Unit 1: Laws of Motion (இயக்க விதிகள்)

Q: Inertia of a body depends on ________.

Answer: mass of the object. Explanation: Inertia is the natural tendency of an object to resist a change in its state of rest or uniform motion. Mass is the quantitative measure of inertia. Greater the mass, greater is its inertia. Therefore, inertia depends solely on mass.

Q: Impulse is equals to ________.

Answer: change of momentum. Explanation: By definition, Impulse J = F × t. From Newton's Second Law, F = Δp / t. Substituting F, we get J = (Δp / t) × t = Δp (Change of momentum).

Q: Newton’s III law is applicable ________.

Answer: both a & b. Explanation: Newton's Third Law states: 'For every action, there is an equal and opposite reaction.' Forces always occur in matched action-reaction pairs in nature, irrespective of whether the bodies are at rest (static equilibrium) or in motion (dynamic state).

Q: Plotting a graph for momentum on the X-axis and time on Y-axis. Slope of momentum-time graph gives ________.

Answer: Force. Explanation: Slope of a curve y vs x is given by dy/dx. Here, momentum is plotted on the axis representing change in momentum Δp, and time Δt. Force F = Δp / Δt. Thus, the slope of the momentum-time curve directly represents Force.

Q: In which of the following sport the turning effect of force is used?

Answer: cycling. Explanation: In cycling, when the rider presses the pedal with their foot, a torque (turning effect of force = Force × perpendicular distance) is produced about the axle. This torque turns the toothed wheel, transferring rotary motion via the chain to propel the cycle forward.

Q: To produce a displacement ________ is required.

Answer: Force. According to Newton's First Law, an object continues in its state of rest unless acted upon by an external unbalanced force.

Q: Passengers lean forward when sudden brake is applied in a moving vehicle. This can be explained by ________.

Answer: Inertia of motion. The upper part of the passenger's body tends to continue moving forward due to inertia of motion while the feet come to rest with the vehicle.

Q: By convention, the clockwise moments are taken as ________ and the anticlockwise moments are taken as ________.

Answer: negative (-), positive (+). Anticlockwise rotation is taken as positive (+), and clockwise rotation is taken as negative (-).

அலகு 1: இயக்க விதிகள் — Tamil Nadu Samacheer Kalvi Book Back Answers & Solutions

Complete book-back solutions for Unit 1 covering Newton's three laws of motion, linear momentum conservation, impulse, gravitational constant, mass vs weight, and rocket propulsion mechanics.

நியூட்டனின் மூன்று இயக்க விதிகள், நேர்க்கோட்டு உந்த மாறாக் கோட்பாடு, கணத்தாக்கு, ஈர்ப்பியல் மாறிலி, நிறை மற்றும் எடை வேறுபாடு, மற்றும் ராக்கெட் ஏவுதல் தத்துவங்களை உள்ளடக்கிய முழுமையான விடைக்குறிப்புகள்.

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📐 Important Formulas & Laws Quick Revision

Linear Momentum (p) (நேர்க்கோட்டு உந்தம் (p)): p = m × v SI Unit: kg m s⁻¹ (or N s). It is a vector quantity having the same direction as velocity.
Newton's Second Law of Motion (நியூட்டனின் இரண்டாம் இயக்க விதி): F = m × a = m × (v - u) / t SI Unit: Newton (N) = 1 kg m s⁻². Defines Force quantitatively.
Impulse (J) (கணத்தாக்கு (J)): J = F × Δt = Δp = m(v - u) SI Unit: kg m s⁻¹ or N s. A large force acting for a very short interval of time.
Newton's Universal Law of Gravitation (நியூட்டனின் பொது ஈர்ப்பியல் விதி): F = G × (m₁ × m₂) / r² G = 6.674 × 10⁻¹¹ N m² kg⁻². Gravitational force is an inverse-square central force.
Acceleration due to gravity (g) (புவிஈர்ப்பு முடுக்கம் (g)): g = (G × M) / R² Standard value on Earth surface: g = 9.8 m s⁻². Independent of the falling body's mass.

