Non-uniform motion

1. Overview

Non-uniform motion occurs when the resultant force acting on an object changes over time, causing the acceleration to be non-constant. In the context of A-Level Physics, this typically involves objects moving through a fluid (a gas or a liquid) where resistive forces—such as air resistance or viscous drag—oppose the motion. Unlike the idealized "free fall" in a vacuum where acceleration remains constant at $g$, real-world motion is governed by the dynamic relationship between velocity and resistive forces. As an object's velocity increases, the resistive force increases, which in turn reduces the resultant force and the acceleration. This process continues until the resistive force perfectly balances the driving force (such as weight), leading to a state of equilibrium known as terminal velocity, where the object continues to move at a constant maximum speed with zero acceleration.

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Key Definitions

• Friction: A force that opposes relative motion between two surfaces in contact. It acts along the interface of the surfaces and converts kinetic energy into thermal energy.
• Drag (Viscous Force): A resistive force experienced by an object moving through a fluid (liquid or gas). It acts in the opposite direction to the object's velocity.
• Air Resistance: A specific form of drag caused by the collisions between a moving object and air molecules.
• Terminal Velocity: The constant maximum velocity reached by an object when the magnitude of the resistive forces is exactly equal to the magnitude of the accelerating force (resultant force is zero).
• Viscosity: A property of a fluid that characterizes its internal resistance to flow. Qualitatively, it is the "thickness" or "stickiness" of a fluid.
• Uniform Gravitational Field: A region where the acceleration of free fall $g$ is constant in magnitude and direction at all points (e.g., near the Earth's surface, $g = 9.81 \text{ m s}^{-2}$).
• Resultant Force: The vector sum of all individual forces acting on an object, which determines the object's acceleration according to $F = ma$.

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Content

3.1 Frictional and Viscous Forces

In introductory kinematics, we often assume "frictionless" environments. In non-uniform motion, we must account for the interaction between the object and its surroundings.

• Contact Friction (Solid-Solid):
  • Occurs when two solid surfaces slide or attempt to slide over each other.
  • At the AS Level, we treat this force as being largely independent of the speed of the object, though it depends on the nature of the surfaces and the normal contact force.
• Fluid Resistance (Drag):
  • This force is fundamentally different from solid friction because it is velocity-dependent.
  • As an object moves through a fluid, it must push fluid particles out of its path.
  • At higher speeds, the object hits more fluid particles per second, and it hits them with greater momentum.
  • Key Principle: Drag force increases as the speed of the object increases.

3.2 Factors Affecting the Magnitude of Drag

While you are not required to use specific coefficients (like the coefficient of viscosity $\eta$), you must be able to explain qualitatively how the following factors change the drag force $D$:

1. Velocity ($v$): For most objects, $D \propto v$ at low speeds (laminar flow) and $D \propto v^2$ at higher speeds (turbulent flow). In all cases, $D$ increases as $v$ increases.
2. Cross-sectional Area ($A$): The larger the area perpendicular to the direction of travel, the more fluid molecules the object must displace, increasing drag.
3. Shape (Streamlining): A "blunt" object creates more disturbance in the fluid than a "streamlined" (aerodynamic) object. Streamlining allows fluid to flow smoothly around the object, reducing the pressure difference between the front and back.
4. Fluid Properties: Drag is greater in fluids with higher density or higher viscosity. For example, a ball falls much slower through honey (high viscosity) than through water (lower viscosity).

3.3 Motion in a Uniform Gravitational Field with Air Resistance

This is the most common exam application. Consider an object of mass $m$ dropped from rest in a uniform gravitational field where air resistance is present.

Stage 1: The Instant of Release ($t = 0$)

• The initial velocity $v = 0$.
• Since drag $D$ depends on velocity, $D = 0$.
• The only force acting is the weight $W = mg$ acting downwards.
• The resultant force $F_{net} = W$.
• The initial acceleration $a = \frac{F_{net}}{m} = \frac{mg}{m} = g$.
• Exam Note: The gradient of a velocity-time graph at $t=0$ must be exactly $9.81 \text{ m s}^{-2}$.

