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    Rest and Motion : Kinematics

    Flashcards for topic Rest and Motion : Kinematics

    Intermediate47 cardsGeneral

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    Card 1

    Front

    How would you properly define a frame of reference, and what are the minimum components needed for a complete frame of reference in classical mechanics?

    Back

    • A frame of reference is a coordinate system used to specify the position of objects and measure their motion

    • Minimum components required:

    1. A set of three mutually perpendicular axes (X-Y-Z) to locate positions in space
    2. A specified origin point (0,0,0) from which all positions are measured
    3. A clock or time-keeping system to measure when events occur
    4. A rigid body to which the coordinate system is attached

    • A complete reference frame allows:

    • Position to be specified using coordinates (x,y,z)
    • Motion to be described as change in coordinates with time
    • Velocity and acceleration to be measured relative to this frame

    • Example: When analyzing motion in a train, we might define a frame attached to the train with:

    • Origin at a specific point in the train
    • X-axis along the train's length
    • Y-axis across the width
    • Z-axis vertically
    • Time measured by a clock in the train

    Note: The choice of frame is arbitrary and depends on what makes the analysis most convenient

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    Card 2

    Front

    What is the fundamental difference between distance and displacement when a particle moves from point A to point B along a curved path?

    Back

    • Distance: The total length of the actual path traveled

    • Scalar quantity (magnitude only)
    • Always positive
    • Path-dependent

    • Displacement: The straight-line vector from initial to final position

    • Vector quantity (magnitude and direction)
    • Direction is from initial to final position
    • Path-independent
    • Can be smaller than the distance traveled

    Example: When walking 3 km along a winding mountain trail that ends just 1 km east of your starting point, your distance traveled is 3 km, but your displacement is only 1 km east.

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    Card 3

    Front

    When a particle follows a curved path ACB from initial point A to final point B, how do you calculate:

    1. The total distance traveled
    2. The displacement

    Back

    Distance calculation: • Sum the lengths of all path segments along the actual trajectory • Mathematically: s=∫path∣dr⃗∣s = \int_{path} |d\vec{r}|s=∫path​∣dr∣ • For discrete segments: s=∑∣Δr⃗i∣s = \sum |Δ\vec{r}_i|s=∑∣Δri​∣

    Displacement calculation: • Vector connecting initial point A to final point B • Mathematically: D⃗=r⃗B−r⃗A\vec{D} = \vec{r}_B - \vec{r}_AD=rB​−rA​ • Magnitude equals the straight-line distance between A and B • Direction points from A to B

    Note: Distance ≥ Displacement (magnitude), with equality only when the path is a straight line.

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    Card 4

    Front

    How does the relationship between a chord and a tangent on a distance-time curve relate to average speed versus instantaneous speed?

    Back

    On a distance-time curve:

    • Average speed between two times (ttt and t+Δtt + \Delta tt+Δt) corresponds to the slope of the chord connecting the two corresponding points on the curve
    • Instantaneous speed at time ttt corresponds to the slope of the tangent to the curve at that point

    Mathematical relationship:

    • As Δt\Delta tΔt approaches zero, the chord approaches the tangent
    • The slope of the chord (ΔsΔt\frac{\Delta s}{\Delta t}ΔtΔs​) approaches the slope of the tangent (dsdt\frac{ds}{dt}dtds​)

    This geometric interpretation shows why: v=lim⁡Δt→0ΔsΔt=dsdtv = \lim_{\Delta t \to 0} \frac{\Delta s}{\Delta t} = \frac{ds}{dt}v=limΔt→0​ΔtΔs​=dtds​

    Note: The tangent represents the "best linear approximation" of the curve at that point, showing the instantaneous rate of change.

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    Card 5

    Front

    What quantitative relationship exists between a linearly increasing speed function and the constant acceleration of an object? How would you find the acceleration value if speed increases from 0 to 6 m/s over 3 seconds?

    Back

    When speed increases linearly with time, acceleration is constant and equals the slope of the speed-time graph:

    • Mathematically: a = Δv/Δt
    • For the given values: a = (6 m/s - 0 m/s)/(3 s - 0 s) = 2 m/s²

    Key insights:

    • The slope of a speed-time graph represents acceleration
    • Constant acceleration produces a straight line on a speed-time graph
    • The steeper the slope, the greater the acceleration

    Example: If a train's speed increases from 10 m/s to 25 m/s in 5 seconds, its acceleration is (25 m/s - 10 m/s)/5 s = 3 m/s².

