9th Sci. Chapter 1: Laws of Motion, Notes -2.

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✍🏻... विज्ञान विषयाच्या माहितीसाठी खालील⬇️ लिंकला🖇️ स्पर्श करा.

👉🖊️.        माहिती विज्ञानाची 🎷  

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1. Basic Concepts of Motion

Distance: Length of the actual path traversed by an object in motion. (Scalar quantity)

Displacement: Minimum distance from the starting point to the final point. (Vector quantity)

Speed: 

Speed = Distance / Time

Scalar quantity, SI unit: m/s

Velocity:

Velocity = Displacement / Time

Vector quantity, SI unit: m/s

Acceleration: Rate of change of velocity.

a = v - u / t

SI Unit: m/s²

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2. Newton’s Laws of Motion:

🥇First Law (Law of Inertia): An object continues to remain at rest or in a state of uniform motion along a straight line unless an external unbalanced force acts on it.

🥈 Second Law: The rate of change of momentum is proportional to the applied force, and the change of momentum occurs in the direction of the force.

Formula: 

F = m × a

Units of Force: SI = Newton (N = kg·m/s²), 

CGS = Dyne (g.cm/s²).

1N = 10⁵ dynes

🥉 Third Law: Every action force has an equal and opposite reaction force, which acts simultaneously.

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3. Equations of Motion

First Equation 

(Velocity–Time Relation): 

v = u + at

Second Equation 

(Displacement–Time Relation): 

s = ut + 1/2 at²

Third Equation 

(Displacement–Velocity Relation): 

v² = u² + 2as

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4. Law of Conservation of Momentum

When no external force acts on two interacting objects, their total initial momentum is equal to their total final momentum:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Recoil Velocity:

 v₂ = - m_{1 /}  m₂   × v₁

(The backward movement of a gun when a bullet is fired).

5. Uniform Circular Motion

Motion of an object moving at a constant speed along a circular path.

Formula: v = 2πr/t

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Problem: An athlete is running on a circular track. He runs a distance of 400 m in 25 s before returning to his original position. What is his average speed and velocity?

Given:

Total distance = 400 m

Total displacement = 0 m (returns to original position)

Total time t = 25 s

Calculation: 

Average speed = Total distance / Total time 

= 400 / 25

Average speed = 16 m/s

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Average velocity = Total displacement / Total time 

= 0/25 

Average velocity = 0 m/s

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Problem: An aeroplane taxies on the runway for 30 s with an acceleration of 3.2 m/s² before taking off. How much distance would it have covered on the runway?

Given: u = 0 m/s, a = 3.2 m/s², t = 30 s

Formula: s = ut + 1/2at²

s = (0 × 30) + 1/2 × 3.2 × (30)²

s = 0 + 1/2 × 3.2 × 900

s = 1.6 × 900

s = 1440 m

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Problem: A kangaroo can jump 2.5 m vertically. What must be the initial velocity of the kangaroo?

Given: s = 2.5 m, v = 0 m/s, a = -9.8 m/s² (deceleration due to gravity)

Formula: v² = u² + 2as

0² = u² + 2 × (-9.8) × 2.5

0 = u² - 49 

u² = 7 m/s

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Problem: A motorboat starts from rest and moves with uniform acceleration. If it attains the velocity of 15 m/s in 5 s, calculate the acceleration and the distance travelled in that time.

Given: u = 0 m/s, 

v = 15 m/s, 

t = 5 s

Acceleration (a) 

a = v - u/t

= 15 - 0 / 5

a = 3 m/s²

Distance (s)

s = ut + 1/2at²

s = (0 × 5) + 1/2 × 3 × (5)²

s = 0 + 1/2 × 3 × 25

= 75/2

s = 37.5 m. 

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Problem: The mass of a cannon is 500 kg, and it recoils with a speed of 0.25 m/s. What is the momentum of the cannon?

Given: Mass (m) = 500 kg, Velocity (v) = 0.25 m/s

Calculation 

momentum = m × v 

= 500 × 0.25

momentum = 125 kg·m/s

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Problem: Two balls have masses of 50 g and 100 g, respectively, and are moving along the same line in the same direction with velocities of 3 m/s and 1.5 m/s, respectively. They collide with each other, and after the collision, the first ball moves with a velocity of 2.5 m/s. Calculate the velocity of the second ball after collision.

Given:

m₁ = 50 g = 0.05 km

m₂ = 100 gm = 0.1 kg 

u₁ = 3 m/s, 

u₂ = 1.5 m/s

v₁ = 2.5 m/s, 

v₂ = ?

Formula:     

m₁u_{1 +} m₂u_{2 =} m₁v_{1 +} m₂v₂ 

(0.05 × 3) + (0.1 × 1.5 = (0.05 × 2.5) + (0.1 × v₂ )

0.15 + 0.15 = 0.125 + 0.1 × v2

0.3 = 0.125 + 0.1 × v2

0.1 × v2 = 0.3 - 0.125

0.1 × v2 = 0.175

v₂ = 0.175/0.1 

v₂ = 1.75 m/s

Velocity of the second ball after collision: v₂ = 1.75 m/s

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