What Determines the Outcome in an Elastic Collision Between Two Croquet Balls?

[anightlikethis]

Homework Statement

A 0.180 kg croquet ball makes an elastic head-on collision with a second ball initially at rest. The second ball moves off with half the original speed of the first ball.(a) What is the mass of the second ball?(b) What fraction of the original kinetic energy (KE/KE) gets transferred to the second ball?

Homework Equations

a)1/2Mava^b+1/2Mbvb^2=1/2Mav2a^2+1/2Mbv2b^2

b)mava+mbvb=mav2a+mbv2b

The Attempt at a Solution

I solved for "v2a" using equation b and got mb2.77777va I then plugged this back into equation a and got (.5)(.18)(va)^2=(.5)(.18)((mb)(2.77777)(va))^2+(.5)(mb)((.5)(va))^2. If I've done this correctly I still don't know how to solve it correctly to find the mass of "ball b"

 

[Mentz114]

You can factorize m_b from the right-hand side of your final equation. I haven't checked if it's correct but your method obviously is.

 

[m2003]

and the fraction of the original kinetic energy transferred to "ball b".

I would first double check the equations being used to ensure they are appropriate for this situation. In this case, the equations used seem to be correct for an elastic collision between two objects.

To solve for the mass of the second ball (ball b), I would use the equation for conservation of momentum, which states that the total momentum before a collision is equal to the total momentum after the collision. In this case, we know that the first ball has a mass of 0.180 kg and is moving at some initial velocity (va), while the second ball has an unknown mass (mb) and is moving at half the initial velocity (0.5va). Therefore, we can write the equation as:

0.180 kg * va = mb * 0.5va

Solving for mb, we get mb = 0.360 kg. This is the mass of the second ball.

To find the fraction of the original kinetic energy transferred to the second ball, we can use the equation for conservation of kinetic energy, which states that the total kinetic energy before a collision is equal to the total kinetic energy after the collision. In this case, we know that the initial kinetic energy is 0.5 * 0.180 kg * (va)^2, and the final kinetic energy is 0.5 * 0.360 kg * (0.5va)^2. Therefore, the fraction of the original kinetic energy transferred to the second ball is:

0.5 * 0.360 kg * (0.5va)^2 / (0.5 * 0.180 kg * (va)^2) = 0.5 * 0.25 = 0.125

So, the fraction of the original kinetic energy transferred to the second ball is 12.5%.