Help elastic head-on collision

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In an elastic head-on collision, both momentum and kinetic energy are conserved. The initial momentum equation is correctly identified as M1v1i + M2v2i = M1v1f + M2v2f, but two unknowns require a second equation. The conservation of kinetic energy must also be applied, leading to the equation (1/2)M1(v1i)^2 + (1/2)M2(v2i)^2 = (1/2)M1(v1f)^2 + (1/2)M2(v2f)^2. By solving these two equations simultaneously, the velocities after the collision and the fraction of kinetic energy transferred can be determined. Understanding these principles is crucial for solving the problem.
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a 3.42g object moving to the right at 16.4 cm/s makes an elastic head-on collision with a 6.84g object that is initially at rest. find the velocity of the first object immediately after collision. Answer in units of cm/s. What fraction of the initial kinetic energy is transferred to the 6.84g object?

I don't understand how to do this problem. I think for the first part is it
M1v1i+M2V2i=M1V1f+M2V2f
and I am lost after that. I completely don't get part 2
 
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Indeed, you have:
P_{\rm{initial}}=P_{\rm{final}}[/itex]<br /> which gives you one equation.<br /> <br /> But, you have two unknowns (v_1 and v_2), so you&#039;ll need another equation. What do you know about elastic collisions?
 
energy and momentum is conserved in elastic collisions?
 
snowbelle said:
energy and momentum is conserved in elastic collisions?

Right. The equation you have listed only deals with momentum.
 
hmm i still don't get it
 
snowbelle said:
hmm i still don't get it

Can you write expressions for the total kinetic energy before and after the collision?
 

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