Meditaliano IMAT Prep
Lesson 1: Physics : Kinematics & Dynamics I
Lesson 1: Introduction to Motion
Welcome to your first physics lesson for the IMAT! Today, we will build the foundation for understanding how things move. We'll explore Kinematics (the description of motion) and Dynamics (the causes of motion). Additionally, we will briefly touch on how Calculus concepts help us understand physics graphs, although you won't need to perform complex derivatives in the exam.
Learning Objective (LO P1.1):
By the end of this lesson, you will be able to analyze motion with constant acceleration using equations and graphs, understand the concept of inertial forces (elevator problems), and apply Newton's Laws to solve for forces and motion.
Part 1: Kinematics - Describing Motion
1.1 Scalars, Vectors, and Physical Quantities
In physics, we use two types of quantities to describe the world:
- Scalars: Quantities that have only magnitude (a size or amount). Examples: distance (5 m), speed (10 m/s), mass (2 kg), time (15 s).
- Vectors: Quantities that have both magnitude and direction. Examples: displacement (5 m, East), velocity (10 m/s, North), force (20 N, downwards).
Key physical quantities we use to describe motion include:
- Position ($\vec{r}$ or $x$): A vector from the origin to an object's location.
- Displacement ($\Delta\vec{r}$ or $s$): The change in position (a vector). $\Delta\vec{r} = \vec{r}_{final} - \vec{r}_{initial}$
- Distance: The total path length traveled (a scalar).
- Velocity ($\vec{v}$): The rate of change of displacement (a vector). Average velocity is $\vec{v}_{avg} = \frac{\Delta\vec{r}}{\Delta t}$.
- Speed: The rate of change of distance (a scalar). Average speed = Total Distance / Total Time.
- Acceleration ($\vec{a}$): The rate of change of velocity (a vector). Average acceleration is $\vec{a}_{avg} = \frac{\Delta\vec{v}}{\Delta t}$.
Challenge an IMAT Question!
Official Paper: 2017 - Q53
worked solution & explanation
Concept 2D Vector Addition. Aggregate independent X (East/West) and Y (North/South) components algebraically before calculating the final magnitude.
Step 1 X-axis (Horizontal): $\Sigma x = +3.0\text{ km (East)} - 6.0\text{ km (West)} = -3.0\text{ km}$ (Net $3.0\text{ km}$ West).
Step 2 Y-axis (Vertical): $\Sigma y = +8.0\text{ km (North)} - 4.0\text{ km (South)} = +4.0\text{ km}$ (Net $4.0\text{ km}$ North).
Step 3 Apply Pythagoras: $d = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5.0\text{ km}$.
Challenge an IMAT Question!
Official Paper: 2012 - Q75
worked solution & explanation
Concept Vectors vs Scalars. A vector quantity absolutely requires both a numerical magnitude and a specific spatial direction.
Step 1 Acceleration, velocity, weight (gravitational force), and electric field all intrinsically point in a specific spatial direction.
Step 2 Electric charge is an intrinsic fundamental property of matter. While it can be positive or negative (polarity), it has NO spatial direction vector. It is a scalar.
1.2 Uniform Motion (Constant Velocity)
This is the simplest type of motion, where the acceleration is zero ($a=0$). The velocity does not change. The only equation you need is:
Equation for Uniform Motion ($a=0$)
$$ s = vt $$
Displacement = constant velocity × time
Challenge an IMAT Question!
Official Paper: 2023 - Q59
worked solution & explanation
Concept Average Velocity is strictly defined as total vector Displacement divided by total Time ($\Delta\vec{x} / \Delta t$). This differs from Average Speed (Scalar Distance / Time).
Step 1 Calculate displacement magnitude. Since East and South form a $90^\circ$ angle, apply Pythagoras: $\Delta x = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ km}$.
Step 2 Convert time into hours to match the requested output unit (km/h). $15\text{ minutes} = 15 / 60\text{ hours} = 0.25\text{ hours}$ (or $1/4\text{ h}$).
Step 3 Apply formula: $\vec{v}_{avg} = \frac{10\text{ km}}{0.25\text{ h}} = 10 \times 4 = 40\text{ km/h}$.
1.3 Uniformly Accelerated Motion in 1D
This is motion along a straight line where acceleration $a$ is constant. For this special case, we use the kinematic equations, often called the "SUVAT" equations.
