A bullet of mass 10 gram is fired from a gun of mass 8 kg with a velocity of 160m/s.Find the velocity of recoil of gun.Find force required to stop the gun in a distance of 25 cm.
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before fire,u1(velocity of gun)=0 and u2(velocity of bullet)=0
after fire ,v2(velocity of bullet)160m/s and v1(velocity of gun)=?
applying law of conservation of linear momentum,
m1*u1+m2*u2=m1*v1+m2*v2
8*0+.01*0=8*v1+0.01*160
0=8*v1+1.6
8*v1=-1.6
v1= -0.2m/s,........where -ve sign indicates the gun goes in opposite direction to the bullet.
now,
v^2-u^2=2*a*s, a= acceleration ,s=0.25m
0-(0.2)^2=2*a*0.25
-0.04=2*a*0.25
-2/25=a
a= -0.08m/s,
F=m*a
F= -8*0.08
F= -0.64N,,,,,,,-ve sign indicates that this force has to be applied against the recoiling of gun.
after fire ,v2(velocity of bullet)160m/s and v1(velocity of gun)=?
applying law of conservation of linear momentum,
m1*u1+m2*u2=m1*v1+m2*v2
8*0+.01*0=8*v1+0.01*160
0=8*v1+1.6
8*v1=-1.6
v1= -0.2m/s,........where -ve sign indicates the gun goes in opposite direction to the bullet.
now,
v^2-u^2=2*a*s, a= acceleration ,s=0.25m
0-(0.2)^2=2*a*0.25
-0.04=2*a*0.25
-2/25=a
a= -0.08m/s,
F=m*a
F= -8*0.08
F= -0.64N,,,,,,,-ve sign indicates that this force has to be applied against the recoiling of gun.