Forest UCM MnAM ElasticCol: Difference between revisions
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:<math>\vec{v}_1 \cdot \vec{v}_1 = \left ( \vec{v}_1^{\;\prime} + \vec{v}_2^{\;\prime}\right ) \cdot \left ( \vec{v}_1^{\;\prime} + \vec{v}_2^{\;\prime}\right )</math> | :<math>\vec{v}_1 \cdot \vec{v}_1 = \left ( \vec{v}_1^{\;\prime} + \vec{v}_2^{\;\prime}\right ) \cdot \left ( \vec{v}_1^{\;\prime} + \vec{v}_2^{\;\prime}\right )</math> | ||
:<math>v_1^2 = \left(v_1^{\prime}\right)^2 + \left(v_2^{\prime}\right)^2 + 2 \vec{v}_1^{\;\prime} \cdot \vec{v}_2^{\;\prime}</math> | :<math>v_1^2 = \left(v_1^{\prime}\right)^2 + \left(v_2^{\prime}\right)^2 + 2 \vec{v}_1^{\;\prime} \cdot \vec{v}_2^{\;\prime}</math> | ||
In order for both the conservation of momentum and energy properties for eleastic collisions to hold | |||
:<math>\vec{v}_1^{\;\prime} \cdot \vec{v}_2^{\;\prime} = 0 </math> | |||
or the balls make an angle of 90 degrees with repect to eachother | |||
[[Forest_UCM_MnAM#Elastic_Collision_of_2_bodies]] | [[Forest_UCM_MnAM#Elastic_Collision_of_2_bodies]] | ||
Revision as of 12:20, 12 September 2014
An Elastic collision conserves both Momentum and Energy
- [math]\displaystyle{ \vec{P}_{\mbox{initial}} = \vec{P}_{\mbox{final}} }[/math]
- AND
- [math]\displaystyle{ E_{\mbox{initial}} = E_{\mbox{final}} }[/math]
Example: problem 3.5
Consider an elastic collision of two equal balls of mass [math]\displaystyle{ m }[/math] where one ball has an initial velocity [math]\displaystyle{ \vec{v}_1 }[/math] and the remaining ball has zero initial velocity.
Determine the angle between the two balls after the collision.
- Conservation of momentum
- [math]\displaystyle{ m_1 \vec{v}_1 + m_2 \vec{v}_2= m_1 \vec{v}_1^{\;\prime} +m_2 \vec{v}_2^{\;\prime} }[/math]
- [math]\displaystyle{ m_1 \vec{v}_1 = m_1 \vec{v}_1^{\;\prime} +m_2 \vec{v}_2^{\;\prime} }[/math]: ball 2 has zero velocity
- [math]\displaystyle{ \vec{v}_1 = \vec{v}_1^{\;\prime} + \vec{v}_2^{\;\prime} }[/math]: balls have equal masses
- Conservation of energy
- [math]\displaystyle{ m_1 v_1^2 + m_2 v_2^2 = m_1 \left(v_1^{\prime}\right)^2 + m_2\left(v_2^{\prime}\right)^2 }[/math]
- [math]\displaystyle{ v_1^2 = \left(v_1^{\prime}\right)^2 + \left(v_2^{\prime}\right)^2 }[/math]
If I look at the dot product of the conservation of momentum equation
- [math]\displaystyle{ \vec{v}_1 = \vec{v}_1^{\;\prime} + \vec{v}_2^{\;\prime} }[/math]: balls have equal masses
- [math]\displaystyle{ \vec{v}_1 \cdot \vec{v}_1 = \left ( \vec{v}_1^{\;\prime} + \vec{v}_2^{\;\prime}\right ) \cdot \left ( \vec{v}_1^{\;\prime} + \vec{v}_2^{\;\prime}\right ) }[/math]
- [math]\displaystyle{ v_1^2 = \left(v_1^{\prime}\right)^2 + \left(v_2^{\prime}\right)^2 + 2 \vec{v}_1^{\;\prime} \cdot \vec{v}_2^{\;\prime} }[/math]
In order for both the conservation of momentum and energy properties for eleastic collisions to hold
- [math]\displaystyle{ \vec{v}_1^{\;\prime} \cdot \vec{v}_2^{\;\prime} = 0 }[/math]
or the balls make an angle of 90 degrees with repect to eachother