A head-on Collision (Example #7) page 205 Cutnell & Jhonson Physics 8e

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SUMMARY

The discussion focuses on an elastic collision problem involving two balls with masses m1 = 0.250 kg and m2 = 0.800 kg. The initial velocity of the first ball is 5.00 m/s, while the second ball is at rest. The kinetic energy formula, KE = 1/2 mv^2, is utilized to analyze the collision, leading to the conclusion that the total kinetic energy before the collision (3.125 J) must equal the total kinetic energy after the collision. The confusion regarding the 1/2 factor and the squared term in the formula is clarified as essential components of kinetic energy calculations.

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gcombina
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One ball has a mass of m1 = 0.250 kg and an initial velocity of 5.00 m/s. The other has a mass of m2=0.800 kg and is initially at rest. No external forces act on the balls. What are the velocities of the balls after the collision? Elastic Collision.
On the book I have this formula.

STEP 1#

1/2 m1vf1^2 + 1/2 m2Vf2^2 = 1/2 m1V01^2 + 0

(I am confused for the 1/2 at the beginning and why is squares?)

WHAT?
 
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gcombina said:
1/2 m1vf1^2 + 1/2 m2Vf2^2 = 1/2 m1V01^2 + 0
(I am confused for the 1/2 at the beginning and why is squares?)
That is the formula of KE and states that the sum of the energy of the balls will be equal to 1/2 0.250*5^2 = 3.125J
 

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