Kinematic Equations (SUVAT)
Where: $s$ = displacement, $u$ = initial velocity, $v$ = final velocity, $a$ = constant acceleration, $t$ = time
- $$ v = u + at $$
- $$ s = ut + \frac{1}{2}at^2 $$
- $$ v^2 = u^2 + 2as $$
- $$ s = \frac{1}{2}(u + v)t $$
1.3.1 Derivation of Kinematic Equations (SUVAT)
These equations are derived from the definitions of velocity and acceleration, assuming acceleration is constant.
1. Deriving $v = u + at$
From the definition of acceleration: $a = \frac{\Delta v}{t} = \frac{v - u}{t}$. Rearranging gives $v - u = at$, so $\mathbf{v = u + at}$.
2. Deriving $s = \frac{1}{2}(u + v)t$
For constant acceleration, average velocity is the simple mean: $v_{avg} = \frac{u + v}{2}$. Displacement is average velocity times time: $\mathbf{s = \frac{1}{2}(u + v)t}$.
3. Deriving $s = ut + \frac{1}{2}at^2$
Substitute (1) into (2): $s = \frac{1}{2}(u + (u + at))t = \frac{1}{2}(2u + at)t$. Simplifying gives $\mathbf{s = ut + \frac{1}{2}at^2}$.
4. Deriving $v^2 = u^2 + 2as$
Rearrange (1) to get $t = \frac{v - u}{a}$. Substitute into (2): $s = \frac{1}{2}(u + v)\left(\frac{v - u}{a}\right)$. Multiplying gives $2as = v^2 - u^2$, so $\mathbf{v^2 = u^2 + 2as}$.
Challenge an IMAT Question!
Official Paper: 2019 - Q53
worked solution & explanation
Concept Two-Phase Kinematics. Break the problem into Phase 1 (Uniform Acceleration) and Phase 2 (Uniform Velocity).
Step 1 Phase 1: Calculate the final velocity reached after acceleration. $v_f = v_i + at = 0 + (4.0 \times 5.0) = 20\text{ m/s}$. This is the max speed.
Step 2 Phase 2: Calculate the distance traveled during the 20s constant speed phase. $d = v \cdot t = 20\text{ m/s} \times 20\text{ s} = 400\text{ meters}$.
1.4 A Note on Calculus in Physics
While the IMAT Mathematics section does not explicitly cover calculus (differentiation and integration), understanding the concepts behind calculus makes physics graphs much more intuitive. You don't need to perform the calculations, but knowing the relationships is powerful.
Differentiation = The Slope (Gradient)
Differentiation finds the rate of change of a quantity. On a graph, this corresponds to the slope.
- Velocity is the rate of change of position: $v = \frac{dx}{dt}$. Therefore, velocity is the slope of the position-time ($x-t$) graph.
- Acceleration is the rate of change of velocity: $a = \frac{dv}{dt}$. Therefore, acceleration is the slope of the velocity-time ($v-t$) graph.
Integration = The Area Under the Curve
Integration is essentially the reverse of differentiation; it represents the accumulation of a quantity. On a graph, this corresponds to the area between the curve and the x-axis.
- Displacement is the accumulation of velocity over time: $\Delta x = \int v \, dt$. Therefore, displacement is the area under the velocity-time ($v-t$) graph.
- Change in velocity is the accumulation of acceleration over time: $\Delta v = \int a \, dt$. Therefore, change in velocity is the area under the acceleration-time ($a-t$) graph.
Intuition Check: Think of a car moving at a constant speed (a flat horizontal line on a v-t graph). The distance traveled is Speed × Time, which is geometrically the area of the rectangle formed by the line.
Challenge an IMAT Question!
Official Paper: 2024 - Q59
worked solution & explanation
Concept Calculus of Kinematics. Velocity is mathematically defined as the first time-derivative of the position function: $v(t) = \frac{dx(t)}{dt}$.
Step 1 Differentiate $x(t) = 4\cos(2\pi t)$. Applying the chain rule: $v(t) = -4 \cdot (2\pi) \cdot \sin(2\pi t) = -8\pi \sin(2\pi t)$.
Step 2 Evaluate at $t = 0.5$ seconds: $v(0.5) = -8\pi \sin(2\pi \cdot 0.5) = -8\pi \sin(\pi)$.
Step 3 From trigonometry, $\sin(\pi)$ exactly equals $0$. Therefore, $v(0.5) = -8\pi \cdot 0 = 0\text{ m/s}$.
