## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 7 - Problems - Page 265: 49

#### Answer

Before the collision, the original speed of the bullet was $346~m/s$

#### Work Step by Step

Let $m_f$ be the total mass of the block and bullet. We can find the rate of deceleration along the horizontal surface: $F_N~\mu_k = m_f~a$ $m_f~g~\mu_k = m_f~a$ $a = g~\mu_k$ $a = (9.80~m/s^2)(0.400)$ $a = 3.92~m/s^2$ We can find the velocity $v_0$ of the block and bullet when they start sliding along the horizontal surface: $v_f^2 = v_0^2 + 2ax$ $v_0^2 = v_f^2 - 2ax$ $v_0 = \sqrt{v_f^2 - 2ax}$ $v_0 = \sqrt{0 - (2)(-3.92~m/s^2)(1.50~m)}$ $v_0 = 3.43~m/s$ By conservation of momentum, the final momentum of the system just after the collision is equal to the initial momentum of the bullet. We can find the original speed $v_b$ of the bullet: $(0.020~kg)~v_b = (2.00~kg+0.020~kg)(3.43~m/s)$ $v_b = \frac{(2.00~kg+0.020~kg)(3.43~m/s)}{0.020~kg}$ $v_b = 346~m/s$ Before the collision, the original speed of the bullet was $346~m/s$.

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