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How far will two children slide after a perfectly inelastic collision? |
While a family is taking a walk on a frozen pond, the two small children (a boy and a girl) manage to run into each other. They become entangled, resulting in a totally inelastic collision. The boy has a mass of 15 kg and was initially running at 1.0 m/s and the girl has a mass of 20 kg and was initially running at 2.0 m/s. Before their collision, the relative angle between their velocities was 45°. Assuming that the coefficient of friction between the resulting boy+girl combination and the ice is 0.15, how far do they slide after the collision before coming to rest?
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Solution
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Boy and girl as |
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Impulse from external influences is neglected during the collision under the assumption that the collision is instantaneous. After the collision, the system experiences external influences from the earth (gravity, |
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We begin by analyzing the collision using momentum. When a problem gives a relative angle, it is important to develop a coordinate system to orient ourselves as we solve. We therefore begin with a picture. We have arbitrarily assigned the boy to move along the x-axis, and the girl to have positive x- and y-velocity components.
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With our picture developed, we can write the equations of constant momentum, since we are assuming that the collision occurs over such a short time that external forces create a negligible impulse.
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{table:align=right|cellspacing=0|cellpadding=1|border=1|frame=box|width=30%} {tr} {td:align=center|bgcolor=#F2F2F2}*[Examples from Momentum]* {td} {tr} {tr} {td} {pagetree:root=Examples from Momentum} {search-box} {td} {tr} {table}{excerpt:hidden=true}How far will two children slide after a perfectly inelastic collision?{excerpt} While a family is taking a walk on a frozen pond, the two small children (a boy and a girl) manage to run into each other. They become entangled, resulting in a [totally inelastic collision]. The boy has a mass of 15 kg and was initially running at 1.0 m/s and the girl has a mass of 20 kg and was initially running at 2.0 m/s. Before their collision, the relative angle between their velocities was 45°. Assuming that the coefficient of friction between the resulting boy+girl combination and the ice is 0.15, how far do they slide after the collision before coming to rest? {composition-setup}{composition-setup} h4. Solution {toggle-cloak:id=sys} *System:* {cloak:id=sys} Boy and girl as [point particles|point particle]. \\ {cloak}\\ \\ \\ \\ \\ {toggle-cloak:id=int} *Interactions:* {cloak:id=int} Impulse from external influences is neglected during the collision under the assumption that the collision is instantaneous. After the collision, the system experiences external influences from the earth (gravity, [conservative|work#nonconservative]) and the ice (normal force and friction, each non-conservative). \\ {cloak}\\ \\ \\ \\ \\ {toggle-cloak:id=mod} *Models:* \\ {cloak:id=mod}[Momentum and External Force] followed by [Mechanical Energy and Non-Conservative Work]. {cloak}\\ \\ \\ \\ \\ {toggle-cloak:id=app} *Approach:* {cloak:id=app} {toggle-cloak:id=diag1} {color:red}{*}Diagrammatic Representation, Part 1{*}{color}\\ {cloak:id=diag1} We begin by analyzing the collision using momentum. When a problem gives a relative angle, it is important to develop a coordinate system to orient ourselves as we solve. We therefore begin with a picture. We have _arbitrarily_ assigned the boy to move along the x-axis, and the girl to have positive x\- and y-velocity components. \\ !pond1.jpg! {cloak:diag1} {toggle-cloak:id=math1} {color:red}{*}Mathematical Representation, Part 1{*}{color}\\ {cloak:id=math1} With our picture developed, we can write the equations of constant momentum, since we are assuming that the collision occurs over such a short time that [external forces|external force] create a negligible [impulse]. \\ {latex}\begin{large}\[ p^{B}_{x,i} + p^{G}_{x,i} = p^{B+G}_{x,f}\] \[p^{B}_{y,i} + p^{G}_{y,i} = p^{B+G}_{y,f} \]\end{large}{latex} |
Rewriting
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in
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terms
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of
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the
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masses
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and
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velocities,
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and
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substituting
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the
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appropriate
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zeros
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gives:
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}\begin{large}\[ m^{B}v^{B}_{x,i} + m^{G}v^{G}_{x,i} = (m^{B}+m^{G})v_{x,f} \] \[ m^{G}v^{G}_{y,i} = (m^{B}+m^{G})v_{y,f} \]\end{large}{latex} |
We
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have
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all
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the
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givens
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we
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need
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to
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solve
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these
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equations
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directly
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for
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the
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final
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velocity
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of
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the
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system:
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}\begin{large}\[ v_{x,f} = \frac{m^{B}v^{B}_{x,i}+m^{G}v^{G}_{x,i}}{m^{B}+m^{G}} = \mbox{1.24 m/s}\] \[ v_{y,f} = \frac{m^{G}v^{G}_{y,i}}{m^{B}+m^{G}} = \mbox{0.808 m/s} \]\end{large}{latex} |
so
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that
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the
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magnitude
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of
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the
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final
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velocity
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is:
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}\begin{large}\[ v_{f} = \sqrt{(\mbox{1.24 m/s})^2+(\mbox{0.808 m/s})^{2}} = \mbox{1.48 m/s} \]\end{large} |
Notice that the final velocity of the entangled boy and girl lies neither along the original velocity of the boy or that of the girl, but between these two initial velocities. From the components of the final velocity vector we can determine the angle between the final velocity and the boy's initial velocity as:
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\begin{large}\[tan(\phi) = \frac{v_{y,f}}{v_{x,f}} = \frac{0.808}{1.24} = 0.653 \]\end{large} |
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\begin{large}\[\phi = 33.16\]\end{large} |
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This final velocity for the collision of the boy and girl is the initial velocity for their post-collision slide. We can now consider a free body diagram for the boy+girl system as they slide across the ice. Assuming the pond is flat, the gravitational potential energy will be constant and the normal force will do no work. Friction will do negative work, as it is directed at 180° from the direction of motion of the system. Since only one force can do work, the question of how far the system slides after the collision is easily addressed using work and energy.
