Kvant Physics Problem 55

Verified: yes
Verdicts: PASS + PASS
Solve time: 2m41s
Source on kvant.digital

Problem

F55. Estimate to what height an arrow shot vertically upward from a bow will rise. The mass of the arrow is 20 g, the length of the bowstring is 1 m. The bowstring is drawn back by 5 cm. Assume the tension of the bowstring to be...

Setup and Assumptions

An arrow of mass

$$m=20,\text{g}=0.020,\text{kg}$$

is shot vertically upward from a bow. The total length of the bowstring is

$$L=1.0,\text{m},$$

and the midpoint of the string is pulled back by

$$x=5.0,\text{cm}=0.050,\text{m}.$$

The required quantity is the maximum height $h$ reached by the arrow.

The statement available here ends with the words “Assume the tension of the bowstring to be...”, but the numerical value of the tension is missing. Denote this constant tension by $T$.

The bowstring is assumed massless, the tension is assumed constant and equal to $T$ during the draw, air resistance is neglected, the arrow moves vertically, and all work done in drawing the bow is converted into the gravitational potential energy of the arrow at the highest point of its flight.

Physical Principles

The solution rests on two principles.

First, the work done by an external force equals the force multiplied by the displacement of its point of application when the force is constant:

$$A=T,\Delta l .$$

Second, conservation of mechanical energy gives

$$A=mgh,$$

because the work stored in the bow becomes the gravitational potential energy of the arrow at the top of its trajectory.

Derivation

Initially the bowstring is straight. Since its total length is $L$, each half of the string has length

$$\frac{L}{2}.$$

After the midpoint is pulled back by a distance $x$, each half becomes the hypotenuse of a right triangle. Its length is

$$\ell=\sqrt{\left(\frac{L}{2}\right)^2+x^2}.$$

The increase in length of one half of the string is

$$\Delta \ell = \sqrt{\left(\frac{L}{2}\right)^2+x^2} -\frac{L}{2}.$$

Since there are two halves, the total increase in string length is

$$\Delta l = 2\left( \sqrt{\left(\frac{L}{2}\right)^2+x^2} -\frac{L}{2} \right).$$

The work done in drawing the string is therefore

$$A = T,\Delta l = 2T\left( \sqrt{\left(\frac{L}{2}\right)^2+x^2} -\frac{L}{2} \right).$$

At the highest point of the flight this work has become gravitational potential energy:

$$mgh = 2T\left( \sqrt{\left(\frac{L}{2}\right)^2+x^2} -\frac{L}{2} \right).$$

Solving for $h$ gives

$$h = \frac{2T}{mg} \left( \sqrt{\left(\frac{L}{2}\right)^2+x^2} -\frac{L}{2} \right).$$

Result

The final symbolic expression is

$$h = \frac{2T}{mg} \left( \sqrt{\left(\frac{L}{2}\right)^2+x^2} -\frac{L}{2} \right).$$

Substituting

$$L=1.0,\text{m}, \qquad x=0.050,\text{m}, \qquad m=0.020,\text{kg}, \qquad g=9.81,\text{m/s}^2,$$

gives

$$\sqrt{\left(\frac{1.0}{2}\right)^2+(0.050)^2} = \sqrt{0.2525} = 0.50249,\text{m}.$$

Hence

$$\Delta l = 2(0.50249-0.50000),\text{m} = 0.00499,\text{m}.$$

The work stored in the bow is

$$A=(0.00499,\text{m}),T.$$

The height becomes

$$h = \frac{0.00499,T} {(0.020,\text{kg})(9.81,\text{m/s}^2)} = 0.0254,T.$$

Thus

$$\boxed{ h \approx 0.0254,T }$$

with $h$ in metres when $T$ is measured in newtons.

Because the numerical value of the tension is missing from the problem statement provided, a numerical height cannot be obtained. If the intended tension is $T=T_0$, then

$$\boxed{ h \approx 0.0254,T_0\ \text{m}. }$$

Sanity Checks

The quantity inside the parentheses has dimensions of length. Multiplying by $T$ gives units of joules, and dividing by $mg$ gives metres. The dimensions are correct.

If the draw distance $x$ tends to zero, the increase in string length tends to zero, the stored energy vanishes, and the height tends to zero. This matches physical expectations.

For small draw distances, the geometric increase in string length is much smaller than the draw distance itself. Using

$$\sqrt{\left(\frac{L}{2}\right)^2+x^2} \approx \frac{L}{2} +\frac{x^2}{L},$$

the stored energy is proportional to $x^2$, which is characteristic of elastic systems for small deformations.

The most common source of error is confusing the draw distance $x=5,\text{cm}$ with the increase in string length. The string length increases by only about $0.50,\text{cm}$, ten times less than the midpoint displacement. Using $Tx$ instead of $T\Delta l$ would overestimate the height by roughly an order of magnitude.