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NCERT Class 9 Science How Forces Affect Motion: Exercise Solutions

Step-by-step working for every question of End-of-chapter exercises in NCERT's Class 9 Science book, chapter “How Forces Affect Motion”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

1. Exercises Q1

   A horizontal force F moves a table across the floor at constant velocity. What is the friction force from the floor on the table?

   1. The table is moving at a constant velocity, which means its speed and direction are not changing.
   2. According to the chapter, an object moving with constant velocity has no net force acting on it.
   3. The applied horizontal force F acts in the forward direction.
   4. The friction force from the floor acts in the opposite direction to the motion.
   5. Since the net force is zero, the friction force must be equal in magnitude to the applied force F.
   6. Therefore, the frictional force exerted by the floor is equal to F.

   Answer: The frictional force exerted by the floor on the table is equal to F.

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2. Exercises Q2

   A ball moves on a smooth frictionless surface. Choose same / increase / decrease: (i) with no net force its velocity will ___ (ii) with a net force along its motion its speed will ___ (iii) with a net force against its motion its speed will ___

   1. (i) According to Newton's first law, an object with no net force acting on it continues moving with constant velocity.
   2. (ii) When a net force acts in the direction of motion, the object accelerates, causing its speed to increase.
   3. (iii) When a net force acts opposite to the direction of motion, the object decelerates, causing its speed to decrease.

   Answer: (i) same; (ii) increase; (iii) decrease

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3. Exercises Q3

   Block P has forces of 5 N and 4 N acting on it in opposite directions; block Q moves at constant velocity. Which is correct? (i) P has a net force, Q does not (ii) P does not, Q does (iii) both have a net force (iv) neither has a net force

   Uses a figure printed in the NCERT book — open the chapter above to see it.

   1. For block P, the forces of 5 N and 4 N are in opposite directions but have different magnitudes.
   2. Since the forces on P are unequal, they are unbalanced, resulting in a non-zero net force.
   3. Block Q is moving with a constant velocity, which means its speed and direction do not change.
   4. According to the chapter, an object moving with constant velocity has no net force acting on it.
   5. Therefore, P experiences a net force while Q does not.

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

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4. Exercises Q4

   In a snake boat race practice, 95 of 100 oarsmen row to push the boat forward but 5 row the wrong way; each applies 200 N. What is the net force on the boat?

   1. Forward force from the 95 oarsmen rowing correctly = 95 × 200 N = 19000 N.
   2. Backward force from the 5 oarsmen rowing the wrong way = 5 × 200 N = 1000 N.
   3. The two forces act in opposite directions, so the net force = 19000 N − 1000 N = 18000 N.
   4. The same result: (95 − 5) × 200 N = 18000 N, in the forward direction.

   Answer: Net force = 18000 N (18,000 N) in the forward direction.

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5. Exercises Q5

   An object with a net force on it accelerates: (i) against the force, proportional to the force (ii) against the force, proportional to its mass (iii) along the force, inversely proportional to the force (iv) along the force, proportional to the force

   1. Newton's second law states that an object accelerates in the direction of the net force acting on it.
   2. The magnitude of this acceleration is directly proportional to the magnitude of the net force.
   3. Therefore, the object moves along the force direction with acceleration proportional to the force.

   Answer: (iv) along the force, proportional to the force

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6. Exercises Q6

   Position-time graphs (Fig. 6.37): A a rising straight line, B a flat line, C a curve getting steeper, D a falling straight line. On which object does a net force act? (i) A (ii) B (iii) C (iv) D

   Uses a figure printed in the NCERT book — open the chapter above to see it.

   1. (i) Graphs A and B show straight lines, indicating constant velocity or rest, which means no net force acts on these objects.
   2. (ii) Graph D shows a straight line with negative slope, indicating constant velocity in the opposite direction, so no net force acts on it.
   3. (iii) Graph C is a curve that gets steeper, showing that the velocity is changing over time.
   4. (iv) A change in velocity implies acceleration, and according to Newton's laws, a net force is required to cause this acceleration.

