# Chapter 15 - Kinetics of a Particle: Impulse and Momentum - Section 15.4 - Impact - Problems - Page 274: 61

$x_{max}=0.839m$

#### Work Step by Step

We can determine the required maximum distance as follows: We know that: $\frac{1}{2}m_Av_{A_\circ}^2-Fs=\frac{1}{2}m_Av_{A_1}^2$ We plug in the known values to obtain: $\frac{1}{2}(15)(10)^2_(0.3)(10)(9.81)(4)=\frac{1}{2}v_{A_2}^2$ This simplifies to: $v_{A_1}=8.744m/s\leftarrow$ According to the conservation of linear momentum $m_Av_{A_1}+m_Bv_{B_1}=m_Av_{A_2}+m_Bv_{B_2}$ We plug in the known values to obtain: $(15)(8.744)+(10)(0)=15v_{A_2}+10v_{B_2}$ This simplifies to: $3v_{A_2}+2v_{B_2}=26.232~~~$[eq(1)] We know that: $e=\frac{v_{B_2}-v_{A_2}}{v_{A_1}-v_{B_1}}$ $\implies 0.6=\frac{v_{B_2}-v_{A_2}}{8.744-0}$ $\implies 5.246=v_{B_2}-v_{A_2}~~~$[eq(2)] After solving eq(1) and eq(2), we obtain: $v_{A_2}=3.148m/s\leftarrow$ and $v_{B_2}=8.394m/s\leftarrow$ Now we apply the conservation of energy for block $B$ $\frac{1}{2}m_Bv_B^2+\frac{1}{2}ks_1^2=\frac{1}{2}m_Bv^2_{{B^{\prime}_2}}+\frac{1}{2}kx_{max}^2$ We plug in the known values to obtain: $\frac{1}{2}(10)(8.394)^2+\frac{1}{2}(1000)(0)^2=\frac{1}{2}(10)(0)^2+\frac{1}{2}(1000)x_{max}^2$ This simplifies to: $x_{max}=0.839m$

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