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viktelen [127]
2 years ago
13

One car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 5.5 m/s

m/s , they then have the same kinetic energy. Part A What were the original speeds of the two cars?

Mathematics
2 answers:
svp [43]2 years ago
8 0

Step-by-step explanation:

Below are attachments containing the solution.

Oksanka [162]2 years ago
6 0

Answer:

v_1 =3.8891 m/s, v_2 = 7.7782 m/s

Step-by-step explanation:

Consider car 1 with mass m_1 and original speed v_1 and car 2 with mass m_2 and original speed v_2. We can consider both cars as punctual masses. In this case, recall that the kinetic energy of a particle of mass m and speed v is given by the expression \frac{mv^2}{2}. 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.

m_1 = 2 m_2, \frac{m_1 v_1^2}{2} =\frac{1}{2}\frac{m_2 v_2^2}{2}.

Replacing the equation m_1 = 2 m_2 in the second one, leads to 4m_2v_1^2=m_2v_2^2, which implies m_2 (4v_1^2-v_2^2)=0=(2v_1+v_2)(2v_1-v_2)Since m_2>0 and assuming that both speeds are positive,  then v_2 =2v_1.

Given that, if both cars increase their speed by 5.5 m/s then they have the same kinetic energy, we have that

\frac{m_1(v_1+5.5)^2}{2}= \frac{m_2(v_2+5.5)^2}{2}. Using the previous result, and expressing everything in terms of m_2 and v_1 we have that

2m_2(v_1+5.5)^2= m_2(2v_1+5.5)^2 (where m_2 cancells out).

Then, we have the following equation 2(v_1+5.5) ^2 = (2v_1+5.5)^2, which by algebraic calculations leads to v_1 = \pm \frac{5.5}{\sqrt[]{2}} = \pm 3.8891. Since we assumed v_1, we have that v_1 = 3.8891.Then, v_2 = 7.7782

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