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concepts/mechanics/u4/.gitkeep
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concepts/mechanics/u4/.gitkeep
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concepts/mechanics/u4/m4-1-linear-momentum.tex
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concepts/mechanics/u4/m4-1-linear-momentum.tex
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\subsection{Linear Momentum}
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This subsection introduces linear momentum as the vector state variable for translational motion. In AP mechanics, momentum is the quantity that naturally leads into impulse and conservation ideas.
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\dfn{Particle and system momentum}{Let $m$ denote the mass of a particle, let $\vec{v}$ denote its velocity measured in a chosen inertial reference frame, and let $\vec{p}$ denote its linear momentum. The \emph{linear momentum} of the particle is
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\[
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\vec{p}=m\vec{v}.
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\]
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Because $m$ is a scalar and $\vec{v}$ is a vector, $\vec{p}$ is a vector in the same direction as $\vec{v}$. Its SI unit is
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\[
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1\,\mathrm{kg\cdot m/s}.
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\]
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For a system of $N$ particles labeled by $i=1,2,\dots,N$, let $m_i$ denote the mass of particle $i$, let $\vec{v}_i$ denote its velocity, and let $\vec{p}_i=m_i\vec{v}_i$ denote its momentum. The total linear momentum of the system is
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\[
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\vec{p}_{\mathrm{sys}}=\sum_{i=1}^N \vec{p}_i=\sum_{i=1}^N m_i\vec{v}_i.
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\]}
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\nt{Momentum is not the same thing as speed. Speed is the scalar $v=|\vec{v}|$, while momentum is the vector $\vec{p}=m\vec{v}$. Two objects can have the same speed but different momenta if their masses are different or if they move in different directions. Momentum also depends on the chosen reference frame because velocity does: an object at rest in one frame has $\vec{p}=\vec{0}$ in that frame, but it can have nonzero momentum in another frame. In one-dimensional motion, a negative momentum component means motion in the negative coordinate direction; it does not mean negative mass or negative speed.}
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\mprop{Component and system relations}{Let
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\[
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\vec{v}=v_x\hat{\imath}+v_y\hat{\jmath}+v_z\hat{k}
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\]
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be the velocity of a particle of mass $m$, and let
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\[
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\vec{p}=p_x\hat{\imath}+p_y\hat{\jmath}+p_z\hat{k}
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\]
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be its momentum. For a system of particles, let
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\[
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M=\sum_{i=1}^N m_i
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\]
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denote the total mass, and let $\vec{v}_{\mathrm{cm}}$ denote the center-of-mass velocity.
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\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
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\item Momentum components are found component-by-component from the velocity:
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\[
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p_x=mv_x,
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\qquad
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p_y=mv_y,
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\qquad
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p_z=mv_z.
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\]
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In one-dimensional motion along the $x$-axis,
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\[
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p_x=mv_x.
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\]
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\item For a single particle, the momentum magnitude is
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\[
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p=|\vec{p}|=m|\vec{v}|=mv.
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\]
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In two dimensions,
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\[
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p=\sqrt{p_x^2+p_y^2},
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\]
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and in three dimensions,
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\[
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p=\sqrt{p_x^2+p_y^2+p_z^2}.
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\]
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\item System momentum is the vector sum of the particle momenta:
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\[
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\vec{p}_{\mathrm{sys}}=\sum_{i=1}^N \vec{p}_i.
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\]
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Therefore its components also add:
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\[
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p_{\mathrm{sys},x}=\sum_{i=1}^N p_{i,x},
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\qquad
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p_{\mathrm{sys},y}=\sum_{i=1}^N p_{i,y},
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\qquad
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p_{\mathrm{sys},z}=\sum_{i=1}^N p_{i,z}.
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\]
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Equivalently,
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\[
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\vec{p}_{\mathrm{sys}}=M\vec{v}_{\mathrm{cm}}.
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\]
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\end{enumerate}}
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\qs{Worked example}{In a laboratory frame, two pucks slide on nearly frictionless ice. Puck 1 has mass $m_1=2.0\,\mathrm{kg}$ and velocity
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\[
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\vec{v}_1=(1.5\hat{\imath}+0.50\hat{\jmath})\,\mathrm{m/s}.
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\]
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Puck 2 has mass $m_2=1.0\,\mathrm{kg}$ and velocity
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\[
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\vec{v}_2=(-1.0\hat{\imath}+4.0\hat{\jmath})\,\mathrm{m/s}.
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\]
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Let $\vec{p}_1$ and $\vec{p}_2$ denote the individual momenta, let $\vec{p}_{\mathrm{sys}}$ denote the total momentum, let $M$ denote the total mass, and let $\theta$ denote the direction of $\vec{p}_{\mathrm{sys}}$ measured counterclockwise from the positive $x$-axis.
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Find $\vec{p}_1$, $\vec{p}_2$, $\vec{p}_{\mathrm{sys}}$, the magnitude $|\vec{p}_{\mathrm{sys}}|$, the direction $\theta$, and the center-of-mass velocity $\vec{v}_{\mathrm{cm}}$.}
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\sol Use $\vec{p}=m\vec{v}$ for each puck.
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For puck 1,
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\[
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\vec{p}_1=m_1\vec{v}_1=(2.0\,\mathrm{kg})(1.5\hat{\imath}+0.50\hat{\jmath})\,\mathrm{m/s}.
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\]
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Therefore,
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\[
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\vec{p}_1=(3.0\hat{\imath}+1.0\hat{\jmath})\,\mathrm{kg\cdot m/s}.
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\]
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For puck 2,
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\[
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\vec{p}_2=m_2\vec{v}_2=(1.0\,\mathrm{kg})(-1.0\hat{\imath}+4.0\hat{\jmath})\,\mathrm{m/s}.
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\]
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So,
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\[
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\vec{p}_2=(-1.0\hat{\imath}+4.0\hat{\jmath})\,\mathrm{kg\cdot m/s}.
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\]
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Now add the momenta component-by-component:
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\[
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\vec{p}_{\mathrm{sys}}=\vec{p}_1+\vec{p}_2.
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\]
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Hence,
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\[
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\vec{p}_{\mathrm{sys}}=(3.0-1.0)\hat{\imath}+(1.0+4.0)\hat{\jmath}.
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\]
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Therefore,
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\[
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\vec{p}_{\mathrm{sys}}=(2.0\hat{\imath}+5.0\hat{\jmath})\,\mathrm{kg\cdot m/s}.
