10.2 Newton's Laws of Motion, Momentum, Impulse & Elastic Collisions

Newton's Laws of Motion, Momentum, Impulse & Elastic Collisions

Dynamics investigates the physical causes of motion, connecting applied forces to changes in momentum and kinetic energy. In the Intermediate Physics Part I syllabus and the UET ECAT, questions from this domain focus on Newton's laws, inertial frames, elevator tension problems, impulse calculations from force-time graphs, one-dimensional collision kinematics, and variable mass systems such as rocket propulsion.

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1. Newton's Laws of Motion and Reference Frames

Newton's First Law (Law of Inertia)

A body continues in its state of rest or of uniform motion in a straight line unless acted upon by a non-zero net external force:

$\sum \mathbf{F} = \mathbf{0} \implies \mathbf{a} = \mathbf{0} \iff \mathbf{v} = \text{constant}$

Inertia is the inherent property of matter that resists changes in its state of rest or motion. Mass is the quantitative measure of inertia: a larger mass possesses greater inertia.

Reference Frames

• Inertial Reference Frame: A reference frame that is stationary or moving with uniform velocity relative to distant stars. Newton's laws are valid strictly in inertial frames without modification.
• Non-Inertial Reference Frame: An accelerated or rotating reference frame. To apply Newton's second law in a non-inertial frame with acceleration $\mathbf{a}_0$, an observer must introduce a fictitious pseudo-force: $\mathbf{F}_{\text{pseudo}} = -m\mathbf{a}_0$

Newton's Second Law of Motion

Newton formulated his second law in terms of momentum: the time rate of change of linear momentum of a body is directly proportional to the applied net force and occurs in the direction of that force:

$\mathbf{F}_{\text{net}} = \frac{d\mathbf{p}}{dt} = \frac{d(m\mathbf{v})}{dt}$

If the mass $m$ remains constant with respect to time:

$\mathbf{F}_{\text{net}} = m\frac{d\mathbf{v}}{dt} = m\mathbf{a}$

The SI unit of force is the Newton ($\text{N}$), where $1\text{ N} = 1\text{ kg}\cdot\text{m/s}^2$, having dimensional formula $[M L T^{-2}]$.

Apparent Weight in an Elevator

Consider a body of mass $m$ resting on a scale inside an elevator moving vertically:

1. Elevator stationary or moving at uniform velocity ($a = 0$): $T - mg = 0 \implies T = mg \quad (\text{Apparent weight} = \text{True weight})$
2. Elevator accelerating upward with acceleration $a$: $T - mg = ma \implies T = m(g + a) \quad (\text{Apparent weight increases})$
3. Elevator accelerating downward with acceleration $a$ ($a < g$): $mg - T = ma \implies T = m(g - a) \quad (\text{Apparent weight decreases})$
4. Elevator cable snaps (Free Fall, $a = g$): $T = m(g - g) = 0 \quad (\text{Apparent weightlessness})$

Newton's Third Law (Action and Reaction)

To every action, there is always an equal and opposite reaction:

$\mathbf{F}_{AB} = -\mathbf{F}_{BA}$

where $\mathbf{F}_{AB}$ is the force exerted on body $A$ by body $B$, and $\mathbf{F}_{BA}$ is the force exerted on body $B$ by body $A$.

ECAT Conceptual Principle: Action and reaction forces never balance or cancel each other out to produce equilibrium because they act on two different interacting bodies, never on the same body.

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2. Linear Momentum and the Impulse-Momentum Theorem

Linear Momentum ($\mathbf{p}$)

Linear momentum is the product of a body's mass and its velocity:

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

Linear momentum is a vector quantity pointing in the direction of the velocity vector. Its SI unit is kilogram-meter per second ($\text{kg}\cdot\text{m/s}$) or Newton-second ($\text{N}\cdot\text{s}$), with dimensional formula $[M L T^{-1}]$.

