gun of mass M fires a bullet of mass m with velocity v relative to gun the average force required to bring gun to rest in 0.5 sec is
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F=Delta P/Delta T
F=MV/0.5
now share option so that proceed further
F=MV/0.5
now share option so that proceed further
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Let velocity of gun backwards be X m/s
According to Newton's third law of motion,
MX=mv or, X = mv/M
Now the gun has to come at rest from velocity X. Therefore the acceleration caused be -a.
Thus, from the formula, v=u+at (where v is final velocity=0, u is initial velocity=X, a is acceleration= -a, and t is time=0.5)
0=X -(a×0.5)
or, X=a×0.5
or, a=X/0.5
or, a=mv/0.5M
or, a=2mv/M
Therefore force required to stop the gun,
F=ma = M×2mv/M (M gets cancelled)
F=2mv (ans)
According to Newton's third law of motion,
MX=mv or, X = mv/M
Now the gun has to come at rest from velocity X. Therefore the acceleration caused be -a.
Thus, from the formula, v=u+at (where v is final velocity=0, u is initial velocity=X, a is acceleration= -a, and t is time=0.5)
0=X -(a×0.5)
or, X=a×0.5
or, a=X/0.5
or, a=mv/0.5M
or, a=2mv/M
Therefore force required to stop the gun,
F=ma = M×2mv/M (M gets cancelled)
F=2mv (ans)
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