Math

QuestionA car with mass mc=1270 kgm_{\mathrm{c}}=1270 \mathrm{~kg} is traveling west through an intersection with a speed of vc=12.86 m/sv_{\mathrm{c}}=12.86 \mathrm{~m} / \mathrm{s} when a truck of mass mt=1990 kgm_{\mathrm{t}}=1990 \mathrm{~kg} traveling south at vt=15,58 m/sv_{\mathrm{t}}=15,58 \mathrm{~m} / \mathrm{s} fails to yield. The vehicles collide, they stick together, and they slide on the asphalt which has a coefficient of kinetic friction of μk=0.500\mu_{k}=0.500. Associate the positive xx direction with east and the positive yy direction with north.
Part (a) Using the symbols provided in the palette below as opposed to their numeric equivalents, enter an expression, in Cartesian unit-vector notation, for the velocity of the vehicles immediately after the collision, as they begin to slide as a composite object. v=\vec{v}= \square
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Part (b) How far, in meters, will the vehicles slide after the collision?

Studdy Solution
Part (a): v=mcvcmc+mtı^mtvtmc+mtȷ^ \vec{v} = \frac{-m_{\mathrm{c}} v_{\mathrm{c}}}{m_{\mathrm{c}} + m_{\mathrm{t}}} \hat{\imath} - \frac{m_{\mathrm{t}} v_{\mathrm{t}}}{m_{\mathrm{c}} + m_{\mathrm{t}}} \hat{\jmath}
Part (b): d=v022μkg d = \frac{v_{0}^2}{2 \mu_{k} g} , where v0=(mcvcmc+mt)2+(mtvtmc+mt)2 v_{0} = \sqrt{\left(\frac{-m_{\mathrm{c}} v_{\mathrm{c}}}{m_{\mathrm{c}} + m_{\mathrm{t}}}\right)^2 + \left(\frac{-m_{\mathrm{t}} v_{\mathrm{t}}}{m_{\mathrm{c}} + m_{\mathrm{t}}}\right)^2}

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