Impulse of a Force ($\mathbf{J}$)

Impulse is defined as the product of an average force and the time interval during which it acts. For a time-varying force:

$\mathbf{J} = \int_{t_1}^{t_2} \mathbf{F} dt = \mathbf{F}_{\text{avg}}\Delta t$

From Newton's second law $\mathbf{F} = \frac{d\mathbf{p}}{dt}$, integrating over the time interval yields the Impulse-Momentum Theorem:

$\mathbf{J} = \int_{t_1}^{t_2} \frac{d\mathbf{p}}{dt} dt = \mathbf{p}_f - \mathbf{p}_i = \Delta\mathbf{p}$

The impulse of the net force acting on a particle equals the resulting change in linear momentum of the particle.

Geometric Interpretation on Force-Time Graph

On a Force versus Time ($F-t$) graph, the definite area under the curve between times $t_1$ and $t_2$ represents the impulse $\mathbf{J}$:

$\text{Area under } F-t \text{ curve} = \int_{t_1}^{t_2} F(t) dt = J = \Delta p$

Impact Force Mitigation

From $F_{\text{avg}} = \frac{\Delta p}{\Delta t}$, for a fixed change in momentum $\Delta p$, the average force experienced is inversely proportional to the duration of impact $\Delta t$. Practical engineering applications include:

• Airbags and vehicle crumple zones that increase collision duration $\Delta t$, reducing the peak force experienced by passengers.
• A cricket fielder pulling their hands backward when catching a fast ball to increase contact time $\Delta t$ and reduce impact force on the hands.

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3. Law of Conservation of Linear Momentum

An isolated system is a collection of interacting bodies upon which no net external force acts ($\sum \mathbf{F}_{\text{ext}} = \mathbf{0}$).

$\frac{d\mathbf{P}_{\text{total}}}{dt} = \sum \mathbf{F}_{\text{ext}} = \mathbf{0} \implies \mathbf{P}_{\text{total}} = \text{constant}$

$\mathbf{p}_{1i} + \mathbf{p}_{2i} = \mathbf{p}_{1f} + \mathbf{p}_{2f} \implies m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2 = m_1 \mathbf{v}_1' + m_2 \mathbf{v}_2'$

Recoil of a Rifle

A bullet of mass $m_b$ is fired horizontally with velocity $v_b$ from a rifle of mass $m_r$ initially at rest:

$0 = m_r v_r + m_b v_b \implies v_r = -\frac{m_b}{m_r} v_b$

The negative sign indicates recoil in the opposite direction. Since $m_r \gg m_b$, the recoil speed $v_r$ is much smaller than the muzzle speed $v_b$.

Ratio of Recoil Kinetic Energies

Because momentum magnitudes are equal ($p_r = p_b$), using $K = \frac{p^2}{2m}$:

$\frac{K_r}{K_b} = \frac{p^2/(2m_r)}{p^2/(2m_b)} = \frac{m_b}{m_r}$

The bullet carries the vast majority of the explosive chemical energy because of its much smaller mass.

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4. Elastic vs. Inelastic Collisions

Collisions are classified based on whether total mechanical kinetic energy is conserved:

1. Elastic Collision: Total linear momentum AND total kinetic energy are both conserved. No mechanical energy is converted into thermal energy, sound, or permanent deformation ($Q = 0$).
2. Inelastic Collision: Total linear momentum is conserved, but total kinetic energy is NOT conserved ($K_f < K_i$). Some kinetic energy is dissipated into internal heat, sound, or material deformation.
3. Perfectly Inelastic Collision: The colliding bodies stick together after impact and move as a single combined mass with a common velocity $V$: $V = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$ Kinetic energy loss is maximized in perfectly inelastic collisions: $\Delta K_{\text{loss}} = \frac{1}{2}\frac{m_1 m_2}{m_1 + m_2}(v_1 - v_2)^2$

The Coefficient of Restitution ($e$)

$e = \frac{\text{Relative velocity of separation}}{\text{Relative velocity of approach}} = \frac{v_2' - v_1'}{v_1 - v_2}$

• Perfectly elastic collision: $e = 1$
• Inelastic collision: $0 < e < 1$
• Perfectly inelastic collision: $e = 0$
• Super-elastic / explosive collision: $e > 1$

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5. One-Dimensional Elastic Collisions: Derivation & Formulas

Consider two smooth spheres of masses $m_1$ and $m_2$ moving along a straight line with initial velocities $v_1$ and $v_2$ ($v_1 > v_2$), colliding head-on to emerge with final velocities $v_1'$ and $v_2'$.

