2015PUPC普林斯顿大学物理竞赛Onsite答案免费下载

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2015PUPC普林斯顿大学物理竞赛完整版真题免费下载

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Question 1

a. The initial energy is

2015PUPC普林斯顿大学物理竞赛真题答案

and the final energy is

2015PUPC普林斯顿大学物理竞赛真题答案

For the ball not to lose contact,

2015PUPC普林斯顿大学物理竞赛真题答案

Using conservation of energy, it follows that

2015PUPC普林斯顿大学物理竞赛真题答案

b. A simple application of conservation of energy yields

2015PUPC普林斯顿大学物理竞赛真题答案

or

2015PUPC普林斯顿大学物理竞赛真题答案

Problem 2

Note that the moment of inertia of a uniformly distributed triangle with mass M and distance from center to vertex L is2015PUPC普林斯顿大学物理竞赛真题答案

a.  We apply conservation of angular momentum (about the center of the triangle), linear momentum, and energy. This gives

2015PUPC普林斯顿大学物理竞赛真题答案

Subtracting L times (1) from (2) gives2015PUPC普林斯顿大学物理竞赛真题答案which we can substitute to get

2015PUPC普林斯顿大学物理竞赛真题答案

2015PUPC普林斯顿大学物理竞赛真题答案

Substituting into (1) gives2015PUPC普林斯顿大学物理竞赛真题答案

Substituting into (2) gives2015PUPC普林斯顿大学物理竞赛真题答案

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

M=3m

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