Activity 8.1: Understanding Inertia
Objective: To demonstrate the concept of inertia and how it affects the motion of objects.
Materials Needed: Carom coins, striker, table.
Procedure:
1. Make a pile of similar carom coins on a table.
2. Attempt a sharp horizontal hit at the bottom of the pile using another carom coin or the striker.
3. Observe the result.
Observation: The bottom coin moves out quickly, while the rest of the coins in the pile fall vertically and remain almost in the same place.
Explanation: This activity demonstrates inertia, which is the tendency of objects to resist changes in their state of motion. The inertia of the coins keeps them in their vertical position even after the bottom coin is struck out.
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Activity 8.2: Demonstrating Inertia with a Coin and Card
Objective: To demonstrate inertia using a coin and a card.
Materials Needed: Coin (five-rupee or one-rupee), stiff card, glass tumbler.
Procedure:
1. Place a coin on a stiff card covering an empty glass tumbler.
2. Give the card a sharp horizontal flick with a finger.
Observation: The card shoots away, and the coin falls vertically into the glass tumbler.
Explanation: The inertia of the coin tries to maintain its state of rest even when the card is flicked away. This activity demonstrates that objects at rest tend to stay at rest unless acted upon by an external force.
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Activity 8.3: Inertia and Spilling Water
Objective: To understand the effect of inertia when an object is in motion.
Materials Needed: Water-filled tumbler, tray.
Procedure:
1. Place a water-filled tumbler on a tray.
2. Hold the tray and turn around as fast as you can.
Observation: The water spills out of the tumbler.
Explanation: The water tends to remain in its state of motion due to inertia while the tray moves in a circular path. This activity shows how inertia affects the motion of liquids and solids differently.
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Activity 8.4: Demonstrating Action and Reaction Forces
Objective: To demonstrate Newton’s third law of motion: “For every action, there is an equal and opposite reaction.”
Materials Needed: Two carts, bag of sand or heavy object, white paint (optional).
Procedure:
1. Request two children to stand on two separate carts facing each other.
2. Give them a bag full of sand or some other heavy object.
3. Ask them to play a game of catch with the bag.
Observation: When one child throws the bag, both carts move in opposite directions.
Explanation: Each child experiences an instantaneous force as a result of throwing the sandbag. This demonstrates Newton’s third law of motion, as the action of throwing the bag results in an equal and opposite reaction force on the carts.
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Activity 8.5: Action and Reaction in Daily Life
Objective: To demonstrate action and reaction forces using a real-life scenario.
Materials Needed: None (optional: discussion and observation in a daily setting).
Procedure:
1. Consider the scenario of walking. When you walk, you push the ground backward with your feet.
2. Observe the reaction force exerted by the ground, which pushes you forward.
Explanation: This everyday activity demonstrates Newton’s third law of motion, as the action of pushing the ground backward results in an equal and opposite reaction force that propels you forward.
In-Text Questions
1. Which of the following has more inertia: (a) a rubber ball and a stone of the same size? (b) a bicycle and a train? (c) a five-rupee coin and a one-rupee coin?
Answer: (a) A stone of the same size. (b) A train. (c) A five-rupee coin.
2. In the following example, try to identify the number of times the velocity of the ball changes: “A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team.” Also identify the agent supplying the force in each case.
Answer: The velocity of the ball changes three times:
1. When the first player kicks the ball.
2. When the second player kicks the ball.
3. When the goalkeeper collects the ball and kicks it.
The agents supplying the force are the first player, the second player, and the goalkeeper.
3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Answer: When the branch of a tree is vigorously shaken, the leaves tend to remain in their state of rest due to inertia, while the branch moves. This causes the leaves to detach from the branch.
4. Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?
Answer: When a moving bus brakes to a stop, your body tends to remain in motion due to inertia, causing you to fall forward. When the bus accelerates from rest, your body tends to remain at rest due to inertia, causing you to fall backward.
End of Chapter Exercises
1. An object experiences a net zero external unbalanced force. Is it possible for the object to be traveling with a non-zero velocity? If yes, state the conditions that must be placed on the object.
Answer: Yes, it is possible for the object to be traveling with a non-zero velocity if it was already in motion. According to Newton’s first law of motion, an object will continue to move with a constant velocity if no net external force acts on it.
2. When a carpet is beaten with a stick, dust comes out of it. Explain.
Answer: When a carpet is beaten with a stick, the carpet moves suddenly but the dust particles tend to remain at rest due to inertia. As a result, the dust gets detached from the carpet and falls off.
3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Answer: It is advised to tie any luggage kept on the roof of a bus with a rope to prevent it from falling off when the bus suddenly starts, stops, or takes a sharp turn. The inertia of the luggage would otherwise cause it to remain in its state of rest or motion, which could result in it being thrown off the roof.
