Define impulse as force × time for which force acts; recall and use the equation impulse = F Δt = Δ(m v)

Momentum – Impulse

Learning Objectives (Cambridge IGCSE 0625)

• Define linear momentum and state that it is a vector quantity.
• State and use the impulse‑momentum theorem  \( \mathbf{I}= \mathbf{F}\Delta t = \Delta\mathbf{p}\).
• Apply the principle of conservation of momentum to solve:

  • one‑dimensional (core) problems, and
  • simple two‑dimensional (supplementary) problems.

• Determine the resultant of two momentum vectors (supplementary requirement).

Quick‑Reference Box (AO1)

Key Concepts

1. Linear Momentum

• \(\mathbf{p}=m\mathbf{v}\) – product of mass (kg) and velocity (m s⁻¹).
• Vector quantity: direction is the same as the velocity vector.
• Units: kg m s⁻¹ (equivalent to N s).

2. Resultant of Two Momentum Vectors (Supplementary)

• When two bodies have momenta \(\mathbf{p}{1}\) and \(\mathbf{p}{2}\) acting at right angles, the total momentum is the vector sum

  \[

  \mathbf{p}{\text{total}}=\mathbf{p}{1}+\mathbf{p}_{2}.

  \]

• For perpendicular vectors, use the Pythagorean theorem:

  \[

  |\mathbf{p}{\text{total}}|=\sqrt{p{1}^{2}+p_{2}^{2}}.

  \]

• Graphically, place the tail of \(\mathbf{p}{2}\) at the head of \(\mathbf{p}{1}\) (head‑to‑tail method) and draw the resultant from the tail of the first to the head of the second.

3. Resultant Force and Change of Momentum

• Newton’s second law in its momentum form:

  \[

  \mathbf{F}_{\text{net}}=\frac{\Delta\mathbf{p}}{\Delta t}.

  \]

• If the mass is constant, this reduces to \(\mathbf{F}=m\mathbf{a}\).

4. Impulse

• Impulse is the effect of a net force acting for a finite time:

  \[

  \mathbf{I}= \int{t{1}}^{t{2}}\mathbf{F}\,dt\approx\mathbf{F}{\text{avg}}\Delta t.

  \]

• Impulse equals the change in momentum:

  \[

  \mathbf{I}= \Delta\mathbf{p}=m\Delta\mathbf{v}=m(\mathbf{v}{f}-\mathbf{v}{i}).

  \]

• Units: N s (identical to kg m s⁻¹).
• Direction of \(\mathbf{I}\) is the same as the direction of the net force.

5. Conservation of Momentum (Core)

• If the net external force on a closed system is zero, the total momentum of the system remains constant:

  \[

  \sum\mathbf{p}{\text{initial}}=\sum\mathbf{p}{\text{final}}.

  \]

• Applicable to collisions and explosions where external forces are negligible during the short interaction.

Impulse–Momentum Theorem

For a body of constant mass:

\[

\mathbf{I}= \Delta\mathbf{p}=m(\mathbf{v}{f}-\mathbf{v}{i}).

\]

If the force is constant, the theorem can be written as:

\[

\mathbf{F}\,\Delta t = m\Delta\mathbf{v}.

\]

Worked Examples

Example 1 – Elastic collision with a wall (1 D, core)

A 0.5 kg ball strikes a rigid wall at 4 m s⁻¹, rebounds directly backwards with the same speed. Find the impulse delivered by the wall.

1. \(p{i}=mv{i}=0.5\times4=2.0\;\text{kg m s}^{-1}\) (to the right).
2. \(p{f}=mv{f}=0.5\times(-4)=-2.0\;\text{kg m s}^{-1}\) (to the left).
3. \(\Delta p = p{f}-p{i}= -2.0-2.0 = -4.0\;\text{kg m s}^{-1}\).
4. \(\displaystyle I = \Delta p = -4.0\;\text{N s}\).

   The negative sign shows the impulse acts opposite to the ball’s original motion.

Example 2 – Inelastic collision (1 D, core)

A 0.8 kg cart moving at 3 m s⁻¹ collides with a stationary 0.6 kg cart; they stick together. Determine the impulse on the 0.8 kg cart.

1. \(p_{\text{initial}} = (0.8)(3)+(0.6)(0)=2.4\;\text{kg m s}^{-1}\).
2. Total mass after collision: \(m_{\text{total}}=1.4\;\text{kg}\).
3. Common final speed: \(\displaystyle v{f}= \frac{p{\text{initial}}}{m_{\text{total}}}= \frac{2.4}{1.4}=1.71\;\text{m s}^{-1}\) (right).
4. Final momentum of the 0.8 kg cart: \(p_{f}=0.8\times1.71=1.37\;\text{kg m s}^{-1}\).
5. Impulse on the 0.8 kg cart: \(\displaystyle I = p{f}-p{i}=1.37-2.4 = -1.03\;\text{N s}\).

   The impulse opposes the original motion, indicating a loss of forward momentum.

Example 3 – Resultant of two perpendicular momenta (2‑D, supplementary)

Two identical pucks (0.5 kg each) glide on a frictionless air‑table. Puck A moves east at 2 m s⁻¹, Puck B moves north at 3 m s⁻¹. They collide and stick together. Find the speed of the combined mass after the collision.

1. Momentum components before collision:

   \[

   p_{x}= (0.5)(2)=1.0\;\text{kg m s}^{-1}\;( \text{east}),\qquad

   p_{y}= (0.5)(3)=1.5\;\text{kg m s}^{-1}\;( \text{north}).

   \]

2. Resultant momentum (vector addition):

   \[

   |\mathbf{p}|=\sqrt{p{x}^{2}+p{y}^{2}}=\sqrt{1.0^{2}+1.5^{2}}=\sqrt{3.25}=1.80\;\text{kg m s}^{-1}.

   \]

3. Combined mass: \(m_{\text{total}}=0.5+0.5=1.0\;\text{kg}\).
4. Speed after collision:

   \[

   v=\frac{|\mathbf{p}|}{m_{\text{total}}}=1.80\;\text{m s}^{-1}.

   \]

This example explicitly satisfies the supplementary requirement to “apply the principle of conservation of momentum in two dimensions”.

Example 4 – Vector‑addition diagram (supplementary)

Sketch the head‑to‑tail diagram for Example 3:

• Draw a horizontal arrow of length proportional to \(p_{x}=1.0\) (east).
• From its head, draw a vertical arrow of length proportional to \(p_{y}=1.5\) (north).
• The resultant arrow from the tail of the first to the head of the second represents the total momentum \(\mathbf{p}\). Its length corresponds to 1.80 kg m s⁻¹ and its direction is \(\tan^{-1}(p{y}/p{x})\approx56.3^{\circ}\) north of east.

Summary Table

Common Misconceptions – Quick Check

• Sign of impulse: The sign only indicates direction relative to the chosen axis; it is not a “positive = good” notion.
• Impulse vs. force: Impulse is the product of force and the time over which it acts. A large force applied briefly can give the same impulse as a small force applied for a long time.
• Momentum conservation: It holds only when external forces are negligible during the interaction (e.g., the short contact time in a collision).
• Resultant vector: Adding vectors head‑to‑tail is essential; simply adding magnitudes is incorrect unless the vectors are collinear.

Suggested Diagram for Teaching