igcse Momentum
Definition of Momentum
Every moving object has linear momentum. It is defined as the product of mass and velocity.
Key Formula
p = m v \; \text{ (units: kg·m/s) }
Vector! Always include a sign/direction for momentum.
Newton’s Second Law (Momentum Form)
The rate of change of momentum of a body is proportional to the net force and occurs in the same direction.
Key Formula
F_\text{net} = \dfrac{\Delta p}{\Delta t}
If mass is constant, this reduces to F = m a.
Impulse
Impulse is the change in momentum produced by a force acting for a time.
[formula]J = F\,\Delta t = \Delta p \quad \text{(units: N·s = kg·m/s)}[/formula>Idea
A bouncing ball undergoes a larger change in momentum than one that is simply caught, so the impulse (and peak force) is larger.
Principle of Conservation of Momentum
In an isolated system, the total momentum remains constant during any interaction (e.g., collision).
- Choose a positive direction (e.g., right).
- Write total p before = total p after.
- Substitute signs for directions and solve.
Do not add speeds; use signed velocities.
Worked Examples
1) Gun Recoil
A bullet of mass 0.03 kg leaves a gun at 1000 m/s. The gun’s mass is 1.5 kg. Find the recoil speed of the gun (take forward as +).
- Total momentum before firing = 0.
- After:
p_\text{bullet} + p_\text{gun} = 0⇒(0.03)(+1000) + (1.5)v_g = 0. v_g = - (0.03×1000)/1.5 = -20 m/s(20 m/s backward).
2) Two Carts (Inelastic)
Cart A (0.80 kg) at +3.0 m/s sticks to Cart B (0.40 kg) at −1.0 m/s. Find their common speed immediately after collision.
p_\text{before} = 0.80×(+3.0) + 0.40×(−1.0) = 2.4 − 0.4 = 2.0 kg·m/s.- Total mass = 1.20 kg ⇒
v = 2.0/1.20 ≈ 1.67 m/s(forward).
Quick Practice
- A 0.20 kg ball changes velocity from +8.0 m/s to −6.0 m/s in 0.040 s. Find the average force.
- Two skaters (45 kg and 60 kg) push off each other and move apart with 1.8 m/s for the lighter one. Find the other’s speed and direction.
- A 2.0 kg trolley receives a 5.0 N force for 0.30 s. Find its impulse and change in speed.
Summary
p = m v(vector);J = \Delta p = F \Delta tF_\text{net} = \dfrac{\Delta p}{\Delta t}(→F=maif m const.)- Isolated system ⇒ total momentum conserved in all interactions.