The Air Track Project

The design and building instructions for a low friction Air Track are presented. The dynamics of a 1D collision are discussed and predictions are compared with measured data. 

Introduction

One of the fundamental concepts explored in introductory physics is the behavior of objects as they move in a straight line and how their interactions can alter direction and speed. Exploring such principles along one dimension greatly simplifies the math. To demonstrate concepts in 1D requires a straight track with negligible friction so that motion is sustained long enough for easy observation. Two methods are often used for low friction, namely free-rolling sleds, or sleds that float on a bed of pressurized air. The latter is called an Air Track, operating much like an air hockey game. It can also be a rather expensive apparatus.

Typical concepts to explore are the conservation of energy and momentum in the absence of friction, which can be used to predict the trajectories after a collision. 

An Air Track achieves low friction using pressurized air from small holes to allow V shaped sleds to float on a square tube.

The Apparatus

The Track

Couplers and End Stops

Track Flatness

Initial trials with square tubing and sections of aluminum angle as sleds showed excellent freedom of motion along the track with low friction. It also revealed regions where the square tube was not straight. The illustration below shows a sled on a highly exaggerated warped track, sliding into a local valley. Because the friction is so low, warping leads to oscillation of the sled within a valley. 

If the radius of curvature of a valley is R, then the period of oscillation $T = 2\pi \sqrt{\frac{R}{g}}$, the same as a pendulum of length R, where g is the Earth’s gravitational constant (9.8 m/s). In some trials, T was measured at 30s, giving R = 225m (738 ft). If this were a single curve over a length of track C, the dip in the center S (Sag) is derived from the equation shown below. For C = 1.83 m (full length) and R = 225 m, then S = 1.9mm. If we wish to have T > 60s then R > 1km and S < 0.4 mm is required. Some means to correct for the distortions in the tube is required to achieve that performance. For comparison, an Air Track sold by PASCO specifies a deviation < 0.04mm over 2m length track. That translates to R > 10 km and T > 3.4 minutes! In this project we can set 1 minute as acceptable stillness.

Track Support

Many methods were explored to achieve the flatness goal.  In the final configuration a 6’ square tube is attached to a 6’ Empire level using 6 V-block clamps, shown in the cross section (the level is hollow, but a more typical I-beam design could be used). The square tube is secured to each V-block with an 8-32 bolt. The clamp bolts (6-32) have some vertical clearance to adjust the height (sag) at each clamp position.  A 6’ straight edge was placed on the apex of the square tube while the clamps were adjusted to eliminate visual gaps. Observations of oscillation of a sled were then used to fine tune the flatness. Carriage bolts were also used as adjustable feet for leveling, one on the right side and two across from one another on the left, as shown below. 

Sleds and Fixtures

Position Detection

To quantify the motion of the sleds, their position must be recorded as a function of time. Options include:

• Ultrasonic Detector
• An ultrasonic pulse is sent toward the object, and the reflected pulse is detected. The distance is calculated from the speed of sound and the time between sending to reception of the pulse.
• Works best with wide objects, making it difficult to differentiate between multiple small objects. Requires temperature sensing to calculate the speed of sound. Resolution about 3mm (HC-SR04).
• Lidar (laser time of flight)
• A pulse of 940nm sent by a laser diode is detected by an avalanche photo-diode array.
• Can work with narrow objects; 1mm resolution (VL53L1X); multiple objects best done with multiple detectors, multiplexed into a datalogger.
• Smart phone video
• The position and time of all objects on the sled can be captured. Time is recorded by frame number (30fps normal, 240fps with slo-mo). Position can be recorded in pixels, calibrated to a known length. 
• Image processing or visual frame by frame analysis can extract distance vs. time for all objects. The procedure is tedious but very simple.

For simplicity, generality, and accuracy, the smart phone video option was used in these measurements.