   Answer: (iii)

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7. Exercises Q7

   A sailor jumps forward from a small boat to the shore. Will the boat move? If so, which way and why?

   Uses a figure printed in the NCERT book — open the chapter above to see it.

   1. Yes, the boat will move when the sailor jumps forward.
   2. The sailor pushes the boat backward with his feet to propel himself toward the shore.
   3. By Newton's third law, the boat pushes the sailor forward with an equal and opposite force; the sailor's push acts on the boat, backwards.
   4. Since there is no other significant horizontal force to balance this push, the boat accelerates in the backward direction.

   Answer: Yes, the boat moves backward because the sailor applies a force on it in the opposite direction to his jump.

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8. Exercises Q8

   Why is a landing mat or sand bed placed where a high jumper lands?

   Uses a figure printed in the NCERT book — open the chapter above to see it.

   1. The high jumper falls from a height and has a certain velocity just before landing.
   2. Upon hitting the ground, the athlete's velocity must reduce to zero to stop the motion.
   3. According to Newton's second law, the force experienced depends on the rate of change of momentum.
   4. A soft landing mat or sand bed increases the time duration over which the athlete comes to rest.
   5. By increasing this time, the magnitude of the force acting on the athlete is reduced.
   6. This reduction in force prevents injury to the athlete during the impact.

   Answer: A landing mat or sand bed increases the time of impact, which reduces the force exerted on the athlete and prevents injury.

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9. Exercises Q9

   A loaded vegetable cart hits an identical empty cart. During the collision: (i) the loaded cart pushes harder (ii) the empty cart pushes harder (iii) neither pushes on the other (iv) both push on each other with equal force

   1. Newton's third law states that when one object exerts a force on another, the second object exerts an equal and opposite force back.
   2. These interaction forces always occur in pairs and act on two different objects simultaneously.
   3. Since the force pair is equal in magnitude, the loaded cart and the empty cart push on each other with the same strength during the collision.

   Answer: (iv) both push on each other with equal force

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10. Exercises Q10

    The acceleration-mass graph for one force (Fig. 6.40) shows 10 m s⁻² at 1 kg falling to 2.5 m s⁻² at 4 kg. Plot the force-mass graph.

    Uses a figure printed in the NCERT book — open the chapter above to see it.

    1. According to Newton's second law, the force acting on an object is the product of its mass and acceleration.
    2. Calculate the force for each data point by multiplying the given mass by the corresponding acceleration value.
    3. The force remains constant at 10 N for all mass values because the product of mass and acceleration is invariant.
    4. Plot the calculated force values against the mass values to obtain a horizontal straight line parallel to the mass axis.

    Answer: The force-mass graph is a horizontal straight line at 10 N, indicating that the force is constant regardless of the mass.

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11. Exercises Q11

    A 10 kg object's velocity rises uniformly from 10 m s⁻¹ at 0 s to 20 m s⁻¹ at 4 s and 30 m s⁻¹ at 8 s (Fig. 6.41). Find the force on it.

    Uses a figure printed in the NCERT book — open the chapter above to see it.

    1. The mass of the object is given as 10 kg.
    2. From the velocity-time graph, the velocity changes from 10 m/s to 20 m/s in 4 seconds.
    3. Calculate the acceleration using the formula a = (v - u) / t.
    4. Substitute the values: a = (20 - 10) / 4 = 10 / 4 = 2.5 m/s².
    5. Use Newton's second law of motion, F = ma, to find the force.
    6. Substitute the values: F = 10 × 2.5 = 25 N.

    Answer: 25 N

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12. Exercises Q12

    A 50 g bullet at 100 m s⁻¹ enters a heavy fixed wooden block and stops after 50 cm. Assuming constant deceleration, estimate the stopping force.

