University Physics I - Classical Mechanics, 2019

4.4. EXAMPLES 85
K cm ′ = 1 2 (m 1 + m 2 )(v cm ′ )2 = 1 (
2 (170 kg) 0.353 m s − 1.5 m ) 2
= 111.8 J (4.26)
s
K conv,i ′ = K′ sys,i − K′ cm = 641.3J− 111.8 J = 529.5J ≃ 529 J (4.27)
This shows explicitly that the convertible energy, as I pointed out earlier in this chapter, is the same
in every reference frame! (The equality is exact, if you keep enough decimals in the calculation.)
Knowing this, we can simplify the calculation of the final kinetic energy, after the explosive separation:
the convertible energy, K
conv,f ′ , will be the same as in the earth reference frame, that is to
say, 21.2 J, and the total kinetic energy will be K
sys,f ′ = K′ cm + K
conv,f ′ = 111.8J +21.2 J = 133 J.
So, in this frame of reference, we have (to three significant figures):
K sys,i ′ = 641 J, K cm ′ = 112 J, K conv,i ′ = 529 J (before the collision)
K sys ′ = K′ cm = 112 J, K′ conv = 0 (right after the collision)
K sys,f ′ = 133 J, K′ cm = 112 J, K′ conv,i =21.2 J (after the separation)
So, even though the total kinetic energy is different in the two reference frames, all the (inertial)
observers will agree as to the amount of kinetic energy “lost” in the collision, as well as the amount
of kinetic energy put back into the system by the players’ pushing on each other.