Elastic Collisions

Two Body Elastic Collisions

Momentum p is defined as the mass of an object m times its velocity v ($p=mv$). If there is no force acting on the masses, then the sum of the momentum of a system is conserved (does not change after a collision). For the two masses confined to one dimension this is written as: $$m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2$$

The kinetic energy of a moving mass is ½mv2. If the collisions are elastic (no sticking to one another) and there are no other forces such as friction, then the sum of the kinetic energy before and after the collision does not change. For 1D motion this means: $$\frac{1}{2} m_1v^2_1 + \frac{1}{2} m_2v^2_2 = \frac{1}{2} m_1{v'}^2_1 + \frac{1}{2} m_2{v'}^2_2$$  

If we know masses $m_1$ and $m_2$ and start velocities $v_1$ and $v_2$, then these two equations can be solved for the velocities after the collision $v’_1$ and $v’_2$. After some algebra and defining $\mu=\frac{m_2}{m_1}$ and $\nu = \frac{v_2}{v_1}$, the solutions are: $$v’_1 = v_1 \frac{1 - \mu + 2\mu \nu}{1 + \mu}$$ $$v’_2 = v_1 \frac{2 - \nu + \mu \nu}{1 + \mu}$$

Note: for $v_1 \ll v_2$ this becomes: $$v'_1 = v_2 \frac{2\mu}{1+\mu}$$ $$v'_2 = v_2 \frac{\mu-1}{1+\mu}$$

Two Body Elastic Collisions (equal mass)

A typical experiment is to collide two equal masses ($\mu=1$). The equations predict for elastic collisions that $v’_1 = v_2$ and $v’_2 = v_1$ meaning that equal masses trade velocities to conserve momentum and energy. If the bodies are moving in opposite directions, the trade in velocity causes the bodies to reverse direction.

To see this on the air track, a spring is attached to the fixture post on a sled to provide an elastic collision:

Two Body Elastic Collisions (equal mass, equal velocities)

Case: $m_1 = m_2 = 78g, v_1 = 19.7 cm/s, v_2 = -19.3 cm/s \rightarrow v’_1 = -17.3 cm/s, v’_2 = 16.3 cm/s$

With approximately equal but opposite velocities before collision, the velocities are essentially reversed after the collision. There is a slight drop in velocity after the collision, indicating some energy loss during impact. 

Two Body Elastic Collisions (equal mass, different velocities)

Cases: 

$m_1 = m_2 = 78g$

Collision A: $v_1 = 28.9 cm/s, v_2 = -10.1 cm/s \rightarrow$

$v’_1 = -8.4 cm/s, v’_2 = 25.9 cm/s$

Collision B: $v_1 = 3.6 cm/s, v_2 = -11.9 cm/s \rightarrow$

$v’_1 = -9.5 cm/s, v’_2 = 2.9 cm/s$

This illustrates the exchange of velocities between equal masses for the case A: $v_1 \gt -v_2$ and B: $v_1 \lt -v_2$. There is a slight drop in velocity after the collision, indicating some energy loss during impact. 

Two Body Elastic Collisions (equal mass, same direction)

Case: 

$m_1 = m_2 = 78g$

$v_1 = -12.8 cm/s, v_2 = -32.9 cm/s \rightarrow$

$v’_1 = -32.2 cm/s, v’_2 = -13.1 cm/s$

This illustrates the exchange of velocities between equal masses for the case $v_2 \gt v_1$. There is a slight drop in velocity after the collision, indicating some energy loss during impact. 

Two Body Elastic Collisions (equal mass, one stationary)

If one of the equal masses is stationary ($v_2 = 0$), the trade in velocities makes the stationary object move and halts the other object:

Cases: 

$m_1 = m_2 = 78g$

Collision A: $v_1 = 46.5 cm/s, v_2 = 0 \rightarrow$

$v’_1 = 0.21 cm/s, v’_2 = 42.0 cm/s$

Collision B: $v_1 = -0.8 cm/s, v_2 = -38.6 cm/s \rightarrow$

$v’_1 = -36.7 cm/s, v’_2 = -1.2 cm/s$

This illustrates the exchange of velocities between equal masses for the case of either $v_1 =0$ or $v_2=0$. There is a slight drop in velocity after the collision, indicating some energy loss during impact. 

