Constant Net Force, Kinematics Derivations, and Gravitational Field Principles

Course Methodology and Momentum Principle Review

• Pedagogical Strategy:

  • Assessment questions focus on applying fundamental principles to new situations rather than memorizing formulas or answers.
  • Problem-solving requires deep conceptual understanding to generate solutions from first principles.

• Analysis of Discrete Momentum Data:

  • Given data points for an object moving along the x-axis:
    • At $t = 0 \,\text{s}$, initial x-momentum $p_{x,0} = 120 \,\text{kg}\cdot\text{m/s}$.
    • At $t = 1 \,\text{s}$, x-momentum $p_{x,1} = 100 \,\text{kg}\cdot\text{m/s}$.
    • At $t = 2 \,\text{s}$, x-momentum $p_{x,2} = 80 \,\text{kg}\cdot\text{m/s}$.
    • At $t = 3 \,\text{s}$, x-momentum $p_{x,3} = 60 \,\text{kg}\cdot\text{m/s}$.
  • Calculating Change in Momentum ($\Delta p_x$) and Net Force ($F_{\text{net}, x}$):
    • From $t = 0 \,\text{s}$ to $t = 1 \,\text{s}$:             $\Delta p_x = p_{x,1} - p_{x,0} = 100 \,\text{kg}\cdot\text{m/s} - 120 \,\text{kg}\cdot\text{m/s} = -20 \,\text{kg}\cdot\text{m/s}$
    • Rate of change of momentum over interval $\Delta t = 1 \,\text{s}$:             $\frac{\Delta p_x}{\Delta t} = \frac{-20 \,\text{kg}\cdot\text{m/s}}{1 \,\text{s}} = -20 \,\text{kg}\cdot\text{m/s}^2 = -20 \,\text{N}$
    • From $t = 1 \,\text{s}$ to $t = 2 \,\text{s}$:             $\Delta p_x = 80 \,\text{kg}\cdot\text{m/s} - 100 \,\text{kg}\cdot\text{m/s} = -20 \,\text{kg}\cdot\text{m/s}$$\frac{\Delta p_x}{\Delta t} = -20 \,\text{N}$
  • Conclusions:
    • The net force acting on the object is constant throughout the entire time interval.
    • By the Momentum Principle, $F_{\text{net}, x} = \frac{\Delta p_x}{\Delta t} = -20 \,\text{N}$.
    • A net force of $-20 \,\text{N}$ corresponds to a magnitude of $20 \,\text{N}$ directed in the negative x-direction.
    • If the net force were zero, the change in momentum $\Delta p$ would be zero, resulting in a constant velocity and constant momentum.

• Curriculum Roadmap:

  • Sections 2.2 to 2.7 & 2.8: Focus on physical systems subjected to a constant net force to derive exact, analytic solutions for predicting future motion.
  • Section 2.3 / 2.4 onward: Returns to examine systems subject to non-constant (varying) net forces.

Impulse and Average Net Force During Collisions

• Golf Ball Collision Example:

  • A golf ball hit off a driver travels at a high speed of approximately $150 \,\text{mph}$.
  • When the ball collides with a rigid steel plate, contact occurs over a finite time interval $\Delta t$.
  • During the collision, the ball flattens, compresses, re-expands, and subsequently oscillates.

• Neglecting Gravitational Force:

  • During high-speed collisions, the contact force exerted by the steel plate on the golf ball far exceeds the force of gravity.
  • Because the gravitational force is negligible in comparison, it is set to zero for the duration of the impact analysis.
  • General Rule: In collision dynamics, internal contact forces between colliding bodies dominate over ambient non-contact forces like gravity.

• Definition of Time-Average Net Force ($F_{\text{avg}}$):

  • The measured force $F(t)$ during impact starts at zero, increases rapidly to a peak, and drops back to zero across interval $\Delta t$
  • $F_{\text{avg}}$ is not an arithmetic mean of force values.
  • $F_{\text{avg}}$ is defined as a constant force value that produces the exact same integral (area under the curve) over time interval $\Delta t$ as the real time-varying force:         $\Delta p_x = \int_{t_i}^{t_f} F_{\text{net}, x}(t) \, dt = F_{\text{avg}, x} \Delta t$
  • The area under a force vs. time graph represents the total change in momentum ($\Delta p$).