Part I: Choose the Correct Answer (1 Mark)

1. Inertia of a body depends on ________.

ஒரு பொருளின் நிலைமம் எதனைச் சார்ந்தது?

(a) weight of the object
(b) acceleration due to gravity of the planet
(c) mass of the object ✓ (Correct Answer)
(d) both a & b

Explanation: Inertia is the natural tendency of an object to resist a change in its state of rest or uniform motion. Mass is the quantitative measure of inertia. Greater the mass, greater is its inertia. Therefore, inertia depends solely on mass.

2. Impulse is equals to ________.

கணத்தாக்கு என்பது எதற்கு சமமானது?

(a) rate of change of momentum
(b) rate of force and time
(c) change of momentum ✓ (Correct Answer)
(d) rate of change of mass

Explanation: By definition, Impulse J = F × t. From Newton's Second Law, F = Δp / t. Substituting F, we get J = (Δp / t) × t = Δp (Change of momentum).

3. Newton’s III law is applicable ________.

நியூட்டனின் மூன்றாம் இயக்க விதி எதற்கு பொருந்தும்?

(a) for a body at rest
(b) for a body in motion
(c) both a & b ✓ (Correct Answer)
(d) only for bodies with equal masses

Explanation: Newton's Third Law states: 'For every action, there is an equal and opposite reaction.' Forces always occur in matched action-reaction pairs in nature, irrespective of whether the bodies are at rest (static equilibrium) or in motion (dynamic state).

4. Plotting a graph for momentum on the X-axis and time on Y-axis. Slope of momentum-time graph gives ________.

உந்தத்தை X-அச்சிலும் காலத்தை Y-அச்சிலும் கொண்டு வரையப்படும் வரைபடத்தின் சாய்வு தருவது ________.

(a) Impulsive force
(b) Acceleration
(c) Force ✓ (Correct Answer)
(d) Rate of force

Explanation: Slope of a curve y vs x is given by dy/dx. Here, momentum is plotted on the axis representing change in momentum Δp, and time Δt. Force F = Δp / Δt. Thus, the slope of the momentum-time curve directly represents Force.

5. In which of the following sport the turning effect of force is used?

கீழ்க்கண்ட எந்த விளையாட்டில் விசையின் திருப்புத்திறன் பயன்படுகிறது?

(a) swimming
(b) tennis
(c) cycling ✓ (Correct Answer)
(d) hockey

Explanation: In cycling, when the rider presses the pedal with their foot, a torque (turning effect of force = Force × perpendicular distance) is produced about the axle. This torque turns the toothed wheel, transferring rotary motion via the chain to propel the cycle forward.

6. The unit of 'g' is m s⁻². It can also be expressed as ________.

'g' -ன் அலகு m s⁻² ஆகும். இதனை ________ எனவும் குறிப்பிடலாம்.

(a) c m s⁻¹
(b) N k g⁻¹ ✓ (Correct Answer)
(c) N m² k g⁻¹
(d) c m² s⁻²

Explanation: From Newton's second law, Force F = m × g ⟹ g = F / m. The unit of Force is Newton (N) and mass is kilogram (kg). Therefore, unit of g = N / kg = N kg⁻¹.

7. One kilogram force (1 kgf) is equal to ________.

ஒரு கிலோகிராம் எடை (1 kgf) என்பது ________-க்கு சமம்.

(a) 9.8 dyne
(b) 9.8 × 10⁴ N
(c) 98 × 10⁴ dyne ✓ (Correct Answer)
(d) 980 dyne

Explanation: 1 kgf = 1 kg × 9.8 m s⁻² = 9.8 N. To convert into CGS units (dyne): 1 N = 10⁵ dyne. So 9.8 N = 9.8 × 10⁵ dyne = 98 × 10⁴ dyne.

8. The mass of a body is measured on planet Earth as M kg. When it is taken to a planet of half the mass of Earth then its value will be ________.

புவியில் M கிலோகிராம் நிறை கொண்ட ஒரு பொருள், புவியின் நிறையில் பாதியளவு கொண்ட மற்றொரு கோளிற்கு எடுத்துச் செல்லப்பட்டால் அதன் நிறை ________.

