11.1.4.2: Explorations

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Exploration 1: Vectors for a Box on an Incline

Exploration 1 represents a free-body diagram for a \(20\text{-N}\) block on a \(30^{\circ}\) frictionless incline (the length of the vectors is given in newtons). The light gray lines represent the traditional \(xy\) axis, and the black lines represent the coordinates along the incline. The blue vector represents the normal force; the green vector represents the weight. You may move the tails of the blue and green vectors to add them and use the red vector to represent their resultant vector by dragging the red vector's tip. Restart.

Determine the resultant force from the diagram.
Determine the acceleration of the block.

Exploration 2: Change the Two Forces Applied

Drag either of the crosshair cursors or the ball (position is given in centimeters and time is given in seconds). The cursors each exert a constant force on the black ball (either attraction or repulsion) if they are within \(10\mathbf{ cm}\) of the ball. When the ball hits a wall, the wall exerts a force on the ball causing it to recoil. The green and red arrows display the forces due to each cursor, and the blue arrow represents the net force. Restart.

For attraction and repulsion, drag the black ball around to see the net force. When you get the ball in a "good" spot, click the "play" button to see the effect of the forces on the ball. Briefly explain how and why the ball moves according to the forces applied.

Exploration 3: Change the Force Applied to Get to the Goal

Drag the crosshair cursor close to the black ball (position is given in meters and time is given in seconds). Notice that the cursor exerts a force, that is, a push or a pull, on the ball depending on the force you select. The ball will bounce off the purple spheres and will bounce off the soft walls around the animation. The animation will end if the ball hits either rectangle. The blue arrow represents the net force. Restart.

Try to get the ball to hit the green rectangle and not the red rectangle.
Given an applied force, how does the ball move?
Does the ball always move the way you expect? Why or why not?

Exploration 4: Set the Force on a Hockey Puck

A \(250\text{-gram}\) hockey puck is acted upon by a single force. It is free to slide on the ice (position is given in meters and time is given in seconds) in any direction. You can set the force vector by changing its magnitude \((0\text{ N}<F<10\text{ N})\) and direction. The force vector is shown in the animation as a red arrow. You also can set the initial velocity components \((-15\text{ m/s} < v < 15\text{ m/s})\). Restart.

When the initial velocity is zero, in what direction does the ball travel for a given force?
When the initial velocity is not zero, in what direction does the ball travel for a given force? Hint: The best way to do this is to pick a nonzero \(v_{0x}\) or \(v_{0y}\), not both. Also turn on the ghosts.
Try \(F = 5\text{ N}\), \(\theta = 270^{\circ}\), \(v_{0x} = 7\text{ m/s}\), and \(v_{0y} = 15\text{ m/s}\). Does this motion look familiar? Turn on the ghosts to help with the answer.

Exploration 5: Space Probe with Multiple Engines

A space probe is designed with four engines that can fire in the +x, -x, +y, and -y directions, respectively (position is given in meters and time is given in seconds). For each of the situations below, first predict the motion of the space probe. Your prediction should be a detailed description of the motion of the probe. Only after you make a prediction, check it by viewing the animation. An example is shown in the first row of the table. Restart.

Situation	Your prediction	Animation
The space probe has a constant velocity in the \(+x\) direction when suddenly an engine exerts a force on the probe in the \(+x\) direction.	The probe will have an acceleration in the \(+x\) direction. Therefore, since it is already traveling in that direction when the engine fires, it will speed up and will continue moving in the \(+x\) direction.	Animation 1
The space probe has a constant velocity in the \(+x\) direction when suddenly an engine exerts a force on the probe in the \(-x\) direction.		Animation 2
The space probe has a constant velocity in the \(+x\) direction when suddenly an engine exerts a force on the probe in the \(+y\) direction.		Animation 3
The space probe has a constant velocity in the \(+x\) direction when suddenly an engine exerts a force on the probe in the \(-y\) direction.		Animation 4
The space probe has a constant velocity in the \(+x\) direction when suddenly an engine exerts a force on the probe in the \(-y\) direction and another engine exerts a force in the \(-x\) direction.		Animation 5
The space probe has a constant velocity in the \(+x\) direction when suddenly an engine exerts a force on the probe in the \(+y\) direction and another engine exerts a force in the \(+x\) direction.		Animation 6
The space probe has a constant velocity in the \(+x\) direction when suddenly all four engines fire simultaneously.		Animation 7

Table \(\PageIndex{1}\)

Exploration authored by and placed into the public domain by Aaron Titus.

