NCERT Solutions: How Forces Affect Motion
Think It Over (Page 94)
Q1: Why does a canoe move forward when the canoeist pushes water backwards with the paddle?
Ans: When the canoeist pushes water backward with the paddle, the water exerts an equal and opposite force on the canoe. According to Newton's third law of motion, every action has an equal and opposite reaction. This reaction force acts in the forward direction on the canoe, causing it to move ahead.
Q2: Suppose the same amount of water is pushed harder in two different canoes, one empty and the other carrying two passengers. In which case will the canoe move faster?
Ans: The empty canoe will move faster because it has less mass. According to Newton's second law of motion, acceleration is inversely proportional to mass. So, for the same applied force, a lighter canoe gains more acceleration and hence moves faster than a heavier canoe carrying passengers.
Pause and Ponder (Page 97)
Q1. 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: Two forces acting on the barbell are:
- Gravitational force (weight): The Earth pulls the barbell downward.
- Upward force applied by the weightlifter: She pushes the barbell upward.
Yes, these forces are balanced when the weightlifter keeps the barbell steady. Since the barbell is stationary, the net force on it is zero, meaning the two forces are equal in magnitude and opposite in direction.
Q2. 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: When the arms tilt to the front direction, the forces are NOT balanced. A net force is acting that caused the arms to move from their initial position. Since the arms moved in a direction associated with player R's push, player R exerted the larger force.
Pause and Ponder (Page 101)
Q3. An object is moving with a constant velocity. Is there a net force acting upon it?Answer: No. According to Newton's first law of motion, if an object is moving with a constant velocity, the net force acting on it is zero. Constant velocity means no change in speed or direction, which means acceleration is zero. By F = ma, zero acceleration means zero net force.
Q4. 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:
(i) Possible. If the object is at rest and no net force acts on it, it will remain at rest (Newton's first law).
(ii) Possible. If the object is already moving and no net force acts on it, it will continue to move with the same constant velocity (Newton's first law).
(iii) Not possible. A constant acceleration requires a net force (F = ma). If the net force is zero, acceleration must also be zero.
Q5. 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: Example: Consider a box being pushed across the floor at a constant velocity. Two forces act on the box: the applied force (forward) and the force of friction (backward). If the magnitude of the applied force is made exactly equal to the force of friction, the two forces are equal and opposite, so the net force on the box is zero. Even though forces are present, the net force is zero - and the box moves with constant velocity (Newton's first law).
Pause and Ponder (Page 106)
Q6. 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: The toy car is moving with constant velocity. By Newton's first law, constant velocity means acceleration is zero. Using F = ma: F = 0.1 kg × 0 = 0 NThe net force acting on the toy car is zero.
Q7. 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: You would require a larger force for the heavier child. By Newton's second law: F = ma. For the same acceleration (a) to be imparted, the force (F) must be proportional to the mass (m). The heavier child has a larger mass, so a larger force must be applied to produce the same acceleration.
Q8. How are glass items packed for transportation using a bubble wrap or hay protected from damage?
Answer: When glass items experience a jolt during transportation, they are in contact with bubble wrap or hay. When a force is applied on the glass item (e.g., due to vibration or a knock), the soft material increases the time over which the velocity of the item changes. By Newton's second law, a longer time for the same change in velocity means a smaller acceleration, and therefore a smaller force on the glass. This smaller force is less likely to break the fragile item.
Pause and Ponder (Page 110)
Q9. Why does a fireperson sometimes struggle when holding the pipe issuing water?
Answer: According to Newton's third law of motion, when water is expelled from the pipe at high speed in the forward direction, the pipe (and the person holding it) experiences an equal and opposite reaction force in the backward direction. This backward reaction force on the pipe can be large if the water comes out at high speed and in large volume, making it difficult for the fireperson to hold the pipe steady.
Q10. Suppose a spacecraft is moving in a region of space where the gravitational force acting upon it is negligible. Suggest how it can change its velocity.
Answer: In a region where gravity is negligible, the spacecraft can change its velocity by firing its rocket engine. The engine expels exhaust gases in one direction; by Newton's third law, these gases exert an equal and opposite force on the spacecraft, causing it to accelerate in the opposite direction.
- To slow down: fire the engine in the direction of motion (exhaust goes forward, spacecraft decelerates).
- To speed up: fire the engine opposite to the direction of motion.
- To change direction: fire the engine at an appropriate angle.
Revise, Reflect, Refine
Q1. 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 is moving at constant velocity, its acceleration is zero. By Newton's first law, the net force on it is zero. The two horizontal forces acting on the table are the applied force F (forward) and friction (backward). For net force to be zero, friction must equal F in magnitude. Therefore, the frictional force exerted by the floor on the table is equal to F (acting in the direction opposite to motion).
Q2. 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. By Newton's first law, with no net force, velocity does not change. (ii) Increase. A net force in the direction of motion produces acceleration in the same direction, increasing speed. (iii) Decrease. A net force opposite to the direction of motion produces deceleration, decreasing speed.
Q3. 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 statements 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.
