## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 7 - Problems - Page 265: 54

#### Answer

After the collision, the speed of the 5.0 kg object is $3.0~m/s$

#### Work Step by Step

Let $m_A = 1.0~kg$ and let $m_B = 5.0~kg$ Let $v_A$ be the initial velocity of the 1.0-kg object. Let $v_B$ be the initial velocity of the 5.0-kg object. Let $v_A'$ be the final velocity of the 1.0-kg object. Let $v_B'$ be the final velocity of the 5.0-kg object. We can use conservation of momentum to find the speed of the $5.0~kg$ object after the collision: $m_A~v_A' + m_B~v_B' = m_A~v_A + m_B~v_B$ $m_B~v_B' = m_A~(v_A-v_A') + m_B~v_B$ $v_B' = \frac{m_A~(v_A-v_A') + m_B~v_B}{m_B}$ $v_B' = \frac{(1.0~kg)~[10.0~m/s-(-5.0~m/s)] + (5.0~kg)(0)}{5.0~kg}$ $v_B' = 3.0~m/s$ After the collision, the speed of the 5.0 kg object is $3.0~m/s$

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