define and use linear momentum as the product of mass and velocity

Cambridge A‑Level Physics 9702 – Momentum and Newton’s Laws

Learning Objectives (aligned with the syllabus)

• State and apply Newton’s three laws of motion (exact textbook wording for the first law).
• Define linear momentum and impulse, and use the relation \(\mathbf{p}=m\mathbf{v}\).
• Distinguish between scalar and vector quantities; add and subtract coplanar vectors using components (including a worked 2‑D example).
• Apply the impulse‑momentum theorem and interpret both momentum–time and force–time graphs.
• State the principle of conservation of linear momentum for a closed system and use it to solve 1‑D and 2‑D collisions (elastic, inelastic, completely inelastic).
• Explain qualitatively how non‑uniform forces (e.g. air resistance, terminal velocity) affect momentum.
• Link Newton’s second law in its momentum form to the work‑energy concepts (AO1‑AO3).
• Design a simple experiment to test momentum conservation and evaluate sources of error.

1. Symbols, Quantities & Units

2. Scalars vs. Vectors

• Scalar quantities have magnitude only (e.g. mass, speed, kinetic energy).
• Vector quantities have magnitude and direction (e.g. velocity, momentum, force).
• Vectors are added tip‑to‑tail or by components:

  \[

  \mathbf{A}+\mathbf{B}= (Ax+Bx)\,\hat{\mathbf{i}}+(Ay+By)\,\hat{\mathbf{j}}

  \]

Worked 2‑D Vector‑Addition Example

A force \(\mathbf{F}1 = 30\;\text{N}\) acts east and a second force \(\mathbf{F}2 = 40\;\text{N}\) acts at \(30^{\circ}\) north of east. Find the resultant \(\mathbf{R}\).

1. Resolve \(\mathbf{F}_2\):

   \[

   F_{2x}=40\cos30^{\circ}=34.6\;\text{N},\qquad

   F_{2y}=40\sin30^{\circ}=20.0\;\text{N}

   \]

2. Add components:

   \[

   R_x = 30 + 34.6 = 64.6\;\text{N},\qquad

   R_y = 0 + 20.0 = 20.0\;\text{N}

   \]

3. Magnitude and direction:

   \[

   |\mathbf{R}|=\sqrt{64.6^2+20.0^2}=67.6\;\text{N},\qquad

   \theta = \tan^{-1}\!\left(\frac{20.0}{64.6}\right)=17^{\circ}\;\text{north of east}

   \]

3. Linear Momentum – Definition

The linear momentum \(\mathbf{p}\) of a particle of mass \(m\) moving with velocity \(\mathbf{v}\) is

\[

\boxed{\mathbf{p}=m\mathbf{v}}

\]

Because \(\mathbf{v}\) is a vector, \(\mathbf{p}\) points in the same direction as the motion.

4. Newton’s Laws in Momentum Form

4.1 First Law (Law of Inertia)

“A body remains at rest or in uniform straight‑line motion unless acted on by a net external force.”

Equivalently: if the net external force is zero, the momentum of the body remains constant.

4.2 Second Law (Momentum Form)

\[

\boxed{\mathbf{F}_{\text{net}}=\frac{d\mathbf{p}}{dt}}

\]

If the mass is constant, \(\displaystyle \frac{d\mathbf{p}}{dt}=m\frac{d\mathbf{v}}{dt}=m\mathbf{a}\).

4.3 Third Law (Action–Reaction)

For every force \(\mathbf{F}{AB}\) exerted by object A on object B there is an equal and opposite force \(\mathbf{F}{BA}=-\mathbf{F}_{AB}\). The pair of forces produces equal and opposite changes in momentum, ensuring that the total momentum of the isolated pair is conserved.