For Block P: Forces 5 N and 4 N act in opposite directions. Net force = 5 - 4 = 1 N. P experiences a net force. For Block Q: It is moving with constant velocity. By Newton's first law, constant velocity means zero acceleration, which means zero net force.
Q4. 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: Force by 95 oarsmen (forward) = 95 × 200 = 19,000 N Force by 5 oarsmen (backward) = 5 × 200 = 1,000 N Net force = 19,000 - 1,000 = 18,000 N in the forward direction.
Q5. 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.
By Newton's second law, F = ma, so a = F/m. The acceleration is in the direction of the net force, and is directly proportional to the net force (for fixed mass).
Q6. 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 only on objects whose acceleration is non-zero.
- Object A: Position-time graph is a straight line with constant slope → constant velocity → acceleration = 0 → no net force.
- Object B: Position-time graph is a horizontal line → object is at rest → acceleration = 0 → no net force.
- Object C: Position-time graph is a curve with increasing slope → velocity is increasing → acceleration ≠ 0 → net force acts.
- Object D: Position-time graph is a curve with decreasing slope → velocity is decreasing → acceleration ≠ 0 → net force acts.
Answer: A net force acts on (iii) Object C and (iv) Object D.
Q7. 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 (towards the shore), by Newton's third law, the sailor's feet apply a force on the boat in the backward direction (away from shore). The boat, being on water with negligible friction, will move backward (away from shore) in response to this equal and opposite reaction force.
Q8. 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: When the athlete falls onto a soft mat or sand, the time taken for their velocity to reduce to zero is increased (compared to falling on a hard surface). By Newton's second law, for the same change in velocity, a longer time results in a smaller acceleration (deceleration). A smaller acceleration means a smaller force is exerted on the athlete's body. This significantly reduces the risk of injury upon landing.
Q9. 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.
According to Newton's third law, during the collision, the force exerted by the loaded cart on the empty cart and the force exerted by the empty cart on the loaded cart are equal in magnitude and opposite in direction - regardless of their masses.
Q10. 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: From Newton's second law: F = ma.
The acceleration-mass graph in Fig. 6.40 shows a hyperbolic curve (a = F/m - inversely proportional relationship for constant F), confirming that the same force is applied throughout.
For the force-mass graph: Since F = ma and F is constant (the same force is applied on objects of different masses), the force does not change regardless of mass.
Therefore, the force-mass graph will be a horizontal straight line (parallel to the mass axis) at the constant value of F.
Q11. 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: From Fig. 6.41, the velocity increases linearly from 0 to 30 m s⁻¹ in 8 seconds (straight line graph).
Acceleration = Change in velocity / Time = (30 - 0) / 8 = 3.75 m s⁻²
Using F = ma: F = 10 kg × 3.75 m s⁻² = 37.5 N
Q12. 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: Given: m = 50 g = 0.05 kg, u = 100 m s⁻¹, v = 0 m s⁻¹, s = 50 cm = 0.5 m
Using v² = u² + 2as: 0 = (100)² + 2 × a × 0.5 0 = 10000 + a a = -10000 m s⁻²
Force = ma = 0.05 × 10000 = 500 N (acting in the direction opposite to motion of the bullet)
Q13. 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: Converting speed: 108 km h⁻¹ = 108 × (1000/3600) = 30 m s⁻¹
Initial velocity of ball u = 0 (stationary), Final velocity v = 30 m s⁻¹
Using F = m(v - u)/t: 800 = 0.4 × (30 - 0) / t 800t = 12 t = 12/800 = 0.015 s
Q14. 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
Using Newton's second law: a = -F/m = -10/2 = -5 m s⁻² (deceleration)
Initial velocity u = 10 m s⁻¹, final velocity v = 0
Using v² = u² + 2as: 0 = (10)² + 2 × (-5) × s 0 = 100 - 10s s = 10 m
The object travels 10 m before coming to rest.
Q15. 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 harrow alone: F = m₁a₁ → m₁ = F/a₁
For trolley alone: F = m₂a₂ → m₂ = F/a₂
Combined mass = m₁ + m₂ = F/a₁ + F/a₂ = F(a₁ + a₂)/(a₁a₂)
Acceleration when combined: a = F / (m₁ + m₂) = F / [F(a₁ + a₂)/(a₁a₂)]
a = (a₁a₂) / (a₁ + a₂)
Q16. 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 the compass needle and the bar magnet experience equal and opposite forces (Newton's third law). However, their masses are very different. The bar magnet is much heavier (larger mass) than the compass needle.
By Newton's second law: a = F/m. For the same force F, the compass needle (very small mass) experiences a very large acceleration and moves noticeably. The bar magnet (large mass) experiences a very small acceleration - too small to be observed.
This is why only the compass needle appears to move, while the bar magnet does not.
FAQs on NCERT Solutions: How Forces Affect Motion
| 1. What are the basic concepts of motion as explained in the article? |  |
| 2. How do forces affect motion according to the content provided? | |
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| 4. Can you explain the difference between distance and displacement as discussed in the article? | |
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