Answer: 13.7 m/s
Explanation:
[tex]$Mass of $T r u c k$ \ \left(m_{T}\right)=3520\mathrm{kg}[/tex]
[tex]V_{T_{y}}=18.5 \mathrm{~m} / \mathrm{s} \\&V_{\text {Tix }}=0 \mathrm{~m} / \mathrm{s}[/tex]
[tex]\mathrm{Mass \ of \ car \ (m_c) = 1480 \ kg}[/tex]
[tex]V_{c i x}=\text { ? ; } V_{c i y}=0 \mathrm{~m} / \mathrm{s}[/tex]
[tex]Final \ velocity $\left(V_{f}\right)=13.6 \mathrm{~m} / \mathrm{s} \ \ \theta=72.6$[/tex]
[tex]v_{f x}=v_{f} \cos \theta=(13.6) \cos 72.6=4.067 \mathrm{~m} / \mathrm{s}[/tex]
[tex]\mathrm {Using \ the \ conservation \ of \ momentum \ along\ the \ $x$-axis}[/tex]
[tex]\begin{aligned}m_{T} v_{T i x}+m_{c} v_{c_{i x}} &=\left(m_{T}+m_{c}\right) v_{f x} \\0+(1480) V_{c_{i x}} &=(3520+1480)(4.067) \\(1480) v_{c i x} &=20335 \\V_{c i x} &=13.7 \mathrm{~m} / \mathrm{sec}\end{aligned}[/tex]
