Billiards collision at right angle
Question
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The following quantities appear in the problem:
Masse \(m\) / Geschwindigkeit \(v\) / Winkel \(\theta\) / Impuls \(p\) /
The following formulas must be used to solve the exercise:
\(\cos\alpha = \dfrac{b}{c} \quad \) \(p = mv \quad \) \(\sum p_{\scriptscriptstyle\rm tot} \stackrel{!}{=} \sum p_{\scriptscriptstyle\rm tot}' \quad \) \(\sin\alpha = \dfrac{a}{c} \quad \)
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Exercise:
Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Ball mathcalA is moving upward along the y axis at .meterpersecond and ball mathcalB is moving to the right along the x axis with speed .meterpersecond. After the collision ased elastic ball mathcalB is moving along the positive y axis. What is the final direction of ball mathcalA and what are their two speeds?
Solution:
First we make a sketch of the situation. The left drawing shows the situation before the collision the right one the situation after the collision. center tikzpicturescale. latex drawdotted ---; drawdotted ---; Billardkugel-.red; Billardkugel-.blue; drawthick-green!!black -.---. nodeabovegreen!!black w; drawthick-green!!black -.---. nodeleftgreen!!black v; tikzpicture tikzpicturescale. latex drawdotted ---; drawdotted ---; Billardkugel..red; Billardkugel..blue; drawthick-green!!black .--. nodeleftgreen!!black v'; drawthick-green!!black :.cm -- +:.cm nodeabovegreen!!black w'; % drawthick colorred .--.---.--.; tikzpicture center We denote the momenta of the balls mathcal A and mathcal B by vec p and vec q respectively and their velocities by vec v and vec w respectively. Conservation of momentum in x- and y-direction leads to the following s: alptot &mustbe ptot' vec p + vec q vec p' + vec q' mmqtyv_xv_y + mmqtyw_xw_y mmqtyv_x'v_y' + mmqtyw_x'w_y' mqtyv + mqtyw mqtyv'cosalphav'sinalpha + mqtyw' Furthermore we have conversation of the kinetic energy: al Etot &mustbe Etot Ekin Ekin fracmv^ + frac mw^ fracmv'^ + fracmw'^ v^ + w^ v'^ + w'^. This is a system of three non-linear s with three unknowns v' w' and alpha v and w are given in the text: w v'cosalpha labelfirst v v'sinalpha + w' labelsecond v^ + w^ v'^ + w'^ labelthird Since we do not have a general procedure to solve such a system we need some creativity. A good trick for such a system is to use the fact that sin^alpha + cos^alpha i.e. we should rearrange the s in a way that we get the squares of sin and cos. We can do this by simply squaring eqreffirst and squaring eqrefsecond after taking w' to the left side: al w^ v'^cos^alpha qtyv-w'^ v'^sin^alpha We can now add these two s: al w^ + qtyv-w'^ v'^ w^ + v^ - vw' + w'^ v'^ v^ + w^ v'^-w'^ + vw' We can eliminate v' by subtracting this from eqrefthird and then solve for w': al w'^ - vw' w' v .. Putting w'v o eqrefthird and solving for v' yields al v^ + w^ v'^ + v^ v' w .. We use eqreffirst to determine alpha and thus the direction of the first ball: al alpha arccosfracwv' arccosfrac.. degree.
Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Ball mathcalA is moving upward along the y axis at .meterpersecond and ball mathcalB is moving to the right along the x axis with speed .meterpersecond. After the collision ased elastic ball mathcalB is moving along the positive y axis. What is the final direction of ball mathcalA and what are their two speeds?
Solution:
First we make a sketch of the situation. The left drawing shows the situation before the collision the right one the situation after the collision. center tikzpicturescale. latex drawdotted ---; drawdotted ---; Billardkugel-.red; Billardkugel-.blue; drawthick-green!!black -.---. nodeabovegreen!!black w; drawthick-green!!black -.---. nodeleftgreen!!black v; tikzpicture tikzpicturescale. latex drawdotted ---; drawdotted ---; Billardkugel..red; Billardkugel..blue; drawthick-green!!black .--. nodeleftgreen!!black v'; drawthick-green!!black :.cm -- +:.cm nodeabovegreen!!black w'; % drawthick colorred .--.---.--.; tikzpicture center We denote the momenta of the balls mathcal A and mathcal B by vec p and vec q respectively and their velocities by vec v and vec w respectively. Conservation of momentum in x- and y-direction leads to the following s: alptot &mustbe ptot' vec p + vec q vec p' + vec q' mmqtyv_xv_y + mmqtyw_xw_y mmqtyv_x'v_y' + mmqtyw_x'w_y' mqtyv + mqtyw mqtyv'cosalphav'sinalpha + mqtyw' Furthermore we have conversation of the kinetic energy: al Etot &mustbe Etot Ekin Ekin fracmv^ + frac mw^ fracmv'^ + fracmw'^ v^ + w^ v'^ + w'^. This is a system of three non-linear s with three unknowns v' w' and alpha v and w are given in the text: w v'cosalpha labelfirst v v'sinalpha + w' labelsecond v^ + w^ v'^ + w'^ labelthird Since we do not have a general procedure to solve such a system we need some creativity. A good trick for such a system is to use the fact that sin^alpha + cos^alpha i.e. we should rearrange the s in a way that we get the squares of sin and cos. We can do this by simply squaring eqreffirst and squaring eqrefsecond after taking w' to the left side: al w^ v'^cos^alpha qtyv-w'^ v'^sin^alpha We can now add these two s: al w^ + qtyv-w'^ v'^ w^ + v^ - vw' + w'^ v'^ v^ + w^ v'^-w'^ + vw' We can eliminate v' by subtracting this from eqrefthird and then solve for w': al w'^ - vw' w' v .. Putting w'v o eqrefthird and solving for v' yields al v^ + w^ v'^ + v^ v' w .. We use eqreffirst to determine alpha and thus the direction of the first ball: al alpha arccosfracwv' arccosfrac.. degree.