Revise, Reflect, Refine
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, acceleration = 0, so net force = 0.
This means frictional force = applied force = F (in the opposite direction)
The frictional force exerted by the floor on the table is equal to F but acts in the opposite/backward direction.
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.
Answer: Velocity will remain the same.
Since no net force acts, by Newton’s 1st Law, the ball continues moving with the same constant velocity.
(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: Velocity will increase.
Force in the direction of motion causes acceleration in the same direction (F = ma), so the ball speeds up.
(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: Velocity will decrease.
The force opposes the motion, causing a negative acceleration (deceleration). By Newton’s 2nd Law, the ball slows down and may eventually stop.
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: (i) P experiences a net force and Q does not experience a net force.
Explanation:
Block P:
- Force towards right = 5 N
- Force towards left = 4 N
- Net force = 5 – 4 = 1 N (towards right)
- ∴ P experiences a net force
Block Q:
- Moving with constant velocity
- By Newton’s 1st Law, constant velocity means acceleration = 0
- Therefore net force = ma = 0 N
- ∴ Q does not experience a net force
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: Net force on the snake boat:
Given:
- Total oarsmen = 100
- Oarsmen rowing correctly (forward) = 95
- Oarsmen rowing wrongly (backward) = 5
- Force by each oarsman = 200 N
Forward force = 95 × 200 = 19000 N
Backward force = 5 × 200 = 1000 N
Net force = 19000 – 1000 = 18000 N (forward)
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: (iv) in the direction of force, with acceleration proportional to the force acting on the object.
Explanation:
By Newton’s 2nd Law: F = ma, so a = F/m
- Object accelerates in the same direction as net force
- Acceleration is directly proportional to force (for fixed mass)
- Acceleration is inversely proportional to mass (for fixed force)
Options (i) and (ii) are wrong — acceleration is in the same direction as force, not opposite.
Option (iii) is wrong — acceleration is directly proportional to force, not inversely.
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: To find where a net force acts, we check whether the object has acceleration (i.e., curved graph or changing slope).
Analysis of the graphs:
- Object A: Straight line with constant slope → constant velocity → no net force
- Object B: Horizontal line → at rest → no net force
- Object C: Curved graph → velocity changing → acceleration present → net force acts
- Object D: Straight line but decreasing position linearly → constant velocity (in opposite direction) → no net force
Correct answer: (iii) Object C
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 will move.
When the sailor jumps forward, he pushes the boat backward.
By Newton’s third law, the boat experiences an equal and opposite force.
So the boat moves backward because of the reaction force from the sailor’s jump.
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: A landing mat or sand bed is used because it increases the time taken by the athlete to come to rest after landing.
When the time of impact increases:
acceleration decreases,
so the force on the athlete becomes much smaller (from Newton’s second law).
This reduces the risk of injury and provides a safe landing.
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: According to Newton’s third law of motion,
when two objects interact, they exert equal and opposite forces on each other.
So, even though one cart is loaded and the other is empty, both carts exert equal forces on each other during the collision.
(iv) The loaded cart and the empty cart exert equal magnitude of force on each other.
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: Force-Mass Graph:
First, let’s find the force from the graph:
From Fig. 6.40, reading values:
Conclusion: Force = constant = 10 N for all masses
The graph is a horizontal straight line at F = 10 N
Reason:
From a = F/m → F = ma
As mass increases, acceleration decreases proportionally
So their product (Force) remains constant
This confirms Newton’s 2nd Law
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: Reading the graph:
From Fig. 6.41:
At t = 0 s → v = 10 m/s
At t = 8 s → v = 30 m/s
Calculating acceleration:
a = (v – u) / t = (30 – 10) / (8 – 0) = 20 / 8 = 2.5 m/s2
Then force:
F = ma = 10 × 2.5 = 25
12. A bullet of mass 50 g moving with a speed of 100 m s1 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 = 50g = 0.05 kg, u = 100 m/s, v = 0, s = 50cm = 0.5 m
Using: v² = u² + 2as
13. An ace footballer converted a penalty shot by kicking the football with a speed of 108 km h1. 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: Given: v = 108 km/h = 30 m/s, u = 0, F = 800 N, m = 0.4 kg
Using: F = ma → a = F/m = 800/0.4 = 2000 m/s²
Using: v = u + at
14. An object of mass 2 kg moving with a constant velocity of 10 m s1 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: Given: m = 2 kg, u = 10 m/s, v = 0
Total opposing force = 7 + 3 = 10 N
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: From given conditions:
When harrow is placed on trolley, total mass = m₁ + m₂
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: When the pole of a bar magnet is brought near a magnetic compass, the bar magnet and the compass needle exert equal and opposite magnetic forces on each other, as stated by Newton’s third law of motion. However, their responses are different because the effect of a force depends on the mass (inertia) of the object it acts on.
The compass needle is very light and has small inertia, so even a small magnetic force produces a noticeable acceleration. This makes the needle turn or move easily.
On the other hand, the bar magnet is much heavier, so its inertia is much larger. The same magnetic force acting on it produces an extremely small acceleration, too small to observe. Hence, the bar magnet appears to remain stationary.
Thus, although the forces are equal in magnitude and opposite in direction, the lighter compass needle moves while the heavier bar magnet does not.
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