Applying Conservation of Momentum to Car Collision

[David112234]

Homework Statement

Homework Equations

conservation of momentum

3. The Attempt at a Solution

initially car_1 has not y momentum so

\begin{equation}
m_1v_1x=(m_1+m_2)v cos(θ)
\end{equation}

and car_2 has no x momentum
\begin{equation}
m_2v_2y=(m_1+m_2)v sin(θ)
\end{equation}

using algebra
\begin{equation}
v cos(θ) = v_x = m_1v_1 /(m_1+m_2)
\end{equation}
\begin{equation}
v sin(θ) = v_y = m_2v_2 /(m_1+m_2)
\end{equation}

so v should equal the vector addition of its components v_x and v_y
or Patagonian theorem, which was my original answer, why is it wrong?

 

[haruspex]

Homework Statement

Homework Equations

conservation of momentum

3. The Attempt at a Solution

initially car_1 has not y momentum so

\begin{equation}
m_1v_1x=(m_1+m_2)v cos(θ)
\end{equation}

and car_2 has no x momentum
\begin{equation}
m_2v_2y=(m_1+m_2)v sin(θ)
\end{equation}

using algebra
\begin{equation}
v cos(θ) = v_x = m_1v_1 /(m_1+m_2)
\end{equation}
\begin{equation}
v sin(θ) = v_y = m_2v_2 /(m_1+m_2)
\end{equation}

so v should equal the vector addition of its components v_x and v_y
or Patagonian theorem, which was my original answer, why is it wrong?

I believe your answer is correct, but it could be written a little more simply. The denominator can be taken outside the square root. Is it possible the software does not recognise the equivalence?

 

[David112234]

I believe your answer is correct, but it could be written a little more simply. The denominator can be taken outside the square root. Is it possible the software does not recognize the equivalence?

got it, thank you!