(a) 4 M
(b) 2 M
(c) M/4
(d) M ✓ (Correct Answer)

Explanation: Mass is the fundamental quantity of matter contained in a body. It remains constant anywhere in the universe. Only weight (W = mg) changes. Therefore, the mass remains M kg.

9. If the Earth shrinks to 50% of its real radius its mass remaining the same, the weight of a body on the Earth will ________.

புவியின் நிறை மாறாமல் அதன் ஆரம் 50% சுருங்கினால், புவியின் மேற்பரப்பில் உள்ள பொருளின் எடை ________.

(a) decrease by 50%
(b) increase by 50%
(c) decrease by 25%
(d) increase by 300% ✓ (Correct Answer)

Explanation: Acceleration due to gravity g = GM / R². If radius shrinks to 50% (R' = R/2 = 0.5R), new g' = GM / (0.5R)² = GM / (0.25 R²) = 4g. The new weight W' = 4W. Percentage increase = [(W' - W) / W] × 100% = [(4W - W) / W] × 100% = 300% increase.

10. To project a rocket which of the following principle(s) is / (are) required?

ராக்கெட் ஏவுதலில் கீழ்க்கண்ட எந்த விதி/விதிகள் பயன்படுகின்றன?

(a) Newton’s third law of motion
(b) Newton’s law of gravitation
(c) law of conservation of linear momentum
(d) both a and c ✓ (Correct Answer)

Explanation: Rocket propulsion is based on Newton’s Third Law of Motion (Action: high velocity gases ejected downwards; Reaction: forward thrust on the rocket) and the Law of Conservation of Linear Momentum (as burning fuel mass decreases, velocity increases). Both principles are required.

Part II: Fill in the Blanks (கோடிட்ட இடங்களை நிரப்புக)

1. To produce a displacement ________ is required.

இடப்பெயர்ச்சியை உருவாக்க ________ தேவை.

Answer: Force (விசை)

2. Passengers lean forward when sudden brake is applied in a moving vehicle. This can be explained by ________.

ஓடும் பேருந்து திடீரென நிறுத்தப்படும் போது பயணிகள் முன்னோக்கிச் சாய்வது ________ மூலம் விளக்கப்படுகிறது.

Answer: Inertia of motion (இயக்கத்திற்கான நிலைமம்)

3. By convention, the clockwise moments are taken as ________ and the anticlockwise moments are taken as ________.

மரபுப்படி, வலஞ்சுழி திருப்புத்திறன் ________ குறியாகவும், இடஞ்சுழி திருப்புத்திறன் ________ குறியாகவும் கொள்ளப்படும்.

Answer: negative (-), positive (+) (எதிர்க்குறி (-), நேர்க்குறி (+))

4. ________ is used to change the speed of car.

மகிழுந்தின் வேகத்தை மாற்ற ________ பயன்படுகிறது.

Answer: Gears (or Accelerator) (பற்சக்கரம் (Gears))

Part III: State True or False with Statement Correction (சரியா? தவறா?)

1. The linear momentum of a system of particles is always conserved.

துகள் அமைப்பின் நேர்க்கோட்டு உந்தம் எப்போதும் மாறாது.

✗ FALSE (தவறு)

Correct Statement: The linear momentum of a system of particles is conserved only when no external force acts on the system (in the absence of an external unbalanced force).

2. Apparent weight of a person is always equal to his actual weight.

ஒரு மனிதனின் தோற்ற எடை எப்போதும் அவரின் உண்மையான எடைக்குச் சமமாக இருக்கும்.

✗ FALSE (தவறு)

Correct Statement: Apparent weight of a person is not always equal to his actual weight; it depends on the acceleration and direction of motion of the lift/reference frame (R = m(g ± a)).

3. Weight of a body is greater at the equator and less at the polar region.

பொருளின் எடையானது நிலநடுக்கோட்டுப் பகுதியில் அதிகமாகவும், துருவப் பகுதியில் குறைவாகவும் இருக்கும்.

