Question

A bullet is fired from a gun the force on the bullet is given by f equals to 600 - 2 into 10 raise to 5 t where is newton's and 3 seconds the force on the bullet becomes zero as soon as it leaves the barrel what is the average impulse imparted by the bullet

Answer 1

F=600−2×105t

0=600−2×105 t

t=600/(2×105)

t=3×10 −3 s

t=3milliseconds.

Time is taken by bullet leave the barrel is 3 milliseconds

F.idt=(6t×105 t).dt

J=[600t−(2×105 2t² )]

J=[600×(3×10 −3 )−(2×10 5 × 2(3×10 −3 ) 2 )]

J=0.9Ns

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