Step 1: Conservation of Momentum

$m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2' \implies m_1(v_1 - v_1') = m_2(v_2' - v_2) \quad \text{--- (Eq. 1)}$

Step 2: Conservation of Kinetic Energy

$\frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 = \frac{1}{2}m_1 v_1'^2 + \frac{1}{2}m_2 v_2'^2 \implies m_1(v_1^2 - v_1'^2) = m_2(v_2'^2 - v_2^2) \quad \text{--- (Eq. 2)}$

Factoring both sides of Eq. 2 as differences of squares:

$m_1(v_1 - v_1')(v_1 + v_1') = m_2(v_2' - v_2)(v_2' + v_2) \quad \text{--- (Eq. 3)}$

Step 3: Dividing Eq. 3 by Eq. 1

$v_1 + v_1' = v_2' + v_2 \implies v_1 - v_2 = -(v_1' - v_2') = v_2' - v_1'$

Fundamental Law of 1D Elastic Collisions: The relative speed of approach prior to collision equals the relative speed of separation after collision.

Step 4: Final Velocity Formulas

Solving the linear system yields the final velocity equations:

$v_1' = \left(\frac{m_1 - m_2}{m_1 + m_2}\right)v_1 + \left(\frac{2m_2}{m_1 + m_2}\right)v_2$

$v_2' = \left(\frac{2m_1}{m_1 + m_2}\right)v_1 + \left(\frac{m_2 - m_1}{m_1 + m_2}\right)v_2$

Analysis of Critical Special Cases

Case 1: Identical Masses ($m_1 = m_2 = m$)

Substituting $m_1 = m_2$ into the equations:

$v_1' = v_2, \quad v_2' = v_1$

Two bodies of equal mass completely exchange their velocities upon head-on elastic collision. If the second body was initially stationary ($v_2 = 0$), the first body stops dead ($v_1' = 0$) and the second body departs with the initial velocity of the first ($v_2' = v_1$).

Case 2: Massive Target at Rest ($m_2 \gg m_1, v_2 = 0$)

Since $m_1 \ll m_2$, approximate $\frac{m_1 - m_2}{m_1 + m_2} \approx -1$ and $\frac{2m_1}{m_1 + m_2} \approx 0$:

$v_1' \approx -v_1, \quad v_2' \approx 0$

A light particle bounces back with virtually unchanged speed, while the massive stationary target remains at rest.

Case 3: Massive Projectile Colliding with Light Stationary Target ($m_1 \gg m_2, v_2 = 0$)

Since $m_2 \ll m_1$, approximate $\frac{m_1 - m_2}{m_1 + m_2} \approx 1$ and $\frac{2m_1}{m_1 + m_2} \approx 2$:

$v_1' \approx v_1, \quad v_2' \approx 2v_1$

The heavy projectile continues moving forward with virtually unchanged speed, while the light stationary target is propelled forward at twice the projectile's speed.

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6. Rocket Propulsion: Variable Mass Dynamics

A rocket operates via the conservation of linear momentum by ejecting burnt fuel gases rearward at high speed. As the rocket moves, its mass decreases continuously, making it a variable-mass system.

The Rocket Equation

Let $m$ be the instantaneous mass of the rocket and $v_{\text{rel}}$ be the speed of exhaust gases relative to the rocket nozzle. The upward thrust force generated by fuel expulsion is:

$F_{\text{thrust}} = -v_{\text{rel}}\frac{dm}{dt} = v_{\text{rel}}\left|\frac{dm}{dt}\right|$

Applying Newton's second law vertically in Earth's gravitational field:

$m a = F_{\text{thrust}} - mg = v_{\text{rel}}\left(-\frac{dm}{dt}\right) - mg$

Dividing through by the instantaneous mass $m$:

$a = \frac{v_{\text{rel}}}{m}\left(-\frac{dm}{dt}\right) - g$

ECAT Key Insight: Because propellant consumption reduces the rocket's mass $m$ in the denominator while the burn rate $\left|\frac{dm}{dt}\right|$ remains constant, the rocket's acceleration $a$ continuously increases throughout powered flight.

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Collision Types and Formulas Comparison Table