Applied Maths - Mechanics 31

Note that for perfectly elastic objects, e = 1.
When ball C impacts obliquely ball A or ball B, there are two component of the impact velocity, one parallel to the tangent of contact, and one perpendicular. The parallel component of the velocity will remain exchanged after the impact since the balls are smooth. Now for the perpendicular component, let say the velocity for ball C before impact is v and the velocity after impact be v; and the velocity of ball A/B after impact be u.
The equations for the subsequent motion are:
mv = mv' + mu
u – v' = ev = v
Solving v' = 0 and u = v. In other words the momentum is completely transfer from C to ball A/B.
http://img81.imageshack.us/img81/4029/ballstqx.jpg
Referring to figure 1 in the above picture, let's first look at the second impact. Since the perpendicular component is completely transferred to B, only the parallel component is unchanged. Since the subsequent velocity of C is perpendicular to PQ, it should be easy to realise that at the very moment of the impact, the centre of C is on the same line PQ.
Likewise, when ball C makes the first impact with A, C will move parallel to the tangent. So it is easy to see a right angle triangle with hypotenuse = 6a and an adjacent side = 2a.
cos@ = 1/3
For the second part of the question, the concept is similar. In order for ball C to miss B, there are two possible extreme as shown in fig 2 and fig 3. For figure 2, cos@ = 1/2 and for figure 3, cos@ = π/2
Therefore the range for @ is cos-1(1/2) < @ < π/2