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The relevant equation is:
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{latex} Notice that the final velocity of the entangled boy and girl lies neither along the original velocity of the boy or that of the girl, but between these two initial velocities. From the components of the final velocity vector we can determine the angle between the final velocity and the boy's initial velocity as: (Add equation and diagram) {cloak:math1} {toggle-cloak:id=diag2} {color:red}{*}Diagrammatic Representation, Part 2{*}{color} {cloak:id=diag2} !pond2.jpg! This final velocity for the collision of the boy and girl is the initial velocity for their post-collision slide. We can now consider a free body diagram for the boy+girl system as they slide across the ice. Assuming the pond is flat, the [gravitational potential energy] will be constant and the normal force will do no [work]. Friction will do negative work, as it is directed at 180° from the direction of motion of the system. Since only one force can do work, the question of how far the system slides after the collision is easily addressed using work and energy. {cloak:diag2} {toggle-cloak:id=math2} {color:red}{*}Mathematical Representation, Part 2{*}{color} {cloak:id=math2} The relevant equation is: {latex}\begin{large} \[ K_{f} = K_{i} + \int F_{f} \:ds = K_{i}-\int \mu_{k}N\:ds\]\end{large}{latex} |
Here
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we
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have
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used
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a
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minus
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sign
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in
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the
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second
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expression
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because
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the
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frictional
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force
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is
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directed
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against
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the
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direction
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of
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motion,
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and
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reduces
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the
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overall
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kinetic
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energy.
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Considering
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the
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free
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body
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diagram
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plus
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the
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fact
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that
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the
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boy
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and
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girl
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are
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not
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accelerating
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in
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the
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z-direction,
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it
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is
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clear
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that
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the
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normal
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force
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and
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the
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weight
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of
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the
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system
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must
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balance.
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Further,
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we
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want
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to
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know
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how
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far
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the
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boy
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and
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girl
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slide
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before
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they
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stop,
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indicating
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that vf should be zero for this part of the problem. Thus, we can write:
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_v{_}{~}f~ should be zero for this part of the problem. Thus, we can write: {latex}\begin{large}\[ 0 = \frac{1}{2}(m^{B}+m^{G})v_{i}^{2} - \mu (m^{B}+m^{G})gs \]\end{large}{latex} |
Again,
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recognizing
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that
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the
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initial
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velocity
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for
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the
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slide
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is
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the
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final
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velocity
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of
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the
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collision,
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we
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can
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solve
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to
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find:
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}\begin{large}\[ s = \frac{v_{i}^{2}}{2 \mu g} = \mbox{0.75745 m} \]\end{large}{latex} {note}Note that because the frictional force is directed opposite the direction of motion, the final answer is positive, as we expect for a distance. {note} {cloak:math2} {cloak:app} | !copyright and waiver^copyrightnotice.png! | RELATE wiki by David E. Pritchard is [licensed|copyright and waiver] under a [Creative Commons Attribution-Noncommercial-Share Alike 3.0 United States License|http://creativecommons.org/licenses/by-nc-sa/3.0/us/]. | |
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Note that because the frictional force is directed opposite the direction of motion, the final answer is positive, as we expect for a distance. |
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