Mass and Velocity in Elastic Collisions

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In an elastic collision involving a 2.0 kg mass that continues moving at one-fourth its original speed after colliding with a stationary body, the mass of the other body can be calculated using momentum and kinetic energy conservation principles. The initial speed of the 2.0 kg body is given as 4.0 m/s, leading to a derived mass of 2/3 kg for the second body. The calculations involve setting up equations for both momentum and kinetic energy conservation. It is emphasized that both conservation laws must be applied to solve for the unknowns correctly. The discussion concludes with a reminder that the net force relates to the change in momentum over time, which is crucial for understanding the collision dynamics.
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I have not been able to figure out anything about this problem. I hope somebody will help me.

A body of mass 2.0 kg makes an elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its original speed. (A): What is the mass of the other body? (B): What is the speed of the two-body mass center of mass if the initial speed of the 2.0 kg body was 4.0 m/s?
 
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I'm only in High School so don't depend on my word for it!

Here's what I'm thinking...
momentum=m*v
p1=p2
.25v(2kg)=.75v(m)

the other mass is: 2/3 kg

So, v=.25*4*2 = 2m/s

think this is correct?
 
In an elastic collision, both kinetic energy and momentum are conserved. If I denote the initial velocity of the 2kg mass by u and the final velocity of the unknown mass by v then,

\frac{1}{2}(4kg)(u^2) = \frac{1}{2}(4kg)(\frac{u}{4})^2 + \frac{1}{2}(m)(v^2) (Kinetic Energy Conservation)

(4kg)u = 4kg({\frac{u}{4}}) + mv (Linear Momentum Conservation)

You have two equations and two unknowns (m and v). You can solve for them easily now.

UrbanXrisis, you have written only one equation--that for linear momentum conservation. Read the question carefully (note that both energy and linear momentum are conserved in an elastic collision).

Now for the second part, I can only tell you that the hint for solving it lies in this very post coupled with the fact that the net force on a system equals the time rate of change of linear momentum.

Hope that helps...

Cheers
Vivek
 
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Are you supposed to get the answers in terms of u?
 
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