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: 8.84

Answer

The minimum speed of the dart is 58.6 m/s

Work Step by Step

To make a complete loop, the minimum speed $v_t$ at the top of the loop is such that the required centripetal force is equal to the force of gravity. Let $m_2$ be the mass of the ball and dart. $\frac{m_2~v_t^2}{r} = m_2~g$ $v_t^2 = g~r$ $v_t = \sqrt{g~r}$ $v_t = \sqrt{(9.80~m/s^2)(2.80~m)}$ $v_t = 5.24~m/s$ The minimum kinetic energy at the bottom of the loop is equal to the sum of the potential energy and the minimum kinetic energy at the top of the loop. $\frac{1}{2}m_2v_2^2 = m_2~gh+\frac{1}{2}m_2~v_t^2$ $v_2^2 = 2gh+v_t^2$ $v_2 = \sqrt{2gh+v_t^2}$ $v_2 = \sqrt{(2)(9.80~m/s^2)(5.60~m)+(5.24~m/s)^2}$ $v_2 = 11.71~m/s$ We can use conservation of momentum to find the minimum initial speed of the dart. $m_1v_1 = m_2v_2$ $v_1 = \frac{m_2v_2}{m_1}$ $v_1 = \frac{(25.00~kg)(11.71~m/s)}{5.00~kg}$ $v_1 = 58.6~m/s$ The minimum speed of the dart is 58.6 m/s
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