1.5 Graphical Analysis of Motion
Applying the calculus concepts we just learned, we can summarize the relationships between graphs:
| Graph | What the Slope Means (Derivative) | What the Area Means (Integral) |
|---|---|---|
| Position vs. Time (x-t) | Instantaneous Velocity ($v$) | (No simple physical meaning) |
| Velocity vs. Time (v-t) | Instantaneous Acceleration ($a$) | Displacement ($s$ or $\Delta x$) |
| Acceleration vs. Time (a-t) | "Jerk" (rate of change of $a$) | Change in Velocity ($\Delta v$) |
Visualizing Constant Acceleration
The diagram below shows the graphs for an object starting from $x=0$ at rest ($u=0$) and moving with constant positive acceleration (like a car speeding up).
1.6 Motion in Two Dimensions (2D) - Projectile Motion
A projectile moves under gravity. We separate its motion into two independent 1D problems:
- Horizontal Motion (x-axis): Constant velocity motion. $a_x = 0$.
$$ v_x = u_x = u \cos(\theta) $$ $$ x = v_x t $$
- Vertical Motion (y-axis): Constant acceleration motion. $a_y = -g$.
$$ v_y = u_y + a_y t = (u \sin(\theta)) - gt $$ $$ y = u_y t + \frac{1}{2}a_y t^2 = (u \sin(\theta))t - \frac{1}{2}gt^2 $$
The time of flight ($t$) links the two motions.
Trajectory of a projectile. The velocity vector (red) changes direction, but its horizontal component $v_x$ (gray) is constant. The vertical component $v_y$ (gray) changes due to the constant downward acceleration $g$ (green).
Challenge an IMAT Question!
Official Paper: 2020 - Q60
worked solution & explanation
Concept 2D Relative Velocity Vectors. The boat's independent velocity ($y$-axis) and the river's flow velocity ($x$-axis) must be added vectorially.
Step 1 Resultant speed is the hypotenuse. Apply Pythagoras: $v_{res} = \sqrt{4.0^2 + 3.0^2} = \sqrt{16 + 9} = \sqrt{25} = 5.0\text{ m/s}$.
Step 2 Determine the angle $\theta$ from the $90^\circ$ line ($y$-axis). In this right triangle, the adjacent side is $4.0$ and the hypotenuse is $5.0$.
Step 3 Use trigonometry: $\cos(\theta) = \frac{\text{Adj}}{\text{Hyp}} = \frac{4.0}{5.0} \implies \theta = \cos^{-1}(4.0/5.0)$.
Challenge an IMAT Question!
Official Paper: 2016 - Q59
1. The magnitude of the ball's acceleration increases as it falls.
2. No vertical forces act on the ball when it is at its maximum height.
3. When the ball is moving upwards it loses kinetic energy and gains potential energy.
Which of the statement(s) is/are correct?
worked solution & explanation
Concept Kinematics of free fall. Neglecting air resistance, a projectile is under the sole influence of Earth's constant gravitational field.
Step 1 Statement 1: False. The acceleration is a constant $9.8\text{ m/s}^2$ (or 10) acting downwards the entire time. It does not increase.
Step 2 Statement 2: False. At max height, velocity is zero, but the force of gravity ($mg$) never ceases to act. It pulls it back down.
Step 3 Statement 3: True. As it rises, it slows down (losing Kinetic Energy) while gaining height (increasing Gravitational Potential Energy).
Part 2: Dynamics - The Cause of Motion
Dynamics explains *why* objects move. It's all about forces.
2.1 Newton's Laws of Motion
- Law of Inertia (First Law): An object maintains a constant velocity (which can be zero) unless a net external force acts on it. If $\Sigma\vec{F} = 0$, then $\vec{a} = 0$ (equilibrium).
- Force and Acceleration (Second Law): The net force on an object is equal to its mass times its acceleration.
$$ \Sigma\vec{F} = m\vec{a} $$
This is the most important equation in classical dynamics. The $\Sigma$ (sigma) means you must sum all forces acting on the object. - Action-Reaction (Third Law): For every action force, there is an equal and opposite reaction force. These forces act on *different* objects.
Challenge an IMAT Question!
Official Paper: 2018 - Q55
worked solution & explanation
Concept Core Newtonian Principles. This tests the relationship between Force, Acceleration, and Velocity.