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\]
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Its magnitude is
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\[
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|\vec{p}_{\mathrm{sys}}|=\sqrt{(2.0)^2+(5.0)^2}\,\mathrm{kg\cdot m/s}=\sqrt{29}\,\mathrm{kg\cdot m/s}.
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\]
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Numerically,
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\[
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|\vec{p}_{\mathrm{sys}}|\approx 5.39\,\mathrm{kg\cdot m/s}.
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\]
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To find the direction, use the component ratio. Since both components of $\vec{p}_{\mathrm{sys}}$ are positive, the vector lies in the first quadrant. Thus,
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\[
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\tan\theta=\frac{p_{\mathrm{sys},y}}{p_{\mathrm{sys},x}}=\frac{5.0}{2.0}=2.5.
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\]
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So,
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\[
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\theta=\tan^{-1}(2.5)\approx 68.2^\circ.
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\]
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Now find the center-of-mass velocity. The total mass is
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\[
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M=m_1+m_2=2.0\,\mathrm{kg}+1.0\,\mathrm{kg}=3.0\,\mathrm{kg}.
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\]
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Using
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\[
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\vec{p}_{\mathrm{sys}}=M\vec{v}_{\mathrm{cm}},
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\]
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we get
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\[
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\vec{v}_{\mathrm{cm}}=\frac{\vec{p}_{\mathrm{sys}}}{M}=\frac{(2.0\hat{\imath}+5.0\hat{\jmath})\,\mathrm{kg\cdot m/s}}{3.0\,\mathrm{kg}}.
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\]
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Therefore,
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\[
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\vec{v}_{\mathrm{cm}}=\left(\frac{2.0}{3.0}\hat{\imath}+\frac{5.0}{3.0}\hat{\jmath}\right)\,\mathrm{m/s}
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\]
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or numerically,
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\[
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\vec{v}_{\mathrm{cm}}\approx (0.667\hat{\imath}+1.67\hat{\jmath})\,\mathrm{m/s}.
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\]
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So the individual and system momenta are
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\[
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\vec{p}_1=(3.0\hat{\imath}+1.0\hat{\jmath})\,\mathrm{kg\cdot m/s},
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\qquad
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\vec{p}_2=(-1.0\hat{\imath}+4.0\hat{\jmath})\,\mathrm{kg\cdot m/s},
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\]
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\[
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\vec{p}_{\mathrm{sys}}=(2.0\hat{\imath}+5.0\hat{\jmath})\,\mathrm{kg\cdot m/s},
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\]
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with magnitude
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\[
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|\vec{p}_{\mathrm{sys}}|\approx 5.39\,\mathrm{kg\cdot m/s},
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\]
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direction
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\[
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\theta\approx 68.2^\circ,
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\]
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and center-of-mass velocity
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\[
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\vec{v}_{\mathrm{cm}}\approx (0.667\hat{\imath}+1.67\hat{\jmath})\,\mathrm{m/s}.
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\]
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This example shows why momentum must be handled as a vector: the total momentum is found by adding components, not by adding speeds.
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122
concepts/mechanics/u4/m4-2-impulse.tex
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concepts/mechanics/u4/m4-2-impulse.tex
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\subsection{Impulse and Momentum Transfer}
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This subsection connects the local momentum law $\vec{F}_{\text{net}}=d\vec{p}/dt$ to finite-time interactions such as hits, kicks, and collisions, where a large force acts for a short time and changes momentum by a measurable amount.
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\dfn{Momentum change and impulse}{Let a body of constant mass $m$ have velocity $\vec{v}$. Its linear momentum is
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\[
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\vec{p}=m\vec{v}.
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\]
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Let $\vec{p}_i$ and $\vec{p}_f$ denote the initial and final momenta over some time interval from $t_i$ to $t_f$. The change in momentum is
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\[
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\Delta \vec{p}=\vec{p}_f-\vec{p}_i.
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\]
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Let $\vec{F}_{\text{net}}(t)$ denote the net external force on the body during that interval. The impulse delivered to the body is
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\[
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\vec{J}=\int_{t_i}^{t_f} \vec{F}_{\text{net}}\,dt.
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\]
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If the net force is constant, then this reduces to
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\[
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\vec{J}=\vec{F}_{\text{net}}\Delta t,
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\qquad
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\Delta t=t_f-t_i.
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\]
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The SI unit of impulse is $\mathrm{N\,s}$, which is equivalent to $\mathrm{kg\,m/s}$.}
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\thm{Impulse-momentum theorem}{Let $\vec{p}(t)$ denote the momentum of a body and let $\vec{F}_{\text{net}}(t)$ denote the net external force on it. Over any time interval from $t_i$ to $t_f$,
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\[
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\vec{J}=\int_{t_i}^{t_f} \vec{F}_{\text{net}}\,dt=\Delta \vec{p}=\vec{p}_f-\vec{p}_i.
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\]
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Thus the net impulse on a body equals its change in momentum. In one dimension along the $x$-axis, let $J_x$, $F_{\text{net},x}$, and $p_x$ denote the corresponding $x$-components. Then
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\[
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J_x=\int_{t_i}^{t_f} F_{\text{net},x}\,dt=\Delta p_x.
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\]}
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\nt{Impulse is a vector, so its direction is the direction of $\Delta \vec{p}$. A body can be moving in one direction while the impulse points in the opposite direction if the interaction slows or reverses the motion. Impulse depends on both force and time: a large force acting briefly can produce the same impulse as a smaller force acting longer. If an average net force $\vec{F}_{\text{avg}}$ over a time interval $\Delta t$ is defined so that it has the same effect as the actual time-varying force, then
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\[
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\vec{J}=\vec{F}_{\text{avg}}\Delta t.
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\]
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On a force-versus-time graph, the signed area under the curve gives impulse. In component form, the signed area under an $F_x(t)$ graph gives $J_x=\Delta p_x$.}
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\pf{Short derivation from $d\vec{p}/dt$}{Start with the local momentum law
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\[
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\vec{F}_{\text{net}}=\frac{d\vec{p}}{dt}.
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\]
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Multiply by $dt$:
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\[
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\vec{F}_{\text{net}}\,dt=d\vec{p}.
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\]
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Now integrate from $t_i$ to $t_f$:
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\[
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\int_{t_i}^{t_f} \vec{F}_{\text{net}}\,dt=\int_{t_i}^{t_f} d\vec{p}.
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\]
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The left side is the impulse $\vec{J}$, and the right side is the total change in momentum:
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\[
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\vec{J}=\vec{p}_f-\vec{p}_i=\Delta \vec{p}.