Two Body Elastic Collisions ($m_2 \approx 2m_1$)

Cases for $m_2 = 2.05 m_1$ were examined for 9 initial velocities $v_1$ and $v_2$. The plots below shows a comparison of the predicted vs. the measured resulting velocities $v’_1$ and $v’_2$ with good correlation. The next sections show videos of the collisions.

Collision 6: Two Body Elastic Collisions (zero momentum)

We can start with 0 momentum for the system given any two masses if $\frac {v_2}{v_1} = -\frac {m_1}{m_2}$, where the larger mass has lower opposite velocity (also expressed as $\nu = \frac{1}{\mu}$). The final velocities are then $v’_1 = -v_1$ and $v’_2 = \frac{v_1}{\mu}$ (the reverse of the initial velocities). If $\mu = 2$ and $\nu = -\frac{1}{2}$ then $v’_1 = -v_1$ and $v'_2 = \frac{1}{2} v_1$.

Case: 

$m_1 = 78g, m_2 = 160g$

$v_1 = 17.6 cm/s, v_2 = -11.5 cm/s \rightarrow$

$v’_1 = -20.1 cm/s, v’_2 = 6.9 cm/s$

This illustrates the case of low total momentum where the initial velocities are essentially reversed for each mass after the collision (it helps to watch each sled individually and view the video twice). 

Collision 3: Two Body Elastic Collisions (stopping velocity)

For a given value of $\mu = \frac{m_2}{m_1}$, the equations show we can make $v’_1 = 0$ if $\nu = \frac{v_2}{v_1}= \frac{\mu -1}{2\mu}$, resulting in $v’_2 = v_1 \frac{\mu + 1}{2\mu}$. So, if $\mu = 2$, this occurs for $\nu = +\frac{1}{4}$, and $v’_2 = \frac{3}{4}v_1$.  

Case: 

$m_1 = 78g, m_2 = 160g$

$v_1 = 35.4 cm/s, v_2 = 8.8 cm/s \rightarrow$

$v’_1 = -0.3 cm/s, v’_2 = 23.9 cm/s$

This illustrates the case of stopping the lower mass and increasing the velocity of the larger mass. 

Collision 9: Two Body Elastic Collisions (stopping velocity)

If we wish to make $v’_2 = 0$, then we must have $\nu = \frac{2}{1-\mu}$, resulting in $v’_1 = v_1 \frac{1+\mu}{1-\mu}$. So, for $\mu=2$, this requires $\nu=-2$, and $v’_1 = -3 v_1$. 

Case: 

$m_1 = 78g, m_2 = 160g$

$v_1 = 10.0 cm/s, v_2 = -23.7 cm/s \rightarrow$

$v’_1 = -32.4 cm/s, v’_2 = -1.5 cm/s$

This illustrates the case of achieving near zero velocity of the larger mass. Ideally for $\frac{m_2}{m_1} = 2, \frac{v_2}{v_1} = -2$ to achieve $v’_2 = 0$. 

Bouncing 2-Ball Effect ($m_2 \gg m_1$)

A well-known physics demonstration is to place a small ball on a more massive ball and drop both at the same time, resulting in a surprisingly high velocity of the smaller ball on the rebound. If the velocity of both balls just before impact is v, and the larger ball rebounds elastically, its velocity is essentially reversed before it strikes the smaller ball. At that moment $\nu=\frac{v_2}{v_1}=-1$, which gives $v’_1 = v \frac{-3\mu+ 1}{\mu+ 1}$ and $v’_2 = v \frac{3 -\mu}{1 +\mu}$. For $m \gg 1$, this approaches $v’_1 = -3v$ and $v’_2 = -v$. The kinetic energy and the final height of the top ball scales as the square of its upward velocity. There is almost a 9x increase in the height compared to just bouncing the small ball on its own.

This effect is explored using the Air Track in the next section.

