A 120kg polar bear cub is sliding on an icy surface toward her mother at 3.0 m/s

Question

A 120kg polar bear cub is sliding on an icy surface toward her mother at 3.0 m/s. Her 268kg mother starts towards her at 2.0 m/s, intending to catch her. What percentage of the original kinetic energy is convertible?
I've seen similar questions, but they use final Kinetic over initial kinetic. And while that's the amount that IS converted, I'm wondering if that's right or if I must use the convertible kinetic energy equation. In my mind the amount that is converted may be less than the amount that is convertible. A 120kg polar bear cub is sliding on an icy surface toward her mother at 3.0 m/s. Her 268kg mother starts towards her at 2.0 m/s, intending to catch her. What percentage of the original kinetic energy is convertible? A 120kg polar bear cub is sliding on an icy surface toward her mother at 3.0 m/s. Her 268kg mother starts towards her at 2.0 m/s, intending to catch her. What percentage of the original kinetic energy is convertible?
I've seen similar questions, but they use final Kinetic over initial kinetic. And while that's the amount that IS converted, I'm wondering if that's right or if I must use the convertible kinetic energy equation. In my mind the amount that is converted may be less than the amount that is convertible.

Explanation / Answer

I am assuming, after the bear catches the cub they move with a common velocity

So, now conserving momentum,

m1*v1 + m2*v2 = (m1 + m2)*v

So, 120*3 + 268*(-2) = (120 + 268)*v

So, v = -0.454 m/s

So, now the total KE of the system = 0.5*(m1 + m2)*v^2

= 0.5*(268 + 120)*0.454^2

= 40 J

Initial KE of the system = 0.5*m1*v1^2 + 0.5*m2*v2^2

= 0.5*120*3^2 + 0.5*268*2^2

= 1076 J

So, percentage of KE that is consertible = (40/1076)*100 = 3.72 percent

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