Answer:
[tex]v_1 =3.8891 m/s, v_2 = 7.7782 m/s[/tex]
Step-by-step explanation:
Consider car 1 with mass [tex]m_1[/tex] and original speed [tex]v_1[/tex] and car 2 with mass [tex]m_2[/tex] and original speed [tex]v_2[/tex]. We can consider both cars as punctual masses. In this case, recall that the kinetic energy of a particle of mass [tex]m[/tex] and speed [tex]v[/tex] is given by the expression [tex]\frac{mv^2}{2}[/tex]. Then, since car 1 has twice the mass of a second car, but only half as much kinetic energy based on the description, we have the following equations.
[tex]m_1 = 2 m_2 [/tex], [tex]\frac{m_1 v_1^2}{2} =\frac{1}{2}\frac{m_2 v_2^2}{2} [/tex].
Replacing the equation [tex]m_1 = 2 m_2 [/tex] in the second one, leads to [tex]4m_2v_1^2=m_2v_2^2[/tex], which implies [tex]m_2 (4v_1^2-v_2^2)=0=(2v_1+v_2)(2v_1-v_2)[/tex]Since [tex]m_2>0[/tex] and assuming that both speeds are positive, then [tex]v_2 =2v_1[/tex].
Given that, if both cars increase their speed by 5.5 m/s then they have the same kinetic energy, we have that
[tex]\frac{m_1(v_1+5.5)^2}{2}= \frac{m_2(v_2+5.5)^2}{2}[/tex]. Using the previous result, and expressing everything in terms of [tex]m_2[/tex] and [tex]v_1[/tex] we have that
[tex]2m_2(v_1+5.5)^2= m_2(2v_1+5.5)^2[/tex] (where [tex]m_2[/tex] cancells out).
Then, we have the following equation [tex]2(v_1+5.5) ^2 = (2v_1+5.5)^2[/tex], which by algebraic calculations leads to [tex]v_1 = \pm \frac{5.5}{\sqrt[]{2}} = \pm 3.8891[/tex]. Since we assumed [tex]v_1[/tex], we have that [tex]v_1 = 3.8891[/tex].Then, [tex]v_2 = 7.7782 [/tex]
Step-by-step explanation:
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