find v of the greater piece by conserving momentum:
55*19 = 55/3 * -19 + 110/3 v
actually, because each term has a factor of 55, we can use
19 = -19/3 + 2v/3
Now, having all the speeds, figure the new KE = 1/2 mv^2 values
An object, with mass 55 kg and speed 19 m/s relative to an observer, explodes into two pieces, one 2 times as massive as the other; the explosion takes place in deep space. The less massive piece stops relative to the observer. How much kinetic energy is added to the system during the explosion, as measured in the observer's reference frame?
2 answers
dang. The smaller piece has v=0. Make the adjustment.