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 $$2mv_y+MV_y=\sum p_{fy}=0$$ $$M=\frac{2mv_y}{-V_y}=2.54\times10^{-3}kg$$
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