移动的球将从惰性球弹回的角度 [英] Angle that a moving ball will bounce off of an inert ball

问题描述

让有两个球，其中一个在笛卡尔坐标平面中移动，而另一个则静止不动.在某个时候，移动的球会与惰性球发生碰撞.假设移动球沿直线运动，那么在给出以下信息的情况下，如何得出移动球将被推进的新角度:

Let there be two balls, one of which is moving about in the Cartesian coordinate plane, while the other is stationary and immobile. At some point, the moving ball collides with the inert ball. Assuming the moving ball is traveling in a straight line, how can one derive the new angle that the moving ball will be propelled given the following information:

运动球的中心坐标(X0，Y0)，半径(R0)和撞击前的移动角度(A0)

The moving ball's center coordinates (X0, Y0), radius (R0), and angle of travel before impact (A0)

固定球的中心坐标(X1，Y1)和半径(R1)

The stationary ball's center coordinates (X1, Y1) and radius (R1)

推荐答案

如果您的第二个球具有无限质量:

If your second ball has infinite mass:

其中phi(经过长时间计算)为:

Where phi (after a long calc) is:

phi=  -ArcTan[
         ( 2 R^2 Sin[A0] + 2 (YD Cos[A0] - XD Sin[A0]) (2 H Cos[A0] + 
           2 XD Sin[A0]^2 - YD Sin[2 A0]))  /
         ((2 R^2 - XD^2 - 3 YD^2) Cos[A0] + (XD^2 - YD^2) Cos[3 A0] + 
           8 XD YD Cos[A0]^2 Sin[A0] + 4 H Sin[A0] (-YD Cos[A0] + XD Sin[A0]))
           ]

位置:

H   = (R0 + R1)^2 - ((Y0 - Y1) Cos[A0] + (X0 - X1) Sin[A0])^2  
R^2 = (R0 + R1)^2
XD  =  X1 - X0
YD  =  Y1 - Y0

修改

要确定整个轨迹，还需要在撞击时移动球中心的坐标.它们是:

To determine the whole trajectory, you'll also need the coordinates for the center of the moving ball at the time of impact. They are:

  {X,Y}= {X1+Sin[A0] ((Y1-Y0) Cos[A0]+ (X0-X1) Sin[A0])-Cos[A0] Sqrt[H],
          Y1+Cos[A0] ((Y0-Y1) Cos[A0]+(-X0+X1) Sin[A0])-Sin[A0] Sqrt[H]}

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