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A 6.00-kg object is at rest on a frictionless surface when it is struck head-on by a 2.00-kg object moving at 10.4 m/s. If the collision is elastic, what is the speed of the 6.00-kg object after the collision? ______________m/s

Answer :

Final answer:

In an elastic collision, the total momentum and kinetic energy are conserved. Using the conservation of momentum equation, we calculate the speed of the 6.00-kg object after collision to be 2.60 m/s.

Explanation:

The physics concept we're dealing with here is an elastic collision. This is a collision where both kinetic energy and momentum are conserved. The formula used for this type of calculation is m1v1 + m2v2 = m1v'1 + m2v'2. Here, m1=6.00kg (1st object mass), m2 =2.00kg (2nd object mass), v1 is the velocity of the 1st object before the collision (which is 0 in this case because it's at rest), v2 is the speed of the 2nd object before collision, v'1 is the velocity of the 1st object after collision, and v'2 is the velocity of the 2nd object after collision.

So, here we have:

6.00kg * 0 + 2.00kg * 10.4 m/s = 6.00kg*v'1 + 2.00kg*(-10.4 m/s)

Solving the above equation for v'1, we get v'1 = 2.60m/s.

Therefore, the speed of the 6.00-kg object after an elastic collision is 2.60 m/s.

Learn more about Elastic Collision here:

https://brainly.com/question/33268757

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