4. A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because
(a) the batsman did not hit the ball hard enough. (b) velocity is proportional to the force exerted on the ball. (c) there is a force on the ball opposing the motion. (d) there is no unbalanced force on the ball, so the ball would want to come to rest.
Answer: (c) there is a force on the ball opposing the motion.
5. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes (Hint: 1 tonne = 1000 kg).
Answer:
• Distance, s = 400 m
• Time, t = 20 s
• Initial velocity, u = 0 m/s
Using the equation of motion: s = ut + (1/2)at²
400 = 0 + (1/2)a(20)²
400 = 200a
a = 2 m/s²
• Mass, m = 7 tonnes = 7000 kg
Using Newton’s second law of motion: F = ma
F = 7000 kg × 2 m/s²
F = 14000 N
6. A stone of 1 kg is thrown with a velocity of 20 m/s across the frozen surface of a lake and comes to rest after traveling a distance of 50 m. What is the force of friction between the stone and the ice?
Answer:
• Mass, m = 1 kg
• Initial velocity, u = 20 m/s
• Final velocity, v = 0 m/s
• Distance, s = 50 m
Using the equation of motion: v² = u² + 2as
0 = (20)² + 2a(50)
0 = 400 + 100a
a = -4 m/s²
Using Newton’s second law of motion: F = ma
F = 1 kg × (-4 m/s²)
F = -4 N
The negative sign indicates that the force of friction is opposite to the direction of motion.
7. A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate: (a) the net accelerating force and (b) the acceleration of the train.
Answer:
• Mass of engine, m₁ = 8000 kg
• Mass of 5 wagons, m₂ = 5 × 2000 kg = 10000 kg
• Total mass, m = m₁ + m₂ = 18000 kg
• Force exerted by engine, F = 40000 N
• Friction force, f = 5000 N
(a) Net accelerating force: F_net = F – f
F_net = 40000 N – 5000 N
F_net = 35000 N
(b) Acceleration: Using Newton’s second law of motion: F_net = ma
35000 N = 18000 kg × a
a = 35000 N / 18000 kg
a = 1.94 m/s²
8. An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 m/s²?
Answer:
• Mass, m = 1500 kg
• Acceleration, a = -1.7 m/s²
Using Newton’s second law of motion: F = ma
F = 1500 kg × (-1.7 m/s²)
F = -2550 N
The negative sign indicates that the force is in the direction opposite to the motion.
9. What is the momentum of an object of mass m, moving with a velocity v?
(a) (mv)² (b) mv² (c) ½ mv² (d) mv
Answer: (d) mv
10. Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
Answer: The friction force that will be exerted on the cabinet is 200 N. Since the cabinet is moving at a constant velocity, the applied force and the friction force must be equal and opposite.
11. According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.
Answer: The student’s logic is incorrect. The two forces do not cancel each other because they act on different objects. When you push the truck, the truck pushes back with an equal and opposite force. However, the truck does not move because the force you exert is not sufficient to overcome the truck’s inertia and the frictional force between the truck’s tires and the ground.
12. A hockey ball of mass 200 g traveling at 10 m/s is struck by a hockey stick so as to return it along its original path with a velocity of 5 m/s. Calculate the magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Answer:
• Mass, m = 200 g = 0.2 kg
• Initial velocity, u = 10 m/s
• Final velocity, v = -5 m/s (opposite direction)
Change in momentum: Δp = mv – mu
Δp = 0.2 kg × (-5 m/s) – 0.2 kg × 10 m/s
Δp = -1 kg m/s – 2 kg m/s
Δp = -3 kg m/s
The magnitude of change of momentum is 3 kg m/s.
13. A bullet of mass 10 g traveling horizontally with a velocity of 150 m/s strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.
Answer:
• Mass, m = 10 g = 0.01 kg
• Initial velocity, u = 150 m/s
• Final velocity, v = 0 m/s
• Time, t = 0.03 s
Using the equation of motion:
v = u + at
0 = 150 + a(0.03)
a = -150 / 0.03
a = -5000 m/s²
Using Newton’s second law of motion: F = ma
F = 0.01 kg × (-5000 m/s²)
F = -50 N
Distance of penetration: Using the equation of motion: v² = u² + 2as
0 = (150)² + 2(-5000)s
0 = 22500 – 10000s
10000s = 22500
s = 22500 / 10000
s = 2.25 m
The distance of penetration is 2.25 m, and the magnitude of the force exerted by the wooden block on the bullet is 50 N.
14. An object of mass 1 kg traveling in a straight line with a velocity of 10 m/s collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.
