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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 in second. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?

1. 1.8 N-s
2. zero
3. 9 N-s
4. 0.9 N-s

Correct answer: 0.9 N-s

Solution

Impulse is the integral of force over time. The force becomes zero when 600 - 2 × 10⁵ t = 0, giving t = 0.003 s. The impulse is ∫F dt = ∫(600 - 2 × 10⁵ t) dt from 0 to 0.003, which evaluates to 0.9 N-s.

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