• Fundamental Principles vs. Approximations:

  • Momentum Principle (General):$\Delta \mathbf{p} = \mathbf{F}_{\text{net}} \Delta t$
  • Relativistic Momentum Relation:$\mathbf{p} = \gamma m \mathbf{v}$
  • Low-Speed Approximation: When speed $v$ is much less than the speed of light $c$ ($v \ll c$), the Lorentz factor $\gamma \approx 1$.
  • Rule of Thumb: $\gamma \approx 1$ can be assumed whenever speed is less than $10\%$ of the speed of light ($v < 0.10 \, c$).

Kinematic Equations under Constant Net Force

• Velocity as a Function of Time:

  • Starting from the Momentum Principle with $\gamma = 1$:         $\mathbf{p}_{\text{final}} = \mathbf{p}_{\text{initial}} + \mathbf{F}_{\text{net}} \Delta t$
  • Dividing through by constant mass $m$ yields the velocity update equation:         $\mathbf{v}_{\text{final}} = \mathbf{v}_{\text{initial}} + \frac{\mathbf{F}_{\text{net}}}{m} \Delta t$
  • Under a constant net force, the velocity graph $\mathbf{v}(t)$ is strictly linear with respect to time.

• Average Velocity under Constant Net Force:

  • Because the velocity-time relationship is strictly linear for constant forces, the time-average velocity equals the arithmetic mean of the initial and final velocities:         $\mathbf{v}_{\text{avg}} = \frac{1}{2} (\mathbf{v}_{\text{initial}} + \mathbf{v}_{\text{final}})$
  • Critical Warning: This arithmetic mean relationship is only valid when net force is constant. It cannot be applied if the net force varies over time.

• Derivation of Position Update Equation:

  • Definition of Average Velocity (General):$\mathbf{r}_{\text{final}} = \mathbf{r}_{\text{initial}} + \mathbf{v}_{\text{avg}} \Delta t$
  • Substituting $\mathbf{v}_{\text{final}} = \mathbf{v}_{\text{initial}} + \frac{\mathbf{F}_{\text{net}}}{m} \Delta t$ into the arithmetic average velocity formula:         $\mathbf{v}_{\text{avg}} = \frac{1}{2} \left( \mathbf{v}_{\text{initial}} + \mathbf{v}_{\text{initial}} + \frac{\mathbf{F}_{\text{net}}}{m} \Delta t \right) = \mathbf{v}_{\text{initial}} + \frac{1}{2} \frac{\mathbf{F}_{\text{net}}}{m} \Delta t$
  • Substituting this expression for $\mathbf{v}_{\text{avg}}$ into the general position definition:         $\mathbf{r}_{\text{final}} = \mathbf{r}_{\text{initial}} + \left( \mathbf{v}_{\text{initial}} + \frac{1}{2} \frac{\mathbf{F}_{\text{net}}}{m} \Delta t \right) \Delta t$
  • Distributing $\Delta t$ yields the complete analytic position prediction equation for constant net force:         $\mathbf{r}_{\text{final}} = \mathbf{r}_{\text{initial}} + \mathbf{v}_{\text{initial}} \Delta t + \frac{1}{2} \frac{\mathbf{F}_{\text{net}}}{m} (\Delta t)^2$

• Summary of Core Constant-Force Equations:

  1. $\mathbf{v}_{\text{final}} = \mathbf{v}_{\text{initial}} + \frac{\mathbf{F}_{\text{net}}}{m} \Delta t$
  2. $\mathbf{v}_{\text{avg}} = \frac{1}{2} (\mathbf{v}_{\text{initial}} + \mathbf{v}_{\text{final}})$ (Constant force only)
  3. $\mathbf{r}_{\text{final}} = \mathbf{r}_{\text{initial}} + \mathbf{v}_{\text{avg}} \Delta t$
  4. $\mathbf{r}_{\text{final}} = \mathbf{r}_{\text{initial}} + \mathbf{v}_{\text{initial}} \Delta t + \frac{1}{2} \frac{\mathbf{F}_{\text{net}}}{m} (\Delta t)^2$

Analytical Application: Fan Cart vs. Fan Truck

• System Setup:

  • A fan cart (car) has mass $m_{\text{car}} = m$.
  • A fan truck has double the mass $m_{\text{truck}} = 2m$.
  • Both objects experience the exact same constant forward pushing force $F_{\text{net}, x} = +F$ generated by pushing air backwards.
  • Both objects start from rest: $v_{\text{initial}, x} = 0$
  • Both objects travel the exact same linear distance $d$: $\Delta x = d$

• Derivation of Travel Time ($\Delta t$):