Collision 8: Bouncing 2-Ball Effect ($m_2 = 2 m_1$)

The moment of impact from the recoiling ball (large mass) and the smaller ball can be simulated by setting $v_1 = -v_2$, with $\frac{m_2}{m_1} = 2$ (for example). Then the resulting velocity of the small mass becomes a high recoil velocity of $v’_1 = -5 \frac{v_1}{3}$ and $v’_2 = \frac{v_1}{3}$.

This illustrates the case of achieving high velocity of the smaller mass after colliding with the large mass. 

Collision 4: Bouncing 2-Ball Effect ($m_2 = 2 m_1$)

To more directly simulate the two-ball effect, a small and large mass can be directed to an elastic wall. For $\frac{m_2}{m_1} = 2$ and $v_1 = -v_2$:

Measured: 

$m_1 = 78g, m_2 = 160g$

$v_1 = 25.9 cm/s, v_2 = 27.5 cm/s \rightarrow$

$v’_1 = -43.8 cm/s, v’_2 = 6.1 cm/s$

Predicted: $v'_1 = -45.9 cm/s, v'_2=7.5 cm/s$

A plot of the velocities of each sled shows A: nearly equal velocity prior to the wall collision ( $v_1 = 25.9 cm/s$, $v_2 = 27.5 cm/s$ ), then B: sled $m_2$ (160g) reflects off the end wall and strikes $m_1$ (78g), C: sending it back with velocity $v’_1$ = -43.8cm/s, while sled $m_2$ recoils at $v’_2 \approx 7.5 cm/s$, and D: reflects off the wall again for a final velocity $v’_2 =  -6.1 cm/s$.

This illustrates the case of achieving high resulting velocity of the smaller mass after colliding with the large mass and wall. 

Wall Collisions

If $m_2 \gg m_1 (\mu \gg 1)$ and $v_2 = 0$ (collision with a large stationary mass) then the equations show $v’_1 \approx - v_1$ and $v’_2 \approx 0$. The smaller mass reverses direction, while the larger one barely moves in the opposite direction. This is the case when the $m_2$ mass is a fixed (elastic) wall.

Multi-Bounce Effects

The following is based on problem in Fundamentals of Physics (Jearl Walker), 12th edition, chapter 9, problem 105.

Mass $m_1$ is stationary to right of an elastic wall. Mass $m_2 \gt m_1$ comes from the right at velocity $-v_2$ and collides with $m_1$ at a distance $d_0$ from the wall, causing $m_2$ to slow down to $-v’_2$ and $m_1$ to move left at velocity $-v’_1$. After elastically reflecting off the wall, $m_1$ returns with velocity $+v’_1$ until it collides again with mass $m_2$ at a distance $d_1$ from the wall, at a time $\tau$ from the first collision. It turns out that the ratio of $\frac{d_1}{d_0}$ is independent of the initial velocity $v_2$.

Reasoning:

The mass $m_1$ travels a distance $d_0 + d_1$ at velocity $v’_1$ in the time $\tau$, so $d_0 + d_1 = v’_1\tau$. Mass $m_2$ travels $d_0 – d_1$ in that same time, so $d_0 – d_1 = v’_2 \tau$. Earlier, it was discussed that $v’_1 = v_2 \frac{2\mu}{1 + \mu}$, and $v’_2 = v_2\frac{\mu - 1}{1 + \mu}$, where $\mu = \frac{m_2}{m_1}$. This gives $\frac{d_1}{d_0} = \frac{1 + \mu}{3\mu – 1}$, which is independent of the initial velocity $v_2$. As$\mu$ becomes very large, this ratio approaches 1/3, independent of the masses.

Measurements on the Air Track indeed shows $\frac{d_1}{d_0}$ to be independent of $v_2$ within 10% (probably due to energy losses in the springs). Six trails were recorded in slow motion to obtain x(t) and v(t) for each sled. The calculations used $\mu = \frac{m_2}{m_1} = 2.051$. An example video (trial 4) is shown below.

