Newton's first and second laws of motion; momentum
Section: Dynamics | Syllabus: Cambridge AS Level Physics (9702)
Newton's First Law (Law of Inertia)
Newton's First Law An object remains at rest or continues to move with uniform velocity unless acted upon by a resultant force. Real-World InertiaWhen a bus driver suddenly applies the brakes, you keep moving forward. While the brakes exert a force on the bus to make it stop, no such force acts directly on you initially, so you continue with your uniform velocity until something (like a seatbelt or handrail) provides a force.
Uniform Motion and Equilibrium
When all forces on an object are balanced, the resultant force is zero. In this state, an object is said to be in translational equilibrium and will either remain stationary or move at a constant velocity.
Concept Check: Forces on a Constant Velocity CarFor a car travelling at constant velocity in a straight line:
- Vertical: The weight acting down is balanced by the normal contact force (or normal reaction force) from the ground.
- Horizontal: The driving force is equal and opposite to the resistive forces (friction and air resistance).
Show a car with four balanced force arrows: Weight (down), Normal Contact Force (up), Driving Force (forward), and Resistive Forces/Drag (backward). Label the resultant force as zero.
The car stays in uniform motion because the sum of all force vectors is zero.
Newton's Second Law
Newton's second law describes the relationship between resultant force and the changes in motion produced.
Newton's Second Law (General Form) The rate of change of momentum of an object is directly proportional to the resultant force acting on it. The change in momentum occurs in the direction of the resultant force.Mathematical Form:
F = (Δ(mv)/Δ t) = (Δ p/Δ t)If mass is constant:
F = ma Defining the Newton (N)One Newton (1 N) is the resultant force that gives a mass of 1 kg an acceleration of 1 m s^-2. This means 1 N = 1 kg m s^-2.
Worked Example: Force, Mass, and AccelerationQuestion: A force of 520 N is applied separately to a car (950 kg) and a shopping trolley (34 kg). Calculate their initial accelerations.
SolutionCar: a = (F/m) = (520/950) = 0.55 m s^-2
Trolley: a = (F/m) = (520/34) = 15.3 m s^-2
Notice: A larger mass requires a much greater force to achieve the same acceleration.
Linear Momentum (p)
Momentum is defined as the product of mass and velocity. It is a vector quantity.
p = mvUnits: kg m s^-1 or N s
Worked Example: Momentum Calculations(a) Bullet: Mass 22g, Velocity 490 m s^-1.
p = 0.022 × 490 = 10.8 kg m s^-1(b) Space Probe: Mass 470kg, Velocity 16,000 m s^-1.
p = 470 × 16,000 = 7.52 × 10^6 kg m s^-1Force and Momentum Interactions
The force required to stop an object depends on its momentum and the time taken to stop it.
Inverse ProportionalityForce is inversely proportional to the time taken (F ∝ (1/Δ t)). If you double the stopping time, you halve the force required.
Worked Example: Catching a Tennis BallQuestion: A 65 g tennis ball moving at 22 m s^-1 is caught and brought to rest in 0.25 s. Calculate the force required.
Solution F = (Δ p/Δ t) = (0.065 × 22/0.25) = 5.72 N Worked Example: Stopping a 44,000 kg TruckA truck moves at 25 m s^-1. Calculate the force to stop it in (a) 10 s and (b) 5 s.
SolutionΔ p = p_final - p_initial = 0 - (44,000 × 25) = -1.1 × 10^6 kg m s^-1.
(a) In 10 s: F = (1.1 × 10^6/10) = 1.1 × 10^5 N
(b) In 5 s: F = (1.1 × 10^6/5) = 2.2 × 10^5 N
Safety Applications
Features like airbags and crumple zones in cars are designed systematically based on Newton's Second Law.
- The change in momentum (Δ p) during a crash is fixed by the initial velocity.
- Airbags and crumple zones increase the time (Δ t) of the collision.
- Since F = (Δ p/Δ t), increasing Δ t significantly reduces the resulting average force on the occupants.
Compare two curves on a Force-Time graph with the same area (impulse). Curve A (stiff collision) is tall and narrow; Curve B (crumple zone) is low and wide, showing how peak force is reduced by spreading momentum change over time.
Worked Example: Box in a VanQuestion: A 100 kg box is in a van travelling at 20 m s^-1. The van takes 5 s to stop. Calculate the frictional force needed to prevent the box from sliding forward.
Solution F = (Δ p/Δ t) = (100 × 20/5) = 400 N Worked Example: Football Kick Contact TimeQuestion: A player kicks a 0.43 kg ball at 31 m s^-1 with an average force of 130 N. How long was the foot in contact with the ball?
Solution Δ t = (m Δ v/F) = (0.43 × 31/130) = 0.10 sVariable Mass Systems
The general form of the 2nd Law (F = (Δ p/Δ t)) is essential when mass is not constant, such as in jet engines or water hoses.
Worked Example: Jet Engine ThrustQuestion: A jet engine releases 5.0 kg of exhaust per second at 65 m s^-1. Calculate the force.
SolutionThrust is the rate of change of momentum.
F = (Δ m/Δ t) × v = 5.0 kg/s × 65 m s^-1 = 325 N Worked Example: Hosepipe ForceQuestion: A hosepipe releases water at 3.0 kg s^-1 at a speed of 4.7 m s^-1. Calculate the force.
Solution F = (Δ m/Δ t) × v = 3.0 × 4.7 = 14.1 NInteractive revision notes, videos and practice questions load below.