A cannon of mass 5.95 x 10^3 kg is rigidly bolted to the earth so it can recoil
ID: 1903403 • Letter: A
Question
A cannon of mass 5.95 x 10^3 kg is rigidly bolted to the earth so it can recoil only by a negligible amount. The cannon fires a 76.2-kg shell horizontally with an initial velocity of 531 m/s. Suppose the cannon is then unbolted from the earth, and no external force hinders its recoil. What would be the velocity of a shell fired from this loose cannon? (Hint: In both cases assume that the burning gunpowder imparts the same kinetic energy to the system.) This i what I did: KE= 1/2 (76.2kg)(531m/s)^2 = 1.07x10^7 J multiplied that number by 2 : 1.07x10^7J(2)= 2.15x10^7J vf2=sqrt(2kf/m2((m2/m1)+1)) =sqrt(2(2.15x10^7J)/(26.2kg((76.2kg/5.95x10^3kg)+1)) =sqrt(4.30x10^7J/76.2kg(1.01281)) =sqrt(4.30x10^7J/77.2kg) = +746 m/s The correct solution from my hw was +528 m/s. Can anyone please explain what I did wrong? I also tried using m1v1=m2v2 5.95 x 10^3 kg x v1 = 76.2-kg x 531 m/s v1= (76.2-kg x 531 m/s) / 5.95 x 10^3 kg v1= 6.80037 m/s 531 m/s-6.80037 m/s = +524.2 m/s Still both ways were wrong.PLEASE HELP!Explanation / Answer
By conservation of enery, we have:
Initial KE =Finall KE
0.5*(5.95 x 10^3+76.2)*V^2 = 0.5*5.95 x 10^3*531^2
V = 531*sqrt(5.95 x 10^3/(5.95 x 10^3+76.2))
=527.63 OR 528 m/s
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