Question #62270

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The mass of the gun is #M=6kg#

The mass of the bullet is #m=0.005kg#

The velocity of the bullet is #=300ms^-1#

The recoil velocity of the gun is

#v_(gun)=-(0.005/6)*300ms^-1#

#=-0.25ms^-1#

The concept behind this question is the conservation of linear momentum . When there is no external force acting on the system "Initial momentum of the system is equal to the final momentum of the system i.e, #P_i=P_f# ".

#P_i# denotes the initial momentum of the system.
#P_f# denotes the final momentum of the system .

Let's get into the question ,

Mass of the gun #M_g#= #6kg#.
Mass of the bullet #M_b#= #5xx10^-3kg#.

Since initially the system is at rest the initial momentum of the system is zero i.e, #P_i=0#

When the gun shots the bullet moves with a velocity #V_b=300ms^-1#

And we need to find the recoil velocity #V_g# of the gun

The final momentum of the system is the sum of the final momentums of the bullet and the gun

Therefore #P_f=M_b*V_b+M_g*V_g#

Since there is no external force acting on the system the initial momentum is equal to the final momentum

#P_i=P_f#
#0=M_b*V_b+M_g*V_g#
Rewriting this equation
#V_g=-(M_b*V_b)/M_g#
#V_g=-(300xx5xx10^-3)/6#
#V_g=0.25ms^-1#

Hence the recoil velocity of the gun is #0.25ms^-1# and the negative sign indicates that the gun move in the opposite direction of the bullet since velocity is a vector quantity ( quantities which also depend on their direction)