✗ FALSE (தவறு)

Correct Statement: Weight of a body is greater at the polar region and less at the equator, because g ∝ 1/R² and the equatorial radius is greater than the polar radius (Re > Rp).

4. Turning a nut with a spanner having a short handle is so easy than one with a long handle.

நீளமான கைப்பிடி கொண்ட திருகுக்குறடைக் (spanner) காட்டிலும், குட்டையான கைப்பிடி கொண்ட திருகுக்குறடினால் திருகுமறையை எளிதாகத் திருப்ப முடியும்.

✗ FALSE (தவறு)

Correct Statement: Turning a nut with a spanner having a long handle is much easier than one with a short handle, because the moment of force (torque τ = F × d) increases with perpendicular distance (d).

5. There is no gravity in the orbiting space station around the Earth. So the astronauts feel weightlessness.

புவியினைச் சுற்றி வரும் விண்வெளி நிலையத்தில் ஈர்ப்பு விசை இல்லாததால் விண்வெளி வீரர்கள் எடையின்மையை உணர்கிறார்கள்.

✗ FALSE (தவறு)

Correct Statement: Astronauts do not experience zero gravity; the space station and astronauts are in a continuous state of free fall towards the Earth with acceleration equal to 'g', resulting in apparent weight R = m(g - g) = 0.

Part IV: Match the Following (பொருத்துக)

Newton's I law (நியூட்டனின் முதல் விதி) ➔ Stable equilibrium of a body
Newton's II law (நியூட்டனின் இரண்டாம் விதி) ➔ Law of force (F = ma)
Newton's III law (நியூட்டனின் மூன்றாம் விதி) ➔ Flying nature of bird
Law of conservation of momentum (நேர்க்கோட்டு உந்த மாறா விதி) ➔ Propulsion of a rocket

Part V: Assertion and Reason (கூற்றும் காரணமும்)

Question 1:

Assertion (A): The value of 'g' decreases as height and depth increase from the surface of the Earth.

Reason (R): 'g' depends on the mass of the object and the Earth.

Correct Answer: (c) Assertion is true, but reason is false.

Assertion is true because g decreases both with altitude (g' = g[1 - 2h/R]) and with depth (g' = g[1 - d/R]). Reason is FALSE because acceleration due to gravity g = GM/R² depends only on the mass of the Earth (M), not on the mass of the falling object (m). Hence, Option (c) is correct.

Question 2:

Assertion (A): To float in an orbiting space station without falling, astronauts must experience zero gravity.

Reason (R): Astronauts are in the state of continuous free fall towards the Earth with acceleration equal to 'g'.

Correct Answer: (d) Assertion is false, but reason is true.

Assertion is false because astronauts do not experience 'zero gravity' (gravity in low-Earth orbit is ~90% of surface value). Reason is TRUE because weightlessness is caused entirely by free fall (a = g), making apparent weight R = m(g - g) = 0. Hence, Option (d) is correct.

Part VI: Short Answer Questions — 2 Marks (சுருக்கமான விடையளி)

Q1. Define inertia. Give its classification.

நிலைமம் என்றால் என்ன? அதன் வகைகளைக் கூறுக.

Answer:

1. Definition (1 Mark): The inherent property of a body to resist any change in its state of rest or the state of uniform motion, unless influenced upon by an external unbalanced force, is called inertia.
2. Classification of Inertia (1 Mark):
(a) Inertia of rest (ஓய்வில் நிலைமம்): Resistance to change the state of rest (e.g. vigorously shaking tree branches drops leaves).
(b) Inertia of motion (இயக்கத்தில் நிலைமம்): Resistance to change the state of motion (e.g. an athlete runs some distance before jumping).
(c) Inertia of direction (திசையில் நிலைமம்): Resistance to change the direction of motion (e.g. sharp turn in car throws passenger outwards).

Q2. Classify force based on direction of action.

விசைகளை அவற்றின் திசைகளின் அடிப்படையில் எவ்வாறு வகைப்படுத்தலாம்?

