Momentum and Conservation of energy

AI Thread Summary
The discussion focuses on the conservation of momentum and energy in a collision scenario involving three objects, A, B, and C. Key equations derived include momentum relationships and kinetic energy equations, specifically relating velocities before and after collisions. Participants explore how to express velocity in terms of Vb, with one user expressing difficulty in finding a solution. Another user provides helpful equations that clarify the relationships between the variables. The exchange concludes with the realization that the provided equations are sufficient to solve for Vb.
wcjy
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Homework Statement
Objects A, B, and C with masses m, 2m, and 2m respectively, slide on a horizontal frictionless surface and object C is connected with a massless spring with spring constant k. Object A travels at a velocity v to the right and objects B and C are stationary initially. During the collisions between A, B, and C, show that the maximum change in the length of the spring is sqrt (4mv^2/9k). Note that the collision between A and B is elastic.
Relevant Equations
EPE = 0.5kx^2
When A hits B,
COLM
mV = -mVa + 2mVb
V = 2Vb - Va
COKE
0.5mv^2 = 0.5mVa^2 + 0.5(2m)Vb^2
V^2 = Va^2 + 2Vb^2

When B hits C
COLM
2mVb=4mVc
Vc = 0.5Vb
COE
0.5(2m)Vb^2 = 0.5kx^2 +0.5(4m)Vc^2
sub Vc = 0.5b
mVb^2 = KX^2

After that I am stuck, cause i can't find V in terms of Vb only
 

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Hello @wcjy , :welcome:

You have $$ v = 2v_b - v_a \qquad \& \qquad v^2 = v_a^2 + 2v_b^2$$ which should be enough to get ##v_b##.
 
BvU said:
Hello @wcjy , :welcome:

You have $$ v = 2v_b - v_a \qquad \& \qquad v^2 = v_a^2 + 2v_b^2$$ which should be enough to get ##v_b##.

oh yea. i think my brain is fried for not seeing that. great thanks!
 
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