Finding kinetic energy retained by an object after an elastic head-on collision

[JuniorJumper]

Hi, I am new to the forums & I have a question about the following problem:

A 0.10 kg object moving initially with a velocity of 0.20 m/s makes an elastic head-on collision with a 0.15 kg object initially at rest. What percentage of the original kinetic energy is retained by the 0.10 kg object?

m1=0.10 kg
vi1=0.20 m/s
v1f=-0.0025 m/s

m2=0.15 kg
vi2=0.00 m/s
v2f=0.198 m/s

I calculated the final velocities of both objects after the collision, but can not decide how to find the kinetic energy retained by m1. I tried finding the kinetic energy lost by m1 (0.002) but then I didnt know where to go from there. I know this must be a simple problem, but I am just not seeing it? Thanks for any guidance :)

 

[physics_31415]

You must use conservation of energy AND conservation of linear momemtum. Once you have calculated the final velocities correctly (your's appear to be incorrect ) then you can calculate their respective kinetic energies (1/2*mv^2). You can then compare each final kinetic energy to the initial system energy (K1/Kinitial and K2/Kinitial) to obtain your percentages (note that they should add to 100%). Also, since energy is conserved the sum of the final kinetic energies will equal the initial kinetic energy (of the incoming object). Hope this helps! Good luck!

 

[thomate1]

I would first like to clarify that the term "elastic head-on collision" refers to a type of collision where both objects involved bounce off each other without any loss of kinetic energy. This is in contrast to an inelastic collision where some of the kinetic energy is lost in the form of heat or sound.

Now, to find the kinetic energy retained by the 0.10 kg object, we can use the formula for kinetic energy: KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity.

In this case, the initial kinetic energy of the 0.10 kg object is KE1 = 1/2 * 0.10 kg * (0.20 m/s)^2 = 0.002 J.

After the collision, the final velocity of the 0.10 kg object is v1f = -0.0025 m/s. Plugging this value into the kinetic energy formula, we get KE1f = 1/2 * 0.10 kg * (-0.0025 m/s)^2 = 0.00000125 J.

Therefore, the kinetic energy retained by the 0.10 kg object is the difference between its initial and final kinetic energies: KE retained = KE1 - KE1f = 0.002 J - 0.00000125 J = 0.00199875 J.

To find the percentage of the original kinetic energy retained, we can use the formula: % retained = (KE retained / KE1) * 100%.

Substituting the values, we get % retained = (0.00199875 J / 0.002 J) * 100% = 99.9375%.

In conclusion, after the elastic head-on collision, the 0.10 kg object retains 99.9375% of its original kinetic energy. I hope this helps clarify the problem for you.