Answer:

Based on direction, forces are classified into two types:
1. Like Parallel Forces (ஒத்த இணை விசைகள்): Two or more forces of equal or unequal magnitude acting along the same direction, parallel to one another.
2. Unlike Parallel Forces (மாறுபட்ட இணை விசைகள்): Two or more forces of equal or unequal magnitude acting along opposite directions, parallel to one another.

Q3. State Newton's Second Law of Motion.

நியூட்டனின் இரண்டாம் இயக்க விதியைக் கூறுக.

Answer:

1. Statement (1 Mark): The force acting on an object is directly proportional to the rate of change of linear momentum of the body and the change in momentum takes place in the direction of the force.
2. Mathematical Form (1 Mark):
F ∝ (change in momentum) / time
F ∝ m(v - u) / t
Since acceleration a = (v - u) / t, we get:
F = m × a
SI Unit: Newton (N) or kg m s⁻².

Q4. Why does a fielder pull his hands back while catching a cricket ball?

மட்டைப்பந்து (கிரிக்கெட்) விளையாட்டில் பந்து பிடிப்பவர் கைகளை பின்னோக்கி இழுப்பது ஏன்?

Answer:

From the impulse equation: Impulse J = F × Δt = Δp (constant change in momentum).
Therefore, Force F = Δp / Δt.
By pulling his hands backwards, the fielder increases the time of contact (Δt) to bring the fast-moving ball to rest.
Since Force is inversely proportional to time (F ∝ 1/Δt), a larger time interval drastically reduces the impulsive force experienced by his hands, preventing severe injury.

Q5. Differentiate Mass and Weight.

நிறை மற்றும் எடை வேறுபடுத்துக.

Answer:

Table of Differences (Draw 2 neat columns):
1. Definition: Mass is the fundamental quantity of matter contained in a body | Weight is the gravitational force exerted on a body by Earth (W = mg).
2. Quantity: Mass is a Scalar quantity (has only magnitude) | Weight is a Vector quantity (has magnitude and directed towards Earth center).
3. Constancy: Remains constant everywhere in the universe | Varies from place to place depending on 'g'.
4. Measurement: Measured using a Physical Balance / Beam Balance | Measured using a Spring Balance.
5. SI Unit: Measured in kilogram (kg) | Measured in Newton (N).

Q6. Define torque and state its unit.

விசையின் திருப்புத்திறன் (டார்க்) என்றால் என்ன? அதன் அலகைக் கூறுக.

Answer:

1. Definition: The rotating or turning effect of a force about a fixed point or fixed axis is called moment of the force or torque (τ).
2. Mathematical Formula: Torque τ = F × d
where F = applied force and d = perpendicular distance between the fixed axis and line of action of force.
3. SI Unit: Newton metre (N m).
4. Nature: It is a vector quantity.

Q7. State the principle of moments.

திருப்புத்திறன்களின் தத்துவத்தைக் கூறுக.

Answer:

1. Statement: When a number of like or unlike parallel forces act on a rigid body and the body is in equilibrium, then the algebraic sum of the moments in the clockwise direction is equal to the algebraic sum of the moments in the anticlockwise direction.
2. Equation:
Moment in clockwise direction = Moment in anticlockwise direction
F₁ × d₁ = F₂ × d₂

Part VII: Long Answer Questions — 5 / 7 Marks (விரிவான விடையளி)

Q1. State and prove the Law of Conservation of Linear Momentum with a neat labeled diagram.

நேர்க்கோட்டு உந்த மாறாக் கோட்பாட்டைக் கூறி, படத்துடன் மெய்ப்பிக்கவும்.

1. Statement of the Principle (1 Mark)

There is no change in the total linear momentum of an isolated system of interacting bodies as long as no external unbalanced force acts on the system.
In other words: Total Momentum before collision = Total Momentum after collision.

2. Description of the System & Collision (1 Mark)

Consider two bodies A and B having masses m₁ and m₂ moving along the same straight line in the same direction with initial velocities u₁ and u₂ respectively.
Assume u₁ > u₂, so that Body A collides with Body B.
During the collision for time 't', Body A exerts force F_B on Body B (Action), and simultaneously Body B exerts an equal and opposite reaction force F_A on Body A (Reaction).
After collision, their velocities become v₁ and v₂ respectively.

