A gun of mass 1000kg, free to recoil in the direction of the horizontal barrel, fires a shot of mass 100kg with a velocity off 400ms. Find the speed of recoil of the gun. If the recoil is resisted by a constant force so that the gun moves back only 12cm, find the magnitude of the force.


Using conservation of linear momentum


V1 = 4ms


Where do I go from here

Guest Nov 21, 2018

Conservation of momentum:   m1v1 + m2v2 = 0


Let subscript 1 represent the shot mass, and subscript 2 the gun. m is mass, v is velocity


100*400 + 1000*v2 = 0  so  v2 =   -40 m/s      (I assume you meant m/s for velocity, not ms !!)

 Note that the negative sign indicates the gun travels in the opposite direction to the shot.


Assuming constant acceleration (deceleration here) of the gun we can use the constant linear acceleration equation 


v2 = u2 + 2as  to find the acceleration, a:


0 = (-40)2 + 2a*(-0.12)   so a = 1600/0.24  m/s2   (Note: I'm assuming direction is positive in the direction of the shot)


Newton's second law says force = mass*acceleration, so F = 1000*1600/0.24  N

Alan  Nov 21, 2018

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