Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 9 - Center of Mass and Linear Momentum - Problems - Page 251: 50b

Answer

The speed of the bullet-block center of mass is $~~4.95~m/s$

Work Step by Step

We can use conservation of momentum to find the velocity $v_b$ of the block: $p_f = p_i$ $(5.20~g)(428~m/s)+(700~g)(v_b) = (5.20~g)(672~m/s)$ $(700~g)(v_b) = (5.20~g)(672~m/s)-(5.20~g)(428~m/s)$ $v_b = \frac{(5.20~g)(672~m/s)-(5.20~g)(428~m/s)}{700~g}$ $v_b = 1.81~m/s$ The resulting speed of the block is $~~1.81~m/s$ We can find the speed of the bullet-block center of mass: $v_{com} = \frac{(5.20~g)(428~m/s)+(700~g)(1.81~m/s)}{5.20~g+700~g}$ $v_{com} = 4.95~m/s$ The speed of the bullet-block center of mass is $~~4.95~m/s$
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