Momentum
Section: Motion, Forces & Energy | Syllabus: Cambridge AS Level Physics 9702
Momentum Momentum Mass × velocity. p = mv p = momentum (kg m/s) | m = mass (kg) | v = velocity (m/s) Momentum is a vector quantity - it has the same direction as the velocity. A stationary object has zero momentum.
Units: kg m/s (same as N s) Worked Example: Calculating Momentum A car of mass 600 kg is moving at 15 m/s. Calculate its momentum. p = mv = 600 × 15 = 9000 kg m/s Figure: Momentum as a Vector Two objects moving to the right on a horizontal surface.
Object 1: a large truck (mass 4000 kg) moving at 2 m/s - momentum arrow labelled "8000 kg m/s →". Object 2: a small car (mass 1000 kg) moving at 8 m/s - momentum arrow labelled "8000 kg m/s →". Caption: "Different masses and speeds can give the same momentum.
Direction of momentum matches direction of velocity." Impulse Impulse Force × time for which the force acts. Impulse = FΔt = Δ(mv) Impulse equals the change in momentum of the object. Units: N s (equivalent to kg m/s) The bigger the force and the longer it acts, the greater the change in momentum.
For the same change in momentum : a longer time means a smaller force is needed. Worked Example: Impulse A force of 30 N acts on a car for 10 s. The car initially has momentum 9000 kg m/s. Calculate the impulse and the final momentum.
Impulse = FΔt = 30 × 10 = 300 N s Final momentum = 9000 + 300 = 9300 kg m/s Safety Applications Seatbelts, airbags, and crumple zones all increase the time taken to stop in a crash. For the same change in momentum, a longer stopping time means a smaller force on the passenger - reducing injury.
Conservation of Momentum Whenever two objects interact, the total momentum before equals the total momentum after - provided no external forces act. Total momentum before = total momentum after If one object gains momentum, the other loses an equal amount.
This applies in one dimension - take care with direction (choose a positive direction and use signs). Figure: Conservation of Momentum in a Collision Before collision: Object A (mass m₁, velocity u₁ to the right) approaches Object B (mass m₂, stationary).
Total momentum = m₁u₁ + 0. After collision: both objects have velocities v₁ and v₂ respectively. Total momentum = m₁v₁ + m₂v₂. An equation beneath: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. Note: "Total momentum is conserved - the arrow lengths (momentum) on each side are equal in total." Worked Example: Collision (Objects Stick Together) A 2 kg trolley moving at 3 m/s collides with and sticks to a stationary 3 kg trolley.
Find their final velocity. Total momentum before = (2 × 3) + (3 × 0) = 6 kg m/s Total mass after = 2 + 3 = 5 kg Final velocity = 6 ÷ 5 = 1.2 m/s (in the original direction) Worked Example: Finding Velocity of a Ball A tennis racket (mass 3.0 kg) hits a stationary ball (mass 0.25 kg).
Before: racket moves at 20 m/s. After: racket moves at 18 m/s. Find the velocity of the ball after the collision. Momentum of racket before = 3.0 × 20 = 60 kg m/s Momentum of racket after = 3.0 × 18 = 54 kg m/s Momentum gained by ball = 60 − 54 = 6.0 kg m/s Velocity of ball = p / m = 6.0 / 0.25 = 24 m/s Resultant Force as Rate of Change of Momentum Resultant Force The change in momentum per unit time.
F = Δp / Δt F = resultant force (N) | Δp = change in momentum (kg m/s) | Δt = time (s) This is an alternative form of F = ma. It shows that a larger force produces a faster rate of change of momentum, and explains why increasing stopping time reduces the force experienced.
Worked Example: Force in a Crash A crash test dummy of mass 70 kg is travelling at 15 m/s and comes to rest in a crash. In Test 1 (no airbag) it stops in 0.03 s. In Test 2 (airbag) it stops in 0.15 s.
Calculate the force in each test. Change in momentum (both tests): Δp = 70 × 15 = 1050 kg m/s Test 1: F = Δp / Δt = 1050 / 0.03 = 35 000 N Test 2: F = Δp / Δt = 1050 / 0.15 = 7 000 N The airbag increases stopping time by 5× and reduces force by 5×.
Interactive revision notes, videos and practice questions load below.