1. The problem statement, all variables and given/known data A 9.0 kg artillery shell is moving to the right at 100 m/s when suddenly it explodes into two fragments, one twice as heavy as the other. Measurements reveal that 900 J of energy are released in the explosion and that the heavier fragment was in front of the lighter fragment. Find the velocity of each fragment relative to the ground. 2. Relevant equations I already have the solution 80 m/s for the heavier and 110 m/s for the lighter but i just want to understand why because i got it wrong on my quiz. 3. The attempt at a solution i used the elastic collision galilean transforms, but came up with the wrong number. 9kg=x+2x so the heaver is 6kg and the lighter is 3 kg then i plugged into mu1'final=(m1-m2)/(m1+m2)*mu1'inital and mu2'final=2m/(m1+m2)*mu1'inital but these were incorrect using 100 m/s as mu1'initial any help?
A change of this sort in the flying shell is not an elastic collision, but is referred to as a "superelastic collision", because the total amount of mechanical energy has increased, due to release from an internal energy source (in this case, the chemical explosion; but releasing a compressed spring initially connected to both masses would have a similar effect). Conservation of linear momentum still applies, but you must now also keep track of the kinetic energy change, as you would in a partially inelastic collision. Keep in mind that the center of mass still moves in the same direction at 100 m/sec before and after the explosion. Consider the "center-of-mass frame" (hence the Galilean relativity reference): before the explosion, the total KE is zero; afterwards, the total KE of the two fragments is 900 J. You know that the two fragments have masses of 3 and 6 kg. You must now solve two equations simultaneously, the one for which the total linear momentum of the fragments is zero in this frame of reference, the second being that the total final KE of the fragments must add to 900 J. You will have the velocities of each fragment in this frame; add back the 100 m/sec velocity of the center-of-mass and you'll have the velocities of the fragments in the "ground" or stationary reference frame.
I think i get what you're saying, and it makes sense but I am for some reason still arriving at the wrong answer. can you check my work. The Consevation of linear momentum is m1v1=m2v2 but if the total has to equal zero then 0=3x+6y where x and y are their respective velocities. making x=-2y then for the kinetic energy KEi=KEf and KEf=900 so 0=(0.5*3*x^2)+(0.5*6*y^2)=900 simplifying x^2+2y^2=600 subsituting the x giving (-2y)^2+ 2y^2=600 giving a y=17.32i? i know i messed up but im not sure exactly where
Oh okay that makes sense! Stupid simple mathematical error on my part for some reason i wrote (-2)^2 = -4. Thanks for all your help!! :)
Good! If you check the speeds in the stationary frame, you'll find that the same 900 J change in kinetic energy has occurred there as well. (Measured values of velocity, momentum, and kinetic energy differ between the frames of reference, but the magnitude of changes in these quantities do not. That's also why magnitudes of accelerations and forces also agree between the reference frames in Galilean relativity.) The center-of-mass frame, though, can often be the easier one in which to make calculations.