Stage 2: Intermediate Motion (Accelerating at a decreasing rate)

• As the object falls, its velocity $v$ increases.
• The air resistance (drag) $D$ increases as $v$ increases.
• The weight $W$ remains constant.
• The resultant force $F_{net} = W - D$ (taking downwards as positive).
• As $D$ grows, the term $(W - D)$ decreases.
• Since $a = \frac{F_{net}}{m}$, the acceleration $a$ decreases.
• Crucial Distinction: The object is still speeding up because $a$ is still positive, but it is speeding up more slowly than before.

Stage 3: Reaching Terminal Velocity ($a = 0$)

• Eventually, the velocity reaches a point where the upward drag force $D$ is equal in magnitude to the downward weight $W$.
• The resultant force $F_{net} = W - D = 0$.
• According to Newton’s First Law, if $F_{net} = 0$, the object no longer accelerates.
• The object has reached terminal velocity ($v_t$). It will continue to fall at this constant speed until it hits the ground or another force acts upon it.

3.4 Graphical Representation of Non-Uniform Motion

The Velocity-Time ($v-t$) Graph:

• Shape: A curve starting at the origin $(0,0)$.
• Initial Gradient: Steepest at the start, representing $g = 9.81 \text{ m s}^{-2}$.
• Curvature: The gradient gradually decreases (becomes less steep) as time passes.
• Asymptote: The curve flattens out into a horizontal line. This horizontal line represents the terminal velocity.

The Acceleration-Time ($a-t$) Graph:

• Shape: A curve starting at $a = 9.81 \text{ m s}^{-2}$ on the y-axis.
• Trend: The acceleration decreases over time.
• Final Value: The curve approaches $a = 0$ as the object reaches terminal velocity.

3.5 The Effect of Mass on Terminal Velocity

If two objects have the same size and shape (and thus experience the same drag force at the same speed) but have different masses:

1. The heavier object has a greater weight $W$.
2. It requires a larger drag force $D$ to balance this weight ($D = W$).
3. Since drag increases with speed, the heavier object must reach a higher speed before the drag is large enough to balance its weight.
4. Conclusion: For objects of the same shape/size, the more massive object will have a higher terminal velocity and will take longer to reach it.

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Worked Example 1 — Calculating Instantaneous Acceleration

A weather balloon of mass $12 \text{ kg}$ is released. At a certain point in its descent, it is traveling at a speed where the upward air resistance is $45 \text{ N}$. Calculate the acceleration of the balloon at this instant.

Step 1: Identify the forces acting on the balloon.

• Weight ($W$) acting downwards.
• Air resistance ($D$) acting upwards.

Step 2: Calculate the weight. $$W = mg$$ $$W = 12 \text{ kg} \times 9.81 \text{ m s}^{-2}$$ $$W = 117.72 \text{ N}$$

Step 3: Calculate the resultant force ($F_{net}$). Taking downwards as the positive direction: $$F_{net} = W - D$$ $$F_{net} = 117.72 \text{ N} - 45 \text{ N}$$ $$F_{net} = 72.72 \text{ N}$$

Step 4: Calculate acceleration using Newton's Second Law. $$F_{net} = ma$$ $$a = \frac{F_{net}}{m}$$ $$a = \frac{72.72 \text{ N}}{12 \text{ kg}}$$ $$a = 6.06 \text{ m s}^{-2}$$

Answer: $a = 6.1 \text{ m s}^{-2}$ (to 2 significant figures).

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Worked Example 2 — Qualitative Analysis of a Parachute

A skydiver falls at terminal velocity. They then open their parachute. Describe and explain the subsequent motion of the skydiver until a new terminal velocity is reached.

1. Initial State: The skydiver is at terminal velocity, so Weight ($W$) = Drag ($D_{old}$). The resultant force is zero, and acceleration is zero.

2. Opening the Parachute:

• Describe: The surface area increases drastically.
• Explain: This causes an immediate and large increase in air resistance ($D$). Now, $D > W$.

3. Deceleration Phase:

• Explain: The resultant force $F_{net} = W - D$ is now negative (upwards).
• Describe: The skydiver decelerates (velocity decreases).
• Explain: As the velocity $v$ decreases, the air resistance $D$ also begins to decrease (since $D \propto v^2$).

4. Reaching New Terminal Velocity:

• Explain: Eventually, $D$ decreases until it once again equals the weight $W$.
• Describe: The resultant force becomes zero again. The skydiver moves at a new, lower, constant terminal velocity.

