136 lines
5.7 KiB
TeX
136 lines
5.7 KiB
TeX
\subsection{Relative Motion and Projectile Motion}
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This subsection combines two central AP kinematics ideas in an inertial frame: relative velocity between different observers, and two-dimensional projectile motion with negligible air resistance. In both settings, vectors are handled component-by-component, and careful notation keeps track of what is being measured.
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\dfn{Relative velocity and the projectile-motion setup}{Let $A$ and $B$ denote moving objects, and let $E$ denote an inertial reference frame such as Earth. If $\vec{v}_{A/E}$ denotes the velocity of object $A$ measured in frame $E$, then the \emph{relative velocity of $A$ with respect to $B$} is the velocity vector of $A$ as measured in the frame moving with $B$, written $\vec{v}_{A/B}$.
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For projectile motion, choose Cartesian axes before writing equations. Let $t$ denote time after launch, let $x(t)$ denote the horizontal coordinate, and let $y(t)$ denote the vertical coordinate measured upward. Let $x_0=x(0)$ and $y_0=y(0)$ denote the initial coordinates, let $v_x(t)$ and $v_y(t)$ denote the velocity components, let $\vec{v}(t)=v_x(t)\hat{\imath}+v_y(t)\hat{\jmath}$ denote the velocity vector, and let $v_{0x}=v_x(0)$ and $v_{0y}=v_y(0)$ denote the initial velocity components. In the AP model, air resistance is neglected and the only acceleration is gravity, so the acceleration vector is
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\[
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\vec{a}=-g\hat{\jmath},
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\]
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where $g$ denotes the positive magnitude of the gravitational acceleration. Thus the component equations are
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\[
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\frac{d^2x}{dt^2}=0,
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\qquad
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\frac{d^2y}{dt^2}=-g.
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\]}
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\thm{Relative-velocity addition and projectile component formulas}{Let $A$, $B$, and $E$ denote objects or frames in classical mechanics. Then the relative-velocity addition law is
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\[
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\vec{v}_{A/E}=\vec{v}_{A/B}+\vec{v}_{B/E}.
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\]
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Equivalently,
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\[
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\vec{v}_{A/B}=\vec{v}_{A/E}-\vec{v}_{B/E}.
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\]
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These equations are vector equations, so they may be applied component-by-component in any chosen axes.
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For projectile motion with the setup in the definition,
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\[
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\frac{d^2x}{dt^2}=0,
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\qquad
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\frac{d^2y}{dt^2}=-g.
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\]
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Integrating once gives the velocity components
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\[
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v_x(t)=v_{0x},
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\qquad
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v_y(t)=v_{0y}-gt.
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\]
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Integrating again gives the position components
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\[
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x(t)=x_0+v_{0x}t,
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\qquad
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y(t)=y_0+v_{0y}t-\frac{1}{2}gt^2.
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\]
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If $v_0$ denotes the initial speed and $\theta$ denotes the launch angle measured above the positive $x$-axis, then
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\[
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v_{0x}=v_0\cos\theta,
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\qquad
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v_{0y}=v_0\sin\theta.
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\]
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The horizontal and vertical motions are independent in the sense that each component has its own equation, but they are linked by the same time variable $t$.}
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\ex{Illustrative relative-motion example}{Let $A$ denote a student walking on a moving walkway, let $B$ denote the walkway, and let $E$ denote the ground frame. Suppose
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\[
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\vec{v}_{A/B}=3.0\hat{\jmath}\,\mathrm{m/s}
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\qquad\text{and}\qquad
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\vec{v}_{B/E}=4.0\hat{\imath}\,\mathrm{m/s}.
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\]
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Then
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\[
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\vec{v}_{A/E}=\vec{v}_{A/B}+\vec{v}_{B/E}=4.0\hat{\imath}+3.0\hat{\jmath}\,\mathrm{m/s}.