Energy Transfer from Pendulum to Sled

A pendulum weight $m_1$ is lifted a height h from its rest position, giving it potential energy $m_1 h g$. When released, its speed increases to $v_1$ at its lowest position as the gravitational energy is converted to kinetic. Using energy conservation we have: $m_1 h g = \frac{1}{2} m_1 {v_1}^2$. If a sled of mass $m_2$ is struck by the pendulum, some fraction of the momentum and energy is transferred to the sled. When this experiment was implemented in the Air Track using a steel ball of mass 56g for the pendulum, striking a spring on a sled of mass 78g, 70% - 80% of the pendulum potential energy was transferred into kinetic energy of the sled (PE to KE), based on measurements from the video. The remaining energy resided with the pendulum and some rocking of the sled. Slight misalignment of the pendulum ball with the sled spring led to far lower PE to KE transfer due to excitation of other types of motion in the sled.

Collision Dynamics

What happens during a collision?

We have seen that if two masses collide elastically, they will recoil off one another consistent with conservation of momentum and energy. This provides a means to calculate the final velocities, but it does not describe the details of the interaction that leads to these velocities.

With the Air Track, a spring is used between the sleds for elastic collisions. The combined kinetic energy of two colliding sleds is transferred to the spring, compressing it until all the kinetic energy is stored in the spring, expressed by $\frac{1}{2} m_1 {v_1}^2 + \frac{1}{2} m_2 {v_2}^2 = \frac{1}{2} kx^2$, where $k$ is the spring constant and $x$ is the total compression width of the spring. The compressed spring provides an equal but opposite force $F = k x$ on each mass as it expands, accelerating each sled to new velocities $v’_1$ and $v’_2$, in accord with $F = m a$ (smaller masses accelerate faster than larger masses).

In the case of any two objects colliding elastically, we can usually invoke a spring to model this interaction. 

The collision process can be observed in detail using the Air Track and slow-motion video.

Measurement of Collision Dynamics

The end portion of the Air Track was fitted with a spring made from stainless steel wire of gages 24, 26 or 28, and a sled of mass 78g or 160g was used to record the collision of the sled and spring at various initial velocities. Using the Slo-Mo video option on a smart phone camera at 240 frames per second, the sled’s position vs. time was measured to obtain x(t) and v(t).

In the video below, note that rather than an “instantaneous” collision, there is a total time $\tau$ for the spring to compress and expand over a width L. Video analysis for this case shows $\tau = 60ms, L = 5mm$:

Measurement of Spring k

To determine k, each spring was placed on a weight scale to measure the force resulting from compressing the spring a distance x. The scale reads in mass units m, so the force is mg, where g = 9.8 kg m/s2. 

The value of k is the slope of mg vs. x. The results for springs made from three gages of 216L stainless steel wire are shown in the table.

Note: To check on this measurement, the k values were plotted vs. $d^4$, where d is the wire diameter, showing reasonable agreement between the measured values for k and its expected dependence on d (https://www.engineersedge.com/spring_comp_calc_k.htm).

Collision Dynamics: Using F = ma

A mass m attached to an elastic object with a spring constant k will oscillate its position x back and forth over time t. This is described using F = ma as: $-k x = m \frac{d^2 x}{dt^2}$. The oscillating behavior motivates a solution of $x = L \sin (\omega t)$, where $\omega$ is the oscillation frequency in radians/sec. Calculus then gives $\frac{d^2 x}{dt^2} = - L \omega^2 \sin(\omega t) = -\frac{k}{m} L \sin (\omega t)$, resulting in $\omega = \sqrt{\frac{k}{m}}$. A period of oscillation T is defined by $\omega = \frac{2\pi}{T}$, so $T = 2\pi \sqrt{\frac{m}{k}}$. As we might expect, the larger the mass or the smaller the k, the longer the oscillation period T.

