Inelastic collision

Question

I'm solving a problem that reads

"Two cars collide with masses $m_1=m_2=140\,$kg with perpendicular velocities $v_1=3.6\,$m/s and $v_2=7.2\,$m/s. After the collision they keep going at a common velocity $v_3.$ Calculate $v_3$.”

Now I used the momentum formula and arrive at $v_3 = \frac{m_1v_1+m_2v_2}{m_1+m_2}$ which gives me $5.4\,$m/s. However the solutions also say the formula above but then in the next step the formula changes to $v_3 = \frac{\sqrt{m_1^2v_1^2+m_2^2v_2^2}}{m_1+m_2}.$ Why did they square the masses and velocity and root the sum? I've also solved these types of problems in the past and the first equation I wrote worked just fine. Does it have to do with the fact the collision is at an angle or is it for another reason?

Answer 1

This is a case where using proper vector notation makes your problem much clearer.

Before the collision, your system's total momentum is

\begin{align} \vec p_0 &= \vec p_1 + \vec p_2 & \text{with}\quad \vec p_1 &= m_1 v_1 \hat x \\ && \text{and}\quad \vec p_2 &= m_2 v_2 \hat y \end{align}

Your inelastic collision doesn't change the total momentum, so the final momentum is also

\begin{align} \vec p_3 &= \vec p_1 + \vec p_2 \\ & = m_1 v_1 \hat x + m_2 v_2 \hat y \\ & = p_x \hat x + p_y \hat y \end{align}

Since the two components are perpendicular, the magnitude of the final component is

$$ \left|\vec p\right| = \sqrt{ (p_x)^2 + (p_y)^2 } $$

You can expand this into your expression for $v_3 = \frac{p}{m_1+m_2}$ if you are confident in your algebra skills.

If you are trying to solve these problems by memorizing "the formulas," you'll be stymied when you run across a situation that is just barely different from what you've encountered before. If you can remember how you get to the formulas you know, then you can follow those rules to come up with your own solutions to novel problems.

Answer 2

Since this is a collision at perpendicular directions, the first formula i think is not applicable here.

The second formula is correct, i think, and here is how i arrived at it:

In the x direction, according to momentum conservation, m2 v2= (m1+m2) v3x. In the y direction, accoridng to momentum conservation, m1 v1= (m1+m2) v3y. This way one can get v3x, v3y.

Then one get v3 using $v3 = \sqrt{v3x^2+v3y^2}$. After plugging in v3x, v3y from above, one get $$v3=\sqrt{\frac{m2^2 v2^2}{(m1+m2)^2}+\frac{m1^2 v1^2}{(m1+m2)^2}}$$ which is just $ \sqrt{m2^2 v2^2+m1^2 v1^2}/(m1+m2)$.

Answer 3

It is true that this is an example of the use of Pythagoras to equate the vector sum of the initial momenta $m\vec v_1$ and $m\vec v_1$ to the final momentum $m\vec v_2$ producing $m\,v_1^2+ m\,v_2^2 = m\,v_3^2$.

However you might have found it strange that

$m_1v_1+m_2v_2 = \sqrt{m_1^2v_1^2+m_2^2v_2^2}$, ie $(m_1v_1+m_2v_2)^2 = m_1^2v_1^2+m_2^2v_2^2$?

The reason is to be found by realising that vectors are involved and a dot product is involved.

$(m_1\vec v_1+m_2\vec v_2)^2 = (m_1\vec v_1)\cdot(m_1\vec v_1)+2(m_1\vec v_1)\cdot(m_2\vec v_2)+(m_2\vec v_2)\cdot(m_2\vec v_2)$.

The middle term has the dot product $v_1\cdot v_2$ which is zero because the initial velocities are orthogonal.