5. Impulse–Momentum Theorem

Integrating Newton’s second law over a time interval \(\Delta t\) gives

\[

\boxed{\mathbf{J}= \int{t1}^{t2}\mathbf{F}\,dt = \Delta\mathbf{p}= \mathbf{p}2-\mathbf{p}_1}

\]

Key graphical interpretations:

• The area under a force–time graph equals the impulse \(\mathbf{J}\).
• The slope of a momentum–time graph equals the net force \(\mathbf{F}_{\text{net}}\).
• The area under a momentum–time graph gives the impulse (since \(\int \mathbf{F}\,dt = \Delta\mathbf{p}\)).

6. Non‑Uniform Motion: Air Resistance & Terminal Velocity

• When a resistive force \( \mathbf{F}{\text{drag}} = -k\mathbf{v}\) (linear drag) or \( -c v^{2}\hat{\mathbf{v}}\) (quadratic drag) acts, the net force is no longer zero and the momentum changes according to \(\mathbf{F}{\text{net}} = m\mathbf{a}\).
• For a falling object, the drag force grows with speed until it balances the weight: \(mg = k v{\text{t}}\) (linear) or \(mg = c v{\text{t}}^{2}\) (quadratic). At this point the net force is zero and the object moves with constant momentum (terminal velocity).
• Even though the momentum of the object changes, the total momentum of the system (object + air) remains conserved because the air receives the opposite momentum change.

7. Conservation of Linear Momentum

For a closed system (no net external force), the vector sum of the momenta of all constituents remains constant:

\[

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

\]

This principle underpins the analysis of collisions and explosions.

8. Collisions – Types, Key Relations & Coefficient of Restitution

Coefficient of Restitution (optional but useful)

\[

e = \frac{\text{relative speed after collision}}{\text{relative speed before collision}}

\qquad (0\le e \le 1)

\]

For a one‑dimensional impact of objects 1 and 2:

\[

e = \frac{v{2f}-v{1f}}{v{1i}-v{2i}}

\]

9. Worked Examples

Example 0 – 2‑D Vector Addition (see Section 2)

Resultant force \(\mathbf{R}=67.6\;\text{N}\) at \(17^{\circ}\) north of east.

Example 1 – One‑Dimensional Elastic Collision

Two carts on a frictionless air track:

• Cart A: \(mA=0.50\;\text{kg},\;v{A}=+2.0\;\text{m s}^{-1}\)
• Cart B: \(mB=0.80\;\text{kg},\;v{B}=0\;\text{m s}^{-1}\)

Find the velocities after the collision.

1. Conservation of momentum:

   \[

   mA vA + mB vB = mA vA' + mB vB' \tag{1}

   \]

2. Conservation of kinetic energy (elastic):

   \[

   \tfrac12 mA vA^{2}+ \tfrac12 mB vB^{2}= \tfrac12 mA {vA'}^{2}+ \tfrac12 mB {vB'}^{2} \tag{2}

   \]

3. Solving (1) and (2) gives

   \[

   v_A' = -0.57\;\text{m s}^{-1},\qquad

   v_B' = +1.43\;\text{m s}^{-1}

   \]

4. Check:

   \[

   \text{Momentum: }0.5(2.0)=0.5(-0.57)+0.8(1.43)=1.0\;\text{kg m s}^{-1}

   \]

   \[

   \text{K.E.: } \tfrac12(0.5)(2.0)^2 = \tfrac12(0.5)(-0.57)^2+\tfrac12(0.8)(1.43)^2 \approx 1.0\;\text{J}

   \]

Example 2 – Completely Inelastic Collision

Same masses, but the carts stick together after impact.

\[

mA vA + mB vB = (mA+mB)v'

\]

\[

v' = \frac{0.5\times2.0}{0.5+0.8}=0.77\;\text{m s}^{-1}

\]

Momentum is conserved; kinetic energy after the collision is

\[

K_f = \tfrac12(1.30)(0.77)^2 = 0.39\;\text{J}

\]

Only 39 % of the initial kinetic energy remains; the rest is dissipated as heat/deformation.

