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

#1**+2 **

Conservation of momentum: m_{1}v_{1} + m_{2}v_{2} = 0

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

100*400 + 1000*v_{2} = 0 so v_{2} = -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

v^{2} = u^{2} + 2as to find the acceleration, a:

0 = (-40)^{2} + 2a*(-0.12) so a = 1600/0.24 m/s^{2} (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