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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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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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\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
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\[
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\vec{p}_i=m_i\vec{v}_i
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\]
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denote its momentum. The total momentum of the system is the vector sum
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\[
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\vec{P}=\sum_{i=1}^N \vec{p}_i=\sum_{i=1}^N m_i\vec{v}_i.
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\]
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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
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\[
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\vec{J}_{\mathrm{ext}}=\int_{t_i}^{t_f} \sum \vec{F}_{\mathrm{ext}}\,dt
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\]
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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:
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\[
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\vec{J}_{\mathrm{ext}}=\vec{0}.
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\]}
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\thm{Conservation of momentum for a system}{For a closed system,
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\[
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\Delta \vec{P}=\vec{P}_f-\vec{P}_i=\vec{J}_{\mathrm{ext}}.
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\]
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Therefore, if the net external impulse on the system is zero,
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\[
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\vec{J}_{\mathrm{ext}}=\vec{0},
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\]
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then the total momentum is conserved:
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\[
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\vec{P}_f=\vec{P}_i.
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\]
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In component form, this means each Cartesian component is conserved separately, such as
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\[
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P_{x,f}=P_{x,i}
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\qquad\text{and}\qquad
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P_{y,f}=P_{y,i}
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\]
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when the external impulse is zero.}
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\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.}
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\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
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\[
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\frac{d\vec{p}_i}{dt}=\vec{F}_{\mathrm{ext},i}+\vec{F}_{\mathrm{int},i}.
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\]
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Now sum over all objects:
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\[
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\sum_{i=1}^N \frac{d\vec{p}_i}{dt}=\sum_{i=1}^N \vec{F}_{\mathrm{ext},i}+\sum_{i=1}^N \vec{F}_{\mathrm{int},i}.
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\]
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Since
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\[
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\sum_{i=1}^N \frac{d\vec{p}_i}{dt}=\frac{d\vec{P}}{dt},
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\]
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and the internal forces cancel in equal-and-opposite pairs by Newton's third law,
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\[
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\sum_{i=1}^N \vec{F}_{\mathrm{int},i}=\vec{0},
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\]
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we obtain
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\[
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\frac{d\vec{P}}{dt}=\sum \vec{F}_{\mathrm{ext}}.
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\]
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Integrate from $t_i$ to $t_f$:
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\[
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\vec{P}_f-\vec{P}_i=\int_{t_i}^{t_f} \sum \vec{F}_{\mathrm{ext}}\,dt=\vec{J}_{\mathrm{ext}}.
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\]
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If $\vec{J}_{\mathrm{ext}}=\vec{0}$, then $\vec{P}_f=\vec{P}_i$, so total momentum is conserved.}
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\qs{Worked example}{On a frictionless horizontal air table, puck $A$ has mass $m_A=0.20\,\mathrm{kg}$ and initial velocity
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\[
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\vec{v}_{A,i}=(6.0\hat{\imath})\,\mathrm{m/s}.
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\]
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Puck $B$ has mass $m_B=0.30\,\mathrm{kg}$ and initial velocity
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\[
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\vec{v}_{B,i}=(4.0\hat{\jmath})\,\mathrm{m/s}.
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\]
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The pucks collide and stick together. Let
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\[
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\vec{v}_f=v_{f,x}\hat{\imath}+v_{f,y}\hat{\jmath}
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\]
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denote their common final velocity.
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Find $\vec{v}_f$, its magnitude and direction, and determine whether kinetic energy is conserved.}
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\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,
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\[
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\vec{P}_i=\vec{P}_f.
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\]
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Because this is a two-dimensional problem, conserve momentum separately in the $x$- and $y$-directions.
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First write the initial momentum components.
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For puck $A$,
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\[
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\vec{p}_{A,i}=m_A\vec{v}_{A,i}=(0.20\,\mathrm{kg})(6.0\hat{\imath}\,\mathrm{m/s})=(1.2\hat{\imath})\,\mathrm{kg\cdot m/s}.
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\]
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For puck $B$,
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\[
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\vec{p}_{B,i}=m_B\vec{v}_{B,i}=(0.30\,\mathrm{kg})(4.0\hat{\jmath}\,\mathrm{m/s})=(1.2\hat{\jmath})\,\mathrm{kg\cdot m/s}.
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\]
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So the total initial momentum is
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\[
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\vec{P}_i=(1.2\hat{\imath}+1.2\hat{\jmath})\,\mathrm{kg\cdot m/s}.
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\]
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After the collision, the pucks stick together, so the total mass is
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\[
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M=m_A+m_B=0.20\,\mathrm{kg}+0.30\,\mathrm{kg}=0.50\,\mathrm{kg}.
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\]
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Their final momentum is
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\[
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\vec{P}_f=M\vec{v}_f=(0.50\,\mathrm{kg})(v_{f,x}\hat{\imath}+v_{f,y}\hat{\jmath}).
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\]
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Now conserve momentum by components.
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In the $x$-direction,
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\[
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P_{x,i}=P_{x,f}.
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\]
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Thus,
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\[
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1.2\,\mathrm{kg\cdot m/s}=(0.50\,\mathrm{kg})v_{f,x},
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\]
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so
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\[
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v_{f,x}=2.4\,\mathrm{m/s}.
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\]
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In the $y$-direction,
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\[
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P_{y,i}=P_{y,f}.
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\]
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Thus,
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\[
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1.2\,\mathrm{kg\cdot m/s}=(0.50\,\mathrm{kg})v_{f,y},
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\]
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so
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\[
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v_{f,y}=2.4\,\mathrm{m/s}.
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\]
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Therefore, the common final velocity is
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\[
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\vec{v}_f=(2.4\hat{\imath}+2.4\hat{\jmath})\,\mathrm{m/s}.
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\]
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Its magnitude is
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\[
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|\vec{v}_f|=\sqrt{(2.4\,\mathrm{m/s})^2+(2.4\,\mathrm{m/s})^2}=3.39\,\mathrm{m/s}.
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\]
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Let $\theta$ denote the direction of $\vec{v}_f$ measured counterclockwise from the positive $x$-axis. Then
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\[
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\tan\theta=\frac{v_{f,y}}{v_{f,x}}=\frac{2.4}{2.4}=1,
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\]
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so
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\[
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\theta=45^\circ.
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\]
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Now check the kinetic energy.
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Initially,
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\[
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K_i=\tfrac12 m_A|\vec{v}_{A,i}|^2+\tfrac12 m_B|\vec{v}_{B,i}|^2
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=\tfrac12(0.20)(6.0)^2+\tfrac12(0.30)(4.0)^2=6.0\,\mathrm{J}.
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\]
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Finally,
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\[
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K_f=\tfrac12 M|\vec{v}_f|^2=\tfrac12(0.50)(3.39)^2\approx 2.88\,\mathrm{J}.
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\]
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Since
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\[
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K_f\neq K_i,
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\]
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kinetic energy is not conserved. That is expected because the pucks stick together, so the collision is inelastic.
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Thus the final velocity is
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\[
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\vec{v}_f=(2.4\hat{\imath}+2.4\hat{\jmath})\,\mathrm{m/s},
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\]
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with magnitude
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\[
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|\vec{v}_f|=3.39\,\mathrm{m/s},
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\]
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directed
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\[
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45^\circ
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\]
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above the positive $x$-axis, while momentum is conserved but kinetic energy is not.
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