Exploration 6: Putted Golf Ball Breaks Toward the Hole

A putted golf ball rolls toward the hole on a green. The animation shows a top view of the ball on the green. Restart. The acceleration vector (orange) of the ball is shown on the animation, and the components of the ball's acceleration are displayed in the data table.

The net force on the golf ball is in the same direction as the acceleration of the golf ball, according to Newton's second law. This means that if you know the mass of the golf ball and the acceleration of the golf ball, you can calculate the net force on the golf ball.

Is the net force on the golf ball in the animation constant during the interval from \(t = 0\) to \(t = 4.8\text{ s}\)?
If not, does its magnitude and/or direction change?
If the mass of a golf ball is \(0.046\text{ kg}\), what is the net force on the golf ball at \(t = 1.0\text{ s}\)?
For practice, calculate the net force on the golf ball at \(t = 2.0\text{ s},\: t = 3.0\text{ s}\), and \(t = 4.0\text{ s}\) as well.

Exploration authored by Aaron Titus.

Exploration 7: Atwood's Machine

Note

You must keep the ratio above \(0.25\) and below \(10\).

when \(M = m\) then: \(a = g\).
when \(M = m\) then: \(a = 0\).
when \(M >> m\) then: \(a = g\).
when \(M >> m\) then: \(a = 0\).
when \(M < m\) then: \(a = 0\).
when \(M < m\) then: \(a = g\).
when \(M < m\) then: \(a < 0\).

Verify your answer(s) to (c) by using the animation and your answer for (b).

Exploration 8: Enter a Formula for the Force Applied

This Exploration allows you to choose initial conditions and forces and then view how that force affects the red ball. You can right-click on the graph to make a copy at any time. If you check the "strip chart" mode box, the top graph will show data for a time interval that you set. Note that the animation will end when the position of the ball exceeds +/-100 m from the origin. Restart.

Remember to use the proper syntax, such as: \(-10+0.5\ast t\), \(-10+0.5\ast t\ast t\), and \(-10+0.5\ast t\wedge 2\). Revisit Exploration 1.3 to refresh your memory.

Differential equations can be difficult to solve analytically. One way around this is to use a numerical method to generate a solution at discrete time steps. The above animation does just that by advancing the position of the red ball from its initial value at time \(t_{0}\) to a new value at \(t_{1} = t_{0} + dt\). This process can be repeated over and over to approximate the solution as a function of time.

Clearly there are pitfalls in the above procedure. If the time step is too large (\(1\) year for example) interesting phenomena can be missed. This is clearly not an informative dataset if something interesting happens during the time interval. On the other hand, if the time step is too small (\(1\) nanosecond for example) the computer may take a very long time to plot a representative set of points so that you can see the motion of the ball.

For each of the following forces, first describe the force (magnitude and direction) and then predict the motion of the ball. How close were you? Don't forget to determine how the initial position and velocity affect the motion of the ball for each of the forces.

\(F_{x}(x, t) = 1\)
\(F_{x}(x, t) = -1\)
\(F_{x}(x, t) = 1\ast\text{step}(3-t)\) This function is a constant until \(t = 3\text{ s}\) when it turns off.
\(F_{x}(x, t) = x\)
\(F_{x}(x, t) = -x\)
\(F_{x}(x, t) = \cos(x)\)
\(F_{x}(x, t) = \cos(t)\)

Physlets were developed at Davidson College and converted from Java to JavaScript using the SwingJS system developed at St. Olaf College.