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A 200-gram object moves in the $+x$ direction ar 4m/s and collides with an identical object moving in the $-y$ direction at 3m/s. If the two objects stick together after the collision, what is the magnitude of their resulting velocity?

This is an example of an inelastic collision.

I was initially tempted to solve this question by finding the resultant velocity of the initial velocities. Not, knowing why I couldn't do that I calculated an incorrect velocity.

I then solved the question by finding the resultant momentum vector. I was glad, but then noticed that many people use the equation $m_1v_1 + m_2v_2 = (m_1 + m_2)v_3$ to solve similar problems. However, this equation did not work for me. Using that equation I get $v_3 =0.5 \mathrm{m}/\mathrm{s}$.

Then I watched videos of people finding the final velocity of a combined object by using the initial velocities (that are in different directions) of the initially separate objects using $m_1v_{xi} + m_1v_{yi} = m_1v_{xf}$ (and also in the $y$-direction). This makes the most sense to me because it is systematic and doesn't leave out any assumptions. However, I still don't know why I can not simply use the two object's respective initial velocities to find the resulting velocity?

The closest I get to convincing myself that I can not just find the resultant velocity using the two initial velocities, is that these velocities are not of massless objects and so I have an intuition that mass and velocity can not be separated.

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  • $\begingroup$ Your intuition is correct: momentum (mv) is conserved, not velocity. Note that velocity (v) is a vector, and your problem combines velocity in two orthogonal directions. This is why you need two equations (in separate directions) to solve for final speed in each direction. This is equivalent to adding momentum vectors. $\endgroup$
    – jpf
    Mar 24, 2021 at 12:02
  • $\begingroup$ Hello! I have edited your question using MathJax (LaTeX) math typesetting. For future questions, you can refer to MathJax basic tutorial and quick reference. Thanks! $\endgroup$
    – jng224
    Mar 24, 2021 at 12:16
  • $\begingroup$ You may be right, but isn't velocity conserved too, if momentum is a product of mass and velocity. $\endgroup$
    – Haley
    Mar 26, 2021 at 10:00

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Momentum conservation is a vector equation, not a scalar equation; it should be $$ m_1 \vec{v}_1 + m_2\vec{v}_2 = (m_1 + m_2)\vec{v}_3. $$ In some problems, where the motion is all along a given line, you can ignore this property. But if the motion is inherently 2D (i.e., it involves both $x$ and $y$ motion), then you can't.

To solve this, you need to decompose this equation into its $x$- and $y$-components and solve for the components of $\vec{v}_3$ individually. For example, the $x$-component equation would be $$ m_1 v_{1x} + m_2v_{2x} = (m_1 + m_2)v_{3x} $$ and similarly for the $y$-components. (Note that this is different from the equation you got out of the video.)

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  • $\begingroup$ I'm with you, mostly. But, my primary question (that I think exposes a fundamental error of mine) is "Why can I not use the two velocities to find the resultant final velocity?" Why can I not connect the velocity vectors, tip-to-tail, and apply Pythagoras' theorem to find the final velocity (after collision)? You pointed out that "Momentum conservation is a vector equation, not a scalar equation..." but so is Pythagoras' theorem when I use velocity (a vector). $\endgroup$
    – Haley
    Mar 25, 2021 at 10:02

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