Question: Car #1 has four times the mass of car #2, but they both have the same kinetic energy. If the speed of car #2 is $v$, what is the speed of car #1? Express answer as decimal number $x v$. Type in only the number in box below. Round to nearest tenth.
1. **State the problem:**
We have two cars, car #1 and car #2. Car #1 has four times the mass of car #2, but both have the same kinetic energy. The speed of car #2 is $v$. We need to find the speed of car #1 expressed as a decimal multiple of $v$.
2. **Recall the kinetic energy formula:**
$$KE = \frac{1}{2} m v^2$$
where $m$ is mass and $v$ is speed.
3. **Set up the equation for equal kinetic energies:**
Let $m_2$ be the mass of car #2 and $v_2 = v$ its speed.
Then car #1 has mass $m_1 = 4 m_2$ and speed $v_1$ (unknown).
Since kinetic energies are equal:
$$\frac{1}{2} m_1 v_1^2 = \frac{1}{2} m_2 v_2^2$$
4. **Simplify the equation:**
Cancel $\frac{1}{2}$ on both sides:
$$m_1 v_1^2 = m_2 v_2^2$$
Substitute $m_1 = 4 m_2$ and $v_2 = v$:
$$4 m_2 v_1^2 = m_2 v^2$$
5. **Cancel $m_2$ from both sides:**
$$\cancel{4 m_2} v_1^2 = \cancel{m_2} v^2$$
which simplifies to:
$$4 v_1^2 = v^2$$
6. **Solve for $v_1^2$:**
$$v_1^2 = \frac{v^2}{4}$$
7. **Take the square root:**
$$v_1 = \sqrt{\frac{v^2}{4}} = \frac{v}{2}$$
8. **Express $v_1$ as a decimal multiple of $v$:**
$$v_1 = 0.5 v$$
**Final answer:**
The speed of car #1 is $0.5 v$.