Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 7 - Linear Momentum - General Problems - Page 196: 75

Answer

The initial speed of the bullet was 340 m/s.

Work Step by Step

We can use $\mu_k$ to find the magnitude of deceleration after the collision: $ma = mg~\mu_k$ $a = (9.80~m/s^2)(0.28)$ $a = 2.74~m/s^2$ We can use the acceleration to find the speed $v$ of the block as it started to slide: $v^2 = 2ax$ $v = \sqrt{2ax} = \sqrt{(2)(2.74~m/s^2)(8.5~m)}$ $v = 6.82~m/s$ We can use conservation of momentum to find the initial speed $v_1$ of the bullet: $m_1~v_1 = (m_1+m_2)~v$ $v_1 = \frac{(m_1+m_2)(v)}{m_1}$ $v_1 = \frac{(0.028~kg+1.35~kg)(6.82~m/s)}{0.028~kg}$ $v_1 = 340~m/s$ The initial speed of the bullet was 340 m/s.
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