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 — phy131studiof17:lectures:chapter9b [2017/08/04 13:41] (current) Line 1: Line 1: + ~~SLIDESHOW~~ + ====== Chapter 9 - 2D Momentum Problems and Center of Mass ====== + + ===== Multi-dimensional collisions ===== + + The equation of conservation of momentum is a vector equation. ​ + + $m_{A}\vec{v}_{A}+m_{B}\vec{v}_{B}=m_{A}\vec{v}'​_{A}+m_{B}\vec{v}'​_{B}$ + + Therefore if we have a collision involving more than one dimension we need to consider conservation of each component of momentum, for example in 2 dimensions + + $m_{A}v_{Ax}+m_{B}v_{Bx}=m_{A}v'​_{Ax}+m_{B}v'​_{Bx}$ + + $m_{A}v_{Ay}+m_{B}v_{By}=m_{A}v'​_{Ay}+m_{B}v'​_{By}$ + + ===== Perfectly Elastic collisions in 2 dimensions ===== + + If a collision is perfectly elastic we can add a 3rd equation ​ + + $m_{A}v_{Ax}+m_{B}v_{Bx}=m_{A}v'​_{Ax}+m_{B}v'​_{Bx}$ + + $m_{A}v_{Ay}+m_{B}v_{By}=m_{A}v'​_{Ay}+m_{B}v'​_{By}$ + + $\frac{1}{2}m_{A}v^{2}_{A}+\frac{1}{2}m_{B}v^{2}_{B}=\frac{1}{2}m_{A}v'​^{2}_{A}+\frac{1}{2}m_{B}v'​^{2}_{B}$ + + + ===== Billiards question ===== + + When we hit a billiard ball straight on all the momentum of the incident ball should be transferred to the 2nd ball. + + {{elastic1.gif}} + + However we know already from experience that if we hit a billiard ball at an angle this does not happen. ​ Can we use our conservation of momentum principle to predict the trajectories of billiard balls after a collision? + + If one ball hits another head on, it should come to rest. But if it comes in at a small angle where should it go? + + + {{billiardclickernew.png}} + + + ===== Billiards (or pucks) ===== + + | Conservation of momentum<​html>&​nbsp;&​nbsp;&​nbsp;&​nbsp;&​nbsp;​ \\ $v_{A}=v'​_{Ax}+v'​_{Bx}$ \\ $0=v'​_{Ay}+v'​_{By}$ |Conservation of energy \\ $v_{A}^2=v'​^{2}_{A}+v'​^{2}_{B}$ | + + {{snookerballs.png}} + + |$v'​_{Bx}=v'​_{B}\cos\theta$ <​html>&​nbsp;&​nbsp;&​nbsp;&​nbsp;&​nbsp;​\\ $v'​_{By}=v'​_{B}\sin\theta$ | $v'​_{Ax}=v'​_{A}\sin\theta$ \\ $v'​_{Ay}=v'​_{A}\cos\theta$ | + + + + Square the conservation of momentum equations + + $v'​^{2}_{Ay}=v'​^{2}_{By}$ + + $v_{A}^2-2v_{A}v'​_{Ax}+v'​^{2}_{Ax}=v'​^{2}_{Bx}$ + + Add these together + + $v_{A}^2+v'​^{2}_{A}-2v_{A}v'​_{Ax}=v'​^{2}_{B}$ + + Use kinetic energy equation to eliminate $v'​_{B}$ + + $v_{A}^2+v'​^{2}_{A}-2v_{A}v'​_{Ax}=v_{A}^{2}-v'​^{2}_{A}$ ​ → $v'​^{2}_{A}=v_{A}v'​_{Ax}$ + + $v'​^{2}_{A}=v_{A}v'​_{A}\sin\theta$ → $v'​_{A}=v_{A}\sin\theta$ + + and conservation of energy equation gives us $v'​_{B}=v_{A}\cos\theta$ + + ===== Snooker video ===== + + <​html>​ +