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Worked Example 3 — Comparing Terminal Velocities

Two spheres, $X$ and $Y$, are dropped from a high altitude. Sphere $X$ is solid lead, and Sphere $Y$ is a hollow plastic shell of the same external diameter. Explain why Sphere $X$ reaches a higher terminal velocity than Sphere $Y$.

Step 1: Compare the forces. Both spheres have the same external dimensions, so at any given velocity $v$, they experience the same air resistance $D$. However, the solid lead sphere $X$ has a much greater mass, and therefore a much greater weight $W_X > W_Y$.

Step 2: Apply the condition for terminal velocity. Terminal velocity is reached when $D = W$.

Step 3: Logical Conclusion. For Sphere $Y$, the drag $D$ only needs to reach a small value to equal $W_Y$. This happens at a relatively low velocity. For Sphere $X$, the drag $D$ must reach a much larger value to equal $W_X$. Since drag increases with velocity, Sphere $X$ must accelerate to a much higher velocity before this balance is achieved. Answer: Sphere $X$ has a higher terminal velocity.

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Key Equations

Physical Quantity	Equation	Symbols	Units	Data Sheet?
Weight	$W = mg$	$m$: mass, $g$: accel. of free fall	$\text{N, kg, m s}^{-2}$	No
Newton's 2nd Law	$F = ma$	$F$: resultant force, $a$: acceleration	$\text{N, kg, m s}^{-2}$	Yes
Resultant Force	$F_{net} = W - D$	$W$: weight, $D$: drag/resistance	$\text{N}$	No
Terminal Velocity	$W = D$	$W$: weight, $D$: drag	$\text{N}$	No

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Common Mistakes to Avoid

• ❌ Wrong: Thinking that acceleration is zero at the moment an object is released.
  • ✓ Right: At $t=0$, velocity is zero, so drag is zero. The resultant force is the full weight, so acceleration is maximum ($g = 9.81 \text{ m s}^{-2}$).
• ❌ Wrong: Stating that "velocity decreases" as an object falls towards terminal velocity.
  • ✓ Right: The acceleration decreases, but the velocity is still increasing. The object only slows down if the resistive force becomes greater than the weight (e.g., when a parachute opens).
• ❌ Wrong: Confusing "balanced forces" with "no forces."
  • ✓ Right: At terminal velocity, forces are balanced ($W = D$). Gravity has not stopped acting; it is simply being countered by the fluid resistance.
• ❌ Wrong: Using $g = 10 \text{ m s}^{-2}$ or $g = 9.8 \text{ m s}^{-2}$.
  • ✓ Right: The Cambridge 9702 syllabus specifically requires $g = 9.81 \text{ m s}^{-2}$.
• ❌ Wrong: Drawing a $v-t$ graph that reaches terminal velocity and then curves back down.
  • ✓ Right: Once terminal velocity is reached, the line must remain perfectly horizontal (constant velocity) unless the physical conditions (like area or fluid density) change.

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Exam Tips

1. The "Chain of Reasoning": When asked to explain the motion of a falling object, always follow this logical sequence:
   • State that $v$ increases.
   • State that $D$ increases because $D$ depends on $v$.
   • State that the resultant force $F = W - D$ decreases.
   • State that since $F = ma$, the acceleration $a$ decreases.
2. Graph Gradients: If you are asked to sketch a $v-t$ graph, use a ruler to draw a faint tangent at $t=0$ and ensure its slope looks like $9.81$. Ensure the curve never actually "dips" unless a parachute is mentioned.
3. Significant Figures: Always check the data provided in the question. If the mass is $12 \text{ kg}$ (2 s.f.) and $g$ is $9.81 \text{ m s}^{-2}$ (3 s.f.), your final answer should typically be given to 2 or 3 s.f.
4. Free-Body Diagrams: When drawing forces on a falling object:
   • Draw arrows starting from the center of gravity.
   • Label them clearly (e.g., "Weight" and "Air Resistance").
   • The length of the arrows matters! Before terminal velocity, the weight arrow must be longer than the drag arrow. At terminal velocity, they must be exactly the same length.
5. Read the Context: If the object is rising (like a bubble in oil), the drag force acts downwards (opposing the upward motion). The logic remains the same, but the direction of the vectors is reversed.