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\]
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So the student moves relative to the ground with speed
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\[
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|\vec{v}_{A/E}|=\sqrt{4.0^2+3.0^2}\,\mathrm{m/s}=5.0\,\mathrm{m/s}.
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\]}
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\nt{In $\vec{v}_{A/B}$, the first label tells \emph{whose} velocity is being described, and the second label tells \emph{which frame} measures it. Reversing the labels changes the meaning. In projectile motion, choose axes first so that the signs of $v_{0x}$, $v_{0y}$, and $-g$ are clear. The horizontal and vertical motions must be evaluated at the same time $t$; they are not separate motions with separate clocks. Common mistakes include adding speeds instead of velocity vectors in relative-motion problems, forgetting that $v_x$ stays constant only when air resistance is neglected, and setting $v_y=0$ for the entire flight instead of only at the top of the path.}
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\qs{Worked example}{A ball is launched from the top of a platform. Let $t$ denote time in seconds after launch. Choose the $x$-axis horizontal and the $y$-axis vertical upward. Let $x(t)$ and $y(t)$ denote the coordinates of the ball in meters. At $t=0$, let $x_0=0$, let $y_0=30.0\,\mathrm{m}$, let $v_{0x}=12.0\,\mathrm{m/s}$, and let $v_{0y}=5.0\,\mathrm{m/s}$. Let $g=10.0\,\mathrm{m/s^2}$.
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Find the time when the ball hits the ground, where the ground is given by $y=0$. Then find the horizontal distance traveled and the velocity vector just before impact.}
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\sol From the projectile component formulas,
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\[
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x(t)=x_0+v_{0x}t=12.0t,
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\]
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and
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\[
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y(t)=y_0+v_{0y}t-\frac{1}{2}gt^2=30.0+5.0t-5.0t^2.
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\]
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The ball hits the ground when $y=0$, so solve
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\[
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30.0+5.0t-5.0t^2=0.
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\]
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Divide by $5.0$:
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\[
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6+t-t^2=0.
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\]
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Rearrange:
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\[
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t^2-t-6=0.
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\]
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Factor:
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\[
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(t-3)(t+2)=0.
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\]
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Thus the two algebraic solutions are $t=3.0\,\mathrm{s}$ and $t=-2.0\,\mathrm{s}$. The negative time is not physically relevant after launch, so the impact time is
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\[
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t=3.0\,\mathrm{s}.
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\]
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The horizontal distance traveled is the $x$-coordinate at this time:
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\[
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x(3.0)=12.0(3.0)\,\mathrm{m}=36.0\,\mathrm{m}.
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\]
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So the ball lands $36.0\,\mathrm{m}$ horizontally from the launch point.
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Next find the velocity components. The horizontal component is constant:
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\[
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v_x(t)=v_{0x}=12.0\,\mathrm{m/s}.
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\]
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The vertical component is
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\[
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v_y(t)=v_{0y}-gt=5.0-10.0t.
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\]
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At impact,
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\[
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v_y(3.0)=5.0-10.0(3.0)\,\mathrm{m/s}=-25.0\,\mathrm{m/s}.
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\]
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Therefore the velocity vector just before impact is
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\[
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\vec{v}(3.0\,\mathrm{s})=12.0\hat{\imath}-25.0\hat{\jmath}\,\mathrm{m/s}.
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\]
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Its negative $y$-component shows that the ball is moving downward at impact. If the impact speed is also desired, then
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\[
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|\vec{v}(3.0\,\mathrm{s})|=\sqrt{12.0^2+(-25.0)^2}\,\mathrm{m/s}=\sqrt{769}\,\mathrm{m/s}\approx 27.7\,\mathrm{m/s}.
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\]
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Thus the ball hits the ground after $3.0\,\mathrm{s}$, lands $36.0\,\mathrm{m}$ from the launch point, and has impact velocity
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\[
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\vec{v}=12.0\hat{\imath}-25.0\hat{\jmath}\,\mathrm{m/s}.
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\]
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