Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 11 - Impulse and Momentum - Exercises and Problems: 47

Answer

The third piece is moving with a speed of 14 m/s at an angle of $45^{\circ}$ north of east.

Work Step by Step

We can use conservation of momentum to find the components of the third piece's momentum. Since the initial momentum of the object is zero, the final momentum is also zero. Let $M$ be the mass of the two smaller pieces. The east component of the third piece's momentum will be equal in magnitude to the piece moving to the west. $(2M)~v_x = (M)(20~m/s)$ $v_x = 10~m/s$ The north component of the third piece's momentum will be equal in magnitude to the piece moving to the south. $(2M)~v_y = (M)(20~m/s)$ $v_y = 10~m/s$ We can find the speed of the third piece. $v = \sqrt{(v_x)^2+(v_y)^2}$ $v = \sqrt{(10~m/s)^2+(10~m/s)^2}$ $v = 14~m/s$ Since the north and east components of the velocity are equal, the third piece is moving at an angle of $45^{\circ}$ north of east. The third piece is moving with a speed of 14 m/s at an angle of $45^{\circ}$ north of east.
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