For the case of a sled of mass m colliding with a spring attached to an infinite mass (wall), the spring begins in a relaxed state with x = 0 until the mass contacts it at velocity v (see the diagram below). The compression proceeds to a distance L until all the kinetic energy is transferred to the spring, reducing the mass velocity to 0. The subsequent expansion of the spring returns the mass to x = 0 and restores the velocity to v in the opposite direction. This is physically equivalent to the mass-spring oscillator operating over half of a period, giving us the time from first contact to compression and back to release: $\tau = \frac{T}{2} = \pi \sqrt{\frac{m}{k}}$. Note that as the spring gets stiffer (larger k), the collision time is shorter. Since the velocity is $\frac{dx}{dt} = \omega L \cos{\omega t}$, then at t = 0, $v = \omega L$, giving the compression width $L = \frac{v}{\omega} = v \sqrt{\frac{m}{k}}$. The compression time is independent of v, while L depends linearly on v. Note that the equation for L also follows from energy conservation: $\frac{1}{2} m v^2 = \frac{1}{2} k L^2$.  

Collision Dynamics: Measurement vs. Calculated Results

Using this analysis, the collision time $\tau$ and the compression width L can be calculated from m, k, and v for each collision experiment. The values of m, k, and v were varied, and the final velocity v’, $\tau$ and L were measured from the video analysis. The final sled velocity magnitude v’ was equal to the initial velocity v within a few percent (see plot below), probably owing to energy losses in the spring. The measured collision times $\tau$ were about 20% smaller than the calculated values on average, due to the difficulty in measuring velocities over short times. This is compounded in the L calculation. In certain cases, the springs would fully compress on impact, leading to an underestimate for larger L.

Collision Dynamics: Computer Simulation of Two-Body in 1D

In general, if $m_1$ or $m_2$ has an attached massless spring with spring constant k, and they interact when their separation equals the length of the spring ($\ell_0$), they will compress the spring until their velocities are 0, followed by expansion and release when their separation again equals $\ell_0$. Using F = ma for each mass, when both masses are touching the spring: $$m_1 a_1= -k (\ell_0 - (x_2 - x_1))$$ $$m_2 a_2= +k (\ell_0 - (x_2 - x_1))$$  

If the separation of the masses ($x_2 – x_1$) exceeds the spring length, then the force on both masses is zero and they continue with a fixed velocity. 

One way of solving these two equations is to do so numerically. The simplest approach is Euler’s method where the above equations are used to calculate $a_1$ and $a_2$, then each is multiplied by a time increment dt to get the velocity increment dv. Then dv is multiplied by the same time increment to get the x increment dx. For sufficiently small dt, a repetition of this process is used to accumulate x, v and a for each mass. Then momentum $mv$, kinetic energy $\frac{1}{2} mv^2$, and the spring energy $\frac{1}{2} k (\ell_0 – (x_2-x_1))^2$ can also be calculated at each time value. This can be done in an Excel spreadsheet where each row is a new time. If dt is sufficiently small, the total energy (kinetic + spring) and momentum will be constant as a natural consequence of F = ma. 

Collision with a large object: $m_1 = 1kg, m_2 = 1000kg, v_1 = 0.03 m/s, v_2 = 0, k = 100 kg/m^2, \ell_0 = 0.01m$:

Increasing k: $m_1 = 1kg, m_2 = 1000kg, v_1 = 0.03m/s, v_2 = 0, k = 300 kg/m^2, \ell_0 = 0.01m$

Note that a higher spring stiffness only reduces the collision time.

Equal mass and velocity collision: $m_1 = m_2  = 1kg, v_1 = -v_2 = 0.03 m/s, k = 100 kg/m^2, \ell_0 = 0.01m$

Small + large mass collision: $m_1 = 1kg, m_2 = 2kg, v_1 = -v_2 = 0.03 m/s, k = 100 kg/m^2, \ell_0 = 0.01m$. Note that two unequal masses stop (v = 0) at different times where the larger mass takes longer.

Collision Dynamics: Observing Different Stop Times for $m_1 \ne m_2$ 

A larger mass takes longer to stop, as evident from the data taken from the video below and the numerical simulation. 

