Physics (10th Edition)

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

Chapter 7 - Impulse and Momentum - Problems - Page 193: 29


a) The speed of the combination is $4.87m/s$ b) The combination can rise to $1.21m$ above its initial position.

Work Step by Step

a) The bullet, with mass $m=2.5g=2.5\times10^{-3}kg$, travels at $v=425m/s$ and strikes the block, whose mass is $M=215g=0.215kg$ and has initial velocity $V=0$. After that, the bullet-block system has mass $M+m=0.218kg$ and rises at velocity $V_f$, which needs to be found. The whole process only travels in one direction, so we do not need to assume the signs for the vectors. We assume there are no external forces, so the total linear momentum is conserved. Therefore, $$M\vec{V}+m\vec{v}=(M+m)\vec{V}_f$$ $$V_f=\frac{MV+mv}{M+m}=4.87m/s$$ b) We take the starting point of the combination to be $h_0=0$ and need to find the highest point $h_f$. At the start, we have $V_0=4.87m/s$ and at $h_f$, $V_f=0$ According to the principle of conservation of mechanical energy, $$\frac{1}{2}(m+M)V_0^2=(m+M)gh_f$$ $$\frac{1}{2}V_0^2=gh_f$$ $$h_f=\frac{V_0^2}{2g}=1.21m$$
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