3. Step-by-Step Mathematical Derivation (2.5 Marks)

Step (i): Force exerted on body B by body A (by Newton's 2nd Law):
F_B = mass of B × acceleration of B = m₂ × (v₂ - u₂) / t ---- (Equation 1)
Step (ii): Reaction force exerted on body A by body B:
F_A = mass of A × acceleration of A = m₁ × (v₁ - u₁) / t ---- (Equation 2)
Step (iii): Applying Newton's Third Law (Action = -Reaction):
F_B = - F_A
m₂ × (v₂ - u₂) / t = - [ m₁ × (v₁ - u₁) / t ]
Step (iv): Multiplying both sides by 't' to eliminate the denominator:
m₂(v₂ - u₂) = - m₁(v₁ - u₁)
m₂v₂ - m₂u₂ = - m₁v₁ + m₁u₁
Step (v): Rearranging initial terms to LHS and final terms to RHS:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

4. Conclusion & Inscription (0.5 Mark)

Total Momentum before collision (m₁u₁ + m₂u₂) = Total Momentum after collision (m₁v₁ + m₂v₂).
Since no external force acts on the system, the total linear momentum remains conserved.
Hence, the Principle of Conservation of Linear Momentum is proved.

Q2. Describe the construction and propulsion mechanism of a Rocket based on Newton's laws and law of conservation of linear momentum.

ராக்கெட் ஏவுதலின் செயல்படும் தத்துவத்தை நியூட்டனின் விதிகள் மற்றும் உந்த மாறாக் கோட்பாட்டின் அடிப்படையில் விரிவாக விளக்குக.

1. Core Working Principles (1 Mark)

Rocket propulsion is strictly based on two fundamental physics laws:
1. Newton's Third Law of Motion (Action and Reaction).
2. Law of Conservation of Linear Momentum.

2. Fuel and Combustion Chamber Construction (1.5 Marks)

Rockets are filled with fuel (liquid or solid propellant) and an oxidizer inside propellant chambers.
When the rocket is fired, the fuel undergoes intense combustion inside the combustion chamber.
Extremely hot exhaust gases are ejected out at supersonic speeds through the nozzle at the rear end.
This downward expulsion of high-velocity gases constitutes the Action force.

3. Thrust Generation & Momentum Conservation (2 Marks)

By Newton's Third Law, the downward ejected gases impart an equal and opposite upward reaction force called Forward Thrust.
This thrust accelerates the rocket upwards.
By Law of Conservation of Linear Momentum, the system momentum remains zero initially:
Momentum of Rocket = - (Momentum of Ejected Gases)
As the rocket ascends, its fuel burns continuously, decreasing the total mass of the rocket.
Since momentum must be conserved, as mass 'm' decreases, the velocity 'v' of the rocket increases continuously until it reaches Escape Velocity.

4. Conclusion (0.5 Mark)

The combination of Newton's third law and momentum conservation enables multistage rockets to overcome Earth's gravitational pull and place satellites into precise orbits.

Q3. State Newton's Universal Law of Gravitation and deduce the relation between 'g' and 'G'.

நியூட்டனின் பொது ஈர்ப்பியல் விதியைக் கூறி, ஈர்ப்பு முடுக்கம் (g) மற்றும் ஈர்ப்பியல் மாறிலி (G) ஆகியவற்றிற்கு இடையேயான தொடர்பை வருவிக்க.

1. Statement of Universal Law of Gravitation (1 Mark)

Every particle of matter in this universe attracts every other particle with a force.
This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between the centers of these masses.
The direction of the force acts along the line joining the masses.

2. Mathematical Formulation (1.5 Marks)

Let m₁ and m₂ be the masses of two bodies separated by a distance 'r'.
According to the law:
F ∝ m₁ × m₂
F ∝ 1 / r²
Combining these: F ∝ (m₁ × m₂) / r²
F = G × (m₁ × m₂) / r²
where G is the Universal Gravitational Constant.
Value of G in SI unit: 6.674 × 10⁻¹¹ N m² kg⁻².

