

Answer: A fireperson struggles to hold the pipe because the fast water coming out produces a backward reaction force on the pipe. This force push the pipe back and fireperson apply effort to control the pipe.

14. Suppose a spacecraft is moving in a region of space where the gravitational force acting upon it is negligible. Suggest how can it change its velocity.

Answer: In space there is no gravety when the rockets are fired they push gas backward. As per the Newton’s 3rd law, Gas goes backward and spacecraft goes forward.

16. Using a horizontal force F, a table is moved across the floor at a constant velocity. How much is the frictional force exerted by the floor on the table?

Answer: A table is moved across the floor with a horizontal force F. It moves at constant velocity, the constant velocity means no acceleration. Newton’s 2nd Law says:

𝐹_net = 𝑚 ⋅ 𝑎

• Here, a = 0
• So the net force = 0

It means the forward force (F) is exactly balanced by the frictional force.

18. For a ball moving on a smooth frictionless surface, choose the appropriate option that will make the following statements physically correct.
(i) If no net force is applied on the ball, the velocity of the ball will remain the same/increase/decrease.
(ii) If a net force is applied on the ball in the direction of its motion, the magnitude of the velocity of the ball will remain the same/ increase/decrease.
(iii) If a net force is applied on the ball in a direction opposite to the direction of its motion, the magnitude of the velocity of the ball will remain the same/increase/decrease.

Answer:

(i) If no net force is applied on the ball, the velocity of the ball will remain the same/increase/decrease.

Answer: The velocity will remain the same.

(ii) If a net force is applied on the ball in the direction of its motion, the magnitude of the velocity of the ball will remain the same/ increase/decrease.

Answer: The velocity will increase.

(iii) If a net force is applied on the ball in a direction opposite to the direction of its motion, the magnitude of the velocity of the ball will remain the same/increase/decrease.

Answer: The velocity will decrease.

19. Two blocks P and Q on a smooth horizontal surface are shown in Fig. 6.36a and Fig. 6.36b. Two forces of magnitudes 4 N and 5 N are acting in opposite directions on block P, while block Q is moving with a constant velocity.
Which of the following statement is correct?
(i) P experiences a net force and Q does not experience a net force.
(ii) P does not experience a net force and Q experiences a net force.
(iii) Both P and Q experience a net force.
(iv) Neither P nor Q experiences a net force.

Answer: (i) P experiences a net force and Q does not experience a net force.

Explanation:

For Block P

• The Net Force = 5N – 4N = 1N
• So the P will experience a net force.

For Block Q

• The constant velocity means there will no acceleration.
• Net Force = 0

So, Q do not experience a net force.

20. While practising for the snake boat race (Vallum kalli in Kerala), 100 oarsmen are rowing a boat together. Out of these, 95 row backwards to propel the boat forward. But by mistake, 5 oarsmen row in the opposite direction. If each oarsman applies a horizontal force of 200 N, what is the net force on the snake boat? (Ignore drag forces, air friction, etc.)

Answer: The net force on the snake boat will be 18,000 N forward.

Explanation:

Situation given,

• Total oarsmen: 100
• 95 row backwards: boat moves forward.
• 5 row forward (wrong way): boat gets backward push.
• Each oarsman applies 200 N force.

The Calculation,

Force by 95 oarsmen = 95 × 200 = 19,000 N (forward).

Force by 5 oarsmen = 5 × 200 = 1,000 N (backward).

Net force = Forward force – Backward force

19,000 − 1,000 = 18,000 N

21. When a net force acts on an object, we observe that the object accelerates: 
(i) opposite to the direction of force, with acceleration proportional to the force acting on the object.
(ii) opposite to the direction of force, with acceleration proportional to the mass of the object.
(iii) in the direction of force, with acceleration inversely proportional to the force acting on the object.
(iv) in the direction of force, with acceleration proportional to the force acting on the object.

Answer: (iv) in the direction of force, with acceleration proportional to the force acting on the object.

Explanation:

The Newton’s 2nd Law

F = m . a

• Force and acceleration are directly linked together.
• Acceleration is always in the same direction as the force.
• Bigger force means bigger acceleration.