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\]
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This is the impulse-momentum theorem.}
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\qs{Worked example}{Choose the positive $x$-axis to the right. A tennis ball of mass $m=0.150\,\mathrm{kg}$ moves horizontally with initial velocity
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\[
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\vec{v}_i=-18.0\,\hat{\imath}\,\mathrm{m/s}.
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\]
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Let $F_x(t)$ denote the net horizontal force on the ball during contact. The force is directed to the right and has the following force-versus-time graph: it increases linearly from $0$ at $t=0$ to $900\,\mathrm{N}$ at $t=4.0\,\mathrm{ms}$, then decreases linearly back to $0$ at $t=8.0\,\mathrm{ms}$.
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Find (a) the impulse delivered to the ball, (b) the ball's final momentum, (c) the ball's final velocity, and (d) the average net force during contact.}
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\sol Let $\Delta t=8.0\times 10^{-3}\,\mathrm{s}$ denote the contact time. Because the force is entirely in the $+x$ direction, the impulse is the area under the triangular $F_x(t)$ graph in the positive direction:
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\[
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\vec{J}=\left(\tfrac12 \right)(\Delta t)(900\,\mathrm{N})\,\hat{\imath}.
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\]
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Substitute $\Delta t=8.0\times 10^{-3}\,\mathrm{s}$:
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\[
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\vec{J}=\left(\tfrac12 \right)(8.0\times 10^{-3}\,\mathrm{s})(900\,\mathrm{N})\,\hat{\imath}=3.6\,\hat{\imath}\,\mathrm{N\,s}.
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\]
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Using $1\,\mathrm{N\,s}=1\,\mathrm{kg\,m/s}$,
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\[
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\vec{J}=3.6\,\hat{\imath}\,\mathrm{kg\,m/s}.
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\]
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Now compute the initial momentum:
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\[
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\vec{p}_i=m\vec{v}_i=(0.150\,\mathrm{kg})(-18.0\,\hat{\imath}\,\mathrm{m/s})=-2.70\,\hat{\imath}\,\mathrm{kg\,m/s}.
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\]
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From the impulse-momentum theorem,
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\[
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\vec{J}=\Delta \vec{p}=\vec{p}_f-\vec{p}_i,
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\]
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so
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\[
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\vec{p}_f=\vec{p}_i+\vec{J}.
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\]
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Substitute the values:
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\[
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\vec{p}_f=(-2.70+3.60)\,\hat{\imath}\,\mathrm{kg\,m/s}=0.90\,\hat{\imath}\,\mathrm{kg\,m/s}.
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\]
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Then the final velocity is
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\[
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\vec{v}_f=\frac{\vec{p}_f}{m}=\frac{0.90\,\hat{\imath}\,\mathrm{kg\,m/s}}{0.150\,\mathrm{kg}}=6.0\,\hat{\imath}\,\mathrm{m/s}.
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\]
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The positive sign shows that the ball leaves moving to the right.
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For the average net force, use
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\[
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\vec{J}=\vec{F}_{\text{avg}}\Delta t.
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\]
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Therefore,
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\[
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\vec{F}_{\text{avg}}=\frac{\vec{J}}{\Delta t}=\frac{3.6\,\hat{\imath}\,\mathrm{N\,s}}{8.0\times 10^{-3}\,\mathrm{s}}=4.5\times 10^2\,\hat{\imath}\,\mathrm{N}.
|
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\]
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So the results are
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\[
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\vec{J}=3.6\,\hat{\imath}\,\mathrm{N\,s},
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\qquad
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\vec{p}_f=0.90\,\hat{\imath}\,\mathrm{kg\,m/s},
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\]
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\[
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\vec{v}_f=6.0\,\hat{\imath}\,\mathrm{m/s},
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\qquad
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\vec{F}_{\text{avg}}=4.5\times 10^2\,\hat{\imath}\,\mathrm{N}.
|
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\]
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The graph interpretation is essential here: the triangular area under $F_x(t)$ gives the impulse directly, and that impulse determines the momentum transfer.
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200
concepts/mechanics/u4/m4-3-momentum-conservation.tex
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concepts/mechanics/u4/m4-3-momentum-conservation.tex
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\subsection{Conservation of Momentum for Systems}
|
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|
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This subsection treats momentum conservation as a statement about a chosen system: internal forces can transfer momentum between parts of the system, but if the net external impulse is zero, the total momentum stays constant.
|
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|
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\dfn{System momentum and momentum-isolated systems}{Consider a system of $N$ objects labeled by an index $i=1,2,\dots,N$. Let $m_i$ denote the mass of object $i$, let $\vec{v}_i$ denote its velocity, and let
|
||||
\[
|
||||
\vec{p}_i=m_i\vec{v}_i
|
||||
\]
|
||||
denote its momentum. The total momentum of the system is the vector sum
|
||||
\[
|
||||
\vec{P}=\sum_{i=1}^N \vec{p}_i=\sum_{i=1}^N m_i\vec{v}_i.
|
||||
\]
|
||||
|
||||
For an interval from time $t_i$ to time $t_f$, let $\sum \vec{F}_{\mathrm{ext}}$ denote the net external force on the whole system, and let
|
||||
\[
|
||||
\vec{J}_{\mathrm{ext}}=\int_{t_i}^{t_f} \sum \vec{F}_{\mathrm{ext}}\,dt
|
||||
\]
|
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denote the net external impulse on the system.
|
||||
|
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A system is called \emph{closed} over that interval if the set of objects in the system does not change during the interaction. A closed system is \emph{isolated for momentum} over that interval if the net external impulse is zero or negligible:
|
||||
\[
|
||||
\vec{J}_{\mathrm{ext}}=\vec{0}.
|
||||
\]}
|
||||
|
||||
\thm{Conservation of momentum for a system}{For a closed system,
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\[
|
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\Delta \vec{P}=\vec{P}_f-\vec{P}_i=\vec{J}_{\mathrm{ext}}.
|
||||
\]
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||||
Therefore, if the net external impulse on the system is zero,
|
||||
\[
|
||||
\vec{J}_{\mathrm{ext}}=\vec{0},
|
||||
\]
|
||||
then the total momentum is conserved:
|
||||
\[
|
||||
\vec{P}_f=\vec{P}_i.
