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: 36

Answer

The amount of kinetic energy acquired by the smaller block is 3300 J The amount of kinetic energy acquired by the larger block is 2200 J

Work Step by Step

Let the masses of the pieces be $m$ and $1.5m$ We can use conservation of momentum to set up an equation. $m~v_A = 1.5~m~v_B$ $v_A = 1.5~v_B$ We can find the kinetic energy $KE_1$ of the smaller block. $KE_1 = \frac{1}{2}mv_A^2 = \frac{1}{2}m(1.5v_B)^2$ $KE_1 = \frac{2.25}{2}m~v_B^2$ We can find the kinetic energy $KE_2$ of the larger block. $KE_2 = \frac{1}{2}(1.5~m)v_B^2 = \frac{1.5}{2}m~v_B^2$ We can find the total kinetic energy $KE$ of the two blocks. $KE = KE_1 + KE_2$ $KE = \frac{2.25}{2}m~v_B^2 + \frac{1.5}{2}m~v_B^2$ $KE = \frac{3.75}{2}m~v_B^2$ We can find the fraction of the total energy acquired by the smaller block. $\frac{KE_1}{KE} = \frac{ \frac{2.25}{2}mv_B^2}{\frac{3.75}{2}mv_B^2} = \frac{2.25}{3.75} = 0.6$ The amount of kinetic energy acquired by the smaller block is $(0.6)(5500~J) = 3300~J$ The amount of kinetic energy acquired by the larger block is $5500~J - 3300~J = 2200~J$
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