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A bullet is fired from gun. The force on bullet is, F = 600-2xx10^(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=600-2xx10^(5)t` F becomes zero as soon as the bullet LEAVES the barrel. `0=600-2xx10^(5)t , 600 = 2xx10^(5)t` `t=3XX10^(-3)s` Impulse `= int_(0)^(1) Fdt` `= int_(0)^(t)(600-2xx10^(5)t)dt=[600t-cancel(2)xx10^(5)(t^(2))/(cancel(2))]^(3xx10^(-3)` `=600xx3xx10^(-3)-10^(5)xx9xx10^(-6)=0.9 Ns` |
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