Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 7 - Linear Momentum - Problems - Page 193: 40

Answer

The ratio of the distances that each block travels is 9. The smaller block travels 9 times farther than the larger block.

Work Step by Step

Let the masses of the two blocks be $m$ and $3m$. We can use conservation of momentum to find the ratio of the two speeds after the explosion: $m~v_1 = 3m~v_2$ $v_1 = 3v_2$ Since the surface is rough, a sliding object will decelerate because of the force of friction. Let $M$ be the mass of this object. $Ma = F_f$ $Ma = Mg~\mu_k$ $a = g~\mu_k$ Note that the acceleration $a$ is the same for both blocks of wood since the acceleration does not depend on the mass. We can find an expression for the distance traveled. In general: $d = \frac{v^2}{2a}$ $d_1 = \frac{v_1^2}{2a}$ $d_2 = \frac{v_2^2}{2a}$ We can then find the ratio of the distances traveled by the two blocks of wood. $\frac{d_1}{d_2} = \frac{(v_1^2)/(2a)}{(v_2^2)/(2a)} = \frac{v_1^2}{v_2^2} = \frac{(3v_2)^2}{v_2^2} = 9$ The ratio of the distances that each block travels is 9. The smaller block travels 9 times farther than the larger block.
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