|
||||
\]
|
||||
In component form, this means each Cartesian component is conserved separately, such as
|
||||
\[
|
||||
P_{x,f}=P_{x,i}
|
||||
\qquad\text{and}\qquad
|
||||
P_{y,f}=P_{y,i}
|
||||
\]
|
||||
when the external impulse is zero.}
|
||||
|
||||
\nt{The most important step is choosing the system boundary correctly. If two objects collide and both objects are included in the system, then the contact forces between them are internal forces and cancel in the total momentum balance. Forces from outside the system, such as a large friction force from the floor or a push from a person, are external and can change the system momentum. In two-dimensional problems, write momentum conservation separately in the $x$- and $y$-directions and solve the resulting scalar equations. Also, momentum conservation does \emph{not} require kinetic energy to be conserved: in an inelastic collision, the total momentum can stay constant even though the kinetic energy changes.}
|
||||
|
||||
\pf{Derivation from Newton's laws}{For each object $i$, let $\vec{F}_{\mathrm{ext},i}$ denote the net external force on that object, and let $\vec{F}_{\mathrm{int},i}$ denote the net internal force on it from the other objects in the system. Then Newton's second law gives
|
||||
\[
|
||||
\frac{d\vec{p}_i}{dt}=\vec{F}_{\mathrm{ext},i}+\vec{F}_{\mathrm{int},i}.
|
||||
\]
|
||||
Now sum over all objects:
|
||||
\[
|
||||
\sum_{i=1}^N \frac{d\vec{p}_i}{dt}=\sum_{i=1}^N \vec{F}_{\mathrm{ext},i}+\sum_{i=1}^N \vec{F}_{\mathrm{int},i}.
|
||||
\]
|
||||
Since
|
||||
\[
|
||||
\sum_{i=1}^N \frac{d\vec{p}_i}{dt}=\frac{d\vec{P}}{dt},
|
||||
\]
|
||||
and the internal forces cancel in equal-and-opposite pairs by Newton's third law,
|
||||
\[
|
||||
\sum_{i=1}^N \vec{F}_{\mathrm{int},i}=\vec{0},
|
||||
\]
|
||||
we obtain
|
||||
\[
|
||||
\frac{d\vec{P}}{dt}=\sum \vec{F}_{\mathrm{ext}}.
|
||||
\]
|
||||
Integrate from $t_i$ to $t_f$:
|
||||
\[
|
||||
\vec{P}_f-\vec{P}_i=\int_{t_i}^{t_f} \sum \vec{F}_{\mathrm{ext}}\,dt=\vec{J}_{\mathrm{ext}}.
|
||||
\]
|
||||
If $\vec{J}_{\mathrm{ext}}=\vec{0}$, then $\vec{P}_f=\vec{P}_i$, so total momentum is conserved.}
|
||||
|
||||
\qs{Worked example}{On a frictionless horizontal air table, puck $A$ has mass $m_A=0.20\,\mathrm{kg}$ and initial velocity
|
||||
\[
|
||||
\vec{v}_{A,i}=(6.0\hat{\imath})\,\mathrm{m/s}.
|
||||
\]
|
||||
Puck $B$ has mass $m_B=0.30\,\mathrm{kg}$ and initial velocity
|
||||
\[
|
||||
\vec{v}_{B,i}=(4.0\hat{\jmath})\,\mathrm{m/s}.
|
||||
\]
|
||||
The pucks collide and stick together. Let
|
||||
\[
|
||||
\vec{v}_f=v_{f,x}\hat{\imath}+v_{f,y}\hat{\jmath}
|
||||
\]
|
||||
denote their common final velocity.
|
||||
|
||||
Find $\vec{v}_f$, its magnitude and direction, and determine whether kinetic energy is conserved.}
|
||||
|
||||
\sol Choose the system to be \emph{puck $A$ + puck $B$}. During the short collision, the table is frictionless, so the net external horizontal impulse on this two-puck system is zero. Therefore,
|
||||
\[
|
||||
\vec{P}_i=\vec{P}_f.
|
||||
\]
|
||||
|
||||
Because this is a two-dimensional problem, conserve momentum separately in the $x$- and $y$-directions.
|
||||
|
||||
First write the initial momentum components.
|
||||
|
||||
For puck $A$,
|
||||
\[
|
||||
\vec{p}_{A,i}=m_A\vec{v}_{A,i}=(0.20\,\mathrm{kg})(6.0\hat{\imath}\,\mathrm{m/s})=(1.2\hat{\imath})\,\mathrm{kg\cdot m/s}.
|
||||
\]
|
||||
|
||||
For puck $B$,
|
||||
\[
|
||||
\vec{p}_{B,i}=m_B\vec{v}_{B,i}=(0.30\,\mathrm{kg})(4.0\hat{\jmath}\,\mathrm{m/s})=(1.2\hat{\jmath})\,\mathrm{kg\cdot m/s}.
|
||||
\]
|
||||
|
||||
So the total initial momentum is
|
||||
\[
|
||||
\vec{P}_i=(1.2\hat{\imath}+1.2\hat{\jmath})\,\mathrm{kg\cdot m/s}.
|
||||
\]
|
||||
|
||||
After the collision, the pucks stick together, so the total mass is
|
||||
\[
|
||||
M=m_A+m_B=0.20\,\mathrm{kg}+0.30\,\mathrm{kg}=0.50\,\mathrm{kg}.
|
||||
\]
|
||||
Their final momentum is
|
||||
\[
|
||||
\vec{P}_f=M\vec{v}_f=(0.50\,\mathrm{kg})(v_{f,x}\hat{\imath}+v_{f,y}\hat{\jmath}).
|
||||
\]
|
||||
|
||||
Now conserve momentum by components.
|
||||
|
||||
In the $x$-direction,
|
||||
\[
|
||||
P_{x,i}=P_{x,f}.
|
||||
\]
|
||||
Thus,
|
||||
\[
|
||||
1.2\,\mathrm{kg\cdot m/s}=(0.50\,\mathrm{kg})v_{f,x},
|
||||
\]
|
||||
so
|
||||
\[
|
||||
v_{f,x}=2.4\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
In the $y$-direction,
|
||||
\[
|
||||
P_{y,i}=P_{y,f}.
|
||||
\]
|
||||
Thus,
|
||||
\[
|
||||
1.2\,\mathrm{kg\cdot m/s}=(0.50\,\mathrm{kg})v_{f,y},
|
||||
\]
|
||||
so
|
||||
\[
|
||||
v_{f,y}=2.4\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
Therefore, the common final velocity is
|
||||
\[
|
||||
\vec{v}_f=(2.4\hat{\imath}+2.4\hat{\jmath})\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
Its magnitude is
|
||||
\[
|
||||
|\vec{v}_f|=\sqrt{(2.4\,\mathrm{m/s})^2+(2.4\,\mathrm{m/s})^2}=3.39\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
Let $\theta$ denote the direction of $\vec{v}_f$ measured counterclockwise from the positive $x$-axis. Then
|
||||
\[
|
||||
\tan\theta=\frac{v_{f,y}}{v_{f,x}}=\frac{2.4}{2.4}=1,
|
||||
\]
|
||||
so
|
||||
\[
|
||||
\theta=45^\circ.
