Can Elastic Collisions Be Solved Quickly with Minimal Steps?

[rachelx46]

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

http://img90.imageshack.us/img90/2593/phys2ul7.png

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?

 

[Andrew Mason]

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

http://img90.imageshack.us/img90/2593/phys2ul7.png

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

 

[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]

Because we are given that m₁= m₂?

 

[pmp!]

Oh my bad... I am sorry, I missed that.