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A bullet is fired from gun. The force on bullet is, F = 600 -2 xx 10^(5) t newton. The force reduces to zero just when bullet leaves barrel. Find the impulse imparted to bullet |
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Answer» Solution :`F = 600 -2 xx 10^(5)`t F BECOMES zero as soon as the bullet leaves the BARREL, `0=600 - 2 xx 10^(5), t, 600 = 2 xx 10^(5)`t `t = 3 xx 10^(-3) s` Impulse `=int_(0)^(t)Fdt` `=int_(0)^(t)(600-2 xx 10^(5)t)dt = [600t -2 xx 10^(5) t^(2)/2]^(3 xx 10^(-3))` `=0.9 Ns` |
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