A cannon of mass 5.60 10^3 kg is rigidly bolted to the earth so it can recoil on

Question

A cannon of mass 5.60 10^3 kg is rigidly bolted to the earth so it can recoil only by a negligible amount. The cannon fires an 82.5-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

Mass of Cannon = 5.6*10^3 Kg
Mass of Shell = 81.5 Kg
initial Velocity of Shell = 550 m/s
Let the Velocity of Cannon = v

Using Momentum Coservation =
5.6*10^3 * v = 550*81.58
v = 8.012 m/s

Now, Kinetic Energy of Cannon = 0.5*81.5*550^2 =

KEshell = Wc + Wrecoil;
0.5*81.5*550^2 = Wc + 0.5* 5.6*10^3 *8.012^2
Wc = 1.214*10^7 m/s

Now if the Cannon is lose, Velocity of Shell =
Wc = 0.5*81.5*v^2
v = sqrt[(1.214*10^7)/(0.5*81.5)]
v = 545.8 m/s
velocity of a shell, v =546 m/s

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