# Anyone heard of this?

so apparently, this problem usually would require, many hours, and quite a few pages to write a proof for this, but today, i was shown a way to solve it in only a few mintues with only a few steps...

an object of mass m collides elastically with a second object of equal mass initially at rest. If the collision is a glancing collision, prove they separate at 90 degrees.

where m1 = mass of object 1, m2 =mass of object 2, v = velocity of m1, x = velocity after collision of object 1, y = velocity of object after collision.

http://img159.imageshack.us/img159/6387/pyhsag3.png [Broken]

http://img90.imageshack.us/img90/2593/phys2ul7.png [Broken]

so then

1/2(m1)v² = 1/2(m1)x² + 1/2(m2)y²
cancel out and simplify
v² = x² + y²

would that be completely valid?

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Andrew Mason
Homework Helper
so apparently, this problem usually would require, many hours, and quite a few pages to write a proof for this, but today, i was shown a way to solve it in only a few mintues with only a few steps...

an object of mass m collides elastically with a second object of equal mass initially at rest. If the collision is a glancing collision, prove they separate at 90 degrees.

where m1 = mass of object 1, m2 =mass of object 2, v = velocity of m1, x = velocity after collision of object 1, y = velocity of object after collision.

http://img159.imageshack.us/img159/6387/pyhsag3.png [Broken]

http://img90.imageshack.us/img90/2593/phys2ul7.png [Broken]

so then

1/2(m1)v² = 1/2(m1)x² + 1/2(m2)y²
cancel out and simplify
v² = x² + y²

would that be completely valid?
Almost. You have to add the condition that $\vec v = \vec x + \vec y$, which is a consequence of conservation of momentum. Since this is a triangle and v² = x² + y², as you have shown, then the angle between $\vec x \text{ and } \vec y$ must be 90 degrees.

This is a very useful thing to know for pool players, who use this principle to determine where the cue ball will go.

AM

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pmp!
Almost. You have to add the condition that $\vec v = \vec x + \vec y$, which is a consequence of conservation of momentum.
AM
Consrvation of momentum states that:
$$m_1\vec v=m_1\vec x + m_2\vec y$$

So why is that your condition is true?

HallsofIvy