Define resultant force as the change in momentum per unit time; recall and use the equation F = Δp / Δt

Momentum – Resultant Force (Cambridge IGCSE Physics 0625 – Topic 1.6)

Learning Objectives

• Define momentum as a vector and write the expression p = mv.
• State and label the definition of resultant (net) force as the rate of change of momentum: F = Δp / Δt.
• Explain impulse, its relationship to momentum change, and how to read it from a force‑time graph.
• Apply the conservation of momentum to collisions of closed systems.
• Use vector addition when combining momenta of several objects.

Key Concepts

1. Momentum

• Momentum p is a vector quantity:
  

  \(\mathbf{p}=m\mathbf{v}\)

• Direction of \(\mathbf{p}\) is the same as the direction of the velocity \(\mathbf{v}\).
• For the SI unit to be kg·m·s⁻¹, mass must be entered in kilograms (kg) and velocity in metres per second (m s⁻¹).

2. Resultant (Net) Force

• The resultant force is defined as the rate at which an object’s momentum changes:
  

  Equation (1): \(\displaystyle \mathbf{F}_{\text{net}}=\frac{\Delta\mathbf{p}}{\Delta t}\)

• If the mass is constant, Equation (1) reduces to Newton’s second law:
  

  Equation (2): \(\displaystyle \mathbf{F}=m\mathbf{a}\)

• Equation (1) is the form required by the syllabus because it remains valid when the mass changes (e.g., rockets).

3. Impulse

• Impulse is the cause of a change in momentum.
  

  Impulse = Force × time = Δp

• For a constant force acting over a time interval Δt:
  

  \(\displaystyle \text{Impulse}=F\Delta t=\Delta p\)

• On a force–time (F‑t) graph, the shaded area under the curve equals the impulse (Δp).

4. Conservation of Momentum (Closed System)

• When no external forces act, the total momentum of a system remains constant.
• For two interacting objects (1 and 2):
  

  \(\displaystyle m{1}v{1}+m{2}v{2}=m{1}v'{1}+m{2}v'{2}\)

• This principle is used for both elastic and inelastic collisions.

5. Vector Addition of Momenta

• Momenta are added tip‑to‑tail, just like any other vectors.
• Example: two carts moving at right angles – draw the two momentum vectors and complete the parallelogram to obtain the resultant momentum.

Derivation of \(F = \Delta p / \Delta t\)

1. Start with the definition of momentum: \(\mathbf{p}=m\mathbf{v}\).
2. For constant mass, the change in momentum over a time interval Δt is
   

   \(\displaystyle \Delta\mathbf{p}=m\Delta\mathbf{v}\).

3. Divide by Δt to obtain the average resultant force:
   

   \(\displaystyle \mathbf{F}_{\text{avg}}=\frac{\Delta\mathbf{p}}{\Delta t}=m\frac{\Delta\mathbf{v}}{\Delta t}=m\mathbf{a}\).

4. Taking the limit as Δt → 0 gives the instantaneous force:
   

   \(\displaystyle \mathbf{F}= \frac{d\mathbf{p}}{dt}\).

Units

Worked Examples

Example 1 – Average Resultant Force

Problem: A 0.15 kg ball’s speed changes from 2.0 m s⁻¹ to 8.0 m s⁻¹ in 0.05 s. Find the average resultant force.

1. Initial and final momenta:
   

   \(pi = m vi = 0.15 \times 2.0 = 0.30\ \text{kg·m·s}^{-1}\)
   

   \(pf = m vf = 0.15 \times 8.0 = 1.20\ \text{kg·m·s}^{-1}\)

2. Change in momentum:
   

   \(\Delta p = pf - pi = 1.20 - 0.30 = 0.90\ \text{kg·m·s}^{-1}\)

3. Average resultant force (use Equation 1):
   

   \(\displaystyle F_{\text{avg}} = \frac{\Delta p}{\Delta t}= \frac{0.90}{0.05}= 18\ \text{N}\)

4. Answer: 18 N in the direction of the ball’s motion.

Example 2 – Impulse and Speed Change

Problem: A constant horizontal force of 5 N acts on a 0.30 kg cart for 0.20 s. Determine the increase in the cart’s speed.

1. Impulse:
   

   \(\displaystyle \text{Impulse}=F\Delta t = 5 \times 0.20 = 1.0\ \text{N·s}\)

2. Impulse = Δp, so
   

   \(\Delta p = 1.0\ \text{kg·m·s}^{-1}\)

3. Using \(\Delta p = m\Delta v\):
   

   \(\displaystyle \Delta v = \frac{\Delta p}{m}= \frac{1.0}{0.30}=3.33\ \text{m s}^{-1}\)

4. Answer: The cart’s speed increases by 3.3 m s⁻¹.

Example 3 – Inelastic Collision (Conservation of Momentum)

Problem: Two carts on a frictionless track stick together after colliding. Cart A (0.5 kg) moves at 2.0 m s⁻¹ to the right; Cart B (0.8 kg) is initially at rest. Find the speed of the combined mass after the collision.

1. Initial total momentum:
   

   \(p{\text{initial}} = mA vA + mB v_B = 0.5 \times 2.0 + 0.8 \times 0 = 1.0\ \text{kg·m·s}^{-1}\)

2. Let \(v'\) be the speed after sticking together. Conservation of momentum gives:
   

   \((mA+mB)v' = p_{\text{initial}}\)

3. Solve for \(v'\):
   

   \(\displaystyle v' = \frac{1.0}{0.5+0.8}= \frac{1.0}{1.3}=0.77\ \text{m s}^{-1}\) (to the right)

4. Answer: The combined carts move at 0.77 m s⁻¹ to the right.

Common Mistakes to Avoid

• Treating momentum as a scalar – always keep the vector direction (use bold symbols or arrows).
• Confusing the units: impulse (N·s) ≠ force (N) and ≠ energy (J).
• Applying \(F = ma\) when the mass is not constant; use \(F = \Delta p / \Delta t\) instead.
• Neglecting unit conversions (g → kg, cm → m) before substituting into formulas.
• Assuming momentum is conserved when external forces act – the system must be closed.

Summary Box

Resultant (net) force is defined as the rate of change of momentum:

\(\displaystyle \mathbf{F}_{\text{net}} = \frac{\Delta\mathbf{p}}{\Delta t}\)  (Equation 1)

Impulse (\(F\Delta t\)) is the cause of that change, and when external forces are absent the total momentum of a closed system remains constant.