define and use force as rate of change of momentum

Momentum and Newton’s Laws of Motion

Learning Objective

Define force as the rate of change of momentum and, using this definition together with Newton’s three laws, solve quantitative problems involving constant‑force motion, variable‑mass systems, elastic and inelastic collisions, and non‑uniform motion (friction, air‑resistance and terminal velocity).

1. Momentum – the Fundamental Quantity

• Definition: \(\displaystyle \mathbf{p}=m\mathbf{v}\)
• Nature: vector; direction is the same as the velocity \(\mathbf{v}\).
• SI unit: kg·m·s⁻¹.

2. Newton’s Three Laws (Qualitative)

1. First law (Inertia) – A body remains at rest or in uniform straight‑line motion unless acted on by a net external force.
2. Second law (Quantitative) – The net external force on a particle equals the time‑rate of change of its momentum: \[ \mathbf{F}_{\text{net}}=\frac{d\mathbf{p}}{dt}. \]
3. Third law (Action–Reaction) – For every force \(\mathbf{F}_{AB}\) exerted by body A on body B there is an equal and opposite force \(\mathbf{F}_{BA}=-\mathbf{F}_{AB}\).

3. Force as the Rate of Change of Momentum

3.1 Constant‑mass systems

If the mass does not change (\(dm/dt=0\)):

\[ \mathbf{F}= \frac{d}{dt}(m\mathbf{v}) = m\frac{d\mathbf{v}}{dt}=m\mathbf{a}. \]

3.2 Variable‑mass systems (e.g. rockets)

When the mass varies with time:

\[ \mathbf{F}= \frac{d}{dt}(m\mathbf{v}) = m\frac{d\mathbf{v}}{dt}+ \mathbf{v}\frac{dm}{dt}. \]

The extra term \(\mathbf{v}\,dm/dt\) accounts for the momentum carried away (or added) with the ejected (or accreted) mass.

3.3 Impulse–Momentum Theorem

\[ \int_{t_1}^{t_2}\mathbf{F}\,dt = \mathbf{p}(t_2)-\mathbf{p}(t_1)=\Delta\mathbf{p}. \]

• Impulse \(\mathbf{J}= \displaystyle\int\mathbf{F}\,dt\) (units kg·m·s⁻¹, same as momentum).
• Especially useful for short‑duration forces such as collisions.

3.4 Constant‑force example

If a constant force \(\mathbf{F}\) acts for a duration \(\Delta t\):

\[ \Delta\mathbf{p}= \mathbf{F}\,\Delta t,\qquad \mathbf{v}_{f}= \mathbf{v}_{i}+ \frac{\mathbf{F}}{m}\,\Delta t. \]

3.5 Variable‑mass (rocket) example

Rocket of instantaneous mass \(m\) ejects exhaust at speed \(\mathbf{u}\) relative to the rocket (opposite to \(\mathbf{v}\)). The thrust is

\[ \mathbf{F}_{\text{thrust}} = -\mathbf{u}\,\frac{dm}{dt}. \]

Including external forces (gravity, drag):

\[ m\frac{d\mathbf{v}}{dt}= -\mathbf{u}\frac{dm}{dt} + \mathbf{F}_{\text{ext}}. \]

4. Conservation of Linear Momentum

For a closed system (\(\sum\mathbf{F}_{\text{ext}}=0\)):

\[ \frac{d}{dt}\Bigl(\sum\mathbf{p}\Bigr)=0 \;\Longrightarrow\; \sum\mathbf{p}_{\text{initial}}=\sum\mathbf{p}_{\text{final}}. \]

This principle underlies all collision problems.

5. Non‑Uniform Motion (Friction, Air‑Resistance & Terminal Velocity)

In many real situations the net force is not constant because it contains a resistive component that depends on speed. Typical forms are:

• Kinetic friction: \(\displaystyle \mathbf{F}_{\text{fr}}=-\mu_k N\,\hat{\mathbf{v}}\) (independent of speed, opposite to motion).
• Linear (viscous) drag: \(\displaystyle \mathbf{F}_{\text{drag}}=-b\,\mathbf{v}\) (proportional to speed).
• Quadratic drag (high‑speed air resistance): \(\displaystyle \mathbf{F}_{\text{drag}}=-c\,v^{2}\,\hat{\mathbf{v}}\).

When a falling object reaches a speed at which the downward weight \(mg\) is exactly balanced by the upward resistive force, the net force becomes zero and the object moves at a constant speed – the terminal velocity \(v_t\). Using \(F_{\text{net}}=0\) gives, for linear drag, \(mg=bv_t\) and for quadratic drag, \(mg=cv_t^{2}\). The same momentum‑force relation (\(\mathbf{F}=d\mathbf{p}/dt\)) applies; the resistive force simply appears as part of \(\mathbf{F}_{\text{net}}\).

6. Practical Applications

• Predicting vehicle speed after a known thrust or braking force.
• Analyzing collisions – elastic (kinetic energy conserved) and inelastic (objects stick together).
• Designing rocket propulsion systems using the \(\mathbf{v}\,dm/dt\) term.
• Estimating terminal velocities of sky‑divers, raindrops, or objects moving through viscous fluids.

