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Grade 11Mechanics

DESCRIBE THE CONCEPT OF RECOIL OF GUN

Profile image of HARSH PAWAR
10 Years agoGrade 11
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1 Answer

Profile image of Gayatri Jayesh Bondriya
ApprovedApproved Tutor Answer10 Years ago
Recoil of a gun

    Consider the gun and bullet in its barrel as an isolated system
    In the begining when bullet is not fired both the gun and bullet are at rest.So the momentum of the before firing is zero
    pi=0
    Now when the bullet is fired ,it moves in the forward direction and gun recoil back in the opposite direction
    Let mb be the mass and vb of velocity of the bullet And mg and vg be the velcoity of the gun after the firing
    Total momentum of the system after the firing would be
    pf=mbvb +mgvg
    since no external force are acting on the system,we can apply the law of conservation of linear momentum to the system
    Therfore
    pf=pi
    or mbvb +mgvg=0
    or vg=-(mbvb/mg)
    The negative sign in above equation shows that velocity of the recoil of gun is opposite to the velocity of the bullet
    Since mass of the gun is very large as compared to the mass of the bullet,the velocity of the recoil is very small as compared to the velocity of the bullet
 
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