    1. Convert the mass of the bullet from grams to kilograms: m = 50 g = 0.05 kg.
    2. Identify the initial velocity u = 100 m/s and final velocity v = 0 m/s.
    3. Convert the stopping distance from centimeters to meters: s = 50 cm = 0.5 m.
    4. Use the equation of motion v² = u² + 2as to find the acceleration.
    5. Substitute the values: 0² = 100² + 2 × a × 0.5.
    6. Simplify the equation: 0 = 10000 + 1a.
    7. Solve for acceleration: a = -10000 m/s².
    8. The negative sign indicates deceleration, so the magnitude of acceleration is 10000 m/s².
    9. Apply Newton's second law to find the force: F = m × a.
    10. Substitute the values: F = 0.05 kg × 10000 m/s².
    11. Calculate the force: F = 500 N.

    Answer: 500 N

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13. Exercises Q13

    A footballer kicks a 0.4 kg ball to 108 km h⁻¹ with an estimated force of 800 N. Find the contact time.

    1. Convert the final velocity from 108 km/h to m/s by dividing by 3.6, which gives 30 m/s.
    2. Use the formula for impulse, which is Force multiplied by time equals the change in momentum.
    3. The change in momentum is the mass times the final velocity, so 0.4 kg times 30 m/s equals 12 kg m/s.
    4. Set up the equation 800 N times t equals 12 kg m/s.
    5. Solve for t by dividing 12 by 800.
    6. The contact time t is 0.015 s.

    Answer: 0.015 s

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14. Exercises Q14

    A 2 kg object moving at a steady 10 m s⁻¹ enters a rough patch with 7 N friction, and an extra 3 N force opposing the motion acts at the same time. How far does it go before stopping?

    1. The total opposing force is the sum of friction and the extra force: 7 N + 3 N = 10 N.
    2. Using Newton's second law, the deceleration is a = F/m = 10 N / 2 kg = 5 m s⁻².
    3. The object stops when its final velocity v becomes 0 m s⁻¹.
    4. Using the equation v² = u² − 2as, we substitute the values: 0² = 10² − 2 × 5 × s.
    5. Solving for distance s gives 100 = 10s, so s = 10 m.

    Answer: 10 m

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15. Exercises Q15

    A force F gives a harrow of mass m₁ acceleration a₁ and a trolley of mass m₂ acceleration a₂. With the harrow on the trolley and the same F, find the acceleration in terms of a₁ and a₂ (ignore friction).

    1. (i) Using Newton's second law, the force F equals m₁a₁ for the harrow and m₂a₂ for the trolley.
    2. (ii) When the harrow is on the trolley, the total mass is the sum m₁ + m₂.
    3. (iii) The new acceleration a is the force F divided by this total mass.
    4. (iv) Substituting m₁ = F/a₁ and m₂ = F/a₂ into the equation for a gives a = F / (F/a₁ + F/a₂).
    5. (v) The force F cancels out, leaving a = 1 / (1/a₁ + 1/a₂), which equals a₁a₂ / (a₁ + a₂).

    Answer: The resulting acceleration is a₁a₂ / (a₁ + a₂).

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16. Exercises Q16

    A bar magnet and a compass needle pull on each other with equal and opposite forces (Newton's third law), yet only the needle moves. Why?

    Uses a figure printed in the NCERT book — open the chapter above to see it.

    1. The bar magnet and the compass needle exert equal and opposite magnetic forces on each other as per Newton's third law.
    2. The compass needle has a very small mass compared to the bar magnet.
    3. According to Newton's second law, the same force produces a larger acceleration on the lighter needle than on the heavier magnet.
    4. The bar magnet is usually held fixed or has enough mass and friction to prevent noticeable motion.
    5. Thus, the needle rotates or moves while the bar magnet remains stationary.

    Answer: The needle moves because it has a much smaller mass, so the equal force causes it a larger acceleration, while the heavier bar magnet does not move significantly.

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