|
||||
\]
|
||||
|
||||
Now check the kinetic energy.
|
||||
|
||||
Initially,
|
||||
\[
|
||||
K_i=\tfrac12 m_A|\vec{v}_{A,i}|^2+\tfrac12 m_B|\vec{v}_{B,i}|^2
|
||||
=\tfrac12(0.20)(6.0)^2+\tfrac12(0.30)(4.0)^2=6.0\,\mathrm{J}.
|
||||
\]
|
||||
|
||||
Finally,
|
||||
\[
|
||||
K_f=\tfrac12 M|\vec{v}_f|^2=\tfrac12(0.50)(3.39)^2\approx 2.88\,\mathrm{J}.
|
||||
\]
|
||||
|
||||
Since
|
||||
\[
|
||||
K_f\neq K_i,
|
||||
\]
|
||||
kinetic energy is not conserved. That is expected because the pucks stick together, so the collision is inelastic.
|
||||
|
||||
Thus the final velocity is
|
||||
\[
|
||||
\vec{v}_f=(2.4\hat{\imath}+2.4\hat{\jmath})\,\mathrm{m/s},
|
||||
\]
|
||||
with magnitude
|
||||
\[
|
||||
|\vec{v}_f|=3.39\,\mathrm{m/s},
|
||||
\]
|
||||
directed
|
||||
\[
|
||||
45^\circ
|
||||
\]
|
||||
above the positive $x$-axis, while momentum is conserved but kinetic energy is not.
|
||||
178
concepts/mechanics/u4/m4-4-collisions.tex
Normal file
178
concepts/mechanics/u4/m4-4-collisions.tex
Normal file
@@ -0,0 +1,178 @@
|
||||
\subsection{Elastic, Inelastic, and Perfectly Inelastic Collisions}
|
||||
|
||||
This subsection classifies collisions by what happens to the system's kinetic energy. In AP mechanics, the first step is still to apply momentum conservation to an isolated system; only elastic collisions add a kinetic-energy conservation equation.
|
||||
|
||||
\dfn{Elastic, inelastic, and perfectly inelastic collisions}{Consider two objects of masses $m_1$ and $m_2$. Let $\vec{v}_{1,i}$ and $\vec{v}_{2,i}$ denote their velocities just before a collision, and let $\vec{v}_{1,f}$ and $\vec{v}_{2,f}$ denote their velocities just after it. Let the total kinetic energy before and after the collision be
|
||||
\[
|
||||
K_i=\tfrac12 m_1|\vec{v}_{1,i}|^2+\tfrac12 m_2|\vec{v}_{2,i}|^2
|
||||
\]
|
||||
and
|
||||
\[
|
||||
K_f=\tfrac12 m_1|\vec{v}_{1,f}|^2+\tfrac12 m_2|\vec{v}_{2,f}|^2.
|
||||
\]
|
||||
|
||||
A collision is called \emph{elastic} if the total kinetic energy is unchanged, so
|
||||
\[
|
||||
K_f=K_i.
|
||||
\]
|
||||
It is called \emph{inelastic} if the total kinetic energy changes, so
|
||||
\[
|
||||
K_f\ne K_i.
|
||||
\]
|
||||
In the usual AP collision problems without stored energy being released during impact, an inelastic collision has $K_f<K_i$.
|
||||
|
||||
A collision is \emph{perfectly inelastic} if the objects stick together after the collision, so they share one final velocity:
|
||||
\[
|
||||
\vec{v}_{1,f}=\vec{v}_{2,f}=\vec{v}_f.
|
||||
\]
|
||||
Every perfectly inelastic collision is inelastic.}
|
||||
|
||||
\nt{Do not decide whether momentum is conserved by asking whether the collision is elastic. Those are different ideas. Momentum conservation depends on the net external impulse on the chosen system. If the system is isolated during the collision, then total momentum is conserved for elastic, inelastic, and perfectly inelastic collisions alike. Kinetic energy supplies an \emph{extra} condition only in the elastic case. In two-dimensional AP problems, conserve momentum separately in the $x$- and $y$-directions.}
|
||||
|
||||
\ex{Illustrative example}{On a frictionless track, cart 1 has mass $m_1=0.40\,\mathrm{kg}$ and initial velocity
|
||||
\[
|
||||
\vec{v}_{1,i}=(3.0\hat{\imath})\,\mathrm{m/s}.
|
||||
\]
|
||||
Cart 2 has mass $m_2=0.20\,\mathrm{kg}$ and initial velocity
|
||||
\[
|
||||
\vec{v}_{2,i}=\vec{0}.
|
||||
\]
|
||||
After the collision, the carts move together with common velocity
|
||||
\[
|
||||
\vec{v}_f=(2.0\hat{\imath})\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
Because the carts move together after impact, the collision is perfectly inelastic. The initial kinetic energy is
|
||||
\[
|
||||
K_i=\tfrac12(0.40)(3.0)^2=1.8\,\mathrm{J},
|
||||
\]
|
||||
while the final kinetic energy is
|
||||
\[
|
||||
K_f=\tfrac12(0.60)(2.0)^2=1.2\,\mathrm{J}.
|
||||
\]
|
||||
Since $K_f<K_i$, the collision is inelastic, as expected. Momentum is still conserved because
|
||||
\[
|
||||
\vec{P}_i=(0.40)(3.0\hat{\imath})=(1.2\hat{\imath})\,\mathrm{kg\cdot m/s}
|
||||
\]
|
||||
and
|
||||
\[
|
||||
\vec{P}_f=(0.60)(2.0\hat{\imath})=(1.2\hat{\imath})\,\mathrm{kg\cdot m/s}.
|
||||
\]}
|
||||
|
||||
\mprop{Practical relations for isolated collisions}{Consider two objects that form an isolated system during a short collision. Let
|
||||
\[
|
||||
\vec{P}_i=m_1\vec{v}_{1,i}+m_2\vec{v}_{2,i}
|
||||
\]
|
||||
denote the total initial momentum and let
|
||||
\[
|
||||
\vec{P}_f=m_1\vec{v}_{1,f}+m_2\vec{v}_{2,f}
|
||||
\]
|
||||
denote the total final momentum.
