Momentum question asking ratio of final 2 velocities

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SUMMARY

The discussion centers on determining the ratio of final velocities after an elastic collision involving two equal masses, where one mass is initially moving at 3V. The relevant equations used include momentum conservation (m1v1 = m1v1' + m2v2') and kinetic energy conservation (Ek = mv^2/2). The calculations yield a final velocity ratio of 3:0, indicating that one mass comes to rest while the other moves at 3V. This conclusion is validated by the principles of elastic collisions and the conservation laws applied.

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hamza2095
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Homework Statement


A cart mass of M moving at 3V (right) collides with a stationary mass of M, Determine the ratio of the two final velocities if the collision is elastic

Homework Equations


p=mv
ek=(mv^2/2)

The Attempt at a Solution


m1v1=m1v1'+m2v2'

Since the masses are all equal and v1 is 3V the equation becomes
3 = v1' + v2'

Since the collision is elastic the total energy afterwards equals the total energy afterwards

Ek=(mv^2/2)

so Ek1=Ek1'Ek2'
and since every term is over 2,and the masses equal the equation simplifies to
v1^2=v1'^2+v2^2

After rearranging the first equation you get v1' = v2'-3, and then plug that into the energy equation to get
9= 2v2'^2 -6v2-9
0=2v2'^2 -6v2

after applying the quadratic equation you get 3 or 0, and since the velocity has to change v2' is 3, subsequently v1' is 0 so the ratio is 3:0, or 0:3

3:0 doesn't make sense to me for some reason but I can't find any mistake in what I did, is this right?
 
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I can't find a mistake either.
Check it by sliding a coin against a similar coin on a flat table (aim for a central collision)

By the way: if an initial speed of 3V is given, your calculations should be in units of V, e.g.
3 MV = MV1'+ MV2'
 
hamza2095 said:
3:0 doesn't make sense to me
It's quite correct. Look for a video of Newton's cradle.
With equal masses and elastic collision, swapping over the two initial speeds to get the final speeds must be right. It would have the same momentum and the same energy.
 

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