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Newton’s Laws of Motion & Collisions

Objectives:

1. Solve numerical problems involving impulse and momentum.
2. Interpret the area under a force–time graph.
3. Interpret and apply Newton’s laws of motion.
4. Compare the concepts of inertia, mass, and force.
5. Deduce the relationship between mass and acceleration.
6. Interpret the conservation of linear momentum in collisions.
7. Solve problems on elastic and inelastic collisions.

Key Laws and Formulas

Newton’s First Law (Law of Inertia): An object remains at rest or in uniform motion unless acted upon by an external force.

Newton’s Second Law: The net force on an object is given by F = ma, where F is force, m is mass, and a is acceleration.

Momentum: The momentum of an object is p = mv, where v is the velocity.

Impulse: Impulse is the product of force and time, Impulse = FΔt, and equals the change in momentum (Δp).

Force–Time Graph: The area under the force–time graph represents the impulse imparted to an object.

Conservation of Linear Momentum: In a closed system with no external forces, the total momentum is conserved. For a collision: m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'.

Elastic Collisions: Both momentum and kinetic energy are conserved. For two colliding bodies, the final velocities can be found using:
v₁' = [(m₁ - m₂)/(m₁ + m₂)]v₁ + [2m₂/(m₁ + m₂)]v₂
v₂' = [2m₁/(m₁ + m₂)]v₁ + [(m₂ - m₁)/(m₁ + m₂)]v₂

Inelastic Collisions: Momentum is conserved but kinetic energy is not. If the objects stick together, then:
m₁v₁ + m₂v₂ = (m₁ + m₂)v'

Worked Examples

Below are 8 worked examples (each in a green box) demonstrating the application of Newton’s laws and collision principles.

Example 1:
Newton’s First Law – A book remains at rest on a table until a force (like a push) is applied. (Concept demonstration)

Example 2:
Newton’s Second Law – A 10 kg object with a net force of 50 N has acceleration: a = 50/10 = 5 m/s².

Example 3:
Impulse – A 2 kg ball moving at 3 m/s (p = 6 kg·m/s) receives a 10 N force for 0.2 s. Impulse = 10×0.2 = 2 N·s, so new momentum = 6 + 2 = 8 kg·m/s.

Example 4:
Force–Time Graph – A constant force of 20 N applied for 5 s gives impulse = 20×5 = 100 N·s, so momentum increases by 100 kg·m/s.

Example 5:
Conservation of Momentum – A 4 kg cart moving at 3 m/s collides with a 2 kg cart at rest. Total momentum = 12 kg·m/s. After collision (if they stick together), combined mass = 6 kg and final velocity = 12/6 = 2 m/s.

Example 6:
Elastic Collision – Two balls (2 kg and 3 kg) moving at 4 m/s and 2 m/s respectively collide elastically. Apply momentum and kinetic energy conservation formulas to solve for final speeds.

Example 7:
Inelastic Collision – A 1000 kg car moving at 10 m/s collides with a 1500 kg car moving at 5 m/s (in opposite directions). Using momentum conservation: (1000×10 + 1500×(-5)) = (1000+1500)v'. Solve for v'.

Example 8:
Force–Time Graph Interpretation – A rocket engine’s force–time graph is used to calculate the impulse. Dividing the area under the curve by the rocket’s mass gives the change in velocity.

JAMB CBT Quiz on Newton’s Laws of Motion & Collisions

Total time: 900 seconds

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This lesson covers the following topics: Newton’s First Law (Inertia) Newton’s Second Law: F = ma Momentum: p = mv and Impulse = FΔt Force–Time Graph: Area under the graph equals impulse Conservation of Linear Momentum: m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂' Collision formulas: Elastic: v₁' = [(m₁ - m₂)/(m₁ + m₂)]v₁ + [2m₂/(m₁ + m₂)]v₂ Inelastic: (m₁ + m₂)v' = m₁v₁ + m₂v₂ The worked examples illustrate numerical problems and real-life applications of Newton’s laws and collision principles.