Step 1 From $F = ma$, if mass is constant and the resultant Force is 'constant and >0', then acceleration ($a$) must be constant and >0.
Step 2 Acceleration is the rate of change of velocity. A positive constant acceleration means velocity is continuously increasing over time.
2.2 Types of Forces
Here are common forces you'll encounter:
- Weight ($W$ or $F_g$): Force of gravity. $W = mg$. Always acts vertically downwards.
- Normal Force ($N$ or $F_N$): A perpendicular contact force from a surface. It acts perpendicular to the surface.
- Friction ($f$): A contact force opposing motion. $f_s \le \mu_s N$ (static), $f_k = \mu_k N$ (kinetic).
- Buoyancy ($F_B$): Upward force from a fluid. $F_B = \rho_{fluid}V_{submerged}g$.
- Tension ($T$): The force transmitted through a string, rope, or cable when pulled.
Forces on a floating object: Weight (W) is balanced by Buoyant Force (FB).
Challenge an IMAT Question!
Official Paper: 2020 - Q55
worked solution & explanation
Concept Newton's Second Law Dynamics ($F_{\text{net}} = ma$). Properly define all opposing forces acting on the free-body diagram.
Step 1 Identify the forces: The engine provides an upward thrust force ($+T$). Gravity provides a constant downward weight force ($-mg$).
Step 2 Write the net force equation: $F_{\text{net}} = T - mg$. Equate to $m \cdot a$: $ma = T - mg$.
Step 3 Solve for acceleration ($a$): $a = \frac{T - mg}{m} = \frac{T}{m} - \frac{mg}{m} = \frac{T}{m} - g$.
2.3 Inertial Frames & Apparent Weight (The Elevator Problem)
Newton's laws are strictly valid in inertial frames (frames that are not accelerating). When we are in an accelerating frame (non-inertial), such as an accelerating elevator or a car braking, we experience what feel like "fictitious forces" or "inertial forces" pushing us in the opposite direction of the acceleration.
Apparent Weight
When you stand on a weighing scale in an elevator, the scale measures the Normal Force ($N$) pushing up on you, not necessarily your true weight ($mg$). This is called your "Apparent Weight".
Three Scenarios:
- Accelerating Upwards ($a$ is up): The floor must push you harder to overcome gravity and accelerate you up.
Equation ($F_{net} = ma$): $N - mg = ma \implies N = m(g + a)$.
Result: You feel heavier. Scale reading > Weight. - Accelerating Downwards ($a$ is down): The floor pushes less, allowing you to accelerate down.
Equation ($F_{net} = ma$): $mg - N = ma \implies N = m(g - a)$.
Result: You feel lighter. Scale reading < Weight. - Free Fall ($a = g$ down): The floor drops away from you at the same rate you fall.
Equation: $N = m(g - g) = 0$.
Result: You feel weightless. Scale reading = 0.
2.4 Application: The Inclined Plane
A classic problem is a block on a ramp. The key is to resolve the weight vector ($mg$) into components parallel and perpendicular to the slope.
- The force pulling the block down the slope is $mg\sin(\theta)$.
- The force pressing the block into the slope is $mg\cos(\theta)$.
- Normal Force: $N = mg\cos(\theta)$.
- Net Force (parallel): $F_{net} = mg\sin(\theta) - f = ma$.
Challenge an IMAT Question!
Official Paper: 2014 - Q53
worked solution & explanation
Concept Newton's Second Law with Friction. Determine the net force by subtracting the opposing kinetic friction from the applied push force.
Step 1 Calculate normal force ($N$): Flat horizontal surface, so $N = mg = 0.40\text{ kg} \times 10\text{ N/kg} = 4.0\text{ N}$.
Step 2 Calculate kinetic friction ($f_k$): $f_k = \mu \cdot N = 0.50 \times 4.0\text{ N} = 2.0\text{ N}$.
Step 3 Calculate net force: $F_{\text{net}} = F_{\text{applied}} - f_k = 10\text{ N} - 2.0\text{ N} = 8.0\text{ N}$.
Step 4 Calculate acceleration: $a = F_{\text{net}} / m = 8.0 / 0.40 = 20.0\text{ m/s}^2$.
2.5 Moment of Force (Torque) and Rotational Equilibrium
Torque ($\tau$) is the rotational equivalent of force. It measures a force's ability to cause rotation around a pivot (fulcrum).
