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: 71

Answer

The speed of the blob is 0.85 m/s and it is moving at an angle of $72.2^{\circ}$ below the positive x-axis.

Work Step by Step

We can find the horizontal component of the momentum of the system. $p_x = (0.040~kg)(4.0~m/s)[cos(45^{\circ})]+(0.030~kg)(-3.0~m/s)+(0.020~kg)(0)$ $p_x = 0.023~N~s$ We can find the horizontal component of the velocity of the 90-gram blob of clay. $m~v_x = p_x$ $v_x = \frac{p_x}{m}$ $v_x = \frac{0.023~N~s}{0.090~kg}$ $v_x = 0.26~m/s$ We can find the vertical component of the momentum of the system. $p_y = (0.040~kg)(-4.0~m/s)[sin(45^{\circ})]+(0.030~kg)(0)+(0.020~kg)(2.0~m/s)$ $p_y = -0.073~N~s$ We can find the vertical component of the velocity of the 90-gram blob of clay. $m~v_y = p_y$ $v_y = \frac{p_y}{m}$ $v_y = \frac{-0.073~N~s}{0.090~kg}$ $v_y = -0.81~m/s$ We can find the speed of the blob of clay. $v = \sqrt{(v_x)^2+(v_y)^2}$ $v = \sqrt{(0.26~m/s)^2+(-0.81~m/s)^2}$ $v = 0.85~m/s$ We can find the angle $\theta$ below the positive x-axis. $tan(\theta) = \frac{v_y}{v_x}$ $\theta = arctan(\frac{0.81~m/s}{0.26~m/s})$ $\theta = 72.2^{\circ}$ The speed of the blob is 0.85 m/s and it is moving at an angle of $72.2^{\circ}$ below the positive x-axis.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

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.