In Example 8.14 (Section 8.5), Ramon pulls on the rope to give himself a speed of 0.70 m/s. What is James’s speed?
Solution 57E Step 1: Data given Mass of James m = 90.0 kg j Mass of Ramon m = 60.r kg Distance between them d = 20.0 m Velocity of Ramon v = 0.r0 m/s We need to find velocity of Ramon Let us consider law of conservation of momentum Hence we have m r r m v j j Here v jvelocity of James Substitute the values to find v j vj= (m v r r j vj= (60.0 kg × 0.70 m/s)/90.0 kg v = (42 m/s kg )/90.0 kg j vj= 0.466 m/s This can approximated to 0.47 m/s Thus we have velocity of James as 0.47 m/s
Textbook: University Physics
Author: Hugh D. Young, Roger A. Freedman
University Physics was written by and is associated to the ISBN: 9780321675460. The full step-by-step solution to problem: 57E from chapter: 8 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. This textbook survival guide was created for the textbook: University Physics, edition: 13. This full solution covers the following key subjects: speed, pulls, himself, James, give. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. Since the solution to 57E from 8 chapter was answered, more than 428 students have viewed the full step-by-step answer. The answer to “In Example 8.14 (Section 8.5), Ramon pulls on the rope to give himself a speed of 0.70 m/s. What is James’s speed?” is broken down into a number of easy to follow steps, and 22 words.