Q: A bullet is fired from a gun. The force on a bullet is, F = 600 – 2 × 105 t newton. The force reduces to zero just when the bullet leaves barrel. Find the impulse imparted to the bullet.
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Sol: F = 600 – 2 × 105 t ,
F becomes zero as soon as the bullet leaves the barrel.
0 = 600 -2 × 105 t
⇒ 600 = 2 × 105 t
t = 3 × 10-3 s
$\large Impulse = \int_{0}^{t} F dt $
$\large Impulse = \int_{0}^{3 \times 10^{-3}} (600- 2 \times 10^5 t) dt $
$\large = [600 t – 2 \times 10^5 \frac{t^2}{2}]_{0}^{3 \times 10^{-3}}$
= 600 × 3 × 10-3 – 105 × 9 × 10-6
= 0.9 Ns.