A 23-kg child is sliding on an icy surface toward her mother at 3.0 m/s. Her 70-kg mother starts toward her at 2.0 m/s, intending to catch her.

Question

What percentage of the original kinetic energy is convertible?

I calculated the convertible kinetic energy as Kconv=1/2*(m1)(m2)/(m1+m2)*(v1+v2)^2 so Kconv=1/2*(17.312 kg)*(1.0m/s)^2 =8.656 J

I calculated the original kinetic energy as Ki=1/2*(m1+m2)*(v1+v2)^2 =1/2*(93 kg)*(1.0 m/s)^2 =46.5 J

Kconv/Ki =8.656/46.5 =18.62%

Why is this answer wrong?

I also tried it assuming the original energy was just the girl alone Ki=1/2(23 kg)(3.0m/s)^2 and this was also wrong.

Answer 1

first off your initial Kinetic energy will be wrong because its 0.5(m1v1^2 + m2v2^2)
also your relative velocity seems iffy to me you wrote (v1+v2)^2

Having trouble with the same problem

Answer 2

Hi again just solved it!

once you fix your kinetic energy initial equation, it is a relative velocity problem. It should come out to be 5.

|v1- -v2| v2 is in the negative direction since the mother is traveling towards the child.