Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson

Chapter 9 - Impulse and Momentum - Exercises and Problems: 25

Answer

The 50-gram blob 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. as: $p_x = (0.020~kg)(3.0~m/s)$ $p_x = 0.060~N~s$ 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 blob 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 blob 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.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.