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: 8.43

Answer

(a) The pendulum rises through a vertical height of 2.94 cm. (b) $K_1 = 866~J$ (c) $K_2 = 1.74~J$

Work Step by Step

(a) We can use conservation of momentum to find the speed of the block after the bullet hits it. $m_2~v_2 = m_1~v_1$ $v_2 = \frac{m_1~v_1}{m_2}$ We can equate the kinetic energy of the block at the bottom of the swing to the potential energy at the top. $PE = K$ $(m_1+m_2)gh = \frac{1}{2}(m_1+m_2)v_2^2$ $gh = \frac{1}{2}(\frac{m_1~v_1}{m_2})^2$ $h = \frac{1}{2g}(\frac{m_1~v_1}{m_2})^2$ $h = \frac{1}{(2)(9.80~m/s^2)}(\frac{(0.0120~kg)(380~m/s)}{6.012~kg})^2$ $h = 0.0294~m = 2.94~cm$ The pendulum rises through a vertical height of 2.94 cm. (b) $K_1 = \frac{1}{2}m_1~v_1^2$ $K_1 = \frac{1}{2}(0.0120~kg)(380~m/s)^2$ $K_1 = 866~J$ (c) $K_2 = \frac{1}{2}(m_1+m_2)~v_2^2$ $K_2 = \frac{1}{2}(m_1+m_2)( \frac{m_1~v_1}{m_2})^2$ $K_2 = \frac{1}{2}(6.012~kg)(\frac{(0.0120~kg)(380~m/s)}{6.00~kg})^2$ $K_2 = 1.74~J$
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