College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 7 - Problems - Page 265: 42

Answer

(a) The horizontal component of velocity of the third piece is $-0.133~m/s$ The vertical component of velocity of the third piece is $-4.07~m/s$ (b) After the explosion, the center of mass is stationary at the origin.

Work Step by Step

The initial momentum of the system is zero. By conservation of momentum, the momentum of the system is zero after the explosion. We can find the x-component of the third piece: $(\frac{M}{4})(5.0~m/s)~cos~37^{\circ}-(\frac{M}{3})(4.0~m/s)~cos~45^{\circ}+(\frac{5M}{12})v_x = 0$ $(\frac{5M}{12})v_x = (\frac{M}{3})(4.0~m/s)~cos~45^{\circ}-(\frac{M}{4})(5.0~m/s)~cos~37^{\circ}$ $5~v_x =(4)(4.0~m/s)~cos~45^{\circ} -(3)(5.0~m/s)~cos~37^{\circ}$ $5~v_x = -0.6658~m/s$ $v_x = -0.133~m/s$ The horizontal component of velocity of the third piece is $-0.133~m/s$ We can find the y-component of the third piece: $(\frac{M}{4})(5.0~m/s)~sin~37^{\circ}+(\frac{M}{3})(4.0~m/s)~sin~45^{\circ}+(\frac{5M}{12})v_y = 0$ $(\frac{5M}{12})v_y = -(\frac{M}{3})(4.0~m/s)~sin~45^{\circ}-(\frac{M}{4})(5.0~m/s)~sin~37^{\circ}$ $5~v_y =-(4)(4.0~m/s)~sin~45^{\circ} -(3)(5.0~m/s)~sin~37^{\circ}$ $5~v_y = -20.34~m/s$ $v_y = -4.07~m/s$ The vertical component of velocity of the third piece is $-4.07~m/s$ (b) Before the explosion, the center of mass was stationary at the origin. By conservation of momentum, after the explosion, the center of mass is stationary at the origin.
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