3. Relation Between 'g' and 'G' (2 Marks)

Consider a body of mass 'm' lying on the surface of the Earth.
Let M = Mass of Earth, R = Radius of Earth.
From Newton's Law of Gravitation, gravitational force on the body:
F = (G × M × m) / R² ---- (Equation 1)
From Newton's Second Law, the force on the body is its weight:
F = m × g ---- (Equation 2)
Equating (1) and (2):
m × g = (G × M × m) / R²
Canceling 'm' on both sides:
g = (G × M) / R²

4. Conclusion & Significance (0.5 Mark)

Acceleration due to gravity 'g' depends only on the mass and radius of the Earth (M and R), and is independent of the mass of the body (m).
Standard value on Earth's surface: g = 9.8 m s⁻².

Part VIII: Solved Problems Corner / Numerical Calculations

Problem 1: Two bodies have a mass ratio of 3:4. The force applied on the bigger mass produces the acceleration of 12 m s⁻². What could be the acceleration of the other body, if the same force acts on it?

இரு உடல்களின் நிறைகளின் விகிதம் 3:4. அதிக நிறையுடைய பொருளின் மீது ஒரு விசை செயல்பட்டு 12 மீ வி⁻² முடுக்கத்தை உருவாக்குகிறது எனில், அதே விசை மற்ற பொருளின் மீது செயல்படும் போது ஏற்படும் முடுக்கம் யாது?

Step-by-Step Working:Step 1: Calculate the applied force using the bigger mass:
   F = m₂ × a₂ = 4m × 12 = 48m N
Step 2: Equate the force to the smaller mass:
   F = m₁ × a₁
   48m = 3m × a₁
Step 3: Solve for a₁:
   a₁ = (48m) / (3m) = 16 m s⁻²

                Boxed Final Answer: Acceleration of the other body a₁ = 16 m s⁻²
              

Problem 2: A ball of mass 1 kg moving with a speed of 10 m s⁻¹ rebounds after a perfectly elastic collision with the floor. Calculate the change in linear momentum of the ball.

1 கிகி நிறையுள்ள ஒரு பந்து 10 மீ வி⁻¹ வேகத்தில் தரை மீது விழுந்து முழு மீட்சியுடன் மீண்டும் எழும்புகிறது எனில், அப்பந்தின் நேர்க்கோட்டு உந்த மாற்றத்தைக் காண்க.

Step-by-Step Working:Step 1: Take the downward direction as negative and upward direction as positive.
Step 2: Initial momentum p_initial = m × u = 1 kg × (-10 m s⁻¹) = -10 kg m s⁻¹.
Step 3: Final momentum p_final = m × v = 1 kg × (+10 m s⁻¹) = +10 kg m s⁻¹.
Step 4: Change in momentum Δp = p_final - p_initial:
   Δp = 10 - (-10) = 10 + 10 = 20 kg m s⁻¹ (or N s).

                Boxed Final Answer: Change in linear momentum Δp = 20 kg m s⁻¹ (or 20 N s)
              

Problem 3: A mechanic unscrew a nut by applying a force of 140 N with a spanner of length 40 cm. What should be the length of the spanner if a force of 40 N is applied to unscrew the same nut?

ஒரு மெக்கானிக் 40 செ.மீ நீளமுள்ள திருகுக்குறடின் மூலம் 140 N விசை கொடுத்து திருகுமறையை கழற்றுகிறார். 40 N விசை கொண்டு அதே திருகுமறையைக் கழற்ற தேவையான திருகுக்குறடின் நீளம் யாது?

Step-by-Step Working:Step 1: The torque required to unscrew the nut remains constant in both cases:
   τ = F₁ × d₁ = F₂ × d₂
Step 2: Substitute the given values:
   140 N × 40 cm = 40 N × d₂
Step 3: Solve for d₂:
   d₂ = (140 × 40) / 40 = 140 cm = 1.4 m.

                Boxed Final Answer: Length of the spanner d₂ = 140 cm (or 1.4 m)
              

Problem 4: The ratio of masses of two planets is 2:3 and the ratio of their radii is 4:7. Find the ratio of their accelerations due to gravity.

