Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 7 - Impulse and Momentum - Problems - Page 195: 61


The mass of the bullet is $2.54\times10^{-3}kg$

Work Step by Step

The arrangement of the bullets is in the figure below. 2 bullets with mass $m=4.5\times10^{-3}kg$ fly at velocity $v=324m/s$. The other bullet with mass $M$ flies at velocity $V=575m/s$. After colliding with one another, they all stop, meaning $\sum p_f=0$ We take a look at the y-component of the motions. - 2 bullets with mass $m$ have $v_y=+324\cos60=+162m/s$ - The other bullet has $V_y=-575m/s$ From conservation of momentum, we have $$2mv_y+MV_y=\sum p_{fy}=0$$ $$M=\frac{2mv_y}{-V_y}=2.54\times10^{-3}kg$$
Small 1581854745
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.