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Problem 96 Easy Difficulty

A proton moving with speed $v_{A 1}$ in the $+x$ -direction makes an elastic, off-center collision with an identical proton originally at rest. After impact, the first proton moves with speed $v_{A 2}$ in the first
quadrant at an angle $\alpha$ with the $x$ -axis, and the second moves with speed $v_{B 2}$ in the fourth quadrant at an angle $\beta$ with the $x$ -axis $(\text { Fig. } 8.13)$ . (a) Write the equations expressing conservation of linear momentum in the $x$ - and $y$ -directions. (b) Square the equations
from part $(a)$ and add them. (c) Now introduce the fact that the collision is elastic. (d) Prove that $\alpha+\beta=\pi / 2 .$ (You have shown that this equation is obeyed in any elastic, off-center collision between objects of equal mass when one object is initially at rest.)

Answer

$\alpha+\beta=90^{\circ}=\frac{\pi}{2} \mathrm{rad}$

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