இரு கோள்களின் நிறைகளின் விகிதம் 2:3 மற்றும் அவற்றின் ஆரங்களின் விகிதம் 4:7 எனில், அவற்றின் ஈர்ப்பு முடுக்கங்களின் விகிதத்தைக் காண்க.

Step-by-Step Working:Step 1: Formula for acceleration due to gravity on planet 1 and planet 2:
   g₁ = GM₁ / R₁²  and  g₂ = GM₂ / R₂²
Step 2: Divide g₁ by g₂:
   g₁ / g₂ = (M₁ / M₂) × (R₂ / R₁)²
Step 3: Substitute the ratios:
   g₁ / g₂ = (2 / 3) × (7 / 4)²
   g₁ / g₂ = (2 / 3) × (49 / 16)
Step 4: Simplify:
   g₁ / g₂ = (1 × 49) / (3 × 8) = 49 / 24

                Boxed Final Answer: Ratio of accelerations due to gravity g₁ : g₂ = 49 : 24
              

Problem 5: A force of 5 N produces an acceleration of 8 m s⁻² on a mass m₁ and an acceleration of 24 m s⁻² on a mass m₂. What acceleration would the same force provide if both the masses are tied together?

5 N விசையானது m₁ என்ற நிறையில் 8 மீ வி⁻² முடுக்கத்தையும், m₂ என்ற நிறையில் 24 மீ வி⁻² முடுக்கத்தையும் உருவாக்குகிறது. இவ்விரு நிறைகளும் ஒன்றாக இணைக்கப்பட்டு அதே விசை செலுத்தப்பட்டால் ஏற்படும் முடுக்கம் யாது?

Step-by-Step Working:Step 1: Calculate mass m₁:
   m₁ = F / a₁ = 5 / 8 kg
Step 2: Calculate mass m₂:
   m₂ = F / a₂ = 5 / 24 kg
Step 3: Find combined mass M = m₁ + m₂:
   M = 5/8 + 5/24 = (15 + 5) / 24 = 20 / 24 = 5 / 6 kg
Step 4: Calculate combined acceleration a = F / M:
   a = 5 / (5/6) = 5 × (6/5) = 6 m s⁻².

                Boxed Final Answer: Combined acceleration a = 6 m s⁻²
              

Problem 6: A body of mass 5 kg is moving with an initial velocity of 20 m s⁻¹. A force of 100 N acts on it for 0.5 s. What is the final velocity of the body?

5 கிகி நிறையுள்ள ஒரு பொருள் 20 மீ வி⁻¹ ஆரம்ப திசைவேகத்தில் செல்கிறது. அதன் மீது 100 N விசை 0.5 வினாடி நேரம் செயல்பட்டால், அப்பொருளின் இறுதி திசைவேகம் யாது?

Step-by-Step Working:Step 1: Calculate acceleration: a = F / m = 100 / 5 = 20 m s⁻².
Step 2: Use the first equation of motion v = u + a × t:
   v = 20 + (20 × 0.5)
   v = 20 + 10 = 30 m s⁻¹.

                Boxed Final Answer: Final velocity v = 30 m s⁻¹
              

Problem 7: A body of mass 10 kg is dropped from a height of 10 m. What is the kinetic energy of the body just before touching the ground? (Take g = 9.8 m s⁻²)

10 கிகி நிறையுள்ள ஒரு பொருள் 10 மீ உயரத்திலிருந்து கீழே போடப்படுகிறது. தரையைத் தொடுவதற்கு முன் அப்பொருளின் இயக்க ஆற்றல் யாது? (g = 9.8 மீ வி⁻²)

Step-by-Step Working:Step 1: According to the Law of Conservation of Energy, the potential energy at height 'h' is completely converted into kinetic energy just before hitting the ground.
Step 2: Total Kinetic Energy = m × g × h
Step 3: Substitute the values:
   K.E = 10 kg × 9.8 m s⁻² × 10 m
   K.E = 980 Joules (J).

                Boxed Final Answer: Kinetic energy K.E = 980 J
              