Elastic Collision of two blocks

AI Thread Summary
In an elastic collision between two blocks, Block A (2.0 kg) moves at 5.0 m/s to the right, while Block B (8.0 kg) moves at 3.0 m/s to the left. The conservation of momentum equation is established, resulting in -7 = vaf + 4.0vbf. To solve for the final velocities of both blocks, the kinetic energy conservation law must also be applied. The discussion emphasizes the need to incorporate both momentum and kinetic energy conservation principles to find the solution. Understanding these concepts is crucial for determining the final velocities post-collision.
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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