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

Answer 1

Hint: When a fast-moving object crashes into a stationary object in a matter of a few seconds, the damage is magnified even though the reaction force is not that high. This is because the change of force in unit time matters, and this quantity is called impulse.

Complete step-by-step answer:
The impulse is defined as a huge force acting in a small interval of time. The impact of change of force with time is even higher than the magnitude if the time of application becomes shorter. The reason for this is that force is defined as the change in momentum and this change in momentum is equal to the impulse.
Therefore,
Impulse, $J = F \times t$
In the question, the force is provided as a function of time
$F\left( t \right) = 600 - 2 \times {10^5}t$
Given that the force of bullet becomes zero initially, $F = 0$
Substituting, we get –
\[
  F\left( t \right) = 600 - 2 \times {10^5}t \\
  0 = 600 - 2 \times {10^5}t \\
   \to 2 \times {10^5}t = 600 \\
   \to t = \dfrac{{600}}{{2 \times {{10}^5}}} = 3 \times {10^{ - 3}}\sec \\
 \]
Since the force is a function of time, t, we cannot directly, multiply with the time. We have to integrate force with respect to time because the change in time is very small in number.
Therefore,
\[
  J = \int {F.dt} \\
  Substituting, \\
  J = \int {600 - 2 \times {{10}^5}t.dt} \\
  Integrating, \\
  J = \int {600.dt - \int {2 \times {{10}^5}t.dt} } \\
   \to J\left( t \right) = 600t - 2 \times {10^5}\dfrac{{{t^2}}}{2} \\
 \]
Substituting the value of time, \[t = 3 \times {10^{ - 3}}\sec \], we get the value of average impulse.
\[
  J\left( t \right) = 600t - 2 \times {10^5}\dfrac{{{t^2}}}{2} \\
   \to J\left( {3 \times {{10}^{ - 3}}} \right) = 600 \times 3 \times {10^{ - 3}} - 2 \times {10^5} \times \dfrac{{{{\left( {3 \times {{10}^{ - 3}}} \right)}^2}}}{2} \\
   \to J\left( {3 \times {{10}^{ - 3}}} \right) = 18 \times {10^2} \times {10^{ - 3}} - {2} \times {10^5} \times \dfrac{{9 \times {{10}^{ - 6}}}}{{{2}}} \\
   \to J\left( {3 \times {{10}^{ - 3}}} \right) = 1800 \times {10^{2 - 3}} - 9 \times {10^{5 - 6}} \\
   \to J\left( {3 \times {{10}^{ - 3}}} \right) = 18 \times {10^{ - 1}} - 9 \times {10^{ - 1}} \\
   \to J\left( {3 \times {{10}^{ - 3}}} \right) = 9 \times {10^{ - 1}} = 0.9Ns \\
 \]
Hence, the average impulse, $J = 0.9Ns$

Note: The integral of the types mentioned in this solution is explained as below: 1. $\int {dx = x + C} $
2. $\int {x.dx = \dfrac{{{x^2}}}{2} + C} $