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    Card 6

    Front

    Given a constant acceleration scenario, which kinematic equation should you choose when you know initial velocity (u), acceleration (a), and displacement (x), but need to find final velocity (v)?

    Back

    When you know initial velocity (u), acceleration (a), and displacement (x), but need to find final velocity (v), you should use:

    v² = u² + 2ax

    This equation is ideal for this specific scenario because:

    1. It directly relates the variables you know (u, a, x) to the one you need to find (v)
    2. It doesn't require time (t), which you don't have
    3. You can solve for v by taking the square root of both sides: v = √(u² + 2ax)

    Example application: A car accelerates from 5 m/s at 2 m/s² over a distance of 100 m. Its final velocity would be: v = √(5² + 2×2×100) = √(25 + 400) = √425 ≈ 20.6 m/s

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    Card 7

    Front

    What fundamental principle allows us to analyze motion in a plane by separating it into two independent one-dimensional motions, and what are the practical implications of this principle?

    Back

    The principle of superposition of motions (or principle of independence of motions) allows us to analyze 2D motion by separating it into independent x and y components.

    Key aspects:

    • Motion along each coordinate axis proceeds independently of motion along the other axis
    • The x-component of motion depends only on the x-component of initial velocity and x-component of acceleration
    • The y-component of motion depends only on the y-component of initial velocity and y-component of acceleration

    Practical implications:

    1. Complex 2D problems can be simplified into two 1D problems
    2. Equations for 1D kinematics apply separately to each component:
      • vx = ux + axt and vy = uy + ayt
      • x = uxt + ½axt² and y = uyt + ½ayt²
    3. Projectile motion analysis becomes straightforward by treating horizontal and vertical motions separately
    4. Vector addition of the components fully reconstructs the complete 2D motion

    This principle is why the trajectory of a projectile is parabolic - horizontal motion proceeds at constant velocity while vertical motion undergoes constant acceleration.

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    Card 8

    Front

    What are the two independent components of a projectile's motion, and how does acceleration affect each component?

    Back

    A projectile's motion consists of two independent components:

    1. Horizontal motion:

      • Constant velocity (u·cosθ)
      • Zero acceleration (ax = 0)
      • Distance: x = (u·cosθ)t
    2. Vertical motion:

      • Changing velocity (initial: u·sinθ)
      • Constant downward acceleration (ay = -g)
      • Height: y = (u·sinθ)t - (1/2)gt²

    This independence of horizontal and vertical components is why projectile motion follows a parabolic path, with gravity only affecting the vertical component.

    Example: A baseball thrown horizontally from a cliff moves at constant horizontal speed while simultaneously accelerating downward due to gravity.

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    Card 9

    Front

    For a projectile launched on a horizontal plane, explain why the time taken to reach maximum height is exactly half the total time of flight.

    Back

    This relationship exists because of the symmetry of the parabolic trajectory:

    Time to maximum height: tmax=usin⁡θgt_{max} = \frac{u\sin θ}{g}tmax​=gusinθ​ (when vertical velocity becomes zero)

    Total time of flight: T=2usin⁡θgT = \frac{2u\sin θ}{g}T=g2usinθ​

    Clearly, tmax=T2t_{max} = \frac{T}{2}tmax​=2T​

    This occurs because:

    • The vertical motion is symmetrical about the highest point
    • Gravitational acceleration is constant
    • The initial upward velocity equals the final downward velocity (at the same height)
    • No air resistance is considered in the model
    Card 10

    Front

    If two reference frames S and S' move with uniform velocity relative to each other, how does acceleration transform between these frames, and what important principle does this demonstrate?

    Back

    Acceleration transformation between frames moving with uniform relative velocity:

    a⃗P,S=a⃗P,S′\vec{a}_{P,S} = \vec{a}_{P,S'}aP,S​=aP,S′​

    Where:

    • a⃗P,S\vec{a}_{P,S}aP,S​ is acceleration of particle P measured in frame S
    • a⃗P,S′\vec{a}_{P,S'}aP,S′​ is acceleration of particle P measured in frame S'

    This demonstrates the principle of Galilean invariance of acceleration:

    • In non-relativistic mechanics, acceleration is the same in all inertial reference frames
    • This is fundamental to Newton's laws, which take the same form in all inertial frames
    • It means that while position and velocity measurements depend on the reference frame, acceleration measurements do not (when frames move with constant velocity)
    • This invariance breaks down at very high speeds approaching the speed of light

    This principle is crucial for understanding that the physical laws governing motion are the same for all observers in inertial reference frames, forming a cornerstone of classical mechanics.

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