This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1869 Excerpt: ...in that time, the diagonal R &. Now, R a multiplied by the weight of A, added to R c multiplied by the weight of C, added to R b', plus R &", multiplied by the weight of B; will be equal to A R multiplied by the weight of A, added to CR multiplied by the weight of C: that is, the joint momentum of A, B, and C, after collision, will be equal to the joint momentum of A and C before collision. If, in the example above given, we suppose radius to be equal to 10 feet, the respective velocities of A and C, before collision, would each be equal to 10 feet per second. And, as A is supposed to weigh 2 pounds, the momentum of A before collision, would be equal to 20. As the weight of C is supposed to be 3 pounds, the momentum of C before collision would be equal to 30. Hence, the joint momentum of A and C before collision would be equal to 50: which is the amount that is to be accounted for after collision. Now, if we suppose the angle to be such that the versed sine would be equal to 6 feet, the respective velocities and momenta of A, C, and B, after collision, would be as follows: The velocity of A would be equal to 4f feet, and the momentum of A would be equal to 9£. The velocity of C would be equal to 3 feet, and the momentum of C would be equal to 11. The velocity of B, in the given direction of A, would be equal to 1£ feet, and the momentum of B equal to 10f. The velocity of B, in the given direction of C, would be equal to 3 feet, and the momentum of B would be equal to 19. And 9J-+ 11 + lOf + 19 = 50. SECTION XVIII. DEVOTED TO THE ENUNCIATION OF THE LAWS WHICH DETERMINE THE RESULTS OF COLLISION IN THE EIGHTEENTH CLASS OF CASES; WHICH CONSISTS OF THOSE IN WHICH THE WEIGHT OF B IS EQUAL TO, OR GREATER THAN, THE JOINT WEIGHT OF A AND C; IN...
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