Chapter 2, Problem 113P

(a)

To determine

To Prove:The experimental magnitude of the free-fall acceleration is given by $g_{\text{exp}}=\frac{2\Delta y}{{\left(\Delta t\right)}^2}$ .

Explanation of Solution

Introduction:

The distance $\Delta y$ travelled by the marble in the measured time interval $\Delta t$ is given by the following equation of motion.

  $\Delta y=v_{0y}\Delta t+\frac{1}{2}{g}_{\text{exp}}{\left(\Delta t\right)}^2$  .....(1)

Here, $v_{0y}$ is the initial velocity of the marble.

The marble is dropped from a point at a negligible distance from the first light gate. Therefore, the distance travelled by the marble in the measured time interval $\Delta t$ can be taken to be equal to the distance between the two light gates and the initial velocity of the marble can be taken to be equal to zero.

Therefore, substitute 0 for $v_{0y}$ in equation (1).

  $\Delta y=\frac{1}{2}{g}_{\text{exp}}{\left(\Delta t\right)}^2$

Simplify for $g_{\text{exp}}$ .

  $g_{\text{exp}}=\frac{2\Delta y}{{\left(\Delta t\right)}^2}$  .....(2)

Conclusion:

Thus, it is proved that the experimental value of the free fall acceleration is given by the expression, $g_{\text{exp}}=\frac{2\Delta y}{{\left(\Delta t\right)}^2}$ .

(b)

To determine

The expected value of $\left(\Delta t\right)$ for the marble to pass through the light gates in this experimental setup assuming that the value of free-fall acceleration is equal to the standard value $g_{\text{exp}}=9.81{\text{ m/s}}^2$ .

Answer to Problem 113P

The expected value of the time $\left(\Delta t\right)$ for the marble to pass through the light gates in this experimental setup is found to be $0.319\text{ s}$ .

Explanation of Solution

Given:

The position of the first light gate (at the edge of the table)measured from the floor

  $y_i=1.00\text{ m}$

The position of the second light gate when measured from the floor

  $y_f=0.500\text{ m}$

The standard value of the free-fall acceleration

  $g_{\text{exp}}=9.81{\text{ m/s}}^2$

Formula used:

The time taken for the marble to fall between the two light gates is given from equation (2) and is written as

  $\Delta t=\sqrt{\frac{2\Delta y}{g_{\text{exp}}}}$  .....(3)

Where, $\Delta y=y_i-y_f$.......(4)

Calculation:

From equations (3) and (4)

  $\Delta t=\sqrt{\frac{2\left(y_i-y_f\right)}{g_{\text{exp}}}}$  .....(5)

Substitute the given values of variables in equation (5).

  $\begin{array}{c}\Delta t=\sqrt{\frac{2\left(y_i-y_f\right)}{g_{\text{exp}}}}\\ =\sqrt{\frac{2\left[\left(1.00\text{ m}\right)-\left(0.500\text{ m}\right)\right]}{\left(9.81{\text{ m/s}}^2\right)}}\\ =0.319\text{ s}\end{array}$

Conclusion:

Thus, the expected value of the time $\left(\Delta t\right)$ for the marble to pass through the light gates in this experimental setup is found to be $0.319\text{ s}$ .

(c)

To determine

The value of the $g_{\text{exp}}$ that will be determined if the first light gate is positioned $\text{0}\text{.5 cm}$ below the table and the percentage difference of this value from the standard value of $g$ .

Answer to Problem 113P

The value of the free fall acceleration determined in the current experimental setup would be $\text{10}{\text{.05 m/s}}^2$ and the percentage difference between this value and the standard value of $g$ is $2.5\%$ .

Explanation of Solution

Given:

The height of the edge of the table from the floor

  $y_i=1.00\text{ m}$

The position of the first light gate (at the edge of the table) measured from the table

  $y_{{}_i}^{\text{'}}=0.5\times {10}^{-2}\text{ m}$

The position of the second light gate when measured from the table

  $y_f=0.500\text{ m}$

The standard value of the free-fall acceleration

  $g=9.81{\text{ m/s}}^2$

Formula used:

The marble is dropped from the edge of the table with a velocity $v_{0y}$ and when it reaches the first light gate, it has a velocity $v_y$ which can be calculated using the expression,

  $v_y^2=v_{0y}^2+2g{y}_i^{\text{'}}$  .....(6)

Assume the origin to be located on the table and the initial velocity is zero.

