Question

A bullet is fired from a gun. The force on the bullet is given by F = 600 - 2 $\times$ ${10}^5$ t, where F is in Newton 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 1

The correct option is (C) 0.9 Ns

Given,

$F=600-2\times {10}^{5}t$

The time when this force becomes zero

$600-2\times {10}^{5}t=0$

Or, $t=3\times {10}^{-3}s$

Time taken by the bullet to leave the barrel is 3 milliseconds.

We know,

Impulse = $F.idt=(600-2\times {10}^{5}t)dt$

Or, $J=[600t-(2\times {10}^5\frac{t^2}{2})]$

Now putting values of t in above equation

$J=[600\times (3\times {10}^{-3})-(2\times {10}^5\times \frac{{(3\times {10}^{-3})}^2}{2})]$

After solving

$J=0.9Ns$

Hence, the impulse is 0.9 Ns.