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: 51a

Answer

The speed of the bullet as it leaves block 1 is $~~721.4~m/s$

Work Step by Step

The speed of the bullet as it leaves block 1 is the same as its speed just before it enters block 2. We can use conservation of momentum to find the speed of the bullet as it leaves block 1: $p_i = p_f$ $(3.50~g)~v = (1803.5~g)~(1.40~m/s)$ $v = \frac{(1803.5~g)~(1.40~m/s)}{3.50~g}$ $v = 721.4~m/s$ The speed of the bullet as it leaves block 1 is $~~721.4~m/s$
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