The time $\Delta t\text{'}$ recorded by the light gates can be obtained from the following expression:

  $\Delta y\text{'}=v_y\Delta t\text{'}+\frac{1}{2}g{\left(\Delta t\text{'}\right)}^2$  ........(7)

Here, $\Delta y\text{'}=y_f-y_i^{\text{'}}$  .....(8)

The value of the free fall acceleration that would be obtained from this experiment (with errors) is given by the following expression

  $g_{\text{exp}}^{\text{'}}=\frac{2\Delta y}{{\left(\Delta t\text{'}\right)}^2}$  .....(9)

The percentage difference between the calculated and the standard values of the free fall acceleration is given by

  $\%\text{ difference}=\frac{g_{\text{exp}}^{\text{'}}-g}{g}\times 100\%$  ......(10)

Calculation:

Calculate the speed of the marble when it passes through the first light gate by substituting the values of the variables in equation (6).

  $\begin{array}{c}{v}_y^2=v_{0y}^2+2g{y}_i^{\text{'}}\\ ={\left(0\text{ m/s}\right)}^2+2\left(9.81{\text{ m/s}}^2\right)\left(0.5\times {10}^{-2}\text{m}\right)\\ v_y=3.31\times {10}^{-2}\text{m/s}\end{array}$

Calculate the distance between the light gates using equation (8).

  $\begin{array}{c}\Delta y\text{'}=y_f-y_i^{\text{'}}\\ =0.500\text{ m}-\left(0.5\times {10}^{-2}\text{m}\right)\\ =0.495\text{ m}\end{array}$

Substitute the values of the variables (calculated and given) in equation (7) and write a quadratic equation in $\left(\Delta t\text{'}\right)$ .

  $\begin{array}{c}\Delta y\text{'}=v_y\Delta t\text{'}+\frac{1}{2}g{\left(\Delta t\text{'}\right)}^2\\ \left(0.495\text{ m}\right)=\left(3.31\times {10}^{-2}\text{m/s}\right)\Delta t\text{'}+\frac{1}{2}\left(9.81{\text{ m/s}}^2\right){\left(\Delta t\text{'}\right)}^2\end{array}$

  $\left(4.905{\text{ m/s}}^2\right){\left(\Delta t\text{'}\right)}^2+\left(3.31\times {10}^{-2}\text{m/s}\right)\Delta t\text{'}-\left(0.495\text{ m}\right)=0$

Solve for $\left(\Delta t\text{'}\right)$ .

  $\begin{array}{c}\Delta t\text{'}=\frac{-\left(3.31\times {10}^{-2}\text{m/s}\right)\pm \sqrt{{\left(3.31\times {10}^{-2}\text{m/s}\right)}^2-4\left(4.905{\text{ m/s}}^2\right)\left(-0.495\text{ m}\right)}}{2\left(4.905{\text{ m/s}}^2\right)}\\ =\frac{-\left(3.31\times {10}^{-2}\text{m/s}\right)\pm \left(3.116\text{ m/s}\right)}{\left(9.81{\text{ m/s}}^2\right)}\end{array}$

Taking the positive root alone

  $\Delta t\text{'}=0.3154\text{ s}$

The person performing the experiment does not take into account the difference in the positioning of the light gates. Hence, the value of free fall acceleration $g_{\text{exp}}^{\text{'}}$ she calculates would be given by equation (9).

Substitute the values of variables in equation (9) and calculate the experimental value of the free fall acceleration.

  $\begin{array}{c}{g}_{\text{exp}}^{\text{'}}=\frac{2\Delta y}{{\left(\Delta t\text{'}\right)}^2}\\ =\frac{2\left(0.500\text{ m}\right)}{{\left(0.3154\text{ s}\right)}^2}\\ =10.05{\text{ m/s}}^2\end{array}$

Calculate the percentage difference between the measured value and the standard value is found using equation (10).

  $\begin{array}{c}\%\text{ difference}=\frac{g_{\text{exp}}^{\text{'}}-g}{g}\times 100\%\\ =\frac{\left(10.05{\text{ m/s}}^2\right)-\left(9.81{\text{ m/s}}^2\right)}{\left(9.81{\text{ m/s}}^2\right)}\times 100\%\\ =2.5\%\end{array}$

Conclusion:

Thus, the value of the free fall acceleration determined in the current experimental setup would be $\text{10}{\text{.05 m/s}}^2$ and the percentage difference between this value and the standard value of $g$ is $2.5\%$ .