a. The initial energy is
and the final energy is
For the ball not to lose contact,
Using conservation of energy, it follows that
b. A simple application of conservation of energy yields
Note that the moment of inertia of a uniformly distributed triangle with mass M and distance from center to vertex L is
a. We apply conservation of angular momentum (about the center of the triangle), linear momentum, and energy. This gives
Subtracting L times (1) from (2) giveswhich we can substitute to get
Substituting into (1) gives
Substituting into (2) gives
b. In order for a second collision to occur, we need to consider v/ relative to V (i.e. consider the motion of the point mass in the frame of the triangle). Clearly if v/ >V , there will be no second collision since the point mass will ’escape’ the triangle’s reach -to see this clearly, draw the circle centered at the triangle’s center (i.e. the circumcircle of the triangle). The only possible point of contact for the second collision is at the bottom of this circle so the point mass must remain stationary relative to the triangle so that the triangle can rotate and collide with the point mass a second time.
Similarly, if v/ <V , then there will be no second collision.
If v'= V , then indeed there will be a second collision since this means the point mass remains stationary relative to the triangle while the triangle will spin with angular velocity w> 0.
Thus, a second collision occurs if V = v/ or 2m =5m − M