Torque Equation
$$ \tau = rF\sin(\theta) $$
Where $r$ is the distance from the pivot, $F$ is the force, and $\theta$ is the angle.
For static equilibrium: Net Force = 0 AND Net Torque = 0.
Challenge an IMAT Question!
Official Paper: 2016 - Q58
worked solution & explanation
Concept Rotational Static Equilibrium. A system in equilibrium has zero net torque around any chosen pivot point ($\Sigma \tau = 0$) and zero net vertical force ($\Sigma F_y = 0$).
Step 1 Set Left support (L) as pivot. Clockwise torques: $300\text{N}$ at $1.0\text{m}$ and $500\text{N}$ at $4.0\text{m}$ ($5.0 - 1.0$). Counter-clockwise: Support R at $5.0\text{m}$.
Step 2 Equate torques: $(300 \cdot 1.0) + (500 \cdot 4.0) = R \cdot 5.0 \implies 300 + 2000 = 5R \implies R = 460\text{ N}$.
Step 3 Use vertical equilibrium to find L: Total upward = Total downward. $L + 460 = 300 + 500 \implies L = 340\text{ N}$.
Challenge an IMAT Question!
Official Paper: 2015 - Q60
worked solution & explanation
Concept Definition of a Force Couple. A couple consists of two forces that are equal in magnitude, opposite in direction (anti-parallel), and displaced by a perpendicular distance $d$.
Step 1 A couple produces pure rotation with zero net translational force.
Step 2 The torque (moment) of a couple is mathematically calculated as: $Torque = \text{One Force (F)} \times \text{Total perpendicular distance between them (d)}$.
Step 3 Diagram E correctly shows anti-parallel forces at the ends and lists the correct formula $F \times d$.
Challenge an IMAT Question!
Official Paper: 2013 - Q54
worked solution & explanation
Concept Complex Rotational Equilibrium. Account for the intrinsic weight of the uniform beam acting exactly at its geometric center.
Step 1 Set pivot at the string ($1.0\text{m}$ from left). Beam's Center of Mass (CM) is at $1.5\text{m}$. Distance from pivot to CM = $1.5 - 1.0 = 0.5\text{m}$.
Step 2 CCW torques (left of pivot): $300\text{N}$ weight is at $0.5\text{m}$, so distance to pivot is $1.0 - 0.5 = 0.5\text{m}$. $\tau_{CCW} = 300 \times 0.5 = 150\text{ Nm}$.
Step 3 CW torques (right of pivot): Beam's weight $100\text{N}$ acts at $0.5\text{m}$ from pivot. $\tau_{\text{beam}} = 100 \times 0.5 = 50\text{ Nm}$. Add $80\text{N}$ weight at distance $x$: $\tau_{CW} = 50 + 80x$.
Step 4 Equate torques: $150 = 50 + 80x \implies 100 = 80x \implies x = 1.25\text{m}$ from the pivot.
Step 5 Locate position: Pivot is $1.0\text{m}$ from left. Position is $1.0 + 1.25 = 2.25\text{m}$ from left end. Distance from right end = $3.0 - 2.25 = 0.75\text{m}$.
Challenge an IMAT Question!
Official Paper: 2011 - Q71
worked solution & explanation
Concept Rotational Equilibrium. Always define distances strictly relative to the pivot point.
Step 1 Pivot is at $1.5\text{m}$ from the left end. The Center of Mass (CM) for a $2.0\text{m}$ uniform bar is at $1.0\text{m}$.
Step 2 CCW Torques (left side): $200\text{N}$ force at left end ($1.5\text{m}$ from pivot) + $1000\text{N}$ CM force ($1.5 - 1.0 = 0.5\text{m}$ from pivot).
Step 3 Calculate CCW: $(200 \times 1.5) + (1000 \times 0.5) = 300 + 500 = 800\text{ Nm}$.
Step 4 CW Torques (right side): Force P at the right end ($2.0\text{m}$). Distance to pivot = $2.0 - 1.5 = 0.5\text{m}$. CW torque = $P \times 0.5$.
Step 5 Equate: $800 = 0.5P \implies P = 1600\text{ N}$.
Part 3: Practice Problems
1. (Kinematics) A train starts from rest and accelerates uniformly at $1.5 \text{ m/s}^2$ for 10 seconds. How far does it travel in this time?
