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: 33

Answer

The 50-gram ball of clay is moving at a speed of 1.7 m/s at an angle $45^{\circ}$ north of east.

Work Step by Step

We can find the east component of momentum. $p_x = (0.020~kg)(3.0~m/s)$ $p_x = 0.060~N~s$ We can find the north component of momentum. $p_y = (0.030~kg)(2.0~m/s)$ $p_x = 0.060~N~s$ We can find the magnitude of the momentum of the 50-gram ball of clay. $p = \sqrt{(p_x)^2+(p_y)^2}$ $p = \sqrt{(0.060~N~s)^2+(0.060~N~s)^2}$ $p = 0.085~N~s$ We can use the momentum to find the speed $v$ of the 50-gram ball of clay. $m~v = p$ $v = \frac{p}{m}$ $v = \frac{0.085~N~s}{0.050~kg}$ $v = 1.7~m/s$ Since the east component and the north component of the momentum are equal, the direction is $45^{\circ}$ north of east.
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