Chapter 7, Problem 75GP

(a)

To determine

To plot: A graph of force versus time for $t=0\sec$ to $t=3\text{m}\sec$

Explanation of Solution

Given:

The force on a bullet varies as $F=580-(1.8\times {10}^5)t$ for the time interval $t=0\sec$ to $t=3\text{ms}$

Calculation:

Force has divided in two components; the first one is a constant part and other than depends on the first power of time. Thus, with the increasing time, the force will be negative because of the dominating second part with the maximum $F=580\text{N}$ at $t=0\sec$ and minimum $F=40\text{N}$ at $t=3\text{ms}$

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Conclusion:

Thus, force versus time graph would be linearly straight in the negative axis.

(b)

To determine

The impulse given by the bullet, using graphical methods.

Answer to Problem 75GP

The impulse given by the bullet is $0.93\text{N-s}$

Explanation of Solution

Given:

The force on a bullet varies as $F=580-(1.8\times {10}^5)t$ for the time interval $t=0s$ to $t=3ms$

Formula used:

The area of trapezium is = $\frac{\text{1}}{\text{2}}\text{(sum of parallel sides)×altitude}$

Calculation:

The impulse is calculated as the area under the graph of force versus time.

Graph represents an inverted trapezium. Thus, the impulse given by the bullet is

  $\begin{array}{l}\text{Impulse}=\left(\frac{580 +40 }{2}\right)\left(3\times {10}^{-3}\right)\\ \text{Impulse}=0.93\text{N-s}\end{array}$

Conclusion:

The impulse given by the bullet is $0.93\text{N-s}$

(c)

To determine

The mass of the bullet.

Answer to Problem 75GP

The mass of the bullet is $4.2\times {10}^{-3}\text{kg}$

Explanation of Solution

Given:

The force on a bullet varies as $F=580-(1.8\times {10}^5)t$ for the time interval $t=0\sec$ to $t=3\text{ms}$ . The impulse given by the bullet is $0.93\text{N-s}$ . Initially, the bullet is at rest. Bullet achieves a final velocity of $220\text{m/s}$ as a result of this impulse.

Calculation:

The impulse is calculated as the change in momentum. Thus, the mass of the bullet is

  $\begin{array}{l}\text{Impulse}=\Delta p=m\left(v_f-v_i\right)\\ 0.92N-s=m\times (220-0)\\ m=4.2\times {10}^{-3}\text{kg}\end{array}$

Conclusion:

Hence, the mass of the bullet is $4.2\times {10}^{-3}\text{kg}$