Question

A bullet is fired from a gun. The force on the bullet is \(F=(600-2\times 10^{5}t)\) N where \(t\) 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

• A. \(9\) N-s
• B. zero
• C. \(0.9\) N-s
• D. \(1.8\) N-s

Hint:

Find the time at which the force becomes zero, then integrate the force over that time to get the impulse.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Impulse is the time integral of force: \(J=\int F\,dt\).

Step 2: Find the Time in the Barrel
The force is zero when the bullet leaves: \(600-2\times 10^{5}t=0\)
\(t=3\times 10^{-3}\) s

Step 3: Calculation
\(J=\int_{0}^{3\times 10^{-3}}(600-2\times 10^{5}t)\,dt=600t-10^{5}t^{2}\)
\(J=1.8-0.9=0.9\) N-s

Step 4: Compare the options
The impulse is \(0.9\) N-s, option (C).

Final Answer:
The impulse imparted to the bullet is 0.9 N-s. \[ \boxed{\text{(C) }\text{0.9 N-s}} \]