|
||||
|
||||
\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
|
||||
\item For any isolated collision,
|
||||
\[
|
||||
\vec{P}_i=\vec{P}_f,
|
||||
\]
|
||||
so
|
||||
\[
|
||||
m_1\vec{v}_{1,i}+m_2\vec{v}_{2,i}=m_1\vec{v}_{1,f}+m_2\vec{v}_{2,f}.
|
||||
\]
|
||||
|
||||
\item In two dimensions, let $P_{x,i}$ and $P_{y,i}$ denote the initial momentum components, and let $P_{x,f}$ and $P_{y,f}$ denote the final momentum components. Then write two separate component equations:
|
||||
\[
|
||||
P_{x,i}=P_{x,f},
|
||||
\qquad
|
||||
P_{y,i}=P_{y,f}.
|
||||
\]
|
||||
|
||||
\item If the collision is perfectly inelastic, then $\vec{v}_{1,f}=\vec{v}_{2,f}=\vec{v}_f$, so momentum conservation gives the shared final velocity directly:
|
||||
\[
|
||||
\vec{v}_f=\frac{m_1\vec{v}_{1,i}+m_2\vec{v}_{2,i}}{m_1+m_2}.
|
||||
\]
|
||||
|
||||
\item If the collision is elastic, then in addition to momentum conservation,
|
||||
\[
|
||||
\tfrac12 m_1|\vec{v}_{1,i}|^2+\tfrac12 m_2|\vec{v}_{2,i}|^2
|
||||
=
|
||||
\tfrac12 m_1|\vec{v}_{1,f}|^2+\tfrac12 m_2|\vec{v}_{2,f}|^2.
|
||||
\]
|
||||
For a one-dimensional elastic collision, this is equivalent to the relative-speed relation
|
||||
\[
|
||||
v_{1,i}-v_{2,i}=-(v_{1,f}-v_{2,f}),
|
||||
\]
|
||||
where each $v$ is an $x$-component.
|
||||
\end{enumerate}}
|
||||
|
||||
\qs{Worked example}{Choose the positive $x$-axis to the right. On a frictionless track, cart 1 has mass $m_1=1.0\,\mathrm{kg}$ and initial velocity
|
||||
\[
|
||||
\vec{v}_{1,i}=(4.0\hat{\imath})\,\mathrm{m/s}.
|
||||
\]
|
||||
Cart 2 has mass $m_2=3.0\,\mathrm{kg}$ and is initially at rest, so
|
||||
\[
|
||||
\vec{v}_{2,i}=\vec{0}.
|
||||
\]
|
||||
After the collision, the carts lock together and move with common final velocity $\vec{v}_f$.
|
||||
|
||||
Find $\vec{v}_f$, determine whether the collision is elastic, inelastic, or perfectly inelastic, and calculate the change in kinetic energy $\Delta K=K_f-K_i$.}
|
||||
|
||||
\sol Because the carts lock together, this is a perfectly inelastic collision by definition. During the short collision the track is frictionless, so the two-cart system is isolated horizontally. Therefore total momentum is conserved:
|
||||
\[
|
||||
\vec{P}_i=\vec{P}_f.
|
||||
\]
|
||||
|
||||
The initial momentum is
|
||||
\[
|
||||
\vec{P}_i=m_1\vec{v}_{1,i}+m_2\vec{v}_{2,i}.
|
||||
\]
|
||||
Substitute the given values:
|
||||
\[
|
||||
\vec{P}_i=(1.0\,\mathrm{kg})(4.0\hat{\imath}\,\mathrm{m/s})+(3.0\,\mathrm{kg})(\vec{0})
|
||||
=4.0\hat{\imath}\,\mathrm{kg\cdot m/s}.
|
||||
\]
|
||||
|
||||
After the collision the carts move together, so their total mass is
|
||||
\[
|
||||
m_1+m_2=4.0\,\mathrm{kg}
|
||||
\]
|
||||
and their final momentum is
|
||||
\[
|
||||
\vec{P}_f=(m_1+m_2)\vec{v}_f.
|
||||
\]
|
||||
Thus,
|
||||
\[
|
||||
4.0\hat{\imath}\,\mathrm{kg\cdot m/s}=(4.0\,\mathrm{kg})\vec{v}_f,
|
||||
\]
|
||||
so
|
||||
\[
|
||||
\vec{v}_f=(1.0\hat{\imath})\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
Now compare the kinetic energies.
|
||||
|
||||
Initially,
|
||||
\[
|
||||
K_i=\tfrac12 m_1|\vec{v}_{1,i}|^2+\tfrac12 m_2|\vec{v}_{2,i}|^2
|
||||
=\tfrac12(1.0)(4.0)^2+\tfrac12(3.0)(0)^2=8.0\,\mathrm{J}.
|
||||
\]
|
||||
|
||||
Finally,
|
||||
\[
|
||||
K_f=\tfrac12 (m_1+m_2)|\vec{v}_f|^2
|
||||
=\tfrac12(4.0)(1.0)^2=2.0\,\mathrm{J}.
|
||||
\]
|
||||
|
||||
Therefore,
|
||||
\[
|
||||
\Delta K=K_f-K_i=2.0\,\mathrm{J}-8.0\,\mathrm{J}=-6.0\,\mathrm{J}.
|
||||
\]
|
||||
|
||||
The negative sign means $6.0\,\mathrm{J}$ of kinetic energy was transformed into other forms of energy during the collision. Since the carts stick together and the kinetic energy decreases, the collision is inelastic, more specifically perfectly inelastic.
|
||||
|
||||
So the results are
|
||||
\[
|
||||
\vec{v}_f=(1.0\hat{\imath})\,\mathrm{m/s},
|
||||
\qquad
|
||||
\Delta K=-6.0\,\mathrm{J},
|
||||
\]
|
||||
and the collision is perfectly inelastic rather than elastic.
|
||||
244
concepts/mechanics/u4/m4-5-recoil-explosions.tex
Normal file
244
concepts/mechanics/u4/m4-5-recoil-explosions.tex
Normal file
@@ -0,0 +1,244 @@
|
||||
\subsection{Recoil, Explosions, and the Center-of-Mass Viewpoint}
|
||||
|
||||
This subsection treats recoil and explosions as momentum-redistribution processes. In AP mechanics, internal energy may be released during the interaction, but if the net external impulse on the chosen system is zero, the total momentum and the center-of-mass motion do not change.