22. The position-time graph for four objects A, B, C and D moving along a straight line are given in Fig. A net force acts on:
(i) Object A
(ii) Object B
(iii) Object C
(iv) Object D

Answer: From the position-time graphs of objects A, B, C, and D:

• In Object A, the graph is a straight line going upward. This means the object is moving with constant speed. Therefore, no net force acts on Object A.
• In Object B, the graph is a horizontal line. This means the object is not moving and remains at rest. Therefore, no net force acts on Object B.
• In Object C, the graph is a straight line going downward. This means the object is speeding up, or accelerating. Therefore, a net force acts on Object C.
• In Object D, the graph is a straight line going downward. This means the object is moving with constant speed in the opposite direction. Therefore, no net force acts on Object D.

23. A sailor jumps out from a small boat to the shore (Fig. 6.38). As the sailor jumps forward, will the boat move? If yes, in which direction and why.

Answer: Yes, the boat will move when the sailor jumps forward to the shore; he pushes the boat backward. This happens because of Newton’s third law of motion, “Every action has an equal and opposite reaction.”

• Action: The sailor pushes the boat backward while jumping forward.
• Reaction: The boat moves backward in the opposite direction.

24. During a high jump event, a landing mat or sand bed is placed for the athlete to fall upon (Fig. 6.39). Explain the reason behind it.

Answer: In a high jump, a soft mat or sand bed is used so the athlete stops slowly after falling. This increases stopping time, reduces the force on the body, and prevents injury.

25. A hand cart loaded with vegetables collides with an identical but empty hand cart. During the collision:
(i) the loaded cart exerts a force of larger magnitude on the empty cart.
(ii) the empty cart exerts a force of larger magnitude on the loaded cart.
(iii) neither cart exerts a force on the other.
(iv) the loaded cart and the empty cart, both exert an equal magnitude of force on each other.

Answer: (iv) the loaded cart and the empty cart, both exert an equal magnitude of force on each other.

Explanation:

As per the Newton’s 3rd Law, When two objects push each other, the force are equal in size but oppooite in direction. So,

• Loaded cart pushes empty cart.
• Empty cart pushes loaded cart
• Both force are equal.

26. The acceleration-mass graph for the acceleration produced by a force on objects of different masses is plotted in Fig. 6.40. Plot the force-mass graph for this case.

Answer: The graph shows acceleration versus mass. It tells us that when mass increases, acceleration decreases, if the force is the same. If we plot force versus mass:

From Newton’s Second Law:

F = m x a

In this case, the applied force is constant. This means, no matter what the mass is, the force value does not change.

So, the force-mass graph will be a straight horizontal line, which shows that the force remains the same for all values of mass.

27. The velocity-time graph of an object of mass 10 kg moving along a straight line is shown in Fig. 6.41. Calculate the force acting on the object by using the graph.

Answer: The time graph of an object of mass 10 kg.

• In graph, initial velocity: At time t = 0, velocity = 10 m/s
• In graph, final velocity: At time t = 8 seconds, velocity = 30 m/s

Find acceleration:

\[ a = \frac{v – u}{t} = \frac{30 – 10}{8} = \frac{20}{8} = 2.5 \, \text{m/s}^2 \]

So, acceleration = 2.5 m/s².

Apply Newton’s Second Law:

F = m x a = 10 x 2.5 = 25 N

The force acting on the object is 25 Newton.

28. A bullet of mass 50 g moving with a speed of 100 m s–1 enters a heavy stationary wooden block and stops after penetrating a distance of 50 cm. Estimate the stopping force acting on the bullet (assume that the bullet undergoes constant acceleration within the block).

Answer: The stopping force on the bullet = 500 N (opposite to motion).

Explanation:

Given

• Mass of bullet = 50 g = 0.05 kg
• Speed of bullet = 100 m/s
• Distance penetrated = 50 cm = 0.5 m
• Bullet stops -> final velocity = 0

Use equation of motion

• v² = u² + 2as

Here:

• v = 0 (final velocity)
• u = 100 m/s (initial velocity)
• s = 0.5 m

• 0 = ( 100 )² + 2.a.0.5
• 0 = 10000 + a
• a = −10000 m/s²

Negative sign means deceleration.