Answer:
• Mass of object, m₁ = 1 kg
• Velocity of object, u₁ = 10 m/s
• Mass of block, m₂ = 5 kg
• Velocity of block, u₂ = 0 m/s
Total momentum before impact: p₁ = m₁u₁ + m₂u₂
p₁ = 1 kg × 10 m/s + 5 kg × 0 m/s
p₁ = 10 kg m/s
After the impact, the combined mass: m = m₁ + m₂
m = 1 kg + 5 kg m = 6 kg
Using the law of conservation of momentum: p₁ = p₂
10 kg m/s = 6 kg × v
v = 10 kg m/s / 6 kg
v = 5/3 m/s
v ≈ 1.67 m/s
The total momentum just before and just after the impact is 10 kg m/s, and the velocity of the combined object is approximately 1.67 m/s.
15. An object of mass 100 kg is accelerated uniformly from a velocity of 5 m/s to 8 m/s in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
Answer:
• Mass, m = 100 kg
• Initial velocity, u = 5 m/s
• Final velocity, v = 8 m/s
• Time, t = 6 s
Initial momentum: p_initial = mu
p_initial = 100 kg × 5 m/s
p_initial = 500 kg m/s
Final momentum: p_final = mv
p_final = 100 kg × 8 m/s
p_final = 800 kg m/s
Force exerted: Using the equation of motion: v = u + at
8 = 5 + a(6)
a = (8 – 5) / 6
a = 3 / 6
a = 0.5 m/s²
Using Newton’s second law of motion: F = ma
F = 100 kg × 0.5 m/s²
F = 50 N
The initial momentum is 500 kg m/s, the final momentum is 800 kg m/s, and the force exerted on the object is 50 N.
16. Akhtar, Kiran, and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result, the insect died. Rahul, while putting an entirely new explanation, said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.
Answer: Rahul is correct. According to Newton’s third law of motion, both the motorcar and the insect experienced the same magnitude of force but in opposite directions. The change in momentum of the insect was significant due to its small mass and the large force exerted by the car. The change in momentum of the car was negligible because of its large mass compared to the insect. Thus, both the insect and the motorcar experienced equal and opposite forces, but the impact was more severe on the insect due to its smaller mass.
17. How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 m/s².
Answer:
• Mass, m = 10 kg
• Height, h = 80 cm = 0.8 m
• Acceleration due to gravity, g = 10 m/s²
Using the equation of motion: v² = u² + 2gh
v² = 0 + 2(10)(0.8)
v² = 16
v = 4 m/s
Momentum transferred to the floor: p = mv
p = 10 kg × 4 m/s
p = 40 kg m/s
Additional Exercises
A1. The following is the distance-time table of an object in motion:
Time (seconds) | Distance (meters) 0 | 0 1 | 1 2 | 8 3 | 27 4 | 64 5 | 125 6 | 216 7 | 343
(a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero?
Answer: The distances are the cubes of the time values (1³, 2³, 3³, etc.), indicating that the acceleration is increasing.
(b) What do you infer about the forces acting on the object?
Answer: The forces acting on the object are increasing because the acceleration is increasing.
A2. Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 m/s². With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.)
Answer:
• Mass, m = 1200 kg
• Acceleration, a = 0.2 m/s²
• Number of persons = 3
Total force, F = ma
F = 1200 kg × 0.2 m/s²
F = 240 N
Force per person: F_person = 240 N / 3
F_person = 80 N
Each person pushes the motorcar with a force of 80 N.
A3. A hammer of mass 500 g, moving at 50 m/s, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?
Answer:
• Mass, m = 500 g = 0.5 kg
• Initial velocity, u = 50 m/s
• Final velocity, v = 0 m/s
• Time, t = 0.01 s
Using the equation of motion: v = u + at
0 = 50 + a(0.01)
a = -50 / 0.01
a = -5000 m/s²
Using Newton’s second law of motion: F = ma
F = 0.5 kg × (-5000 m/s²)
F = -2500 N
The force of the nail on the hammer is 2500 N in the opposite direction.
A4. A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also, calculate the magnitude of the force required.
Answer:
• Mass, m = 1200 kg
• Initial velocity, u = 90 km/h = 25 m/s
• Final velocity, v = 18 km/h = 5 m/s
• Time, t = 4 s
Acceleration, a: Using the equation of motion: v = u + at
5 = 25 + a(4)
a = (5 – 25) / 4
a = -20 / 4
a = -5 m/s²
Change in momentum: Initial momentum, p_initial = mu
p_initial = 1200 kg × 25 m/s
p_initial = 30000 kg m/s
Final momentum, p_final = mv
p_final = 1200 kg × 5 m/s
p_final = 6000 kg m/s
Change in momentum, Δp: Δp = p_final – p_initial
Δp = 6000 kg m/s – 30000 kg m/s
Δp = -24000 kg m/s
Magnitude of the force required: Using Newton’s second law of motion: F = ma
F = 1200 kg × (-5 m/s²)
F = -6000 N
The magnitude of the force required is 6000 N in the opposite direction of motion.