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: 14

Answer

(a) The speed and direction of Piece 1 is $8.00\times 10^3~m/s$ away from the Earth. The speed and direction of Piece 2 is $5.20\times 10^3~m/s$ away from the Earth. (b) The explosion supplied $7.0 \times 10^8~J$ of energy.

Work Step by Step

(a) Let $v$ be the original speed. Let $v_r$ be the relative speed of the two pieces. Since the two parts have equal mass, then one part must move with a speed of $v+v_r/2$ and the other piece must move with a speed of $v-v_r/2$. Then the original momentum is conserved, and the relative speed between the two pieces is $v_r$. Piece 1: $v+v_r/2 = 6.60\times 10^3~m/s + 1.40\times 10^3~m/s$ $v+v_r/2 = 8.00\times 10^3~m/s$ The speed and direction of Piece 1 is $8.00\times 10^3~m/s$ away from the Earth. Piece 2: $v-v_r/2 = 6.60\times 10^3~m/s - 1.40\times 10^3~m/s$ $v-v_r/2 = 5.20\times 10^3~m/s$ The speed and direction of Piece 2 is $5.20\times 10^3~m/s$ away from the Earth. (b) Before the explosion: $KE_0 = \frac{1}{2}(725~kg)(6.60\times 10^3~m/s)^2$ $KE_0 = 1.58\times 10^{10}~J$ After the explosion: $KE = \frac{1}{2}(725~kg/2)(8.00\times 10^3~m/s)^2 + \frac{1}{2}(725~kg/2)(5.20\times 10^3~m/s)^2$ $KE = 1.65\times 10^{10}~J$ $KE-KE_0 = 1.65\times 10^{10}~J - 1.58\times 10^{10}~J$ $KE-KE_0 = 7.0 \times 10^8~J$ The explosion supplied $7.0 \times 10^8~J$ of energy.
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