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class 9 science chapter 6 notes and book solutions

Chapter 6: How Forces Affect Motion

Class 9 Science | NCERT Exploration

Part 1 — Colourful Short Notes

1. Force

A force is a push or pull that can change an object’s speed, direction of motion or shape.

  • Force can start the motion of an object at rest.
  • Force can speed up, slow down or change the direction of a moving object.
  • Force can change the shape of an object, such as when we squeeze a lemon.
  • Force has both magnitude and direction.
SI unit of force: newton (N)

A spring balance can be used to measure the magnitude of a force.

2. Balanced and Unbalanced Forces

Balanced forces: Equal forces acting in opposite directions produce zero net force. They do not change an object’s velocity.

Unbalanced forces: Forces that do not cancel each other produce a non-zero net force and change an object’s velocity.

  • Same direction: add the forces.
  • Opposite directions: subtract the smaller force from the larger one.
  • The net force acts in the direction of the larger force when two opposite forces are unequal.
Example: Forces of 10 N and 6 N acting to the right give a net force of 16 N to the right. If they act in opposite directions, the net force is 4 N towards the 10 N force.

3. Force of Friction

Friction is a force that opposes the relative motion or attempted motion between surfaces in contact.

  • Friction often acts opposite to the direction of motion.
  • Rougher surfaces generally produce greater friction than smoother surfaces.
  • Friction can slow a moving object and eventually stop it.
  • Friction is also useful: it helps us walk, grip objects and allows tyres to move safely on roads.
  • Air also exerts friction on moving objects; this is called air resistance or drag.
Example: A box begins to move when the applied force becomes greater than the opposing frictional force.

4. Newton’s First Law of Motion

An object at rest remains at rest, and an object in motion continues to move with constant velocity, unless a net force acts upon it.

  • When net force is zero, acceleration is zero.
  • An object already moving can continue at constant velocity without a continuous push if no net force acts on it.
  • The tendency to resist a change in the state of rest or motion is called inertia.
Net force = 0 → acceleration = 0

Constant velocity means constant speed in a constant direction.

5. Newton’s Second Law of Motion

A net force produces acceleration in the direction of that force. For a given mass, greater net force produces greater acceleration. For a given force, greater mass produces smaller acceleration.

F = ma

Here, F = net force in newtons, m = mass in kilograms, and a = acceleration in m/s2.

  • 1 N = 1 kg m/s2.
  • Weight is the gravitational force acting on an object.
Weight: W = mg

Near Earth’s surface, g is approximately 9.8 m/s2. For quick estimates, g may be taken as 10 m/s2.

Momentum: Momentum is the product of mass and velocity.

Momentum = mass × velocity, or p = mv

Increasing the time taken to stop a moving object reduces the average force needed to stop it for the same change in momentum. This is why airbags, cushioning and landing mats improve safety.

6. Newton’s Third Law of Motion

Whenever one object exerts a force on a second object, the second object simultaneously exerts an equal and opposite force on the first object.

  • Action and reaction forces are equal in magnitude and opposite in direction.
  • They act on two different objects, so they do not cancel each other on a single object.
  • These forces occur in contact and non-contact interactions.
Examples: Walking, rowing a canoe, a rocket launching, a gun recoiling and a person jumping from a boat.

7. Forces Acting on a System of Objects

Connected objects can often be treated as one system. Internal forces act between parts of the system, while external forces act on the system from outside.

For two connected objects of masses m1 and m2 pulled by an external force F on a frictionless surface:

a = Fm1 + m2

Only the net external force determines the acceleration of the whole system.

Force and motion: a simple illustration Applied force Friction Block

An object’s acceleration depends on the net force acting on it.

Part 2 — NCERT Book Solutions

Revise, Reflect, Refine — all 16 end-of-chapter questions

Question 1

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: Since the table moves at constant velocity, its acceleration is zero and the net force is zero. Therefore, the frictional force has magnitude F and acts opposite to the applied force.

Question 2

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) Remain the same.
  • (ii) Increase.
  • (iii) Decrease.

Question 3

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: Option (i). Block P experiences a net force of 5 − 4 = 1 N in the direction of the 5 N force. Block Q moves at constant velocity, so its net force is zero.

Question 4

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:

Forward force = 95 × 200 = 19,000 N.

Backward force = 5 × 200 = 1,000 N.

Net force = 19,000 − 1,000 = 18,000 N forward.

Question 5

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: Option (iv). Acceleration is in the direction of the net force and is directly proportional to the net force for a fixed mass.

Question 6

The position-time graph for four objects A, B, C and D moving along a straight line are given in Fig. 6.37. A net force acts on:

(i) Object A

(ii) Object B

(iii) Object C

(iv) Object D

Answer: A net force acts on an object whose velocity changes, which is shown by a curved position-time graph. A straight sloping line represents constant velocity, while a horizontal line represents rest. Check the curves in Fig. 6.37 in your textbook to select the corresponding object(s), because the graph itself is not reproduced in the supplied text.

Question 7

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 moves backwards as the sailor jumps forward. The sailor pushes the boat backwards, and the boat pushes the sailor forwards with an equal and opposite force, according to Newton’s third law.

Question 8

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: The mat increases the time taken to stop the athlete. This reduces the average force on the body and helps prevent injury.

Question 9

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: Option (iv). According to Newton’s third law, both carts exert equal and opposite forces on each other during the collision.

Question 10

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: Since F = ma, the force is constant for this graph because the acceleration varies inversely with mass. The force-mass graph is a horizontal straight line, parallel to the mass axis, at the constant value of F.
F Mass (kg) Constant force 0

Question 11

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: Find the acceleration from the slope of the velocity-time graph, then use F = ma.

Acceleration: a = (v − u) / t, using the initial and final values read from Fig. 6.41.

Force: F = 10a newtons.

The exact numerical force depends on the graph’s endpoints. Read the velocity values and corresponding times carefully from the original figure before entering the final number.

Question 12

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:

Mass, m = 50 g = 0.05 kg; initial speed, u = 100 m/s; final speed, v = 0; distance, s = 0.5 m.

Using v2 = u2 + 2as:

0 = 1002 + 2a(0.5)

a = −10,000 m/s2

Using F = ma:

F = 0.05 × (−10,000) = −500 N.

The stopping force has magnitude 500 N, opposite to the bullet’s motion.

Question 13

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:

Final speed = 108 ÷ 3.6 = 30 m/s. Assuming the ball was initially at rest, u = 0.

Using F = ma, a = 800 ÷ 0.4 = 2000 m/s2.

Using v = u + at:

30 = 0 + 2000t

t = 30 ÷ 2000 = 0.015 s.

Time of contact = 0.015 s.

Question 14

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:

Total opposing force = 7 + 3 = 10 N.

Acceleration, a = F/m = −10/2 = −5 m/s2.

Using v2 = u2 + 2as, with v = 0 and u = 10 m/s:

0 = 100 − 10s, so s = 10 m.

The object travels 10 m before stopping.

Question 15

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 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:

For the harrow, F = m1a1, so m1 = F/a1.

For the trolley, F = m2a2, so m2 = F/a2.

When both are pulled together, their total mass is m1 + m2.

a = F/(m1 + m2)

Substituting the expressions for the masses:

a = F/(F/a1 + F/a2)

Result: a = a1a2a1 + a2.

Question 16

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: Both objects exert equal and opposite forces on each other. However, the compass needle has a very small mass and is free to rotate, so it moves noticeably. The bar magnet has a much larger mass and is usually held or supported, so its motion is too small to notice.

Related Visual References

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StudyCBSE.online | Class 9 Science | Chapter 6: How Forces Affect Motion