Elastic Collision of two blocks

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SUMMARY

The discussion focuses on an elastic collision problem involving two blocks: Block A with a mass of 2.0 kg moving at 5.0 m/s and Block B with a mass of 8.0 kg moving at -3.0 m/s. The conservation of momentum equation is established as (2.0)(5.0) + (8.0)(-3.0) = (2.0)vaf + (8.0)vbf, leading to -14 = (2.0)vaf + (8.0)vbf. To solve for the final velocities (vaf and vbf), the kinetic energy conservation law must also be applied, as indicated by the user "marco".

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chiurox
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Homework Statement



Block A of mass, mA = 2.0 kg, is moving on a frictionless surface with a velocity of 5.0 m/s to the right and another block B of mass, mB = 8.0 kg, is moving with a velocity of 3.0 m/s to the left, as shown in the diagram below. The two block eventually collide.

a. If the collision is elastic, what are the final velocities of the two blocks?

The Attempt at a Solution



(2.0)(5.0) + (8.0)(-3.0) = (2.0)vaf + (8.0)vbf
10 – 24 = (2.0)vaf + (8.0)vbf
-14 = (2.0)vaf + (8.0)vbf divide the equation by 2
-7 = vaf + 4.0vbf //equation 1
So, I don't know now how to solve for these two final variables.
 
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chiurox said:

Homework Statement



Block A of mass, mA = 2.0 kg, is moving on a frictionless surface with a velocity of 5.0 m/s to the right and another block B of mass, mB = 8.0 kg, is moving with a velocity of 3.0 m/s to the left, as shown in the diagram below. The two block eventually collide.

a. If the collision is elastic, what are the final velocities of the two blocks?

The Attempt at a Solution



(2.0)(5.0) + (8.0)(-3.0) = (2.0)vaf + (8.0)vbf
10 – 24 = (2.0)vaf + (8.0)vbf
-14 = (2.0)vaf + (8.0)vbf divide the equation by 2
-7 = vaf + 4.0vbf //equation 1
So, I don't know now how to solve for these two final variables.

you need to put also the kinetic energy conservation law.

bye marco
 

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