|
||||
|
||||
\dfn{Recoil/explosion interactions and the center-of-mass viewpoint}{Consider a closed system of particles labeled by $i=1,2,\dots,N$. Let $m_i$ denote the mass of particle $i$, let $\vec{v}_i$ denote its velocity, let
|
||||
\[
|
||||
\vec{p}_i=m_i\vec{v}_i
|
||||
\]
|
||||
denote its momentum, and let
|
||||
\[
|
||||
\vec{P}=\sum_{i=1}^N \vec{p}_i=\sum_{i=1}^N m_i\vec{v}_i
|
||||
\]
|
||||
denote the total momentum. Let
|
||||
\[
|
||||
M=\sum_{i=1}^N m_i
|
||||
\]
|
||||
denote the total mass, and let $\vec{v}_{\mathrm{cm}}$ denote the center-of-mass velocity. Then
|
||||
\[
|
||||
\vec{P}=M\vec{v}_{\mathrm{cm}}.
|
||||
\]
|
||||
|
||||
A \emph{recoil} or \emph{explosion} interaction is a short internal interaction in which parts of the system push apart or are driven apart by released internal energy. If the net external impulse on the system during the interaction is zero or negligible, then
|
||||
\[
|
||||
\vec{P}_f=\vec{P}_i,
|
||||
\]
|
||||
so equivalently
|
||||
\[
|
||||
\vec{v}_{\mathrm{cm},f}=\vec{v}_{\mathrm{cm},i}.
|
||||
\]
|
||||
If the system is initially at rest, then $\vec{P}_i=\vec{0}$ and $\vec{v}_{\mathrm{cm}}=\vec{0}$ both before and after the interaction.}
|
||||
|
||||
\nt{In recoil and explosion problems, the internal forces between the parts of the system can be very large, but they occur in equal-and-opposite pairs and therefore only redistribute momentum within the system. They can change the individual velocities and can increase the total kinetic energy if internal energy is released, but they do not change the total momentum of an isolated system. From the center-of-mass viewpoint, the whole event is just internal rearrangement: the center of mass continues to remain at rest or to move with constant velocity if the external impulse is zero.}
|
||||
|
||||
\ex{Illustrative example}{A student of mass $m_s=60.0\,\mathrm{kg}$ stands at rest on a frictionless skateboard and throws a backpack of mass $m_b=5.0\,\mathrm{kg}$ horizontally backward with velocity
|
||||
\[
|
||||
\vec{v}_b=(-8.0\hat{\imath})\,\mathrm{m/s}.
|
||||
\]
|
||||
Let $\vec{v}_s=v_s\hat{\imath}$ denote the student's recoil velocity after the throw.
|
||||
|
||||
Because the student-backpack system starts at rest and the external horizontal impulse is negligible,
|
||||
\[
|
||||
\vec{P}_f=\vec{0}.
|
||||
\]
|
||||
Thus,
|
||||
\[
|
||||
m_s\vec{v}_s+m_b\vec{v}_b=\vec{0}.
|
||||
\]
|
||||
Substitute the values:
|
||||
\[
|
||||
(60.0\,\mathrm{kg})v_s\hat{\imath}+(5.0\,\mathrm{kg})(-8.0\hat{\imath}\,\mathrm{m/s})=\vec{0}.
|
||||
\]
|
||||
So,
|
||||
\[
|
||||
60.0v_s-40.0=0,
|
||||
\]
|
||||
which gives
|
||||
\[
|
||||
v_s=0.667\,\mathrm{m/s}.
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
\vec{v}_s=(0.667\hat{\imath})\,\mathrm{m/s}.
|
||||
\]
|
||||
The student recoils forward while the backpack moves backward, and the total momentum remains zero.}
|
||||
|
||||
\mprop{Useful recoil and center-of-mass relations}{Let a system of total mass $M$ have velocity $\vec{V}_0$ just before a recoil or explosion event. Let the final pieces have masses $m_1,m_2,\dots$ and velocities $\vec{v}_1,\vec{v}_2,\dots$. If the net external impulse during the short interaction is zero or negligible, then:
|
||||
|
||||
\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
|
||||
\item Total momentum is conserved:
|
||||
\[
|
||||
\sum \vec{p}_{f}=\sum \vec{p}_{i},
|
||||
\qquad
|
||||
\sum m_k\vec{v}_k=M\vec{V}_0.
|
||||
\]
|
||||
|
||||
\item The center-of-mass velocity stays constant:
|
||||
\[
|
||||
\vec{v}_{\mathrm{cm}}=\frac{\vec{P}}{M}=\vec{V}_0.
|
||||
\]
|
||||
If the system starts from rest, then $\vec{v}_{\mathrm{cm}}=\vec{0}$ and
|
||||
\[
|
||||
\sum m_k\vec{v}_k=\vec{0}.
|
||||
\]
|
||||
|
||||
\item For a two-piece explosion from rest,
|
||||
\[
|
||||
m_1\vec{v}_1+m_2\vec{v}_2=\vec{0},
|
||||
\]
|
||||
so the two final momenta are equal in magnitude and opposite in direction:
|
||||
\[
|
||||
m_1\vec{v}_1=-m_2\vec{v}_2.
|
||||
\]
|
||||
|
||||
\item In two dimensions, conserve momentum component-by-component:
|
||||
\[
|
||||
\sum p_{x,f}=\sum p_{x,i},
|
||||
\qquad
|
||||
\sum p_{y,f}=\sum p_{y,i}.
|
||||
\]
|
||||
This is usually the most direct way to find unknown fragment velocities.
|
||||
\end{enumerate}}
|
||||
|
||||
\qs{Worked example}{A firework shell of total mass $M=4.0\,\mathrm{kg}$ is moving with velocity
|
||||
\[
|
||||
\vec{V}_0=(10.0\hat{\imath}+6.0\hat{\jmath})\,\mathrm{m/s}
|
||||
\]
|
||||
just before it explodes into two fragments. Fragment 1 has mass $m_1=1.5\,\mathrm{kg}$ and velocity
|
||||
\[
|
||||
\vec{v}_1=(18.0\hat{\imath}+2.0\hat{\jmath})\,\mathrm{m/s}
|
||||
\]
|
||||
immediately after the explosion. Fragment 2 has mass $m_2=2.5\,\mathrm{kg}$ and velocity
|
||||
\[
|
||||
\vec{v}_2=v_{2,x}\hat{\imath}+v_{2,y}\hat{\jmath}.
|
||||
\]
|
||||
Assume the net external impulse during the explosion is negligible.
|
||||
|
||||
Find $\vec{v}_2$, find the center-of-mass velocity after the explosion, and determine whether the explosion could have increased the total kinetic energy.}
|
||||
|
||||
\sol Because the explosion is brief and the net external impulse is negligible, the shell-plus-fragments system conserves momentum:
|
||||
\[
|
||||
\vec{P}_i=\vec{P}_f.
|
||||
\]
|
||||
|
||||
First compute the initial total momentum. Since the shell has total mass $M=4.0\,\mathrm{kg}$ and velocity $\vec{V}_0$ just before the explosion,
|
||||
\[
|
||||
\vec{P}_i=M\vec{V}_0.
|
||||
\]
|
||||
Substitute the given values:
|
||||
\[
|
||||
\vec{P}_i=(4.0\,\mathrm{kg})(10.0\hat{\imath}+6.0\hat{\jmath})\,\mathrm{m/s}.