Example 3 – Two‑Dimensional Elastic Collision (Billiard Balls)

Ball 1 (\(m1=0.20\;\text{kg}\)) moves at \(5.0\;\text{m s}^{-1}\) along +x and strikes stationary ball 2 (\(m2=0.20\;\text{kg}\)). After impact ball 1 moves at \(3.0\;\text{m s}^{-1}\) at \(30^{\circ}\) above +x. Find the speed and direction of ball 2.

1. Resolve ball 1 after the collision:

   \[

   v_{1x}' = 3.0\cos30^{\circ}=2.60\;\text{m s}^{-1},\qquad

   v_{1y}' = 3.0\sin30^{\circ}=1.50\;\text{m s}^{-1}

   \]

2. Momentum conservation in components:

   \[

   \begin{aligned}

   x:&\; m1 v{1x}=m1 v{1x}'+m2 v{2x}'\\

   y:&\; 0=m1 v{1y}'+m2 v{2y}'

   \end{aligned}

   \]

   Substituting numbers (both masses = 0.20 kg):

   \[

   \begin{aligned}

   0.20(5.0)&=0.20(2.60)+0.20 v_{2x}'\\

   0&=0.20(1.50)+0.20 v_{2y}'

   \end{aligned}

   \]

3. Solve:

   \[

   v_{2x}' = 2.20\;\text{m s}^{-1},\qquad

   v_{2y}' = -1.50\;\text{m s}^{-1}

   \]

4. Magnitude and direction of ball 2:

   \[

   v_2'=\sqrt{(2.20)^2+(1.50)^2}=2.7\;\text{m s}^{-1},\qquad

   \theta = \tan^{-1}\!\left(\frac{-1.50}{2.20}\right) = -34^{\circ}

   \]

   (34° below the +x‑axis).

10. Common Misconceptions

• Momentum = force. Momentum is a property of a moving object; force is the *rate of change* of momentum.
• Only speed matters. Momentum is a vector; direction determines sign and must be considered.
• Mass can be ignored. Momentum scales linearly with mass; a heavy slow object can have the same momentum as a light fast one.
• Momentum is always conserved. It is conserved only when the net external force on the system is zero.
• Energy is always conserved in collisions. Kinetic energy is conserved only in elastic collisions; inelastic collisions convert kinetic energy into other forms.
• Impulse is a different physical quantity. Impulse is numerically equal to the change in momentum; they share the same units.
• Vector addition is optional. In two‑dimensional problems you must resolve momenta into components before adding.
• Air resistance does not affect momentum. Drag provides an external force, so the object's momentum changes; the total momentum of object + air remains conserved.

11. Practical Activity – Testing Momentum Conservation

1. Set up a low‑friction air track with two gliders of known masses \(m1\) and \(m2\).
2. Measure initial velocities using photogate timers; record \(v{1i}\) and \(v{2i}\).
3. Cause collisions:

   • Elastic – use a spring‑loaded bumper.
   • Inelastic – attach Velcro so the gliders stick together.

4. Measure final velocities \(v{1f}, v{2f}\) (or common velocity for the stuck pair).
5. Calculate total momentum before and after:

   \[

   p{\text{initial}} = m1 v{1i}+m2 v_{2i},\qquad

   p{\text{final}} = m1 v{1f}+m2 v_{2f}

   \]

6. Discuss agreement and evaluate sources of error:

   • Residual air‑track friction.
   • Timing uncertainty of photogates.
   • Imperfect elasticity or incomplete sticking.
   • Unaccounted air resistance (demonstrates Section 6).

12. Summary

Linear momentum \(\mathbf{p}=m\mathbf{v}\) links an object’s mass and its motion. Newton’s first law states that without a net external force the momentum stays constant; the second law quantifies the change (\(\mathbf{F}=d\mathbf{p}/dt\)), leading to the impulse‑momentum theorem. Newton’s third law guarantees that internal forces produce equal and opposite momentum changes, so the total momentum of a closed system is conserved. This conservation law, together with the distinction between elastic and inelastic collisions (and the coefficient of restitution), provides a powerful framework for solving both one‑ and two‑dimensional problems, interpreting graphs, and designing experiments that test fundamental dynamics.

Suggested Diagram