Newton’s Laws of Motion & Collisions
Objectives:
- Solve numerical problems involving impulse and momentum.
- Interpret the area under a force–time graph.
- Interpret and apply Newton’s laws of motion.
- Compare the concepts of inertia, mass, and force.
- Deduce the relationship between mass and acceleration.
- Interpret the conservation of linear momentum in collisions.
- 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.
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