How Do You Calculate Change in Kinetic Energy for an Elastic Collision?

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To calculate the change in kinetic energy for an elastic collision, use the formula KE = 1/2mv^2. In the given scenario, a rubber ball with a mass of 0.015 kg and a velocity of 6.26 m/s has the same speed before and after bouncing off the floor. The change in kinetic energy is determined by subtracting the kinetic energy before the collision from the kinetic energy after, resulting in 0 joules. This outcome is expected because, in elastic collisions, kinetic energy is conserved, meaning there is no change in energy. Therefore, the change in kinetic energy for the rubber ball in this case is zero.
psruler
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Hi!

How do i find change in kinetic energy?

Here is the problem:
The rubber ball is dropped 2m elastically off the floor, so its speed immediately after bouncing is the same as immediately b4.
If the rubber ball has a mass of 0.015kg and its velocity is 6.26m/s, what is its change in kinetic energy from immediately b4 the collision to immediately after the collision? Breifly justify your answer.

Do i use this equation, KE = 1/2mv^2?

THANKS!
 
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Yup. Change in kinetic energy is just the kinetic energy after minus the kinetic energy before.

So KE = 1/2*mv^2, now if m is the same for both and v is the same for both...

cookiemonster
 
So is my answer, 0.29N?
 
Hold on now!

1/2(.015kg)(6.26m/s)^2 - 1/2(.015)(6.26m/s)^2 = ?

Also, what units do you get? Here's a hint: it's not Newtons! It's kinetic energy, so its units will probably be energy.

cookiemonster
 
I got 0 joules.
 
Yup. And this makes sense, too, since this collision is elastic. Elastic collisions are collisions in which energy is conserved, so energy before had better equal energy after (and their difference had better equal zero)!

cookiemonster
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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