Question

A bullet is fired from a gun. The force on the bullet is given by $F=600-(2\times10^5)t$ 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. 1.8 Ns
• D. 0.9 Ns

Answer 1

The Correct Option is D

To find the average impulse imparted to the bullet, we must first understand the relationship between force and impulse. The impulse imparted to an object is the integral of force with respect to time over the time interval the force is applied.

The force on the bullet as a function of time is given by:

F(t) = 600 - (2 \times 10^5) t

The impulse J imparted to the bullet during the time interval from t = 0 to t = T (where T is when the force becomes zero) is given by:

J = \int_0^T F(t) \, dt

First, we find the time T when the force becomes zero. Set F(T) = 0:

600 - (2 \times 10^5) T = 0

Solving for T:

(2 \times 10^5) T = 600

T = \frac{600}{2 \times 10^5} = 0.003 \, \text{seconds}

Now, we calculate the impulse:

J = \int_0^{0.003} (600 - (2 \times 10^5)t) \, dt

This integral can be split and solved as follows:

J = \int_0^{0.003} 600 \, dt - \int_0^{0.003} (2 \times 10^5)t \, dt

The first part of the integral is:

\int_0^{0.003} 600 \, dt = 600 \cdot (0.003 - 0) = 1.8

The second part of the integral is:

\int_0^{0.003} (2 \times 10^5)t \, dt = \left[ (2 \times 10^5) \cdot \frac{t^2}{2} \right]_0^{0.003}

= 1 \times 10^5 \cdot (0.003)^2 = 0.9

Thus, the total impulse J is:

J = 1.8 - 0.9 = 0.9 \, \text{Ns}

Therefore, the average impulse imparted to the bullet is 0.9 Ns.