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

Explanation:

Below are attachments containing the solution.

Answer 2

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

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
`(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`,
`(m_1 v_1^2)/(2) =(1)/(2)(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

`(m_1(v_1+5.5)^2)/(2)= (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`