Find Magnitude of velocity in a collision?

Question

A baseball and a ball of clay are approaching each other. The mass of the baseball is $145 \ \mathrm{ g }$ and it is moving due west $5 \ \left.\mathrm{ m }\middle/\mathrm{ s }\right.$ at $180^\circ$. The ball of clay is $290 \ \mathrm{ g }$ and it is moving northwest at $135^\circ$ $4 \ \left.\mathrm{ m }\middle/\mathrm{ s }\right.$. What is the magnitude of their combined velocity after they collide and stick together?

I first tried to find velocity in the $x$-direction, then velocity in the $y$-direction, and use the Pythagorean theorem to find total magnitude. I got $3.14 \ \left.\mathrm{ m }\middle/\mathrm{ s }\right.$. Why is this wrong? What should I be doing instead?

I'm pretty sure it has to do with the angle, but what? The only practice I have done was where one object moved straight up.

Answer 1

Use the law of conservation of momentum. In particular, the combined velocity $v$ can be found using $$145 (-5 \hat{i}) + 290 (-2\sqrt{2} \hat{i} + 2\sqrt{2} \hat{j}) = (145+290) v.$$

Answer 2

The momentum $\mathbf{p}$ of an object is given by $m \mathbf{v}$, where $m$ is the mass of that object, and $\mathbf{v}$ is the velocity of that object. Based on your description, the momentum of the baseball is $$ \mathbf{p}_1 = 145 \langle -5, 0 \rangle $$ and the momentum of the clay is $$ \mathbf{p}_2 = 290 \left\langle -4\frac{\sqrt{2}}{2}, 4\frac{\sqrt{2}}{2} \right\rangle = 580 \langle -\sqrt{2}, \sqrt{2} \rangle $$ (where I have rather lazily elided the units).

Momentum and mass are conserved in the collision of the two objects, thus the resulting momentum is $$ \mathbf{p} = \mathbf{p}_1 + \mathbf{p}_2 = 145 \langle -5, 0 \rangle + 580 \langle -\sqrt{2}, \sqrt{2} \rangle = \langle -725-580 \sqrt{2}, 580\sqrt{2} \rangle. $$ Since mass is conserved, it follows that the resulting velocity is $$ \mathbf{v} = \frac{\mathbf{p}}{m} = \frac{\langle -725-580 \sqrt{2}, 580\sqrt{2} \rangle}{145+290} \approx \langle 3.55, 1.89 \rangle.$$ Note that this is a bit north of due west, which seems reasonable. A little bit of trigonometry could give a more exact heading if we really need it. From this, it follows that the magnitude of the velocity is $$\|\mathbf{v}\| = \frac{\mathbf{p}}{m} = \frac{\sqrt{ (725+580 \sqrt{2})^2 + (580\sqrt{2})^2 }}{145+290} \approx 4.02\ \frac{\text{m}}{\text{s}}.$$