## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson

# Chapter 11 - Impulse and Momentum - Exercises and Problems: 64

#### Answer

After the collision, the 100-g ball is moving to the left at a speed of 0.8 m/s After the collision, the 400-g ball is moving to the right at a speed of 2.2 m/s

#### Work Step by Step

Let $m_A = 0.10~kg$ and let $m_B = 0.40~kg$ Let $v_A$ be the initial velocity of the 100-g ball. Let $v_B$ be the initial velocity of the 400-g ball. Let $v_A'$ be the final velocity of the 100-g ball. Let $v_B'$ be the final velocity of the 400-g ball. We first use conservation of momentum to set up an equation. $m_A~v_A + m_B~v_B = m_A~v_A' + m_B~v_B'$ Since the collision is elastic, we then set up another equation. $v_A - v_B = v_B' - v_A'$ $v_A' = v_B' - v_A + v_B$ Next, we use this expression for $v_A'$ in the first equation. $m_A~v_A + m_B~v_B = m_A~v_B' - m_Av_A + m_Av_B+ m_B~v_B'$ $v_B' = \frac{2m_A~v_A+m_B~v_B -m_Av_B}{m_A+m_B}$ $v_Bâ€™ = \frac{(2)(0.10~kg)(4.0~m/s)+(0.40~kg)(1.0~m/s)- (0.10~kg)(1.0~m/s)}{(0.10~kg)+(0.40~kg)}$ $v_B' = 2.2~m/s$ We can use $v_B'$ to find $v_A'$. $v_A' = v_B' - v_A + v_B$ $v_A' = 2.2~m/s - 4.0~m/s + 1.0~m/s$ $v_A' = -0.8~m/s$ After the collision, the 100-g ball is moving to the left at a speed of 0.8 m/s. After the collision, the 400-g ball is moving to the right at a speed of 2.2 m/s.

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