Question

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

Answer 1

Step-by-step explanation:

Below are attachments containing the solution.

Answer 2

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, $v_2 = 7.7782$