Define momentum as mass × velocity; recall and use the equation p = m v
1.6 Momentum
Learning Objectives
- Define linear momentum and recognise it as a vector quantity.
- Recall and use the fundamental relation \(p = mv\).
- Explain the link between momentum, impulse and resultant force (including the connection to \(F = ma\)).
- Interpret momentum–time graphs (area = impulse).
- Apply the principle of conservation of momentum in one‑ and two‑dimensional situations.
- Plan, carry out and evaluate a simple experimental investigation of momentum, including error analysis.
1. Definition and Vector Nature
Linear momentum (\(\mathbf{p}\)) of a particle is the product of its mass (\(m\)) and its velocity (\(\mathbf{v}\)). Both magnitude and direction matter, so momentum is a vector that points in the same direction as the velocity.
Formula: \(\displaystyle \mathbf{p}=m\mathbf{v}\)
- When several momenta act together, add them vectorially:
- Component method – add the \(x\) and \(y\) components separately.
- Tip‑to‑tail method – place the tail of the second vector at the tip of the first; the resultant runs from the free tail to the free tip.
Tip‑to‑Tail Example (perpendicular momenta)
Cart A: \(mA = 2.0\;\text{kg},\; \mathbf{v}A = 3.0\;\text{m s}^{-1}\,\hat{\mathbf{i}}\) → \(\mathbf{p}_A = 6.0\;\text{kg m s}^{-1}\,\hat{\mathbf{i}}\)
Cart B: \(mB = 1.0\;\text{kg},\; \mathbf{v}B = 4.0\;\text{m s}^{-1}\,\hat{\mathbf{j}}\) → \(\mathbf{p}_B = 4.0\;\text{kg m s}^{-1}\,\hat{\mathbf{j}}\)
Resultant momentum:
\[
\mathbf{p}_{\text{tot}} = (6.0\,\hat{\mathbf{i}} + 4.0\,\hat{\mathbf{j}})\;\text{kg m s}^{-1},
\qquad
|\mathbf{p}_{\text{tot}}| = \sqrt{6.0^2 + 4.0^2}=7.2\;\text{kg m s}^{-1}
\]
2. Units and Symbols
| Quantity | Symbol | SI Unit |
|---|---|---|
| Mass | m | kg |
| Velocity | \(\mathbf{v}\) | m s⁻¹ |
| Momentum | \(\mathbf{p}\) | kg m s⁻¹ (N s) |
| Force | \(\mathbf{F}\) | N (kg m s⁻²) |
| Impulse | \(\mathbf{I}\) | N s (kg m s⁻¹) |
3. Relationship with Force – Impulse
The change in momentum of an object is produced by a resultant force acting over a time interval.
Resultant force: \(\displaystyle \mathbf{F}_{\text{res}} = \frac{\Delta\mathbf{p}}{\Delta t}\)
Impulse: \(\displaystyle \mathbf{I}= \mathbf{F}_{\text{res}}\Delta t = \Delta\mathbf{p}\)
- Impulse has the same units as momentum (N s).
- Combining with \(F = ma\) gives the familiar form \(\displaystyle F = \frac{\Delta p}{\Delta t}\), emphasising that force is the rate of change of momentum.
4. Momentum–Time Graphs
On a momentum–time graph the area under the curve between two times equals the impulse delivered during that interval.
Graphical Illustration (placeholder)
5. Conservation of Momentum (One‑Dimensional)
In the absence of external forces, the total momentum of a closed system remains constant.
Mathematical statement (1 D): \(\displaystyle m1v{1i}+m2v{2i}=m1v{1f}+m2v{2f}\)
- Valid for both elastic and perfectly inelastic collisions.
- Often used to find an unknown final speed when the masses are known.
Worked Example – Perfectly Inelastic Collision
A 0.8 kg cart moving at \(4.0\;\text{m s}^{-1}\) collides head‑on with a stationary 1.2 kg cart and they stick together. Find the speed of the combined system.
- Initial momentum: \(p_i = (0.8)(4.0) + (1.2)(0) = 3.2\;\text{kg m s}^{-1}\).
- Total mass after sticking: \(m_{\text{tot}} = 0.8 + 1.2 = 2.0\;\text{kg}\).
- Conservation: \(pi = m{\text{tot}}vf \;\Rightarrow\; vf = \dfrac{3.2}{2.0}=1.6\;\text{m s}^{-1}\) (in the original direction of the 0.8 kg cart).
6. Conservation of Momentum (Two‑Dimensional) – Supplement
When motion occurs in a plane, momentum must be conserved in each perpendicular direction independently.
Vector form: \(\displaystyle \sum \mathbf{p}{\text{initial}} = \sum \mathbf{p}{\text{final}}\)
2‑D Example – Perpendicular Collision
Cart A (mass = 1.0 kg) moves east at \(2.0\;\text{m s}^{-1}\). Cart B (mass = 2.0 kg) moves north at \(1.0\;\text{m s}^{-1}\). They collide and stick together. Find the speed of the combined mass after the collision.
