Momentum and Conservation of energy

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SUMMARY

The discussion focuses on the principles of momentum and conservation of energy in a collision scenario involving three objects A, B, and C. Key equations derived include the conservation of linear momentum (COLM) and conservation of energy (COE) formulas, specifically mV = -mVa + 2mVb and 0.5mv^2 = 0.5mVa^2 + 0.5(2m)Vb^2. The participants clarify the relationships between the velocities of the objects, particularly V in terms of Vb, leading to the conclusion that V = 2Vb - Va and V^2 = Va^2 + 2Vb^2. This exchange highlights the importance of correctly applying these fundamental physics concepts to solve for unknown variables.

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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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