7. Worked Examples

Example 1 – Elastic Collision (head‑on)

Problem: A 0.150 kg ball moving at \(8.0\ \text{m s}^{-1}\) collides head‑on with a 0.250 kg ball initially at rest. The impact lasts \(0.020\ \text{s}\) and the average force on each ball is \(120\ \text{N}\). Find the final velocities assuming a perfectly elastic collision.

1. Impulse on ball 1: \(J = F\Delta t = 120\times0.020 = 2.4\ \text{N·s}\).
2. Momentum change for ball 1: \(2.4 = m_1(v_{1f}-8.0)\).
3. Impulse on ball 2 (opposite direction): \(-2.4 = m_2(v_{2f}-0)\).
4. Solving the two equations gives \[ v_{1f}= -2.0\ \text{m s}^{-1},\qquad v_{2f}= 5.6\ \text{m s}^{-1}. \]
5. Check the elastic‑collision condition (kinetic‑energy conservation) – it is satisfied within rounding error.

Example 2 – Completely Inelastic Collision (carts sticking together)

Problem: A 1.20 kg cart moving at \(3.0\ \text{m s}^{-1}\) collides with a stationary 0.80 kg cart. The carts lock together after the impact, which lasts \(0.050\ \text{s}\). The average force during the collision is \(40\ \text{N}\). Find the common final speed.

1. Impulse on the system: \(J = F\Delta t = 40\times0.050 = 2.0\ \text{N·s}\).
2. Initial total momentum: \(p_i = (1.20)(3.0)+(0.80)(0)=3.6\ \text{kg·m s}^{-1}\).
3. Internal forces are equal and opposite, so the net external impulse on the combined system is zero; therefore \(p_f = p_i\).
4. Combined mass \(m_{\text{tot}} = 1.20+0.80 = 2.00\ \text{kg}\). \[ v_f = \frac{p_f}{m_{\text{tot}}}= \frac{3.6}{2.0}=1.8\ \text{m s}^{-1}. \]
5. Kinetic energy is not conserved: \[ K_i=\tfrac12(1.20)(3.0)^2=5.4\ \text{J},\qquad K_f=\tfrac12(2.00)(1.8)^2=3.24\ \text{J}. \] The loss (≈2.2 J) appears as deformation, heat, sound, etc.

Example 3 – Rocket Thrust Calculation

Problem: A rocket of mass \(m=500\ \text{kg}\) ejects propellant at a relative speed \(u=2500\ \text{m s}^{-1}\) and a mass‑flow rate \(|dm/dt| = 5\ \text{kg s}^{-1}\). Neglect gravity and drag. Find the thrust and the initial acceleration.

1. Thrust from the momentum term: \[ F_{\text{thrust}} = u\left|\frac{dm}{dt}\right| = 2500\times5 = 1.25\times10^{4}\ \text{N}. \]
2. Instantaneous acceleration (treating the mass as constant at the instant): \[ a = \frac{F_{\text{thrust}}}{m}= \frac{1.25\times10^{4}}{500}=25\ \text{m s}^{-2}. \]

8. Summary Table of Key Relations

Quantity	Symbol	Definition / Equation	SI Unit
Momentum	\(\mathbf{p}\)	\(\mathbf{p}=m\mathbf{v}\)	kg·m·s⁻¹
Force (general)	\(\mathbf{F}\)	\(\mathbf{F}=d\mathbf{p}/dt\)	N (kg·m·s⁻²)
Constant‑mass form		\(\mathbf{F}=m\mathbf{a}\)
Variable‑mass form		\(\mathbf{F}=m\mathbf{a}+\mathbf{v}\,dm/dt\)
Impulse	\(\mathbf{J}\)	\(\mathbf{J}= \int\mathbf{F}\,dt = \Delta\mathbf{p}\)	kg·m·s⁻¹
Conservation of linear momentum		\(\displaystyle\sum\mathbf{p}_{\text{initial}}=\sum\mathbf{p}_{\text{final}}\) (closed system)
Linear friction	\(\mathbf{F}_{\text{fr}}\)	\(-\mu_k N\,\hat{\mathbf{v}}\)	N
Linear drag	\(\mathbf{F}_{\text{drag}}\)	\(-b\,\mathbf{v}\)	N
Quadratic drag	\(\mathbf{F}_{\text{drag}}\)	\(-c\,v^{2}\,\hat{\mathbf{v}}\)	N

9. Suggested Diagram

10. Quick Checklist for Solving Momentum Problems

• Identify whether the system is closed (no external forces) or open.
• Determine if any mass changes during the process.
• Choose the appropriate form of Newton’s second law:
  • Constant mass → \(\mathbf{F}=m\mathbf{a}\).
  • Variable mass → \(\mathbf{F}=m\mathbf{a}+\mathbf{v}\,dm/dt\).
• If a time interval is given, apply the impulse–momentum theorem.
• For collisions, write the conservation of momentum equation; add the kinetic‑energy condition only for elastic collisions.
• When resistive forces are present, include them in \(\mathbf{F}_{\text{net}}\) (friction, linear or quadratic drag) and check whether a terminal velocity condition (\(\mathbf{F}_{\text{net}}=0\)) applies.
• Check sign conventions and units at every step.