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Question: A bullet is fired from a gun. The force on the bullet is given by \(F = 600 - 2 \times 10^{5}t\), wh...

A bullet is fired from a gun. The force on the bullet is given by F=600−2×105tF = 600 - 2 \times 10^{5}t, where F is in newtons and t in seconds. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet

A

9 Ns

B

Zero

C

0.9 Ns

D

1.8 Ns

Answer

0.9 Ns

Explanation

Solution

F=600−2×105t=0F = 600 - 2 \times 10^{5}t = 0 ⇒ t=3×10−3sec⁡t = 3 \times 10^{- 3}\sec

Impulse I=∫0tFdt=∫03×10−3(600−2×103t)dtI = \int_{0}^{t}{Fdt} = \int_{0}^{3 \times 10^{- 3}}{(600 - 2 \times 10^{3}t)dt}

=[600t−105t2]03×10−3=0.9N×sec⁡= \lbrack 600t - 10^{5}t^{2}\rbrack_{0}^{3 \times 10^{- 3}} = 0.9N \times \sec