Chapter 6, Problem 19Q

To determine

The faster arrow will penetrate how much farther than the slower arrow.

Answer to Problem 19Q

Solution:The faster arrow penetrates 4 times farther than the slower arrow.

Explanation of Solution

Given:

The masses of two arrows are equal ( $m_1=m_2$ ). The speed of one arrow is twice the other. The hay exerts a constant “frictional” force on the arrows.

Formula used: From the Work-Energy Principle:

  $\Delta K+\Delta U=W_{\text{nc}}$  (1)

Where:

• $\Delta K$ is the change in kinetic energy.
• $\Delta U$ is the change in potential energy.
• $W_{\text{nc}}$ is the work done by non-conservative forces. In this case is friction.

Calculation:

Equation (1) can be written as:

  $\frac{1}{2}m{v}_f^2+mg{h}_f=\frac{1}{2}m{v}_i^2+mg{h}_i+W_f$  (2)

But, the arrows before and after impact are at the same height, therefore:

  $\frac{1}{2}m{v}_f^2=\frac{1}{2}m{v}_i^2+W_f$  (3)

Now, the work done by friction $\left(W_f\right)$ is:

  $W_f=F_{f}x\cos (180°)=-F_f.\Delta x$  (4)

Inserting (4) into (3) we have:

  $\frac{1}{2}m{v}_f^2=\frac{1}{2}m{v}_i^2-F_f\Delta x$  (5)

When the arrows hit the hay, the final speed is zero:

  $v_f=0$  (6)

Inserting (6) into (5):

  $0=\frac{1}{2}m{v}_i^2-F_f.\Delta x$  (7)

Isolating $\Delta x$ from equation (7):

  $\Delta x=\frac{m{v}_i^2}{2F_f}$  (8)

Now, for arrow 1:

  $\Delta x_1=\frac{m{v}_1^2}{2F_f}$  (9)

Similarly, for arrow 2:

  $\Delta x_2=\frac{m{v}_2^2}{2F_f}$  (10)

But:

  $v_2=2v_1$  (11)

Replacing (11) into (10):

  $\Delta x_2=\frac{m{\left(2v_1\right)}^2}{2F_f}=\frac{2m{v}_1^2}{F_f}$  (12)

Dividing (12) and (9):

  $\frac{\Delta x_2}{\Delta x_1}=\frac{\frac{2m{v}_1^2}{F_f}}{\frac{m{v}_1^2}{2F_f}}=4$  (13)

Or

  $\Delta x_2=4\Delta x_1$  (14)

Conclusion:

According to equation (14),the faster arrow (arrow 2) penetrates 4 times farther than the slower arrow (arrow 1).