Explanation: Use the kinematic equation $s = ut + \frac{1}{2}at^2$. Given: $u=0$ (starts from rest), $a=1.5$, $t=10$. $s = (0)(10) + \frac{1}{2}(1.5)(10)^2 = 0 + 0.5 \times 1.5 \times 100 = 75$ m.
2. (Vectors) Which of the following is a list of only vector quantities?
Explanation: Vectors have both magnitude and direction. Velocity (speed in a direction), acceleration (rate of change of velocity), and displacement (distance in a direction) are all vectors. Speed, mass, distance, time, temperature, and energy are scalars.
3. (Projectile Motion) A cannon fires a ball with an initial speed of 100 m/s at an angle of 30° above the horizontal. What is the initial vertical component of its velocity?
Explanation: The vertical component of initial velocity is $v_{y0} = v_0 \sin(\theta)$. Given: $v_0 = 100$ m/s, $\theta = 30^\circ$. $v_{y0} = 100 \times \sin(30^\circ) = 100 \times 0.5 = 50$ m/s.
4. (Newton's II Law) A net force of 50 N is applied to a 10 kg box. What is the acceleration of the box?
Explanation: Use Newton's Second Law, $F_{net} = ma$. Rearrange to find acceleration: $a = F_{net} / m$. $a = 50 \text{ N} / 10 \text{ kg} = 5 \text{ m/s}^2$.
5. (Newton's I Law) An object is moving at a constant velocity of 20 m/s to the right. What is the net force acting on the object?
Explanation: According to Newton's First Law, an object moving at a constant velocity has zero acceleration. From Newton's Second Law, $F_{net} = ma$. If $a=0$, then $F_{net}$ must also be 0.
6. (Kinematics) A 1000 kg car traveling at 20 m/s brakes to a stop over a distance of 50 m. What is the magnitude of the braking force?
Explanation: First find acceleration using $v^2 = u^2 + 2as$. $0^2 = 20^2 + 2(a)(50) \implies 0 = 400 + 100a \implies a = -4 \text{ m/s}^2$. Now use $F=ma$: $F = (1000 \text{ kg})(-4 \text{ m/s}^2) = -4000$ N. The magnitude is 4000 N.
7. (Free Fall) A ball is dropped from a 20 m high building. How long does it take to hit the ground? (Use $g=10 \text{ m/s}^2$)
Explanation: Use $s = ut + \frac{1}{2}at^2$. Here $s=20, u=0, a=g=10$. $20 = (0)t + \frac{1}{2}(10)t^2 \implies 20 = 5t^2 \implies t^2=4 \implies t=2$ s.
8. (Conceptual) A projectile is fired at an angle. At the highest point of its trajectory, which statement is true?
Explanation: At the peak of its trajectory, the projectile momentarily stops moving upwards before starting to fall, so its vertical velocity is zero. However, its horizontal velocity remains constant throughout the flight, and its acceleration is always constant at $g$ downwards.
9. (Conceptual) A book is resting on a table. According to Newton's First Law, why does it not move?
Explanation: The book is at rest, so its acceleration is zero. Newton's First Law (and Second Law) states that this means the net force must be zero. The force of gravity pulling it down is perfectly balanced by the normal force from the table pushing it up.
10. (Conceptual) Which statement best describes projectile motion in the absence of air resistance?
Explanation: The fundamental principle for solving projectile motion problems is to separate the motion into two independent components: the horizontal motion (which has zero acceleration) and the vertical motion (which has constant downward acceleration due to gravity).
11. (Kinematics) A car starts from rest and accelerates uniformly at $2 \text{ m/s}^2$ over a distance of 100 m. What is its final speed?
Explanation: Use the time-independent kinematic equation $v^2 = u^2 + 2as$. Given: $u=0, a=2, s=100$. $v^2 = 0^2 + 2(2)(100) = 400$. $v = \sqrt{400} = 20$ m/s.
12. (Conceptual - Dynamics) Which statement best describes the relationship between mass and inertia?
Explanation: Inertia is the qualitative concept that objects resist changes in their motion. Mass is the number we assign to an object to quantify how much inertia it has. A more massive object has more inertia.
13. (Projectile Motion) A stone is thrown horizontally at 15 m/s from a 45 m high cliff. How far from the base of the cliff does it land? (Use $g=10 \text{ m/s}^2$)
Explanation: First, find the time of flight from the vertical motion. The initial vertical velocity is $u_y = 0$. Using $s = ut + \frac{1}{2}at^2$, we have $45 = (0)t + \frac{1}{2}(10)t^2 \implies 45 = 5t^2 \implies t^2 = 9 \implies t = 3$ s. Second, find the horizontal distance (range) using the constant horizontal velocity: $d = v_x \times t = 15 \text{ m/s} \times 3 \text{ s} = 45$ m.
