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Q.

A bullet is fired from a gun. The force on the bullet is given by F = 600–2 × 10⁵t, where F is in newton and 't' is in second. The force on the bullet becomes zero as soon as it leaves the barrel. The average impulse imparted to the bullet is

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a

0.9Ns

b

1.8Ns

c

9Ns

d

Zero

answer is C.

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Detailed Solution

$\begin{array}{l} F = 600 - 2 \times 10^{5} t = 0 \\ 600 = 2 \times 10^{5} t \\ t = \frac{300}{105} = 3 \times 10^{- 3} \sec \\ \int dp = \int Fdt; \\ P = \int_{0}^{t} 600 dt - 2 \times 10^{5} \int_{0}^{t} tdt \end{array}$
$\begin{array}{l} = [600 t]_{0}^{3 \times 10^{- 3}} - 2 \times 10^{+ 5} {[\frac{t^{2}}{2}]}_{0}^{3 \times 10^{- 3}} \\ = 1800 \times 10^{- 3} - 9 \times 10^{- 6} \times 10^{5} \\ = 1.8 - 0.9 = 0.9 NS \end{array}$

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A bullet is fired from a gun. The force on the bullet is given by F = 600–2 × 105 t, where F is in newton and 't' is in second. The force on the bullet becomes zero as soon as it leaves the barrel. The average impulse imparted to the bullet is