- Initial momenta:
\[
\mathbf{p}_A = (1.0)(2.0)\,\hat{\mathbf{i}} = 2.0\,\hat{\mathbf{i}},\qquad
\mathbf{p}_B = (2.0)(1.0)\,\hat{\mathbf{j}} = 2.0\,\hat{\mathbf{j}}
\]
- Total momentum vector:
\[
\mathbf{p}_{\text{tot}} = 2.0\,\hat{\mathbf{i}} + 2.0\,\hat{\mathbf{j}}\;\text{kg m s}^{-1}
\]
- Total mass after sticking: \(m_{\text{tot}} = 1.0 + 2.0 = 3.0\;\text{kg}\).
- Resultant speed:
\[
vf = \frac{|\mathbf{p}{\text{tot}}|}{m_{\text{tot}}}
= \frac{\sqrt{2.0^2 + 2.0^2}}{3.0}
= \frac{2.83}{3.0}
= 0.94\;\text{m s}^{-1}
\]
Direction: \(\tan\theta = \frac{py}{px}=1\) → \(\theta = 45^\circ\) north of east.
7. Practical Investigation – Momentum of Colliding Carts
Objective: Verify conservation of linear momentum using two low‑friction carts on a track.
Equipment: air‑track (or low‑friction track), two carts with known masses, motion sensors or photogates, data logger, set of rubber bumpers (elastic) and Velcro pads (inelastic).
Method (summary):
- Measure and record the mass of each cart (to 0.01 kg).
- Calibrate the photogates and set the data logger.
- Launch cart A at a known speed toward stationary cart B; record the speed of each cart before impact.
- Allow the carts to collide (choose elastic or inelastic bumper) and record the final speeds of both carts.
- Repeat for at least five trials for each bumper type.
Data analysis:
- Calculate initial and final momenta for each trial using \(p = mv\).
- Compute the percentage difference:
\[
\%\,\text{diff}= \left|\frac{p{\text{initial}}-p{\text{final}}}{p_{\text{initial}}}\right|\times100\%
\]
- Error analysis:
- Random errors – timing jitter of photogates, slight variations in launch speed; treat with standard deviation of the set of trials.
- Systematic errors – friction of the track, mis‑alignment of the photogates, mass measurement bias; discuss how each would affect the momentum balance (e.g., friction reduces final momentum).
- Suggest improvements: use a vacuum‑track, increase photogate sampling rate, verify cart alignment with a spirit level.
Safety notes:
- Secure the track to the bench to prevent tipping.
- Use low‑mass carts and gentle bumpers; never place fingers near moving carts or photogate beams.
- Turn off the data logger when not in use to avoid accidental exposure to laser beams.
8. Worked Numerical Examples
8.1 Basic Momentum Calculation
Given a 2.0 kg object moving at 3.0 m s⁻¹, calculate its momentum.
\(p = mv = (2.0)(3.0)=6.0\;\text{kg m s}^{-1}\)
8.2 Inelastic Collision (from Section 5)
Result: \(v_f = 1.6\;\text{m s}^{-1}\) (direction of the moving cart).
8.3 Elastic Collision in One Dimension
Two carts: \(m1=1.5\;\text{kg},\;v{1i}=2.0\;\text{m s}^{-1}\) and \(m2=2.5\;\text{kg},\;v{2i}=-1.0\;\text{m s}^{-1}\). Using the elastic‑collision formulas:
\[
v{1f}= \frac{(m1-m2)}{m1+m2}v{1i}+\frac{2m2}{m1+m2}v{2i}= -0.2\;\text{m s}^{-1}
\]
\[
v{2f}= \frac{2m1}{m1+m2}v{1i}+\frac{(m2-m1)}{m1+m2}v{2i}= 1.8\;\text{m s}^{-1}
\]
8.4 Two‑Dimensional Collision (Section 6)
Result from the example: combined speed \(v_f = 0.94\;\text{m s}^{-1}\) at \(45^\circ\) north of east.
9. Practice Questions
- Calculate the momentum of a 0.5 kg ball moving at 10 m s⁻¹.
- What is the momentum of a 3.0 kg object that is at rest?
- A 1.5 kg cart moving at 2.0 m s⁻¹ collides elastically with a 2.5 kg cart moving at –1.0 m s⁻¹. Determine the velocities after the collision.
- In a laboratory set‑up, a 0.8 kg cart strikes a stationary 1.2 kg cart and they stick together. The measured final speed is 1.5 m s⁻¹. Calculate the percentage discrepancy between the experimental result and the theoretical value obtained from conservation of momentum.
- Two carts move in the xy‑plane: Cart A (1.0 kg) east at 3.0 m s⁻¹, Cart B (2.0 kg) north at 2.0 m s⁻¹. They collide and stick. Find the speed and direction of the combined mass after the collision.
10. Links to Other Syllabus Topics
- Force = ma (Section 1.4): Substituting \(a = \Delta v/\Delta t\) into \(F = ma\) yields \(F = \Delta p/\Delta t\), the impulse–momentum relationship.
- Energy, Work & Power (Section 1.7): Compare kinetic energy \(\frac12 mv^2\) with momentum \(mv\) to see that they describe different aspects of motion (energy depends on speed squared, momentum on speed linearly).
- Vectors (Section 1.2): Momentum addition uses the same component techniques introduced for displacement and force vectors.
- Newton’s Laws (Section 1.3): The second law in its original form, \(\mathbf{F} = \dfrac{d\mathbf{p}}{dt}\), underpins the impulse–momentum connection.
Illustration