14. (Newton's II Law) A 5 kg block is pushed with a force of 30 N to the right. A friction force of 10 N opposes the motion. What is the block's acceleration?
Explanation: The net force is the sum of the forces: $F_{net} = F_{push} - F_{friction} = 30 \text{ N} - 10 \text{ N} = 20$ N to the right. Using $F_{net} = ma$, we have $20 \text{ N} = (5 \text{ kg}) \times a \implies a = 20/5 = 4 \text{ m/s}^2$.
15. (Conceptual - Kinematics) In one-dimensional motion, a negative acceleration always means that:
Explanation: Acceleration is the rate of change of velocity. A negative acceleration means the velocity is changing in the negative direction. This could mean slowing down if velocity is positive, or speeding up if velocity is negative. The most accurate and general statement is that the velocity is becoming more negative.
16. (Free Fall) A ball is thrown straight up. At the very top of its path, its velocity and acceleration are:
Explanation: At the highest point, the ball momentarily stops moving to change direction, so its instantaneous velocity is zero. However, the force of gravity is still acting on it, so its acceleration is constant throughout the flight, equal to $g$ and directed downwards.
17. (Friction) A 10 kg block rests on a horizontal surface. The coefficient of kinetic friction is 0.2. A horizontal force of 50 N is applied. What is the acceleration of the block? (Use $g=10 \text{ m/s}^2$)
Explanation: First, calculate the normal force. On a flat surface, $F_N = mg = 10 \text{ kg} \times 10 \text{ m/s}^2 = 100$ N. Next, calculate the kinetic friction force: $f_k = \mu_k F_N = 0.2 \times 100 \text{ N} = 20$ N. The net force is $F_{applied} - f_k = 50 \text{ N} - 20 \text{ N} = 30$ N. Finally, calculate acceleration: $a = F_{net}/m = 30 \text{ N} / 10 \text{ kg} = 3 \text{ m/s}^2$.
18. (Conceptual - Friction) A person is trying to push a heavy crate, but it is not moving. Which statement about the static friction force is true?
Explanation: Static friction is an adaptive force. As long as the object is not moving, the static friction force is exactly equal and opposite to the applied force, resulting in a net force of zero. It only reaches its maximum value just before the object begins to slide.
19. (Graphs) The area under a velocity-time graph represents what physical quantity?
Explanation: Since velocity is displacement/time (v = s/t), or more formally $v = ds/dt$, the integral of velocity with respect to time (area under the graph) gives displacement (s).
20. (Inclined Plane) A 5 kg block slides down a frictionless ramp angled at 30°. What is its acceleration? (Use $g=10 \text{ m/s}^2$)
Explanation: The net force down the slope is $F_{net} = mg\sin(\theta)$. From $F_{net}=ma$, we get $ma = mg\sin(\theta)$, so $a = g\sin(\theta) = 10 \times \sin(30^\circ) = 10 \times 0.5 = 5 \text{ m/s}^2$. Note that the mass (5 kg) is irrelevant.
21. (Torque) To balance a seesaw, a 40 kg child sits 2 m from the pivot. Where must a 20 kg child sit on the other side?
Explanation: For rotational equilibrium, clockwise torque must equal counter-clockwise torque. $\tau_{cw} = \tau_{ccw} \implies r_1 F_1 = r_2 F_2$. The forces are the weights ($F=mg$). $(2 \text{ m})(40\text{g}) = (r_2)(20\text{g})$. The 'g' cancels out. $80 = 20r_2 \implies r_2 = 4$ m.
22. (Inertial Forces / Elevator) A 60 kg person stands on a weighing scale inside an elevator. The elevator accelerates upwards at $2 \text{ m/s}^2$. What does the scale read? (Use $g=10 \text{ m/s}^2$)
Explanation: The person is accelerating upwards. The scale measures the Normal Force ($N$). The net force upward is $F_{net} = N - W$. Using Newton's 2nd Law ($F_{net} = ma$), we have $N - mg = ma \implies N = m(g + a)$.
Calculation: $N = 60(10 + 2) = 60(12) = 720$ N. The scale reads greater than the true weight (600 N).