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Question

A bullet is fired from a gun. The force on the bullet is given by F=6002×105t, where F is in N and t in sec. The force on bullet becomes zero as soon as it leaves barrel. What is average impulse imparted to bullet?

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Solution

F=6002×105t
0=6002×105t
t=600/(2×105)
t=3×103s
t=3milliseconds.
Time is taken by bullet leave the barrel is 3 milliseconds
F.idt=(6t×105t).dt
J=[600t(2×105t²2)]
J=[600×(3×103)(2×105×(3×103)22)]
J=0.9Ns

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