  • Using the 1D position equation:         $\Delta x = v_{\text{initial}, x} \Delta t + \frac{1}{2} \frac{F_{\text{net}, x}}{m} (\Delta t)^2$
  • Substitute known values ($\Delta x = d$, $v_{\text{initial}, x} = 0$, $F_{\text{net}, x} = F$):         $d = \frac{1}{2} \frac{F}{m} (\Delta t)^2$
  • Solving for $\Delta t$:         $(\Delta t)^2 = \frac{2 m d}{F} \implies \Delta t = \sqrt{\frac{2 m d}{F}}$
  • Proportionality Analysis for Time:$\Delta t \propto \sqrt{m}$
  • Evaluating the truck ($2m$) relative to the car ($m$):         $\Delta t_{\text{truck}} = \sqrt{2} \cdot \Delta t_{\text{car}}$
  • Conclusion: The car reaches the finish line first because it has a smaller time interval. The truck takes a factor of $\sqrt{2}$ longer.

• Derivation of Final Velocity ($v_{\text{final}, x}$):

  • Substitute the time equation $\Delta t = \sqrt{\frac{2 m d}{F}}$ into the velocity update formula:         $v_{\text{final}, x} = 0 + \frac{F}{m} \left( \sqrt{\frac{2 m d}{F}} \right)$
  • Bringing the factor $\frac{F}{m}$ inside the radical:         $v_{\text{final}, x} = \sqrt{\left( \frac{F}{m} \right)^2 \left( \frac{2 m d}{F} \right)} = \sqrt{\frac{2 F d}{m}}$
  • Proportionality Analysis for Final Velocity:$v_{\text{final}, x} \propto \frac{1}{\sqrt{m}}$
  • Evaluating the truck ($2m$) relative to the car ($m$):         $v_{\text{truck}} = \frac{1}{\sqrt{2}} v_{\text{car}}$
  • Conclusion: The car attains a higher final velocity at the finish line by a factor of $\sqrt{2}$ compared to the truck.

Gravitational Field and Gravitational Force

• Gravitational Field Definition:

  • Near Earth's surface, the gravitational force exerted on an object of mass $m$ is given by:         $\mathbf{F}_{\text{grav}} = m \mathbf{g}$
  • $\mathbf{g}$ represents the gravitational field produced by surrounding objects (e.g., Earth).
  • Magnitude near Earth's surface: $g \approx 9.8 \,\text{N/kg}$ (or $9.8 \,\text{m/s}^2$).
  • Direction: Directed vertically downward toward the center of Earth.

• Pedagogical Clarification on Terminology:

  • Referring to $g$ as "acceleration due to gravity" is physically imprecise.
  • The correct term is the gravitational field of the source object.
  • An object only accelerates at $g$ if gravity is the sole force acting on it (free fall in a vacuum).

• Conceptual Application Questions:

  • Question 1: Comparing gravitational forces on Ball A (mass $m$) and Ball B (mass $2m$):
    • $\mathbf{F}_{\text{grav}, A} = m \mathbf{g}$
    • $\mathbf{F}_{\text{grav}, B} = 2 m \mathbf{g}$
    • Result: Ball B experiences twice as much gravitational force as Ball A.
  • Question 2: Dropping Ball A ($m$) and Ball B ($2m$) from rest over distance $d$:
    • While the force on Ball B is double ($2mg$ vs. $mg$), its inertia/mass is also double ($2m$ vs. $m$).
    • The resulting acceleration for both objects in free fall is identical:             $a = \frac{F_{\text{net}}}{m} = \frac{m g}{m} = g$

Questions & Discussion

• Question: What is the net force acting on an object if its momentum changes from $120 \,\text{kg}\cdot\text{m/s}$ to $100 \,\text{kg}\cdot\text{m/s}$ and then to $80 \,\text{kg}\cdot\text{m/s}$ in equal 1-second intervals?

  • Response: The change in momentum per second is $-20 \,\text{kg}\cdot\text{m/s}^2$. Therefore, the net force is constant at $-20 \,\text{N}$ (directed in the negative x-direction).

• Question: If force is constant, is net force zero?

  • Response: No. A constant non-zero net force causes a constant rate of change in momentum ($\frac{\Delta p}{\Delta t} = F_{\text{net}}$). If the net force were zero, the momentum would remain entirely unchanged over time.

• Question: Is gravitational force on a heavier ball larger, and does it fall faster?

  • Response: The gravitational force $F_{\text{grav}} = mg$ is twice as large on a ball of mass $2m$ compared to mass $m$. However, because force scales linearly with mass, the acceleration $a = \frac{F}{m} = g$ remains identical for both balls in the absence of air resistance.