If a 10 kg round fired from a gun of 2 ton.the initial velocity of round is 340 m/sec then what is the recoiling velocity of gun?and how much the force required to stop the recoil of gun in 1 mtr?
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bullet,
m = 10kg
u = 0m/s
p1 = mu = 0kg.m/s
Gun,
m = 1814.37
u = 0
p2 = mu = 0kg.m/s
AFTER FIRING,
bullet ,
m = 10kg
v = 340m/s
P1 = mv = 3400kg.m/s
Gun,
m = 1814.17kg
v = v m/s
P2 = mv = 1814.17v kg.m/s
p1 + p2 = P1 + P2 (law of conservation of momentum)
0 = 3400 + 1814.17v
-3400/1814.17 = v
v = -1.873m/s
WORK DONE = CHANGE IN ENERGY ( IN THIS CASE KINETIC ENERGY)
w = 1/2m(v² - u²)
= 1/2×10×(0-(340)²)
= 5×-(340)²
= -578000J
W = FS
-578000 = F×1m ( since it has to be stopped in 1meter)
FORCE REQUIRED = -578000Newtons
m = 10kg
u = 0m/s
p1 = mu = 0kg.m/s
Gun,
m = 1814.37
u = 0
p2 = mu = 0kg.m/s
AFTER FIRING,
bullet ,
m = 10kg
v = 340m/s
P1 = mv = 3400kg.m/s
Gun,
m = 1814.17kg
v = v m/s
P2 = mv = 1814.17v kg.m/s
p1 + p2 = P1 + P2 (law of conservation of momentum)
0 = 3400 + 1814.17v
-3400/1814.17 = v
v = -1.873m/s
WORK DONE = CHANGE IN ENERGY ( IN THIS CASE KINETIC ENERGY)
w = 1/2m(v² - u²)
= 1/2×10×(0-(340)²)
= 5×-(340)²
= -578000J
W = FS
-578000 = F×1m ( since it has to be stopped in 1meter)
FORCE REQUIRED = -578000Newtons
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