A cannon of mass 5.71 x 103 kg is rigidly bolted to the earth so it can recoil only by a negligible amount. The cannon fires a 73.5-kg shell horizontally with an initial velocity of 547 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

541.14 m/s

Explanation:

We are given that

Mass of cannon=

Mass of shell,

Initial velocity of shell,v=547 m/s

We have to find the velocity of shell fired from this loose cannon.

According to law of conservation of momentum

Initial momentum of system=0

When the cannon is bolted to the ground then only shell moves and kinetic energy of system equals to kinetic energy of shell

Kinetic energy of shell,

K.E of shell=

K.E of shell=

K.E of shell=

2K.E of shell=

Velocity of shell fired from this loose cannon,v_2=

Hence, the velocity of shell fired from this loose cannon would be 541.14 m/s

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