• F = m . a
• F = 0.05 × 10000
• = 500 N

29. An ace footballer converted a penalty shot by kicking the football with a speed of 108 km h–1. The estimated force they imparted was 800 N. The mass of the football was 0.4 kg. Calculate the time of contact between their foot and the ball.

Answer: The time of contact between their foot and the ball is 0.015 seconds

Given

• Speed of football = 108 km/h
• Mass of football = 0.4 kg
• Force applied = 800 N

Convert speed to m/s

\[ 108 \ \text{km/h} = \frac{108 \times 1000}{3600} = 30 \ \text{m/s} \]

So, ball speed = 30 m/s.

• Use the impulse formula.
• Impulse = Change in momentum = Force × Time

F.t = m.v

Here:

• F = 800 N
• m = 0.4 kg
• v = 30 m/s

\[ 800 \cdot t = 0.4 \cdot 30 \]\[ 800 \cdot t = 12 \]\[ t = \frac{12}{800} = 0.015 \ \text{s} \]

30. An object of mass 2 kg moving with a constant velocity of 10 m s–1 encounters a rough patch where the force of friction on the object is 7 N. At the same time, an additional constant force of 3 N opposing the motion is applied on the object. After entering the rough patch, how much distance does the object travel before coming to rest?

Answer: The object travels 10 metres before coming to rest.

Given

• Mass of object = 2 kg
• Initial velocity = 10 m/s
• Friction force = 7 N
• Extra opposing force = 3 N
• Final velocity = 0 (object stops)

Net opposing force

• F_net = 7 + 3 = 10 N

Acceleration (deceleration)

• F = m . a
• 10 = 2 . a
• a = 5m/s²

Use equation of motion

• v² = u² + 2as

Here:

• v = 0
• u = 10
• a = -5

• 0 = (10)² + 2 . (-5) . s
• 0 = 100 – 10s
• 10s = 100
• s = 10m

30. A tractor pulls a harrow (a ploughing tool) of mass m1 with a net force F resulting in an acceleration of a1 . The same tractor pulls a trolley of mass m2 with a force F producing an acceleration of a2. If the tractor Fig. 6.42: A bar magnet and a magnetic compass now pulls the trolley with the harrow placed on it (with the same force F ), then obtain an expression for the resulting acceleration in terms of a1 and a2 . Ignore friction.

Answer:

Situation given,

• Tractor pulls harrow: acceleration a₁.
• Tractor pulls trolley: acceleration a₂.
• Same force F is used in both cases.
• Now tractor pulls harrow and trolley together.

For harrow:

\[ F = m_{1} \cdot a_{1} \;\;\Rightarrow\;\; m_{1} = \frac{F}{a_{1}} \]

For trolley:

\[ F = m_{2} \cdot a_{2} \;\;\Rightarrow\;\; m_{2} = \frac{F}{a_{2}} \]

Together:

\[ \text{Total mass} = m_{1} + m_{2} = \frac{F}{a_{1}} + \frac{F}{a_{2}} \]

Acceleration when combined:

\[ a = \frac{F}{m_{1} + m_{2}} = \frac{F}{\frac{F}{a_{1}} + \frac{F}{a_{2}}} \]

Simplify

\[ a = \frac{1}{\frac{1}{a_{1}} + \frac{1}{a_{2}}} \]

The acceleration of harrow and trolley together is:

\[ a = \frac{1}{\frac{1}{a_{1}} + \frac{1}{a_{2}}} \]

32. When the pole of a bar magnet is brought close to a magnetic compass, the bar magnet and the compass needle (which is also a magnet) exert a magnetic force on each other. As per Newton’s third law of motion, both the forces are equal in magnitude and opposite in direction. However, the compass needle moves, whereas the bar magnet does not move (Fig. 6.42). Explain why.

Answer: When the pole of a bar magnet is brought near a compass, both the magnet and the compass needle pull on each other with equal force.

But the compass needle moves and the bar magnet does not. The reason is simple:

• The compass needle is small and light. It is free to rotate, so even a small force makes it move.
• The bar magnet is big and heavy. It is fixed in place, so the same force cannot move it.

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