
#1
Nov612, 06:11 AM

P: 147

Here's an interesting puzzle:
The whole scenario takes place in one dimension of space. Ball B is at rest. Ball A moves with momentum [tex]p_a[/tex] Ball A makes a perfectly elastic headon collision with Ball B. Ball B moves off with momentum [tex]p_b[/tex] Prove that [tex]\frac{p_b}{p_a}<2[/tex] 



#2
Nov612, 06:58 AM

P: 3,546

[tex]{p_a} = {p_a}' + {p_b}[/tex] [tex]1 \frac{{p_a}'}{p_a} = \frac{p_b}{p_a}[/tex] Energy conservation (Ball B had intially no KE, so speed of ball A cannot increase): [tex]\frac{{p_a}'}{p_a} < 1[/tex] 



#3
Nov712, 12:04 AM

Homework
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Thanks ∞
P: 9,179

Using energy conservation more completely: [tex]\frac{{p_a}'^2}{m_a}+\frac{{p_b}^2}{m_b}=\frac{{p_a}^2}{m_a}[/tex] Combining with momentum eqn. we get [tex]\frac{p_b}{p_a}=\frac{2m_b}{m_a+m_b}<2[/tex] 



#4
Nov712, 08:09 AM

P: 3,546

Transfer of momentum problem?[tex]1 \frac{{p_a}'}{p_a} = \frac{p_b}{p_a}[/tex] [tex]\frac{{p_a}'}{p_a} < 1[/tex] This leads to: [tex]0 < \frac{p_b}{p_a} < 2[/tex] 


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