University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 8 - Momentum, Impulse, and Collision - Problems - Exercises - Page 266: 8.45

Answer

The tension in the wire just after the collision is 13.6 N.

Work Step by Step

We can use conservation of momentum to find the speed of the ornament. $m_2~v_2 = m_1~v_1$ $v_2 = \frac{m_1~v_1}{m_2}$ $v_2 = \frac{(0.200~kg)(12.0~m/s)}{1.00~kg}$ $v_2 = 2.40~m/s$ We can use the speed to find the tension in the wire. $\sum F = \frac{mv^2}{R}$ $T-mg = \frac{mv^2}{R}$ $T = \frac{mv^2}{R}+mg$ $T = \frac{(1.00~kg)(2.40~m/s)^2}{1.50~m}+(1.00~kg)(9.80~m/s^2)$ $T = 13.6~N$ The tension in the wire just after the collision is 13.6 N.
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