A cannon of mass 5.65 103 kg is rigidly bolted to the earth so it can recoil onl

Question

A cannon of mass 5.65 103 kg is rigidly bolted to the earth so it can recoil only by a negligible amount. The cannon fires an 85.5-kg shell horizontally with an initial velocity of +550 m/s. Suppose the cannon is then unbolted from the earth, and no external force hinders its recoil. What would be the velocity of an identical shell fired by this loose cannon? (Hint: In both cases assume that the burning gunpowder imparts the same kinetic energy to the system. Enter your answer to the nearest whole number. Indicate the direction with the sign of your answer.)

Explanation / Answer

in first case the speed of cannon = 0

so energy given by birning poweder = 0.5*85*550^2 J

case 2 now canon set to loose

so now it will have some velocity backward due to recoil which wil reduce the velocity of shell so

suppose canon velocity = Vc and shell velocity Vs   

energy by gunpoweder

= 1/2* 5.65*10^3 *Vc^2 + 1/2*85*Vs^2 =  0.5*85*550^2 ...........1

and momentum conservation

Pi = Pf

0 = 5.65*10^3* Vc - 85*Vs ...........2

Vc = 85*Vs / (5.65*10^3) put his into 1 so

1/2* 5.65*10^3 *Vc^2 + 1/2*85*Vs^2 =  0.5*85*550^2

0.5*5.65*10^3* [ 85*Vs / (5.65*10^3)]^2 + 0.5 *85*Vs^2 = 0.5*85*550^2

from here we got

Vs = + 545.909 m/s

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