Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 9 - Linear Momentum and Collisions - Conceptual Questions - Page 189: 1

Answer

$2.49\times 10^5\frac{mi}{h}$

Work Step by Step

We know that $p_{ball}=m_{ball}v_{ball}$ This can be rearranged as: $v_{ball}=\frac{p_{ball}}{m_{ball}}$ but given that $p_{ball}=p_{car}$ $\implies v_{ball}=\frac{p_{car}}{m_{ball}}$ We plug in the known values to obtain: $v_{ball}=\frac{15800}{0.412}=1.11\times 10^6\frac{m}{s}$ Now we convert this speed into miles per hour $v_{ball}=(\frac{11.6\times 10^6m}{s})(\frac{3600s}{h})(\frac{1mi}{1609m})$ $v_{ball}=2.49\times 10^5\frac{mi}{h}$
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