Physics (10th Edition)

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

Chapter 7 - Impulse and Momentum - Problems - Page 194: 41


a) $\theta=73^o$ b) $v_f=4.27m/s$

Work Step by Step

- The 50-kg skater's momentum before the collision is $p_1=mv=50\times3=150kg.m/s$ - The 70-kg skater's momentum before the collision is $p_2=mv=70\times7=490kg.m/s$ We take due east to be the $+x$ direction and due south to be the $+y$ direction. So, in unit-vector notation, we have $\vec{p_1}=150i$ and $\vec{p_2}=490j$ According to the principle of conservation of linear momentum, if we call the momentum of the system after the collision $\vec{p_f}$, we have $$\vec{p_f}=\vec{p_1}+\vec{p_2}=150i+490j (kg.m/s)$$ After the collision, the system of 2 skaters has mass $M=50+70=120kg$, so the velocity of the system is $$\vec{v_f}=\frac{\vec{p_f}}{M}=1.25i+4.08j (m/s)$$ a) We have $$\tan\theta=\frac{4.08}{1.25}=3.264$$ $$\theta=73^o$$ b) The speed is $v_f=\sqrt{1.25^2+4.08^2}=4.27m/s$
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