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
\vec{P}_i=(40.0\hat{\imath}+24.0\hat{\jmath})\,\mathrm{kg\cdot m/s}.
|
||||
\]
|
||||
|
||||
Next compute the momentum of fragment 1:
|
||||
\[
|
||||
\vec{p}_1=m_1\vec{v}_1=(1.5\,\mathrm{kg})(18.0\hat{\imath}+2.0\hat{\jmath})\,\mathrm{m/s}.
|
||||
\]
|
||||
So,
|
||||
\[
|
||||
\vec{p}_1=(27.0\hat{\imath}+3.0\hat{\jmath})\,\mathrm{kg\cdot m/s}.
|
||||
\]
|
||||
|
||||
Let $\vec{p}_2=m_2\vec{v}_2$ denote the momentum of fragment 2. Since
|
||||
\[
|
||||
\vec{P}_f=\vec{p}_1+\vec{p}_2,
|
||||
\]
|
||||
momentum conservation gives
|
||||
\[
|
||||
\vec{p}_2=\vec{P}_i-\vec{p}_1.
|
||||
\]
|
||||
Thus,
|
||||
\[
|
||||
\vec{p}_2=(40.0\hat{\imath}+24.0\hat{\jmath})-(27.0\hat{\imath}+3.0\hat{\jmath}).
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
\vec{p}_2=(13.0\hat{\imath}+21.0\hat{\jmath})\,\mathrm{kg\cdot m/s}.
|
||||
\]
|
||||
|
||||
Now divide by $m_2=2.5\,\mathrm{kg}$ to find the velocity of fragment 2:
|
||||
\[
|
||||
\vec{v}_2=\frac{\vec{p}_2}{m_2}=\frac{(13.0\hat{\imath}+21.0\hat{\jmath})\,\mathrm{kg\cdot m/s}}{2.5\,\mathrm{kg}}.
|
||||
\]
|
||||
Hence,
|
||||
\[
|
||||
\vec{v}_2=(5.2\hat{\imath}+8.4\hat{\jmath})\,\mathrm{m/s}.
|
||||
\]
|
||||
|
||||
Now find the center-of-mass velocity after the explosion. The total momentum after the explosion is still
|
||||
\[
|
||||
\vec{P}_f=\vec{P}_i=(40.0\hat{\imath}+24.0\hat{\jmath})\,\mathrm{kg\cdot m/s}.
|
||||
\]
|
||||
Using
|
||||
\[
|
||||
\vec{v}_{\mathrm{cm}}=\frac{\vec{P}}{M},
|
||||
\]
|
||||
we get
|
||||
\[
|
||||
\vec{v}_{\mathrm{cm},f}=\frac{\vec{P}_f}{M}=\frac{(40.0\hat{\imath}+24.0\hat{\jmath})\,\mathrm{kg\cdot m/s}}{4.0\,\mathrm{kg}}.
|
||||
\]
|
||||
So,
|
||||
\[
|
||||
\vec{v}_{\mathrm{cm},f}=(10.0\hat{\imath}+6.0\hat{\jmath})\,\mathrm{m/s}.
|
||||
\]
|
||||
This is exactly the same as the pre-explosion velocity $\vec{V}_0$, which is the center-of-mass viewpoint: the fragments fly apart, but the center of mass continues with the same velocity because the external impulse is negligible.
|
||||
|
||||
Finally, check whether the total kinetic energy could have increased.
|
||||
|
||||
Before the explosion,
|
||||
\[
|
||||
K_i=\tfrac12 M|\vec{V}_0|^2=\tfrac12 (4.0)\left[(10.0)^2+(6.0)^2\right].
|
||||
\]
|
||||
Thus,
|
||||
\[
|
||||
K_i=2.0(136)=272\,\mathrm{J}.
|
||||
\]
|
||||
|
||||
After the explosion,
|
||||
\[
|
||||
K_f=\tfrac12 m_1|\vec{v}_1|^2+\tfrac12 m_2|\vec{v}_2|^2.
|
||||
\]
|
||||
For fragment 1,
|
||||
\[
|
||||
|\vec{v}_1|^2=(18.0)^2+(2.0)^2=328,
|
||||
\]
|
||||
so
|
||||
\[
|
||||
K_1=\tfrac12 (1.5)(328)=246\,\mathrm{J}.
|
||||
\]
|
||||
For fragment 2,
|
||||
\[
|
||||
|\vec{v}_2|^2=(5.2)^2+(8.4)^2=27.04+70.56=97.60,
|
||||
\]
|
||||
so
|
||||
\[
|
||||
K_2=\tfrac12 (2.5)(97.60)=122\,\mathrm{J}.
|
||||
\]
|
||||
Therefore,
|
||||
\[
|
||||
K_f=246\,\mathrm{J}+122\,\mathrm{J}=368\,\mathrm{J}.
|
||||
\]
|
||||
|
||||
Since
|
||||
\[
|
||||
K_f>K_i,
|
||||
\]
|
||||
the total kinetic energy increased by
|
||||
\[
|
||||
\Delta K=368\,\mathrm{J}-272\,\mathrm{J}=96\,\mathrm{J}.
|
||||
\]
|
||||
That is possible because the explosion released internal energy. Momentum stayed constant because the system was isolated during the brief explosion, but kinetic energy did not have to remain constant.
|
||||
|
||||
So the results are
|
||||
\[
|
||||
\vec{v}_2=(5.2\hat{\imath}+8.4\hat{\jmath})\,\mathrm{m/s},
|
||||
\]
|
||||
and
|
||||
\[
|
||||
\vec{v}_{\mathrm{cm},f}=(10.0\hat{\imath}+6.0\hat{\jmath})\,\mathrm{m/s}.
|
||||
\]
|
||||
The explosion changed the fragment velocities and increased the total kinetic energy, but it did not change the total momentum or the motion of the center of mass.
|
||||
Reference in New Issue
Block a user