Q85P - University Physics with Modern Physics

Question

Question: In a physics lab experiment, you release a small steel ball at various heights above the ground and measure the ball’s speed just before it strikes the ground. You plot your data on a graph that has the release height (in meters) on the vertical axis and the square of the final speed (in ) on the horizontal axis. In this graph your data points lie close to a straight line. (a) Using and ignoring the effect of air resistance, what is the numerical value of the slope of this straight line? (Include the correct units.) The presence of air resistance reduces the magnitude of the downward acceleration, and the effect of air resistance increases as the speed of the object increases. You repeat the experiment, but this time with a tennis ball as the object being dropped. Air resistance now has a noticeable effect on the data. (b) Is the final speed for a given release height higher than, lower than, or the same as when you ignored air resistance? (c) Is the graph of the release height versus the square of the final speed still a straight line? Sketch the qualitative shape of the graph when air resistance is present.

 

Step-by-Step Solution

Answer

Answer

 

(a) The numerical value of the slope of this straight line is $0.0510 {\text{s}}^{\text{2}}\text{/m}$.

(b) The final speed for a given release height is lower when I ignored the air resistance.

(c)

Step 1: Significance of the kinematic equation

The kinematic equation mainly defines motion of a particle with respect to the time. The kinematic equation gives the distance travelled by an object that is the division between the difference between the square of final and initial velocity and the two times of the acceleration.

Step 2: (a) Determination of the numerical value of the slope

The equation of the distance of the small steel ball is expressed as:

 $v_f^2=v_i^2+2g\Delta y$

Here,$v_f$  is the final speed, is the initial speed, $v_i$is the distance travelled by the steel ball and$\Delta y$ is the acceleration due to gravity.

 

As initially the ball was at rest, then the initial velocity of the ball is zero.

 

Substitute 0  for v in the above equation.

 $$\begin{gathered} v_f^2=2g\Delta y \\ \Delta y=\frac{v_f^2}{2g} \\ =\frac{1}{2g}v \end{gathered}$$

The above equation is the equation of the straight line having the slope $\frac{1}{2g}$.

 

The equation of the slope is expressed as:

$s = \frac{1}{2g}$ 

Here,g is the acceleration due to gravity with value 9.8 m/s².

 

Substitute 9.8m/s² for g in the above equation.

 $$\begin{gathered} s=\frac{1}{2\left(9.8 {\text{m/s}}^{\text{2}}\right)} \\ =\frac{1}{\left(19.6 {\text{m/s}}^{\text{2}}\right)} \\ =0.0510 {\text{s}}^{\text{2}}\text{/m} \end{gathered}$$

Thus, the numerical value of the slope of this straight line is $0.0510 {\text{s}}^{\text{2}}\text{/m}$.

Step 3: (b) Determination of the final speed

The acceleration is less than the value 9.8 m/s² of when the air resistance is present. Hence, as the acceleration is less, then the ball’s final speed will also be less.

 

Thus, the final speed for a given release height is lower when I ignored the air resistance.

 

Step 4: (c) Determination of the qualitative shape of the graph

When the air resistance is present, then there will not be any constant acceleration. Hence, the graph will not behave like a straight line. The final velocity will be less in the presence of the air resistance. 

 

The graph of $∆y$ versus $v_f^2$ has been provided below:

According to the graph, the line is not a straight line. Hence, with the increase in the final velocity, the distance also increases.