Video (240 fps) of $m_1 = 78g$ with spring constant of $k = 28 kg/s^2$ and $m_2 = 160g$. The data and simulation are in reasonable agreement.

Inelastic Collisions

Two Body Inelastic Collisions

Two objects moving in 1D that collide can sometimes stick together on impact. This is an inelastic collision, where momentum is still conserved, but some of the energy is lost to friction and heat. If two bodies of mass $m_1$ and $m_2$ and corresponding velocities $v_1$ and $v_2$ combine into a single body, they produce a mass $m’ = m_1 + m_2$ moving at some velocity $v’$. Conservation of momentum dictates $m_1v_1 + m_2v_2 = m’v’$. Using $\mu = \frac{m_2}{m_1}, \nu = \frac{v_2}{v_1}$ and $\nu’ = \frac{v’}{v_1}$ gives us $\nu’ = \frac{1 + \mu\nu}{1 + \mu}$. The kinetic energy before and after the collision are $KE = \frac{1}{2}(m_1{v_1}^2 + m_2{v_2}^2$ and $KE’ = \frac{1}{2} m’{v’}^2$. The energy lost to heat is $LE = KE – KE’$.

To see this on the Air Track, a hook and loop patch is attached to the fixture post on the sleds, so they adhere when they contact, as shown in the video:

Various initial velocities $v_1$ and $v_2$ were used for $m_1 = 160g$ and $m_2 = 78g$, using Hook & Loop to adhere colliding sleds. The position and velocities vs. time were derived from slow-motion video (240fps) and are plotted on the left for 10 cases.  The resulting velocity v’ of the joined sleds is also plotted vs. the predictions from conservation of momentum, showing reasonable agreement. Note the slight recoil in the video below during impact resulting from some elasticity in the Hook & Loop.

Video of inelastic collisions 1 and 4 (240fps):

Case of Zero Momentum

If any two masses move toward one another so that the total momentum is 0 $(\mu \nu = -1)$, an inelastic collision results in a zero final velocity (v’ = 0) to conserve momentum. This means LE = KE or 100% energy loss to heat. 

An interesting demonstration of heat generated by inelastic collisions can be seen by hitting two steel ball bearings together with a sheet of paper between them. The resulting heat energy can create a charred hole: https://youtu.be/gTh5aABIwoY

Two Body Inelastic Collisions (small $v_2$)

If a large mass at high speed inelastically collides with a slower small mass, then $v' \approx \frac{v_1}{1+\mu}$. In the video below, $\mu = 0.49$, and the measured $\frac{v’}{v_1} = 0.67$.

Gravitational Acceleration

Tilted Ramp

Falling objects accelerate due to a gravitational force F = mg between the object of mass m and Earth (or any other mass). The acceleration rate $g = 980 cm/s^2$. If the falling distance is denoted x then Newton’s second law gives us $x(t) = x_0 + v_0t + \frac{1}{2} g t^2$ and $v(t) = v_0 + g t$, where $x_0$ and $v_0$ are the initial position and velocity.

If an object slides down a frictionless ramp tilted an angle $\theta$ from the horizontal then the gravitational force along the track is $F = mg \sin{\theta}$. The equations of motion are the same as for a vertically falling object, except that g is replaced by $g \sin{\theta}$. 

The Air Track can be used to measure x(t) and v(t) of sliding masses on a tilted track. Fitting the data to linear and quadratic functions for various $\theta$, we can test these predictions and determine the value of g.

Note: The gravitational acceleration g’ on less massive celestial bodies can therefore be simulated on a tilted ramp using $g’ = g \sin{\theta}$. For our moon, $\theta =  9.5^\circ$, while for Mercury and Mars, which have similar g’, $\theta =  22.3^\circ$.

Tilted ramp example data

The Air Track was tilted by placing a spacer of known thickness h under the right hand footing a distance L from the left footing. This creates a tilt angle $\theta$ given by $\sin{\theta} = \frac{h}{L}$. 

Using video capture, the sled position x along the track was measured as a function of time. Five ramp angles and two mass values were used in the experiment. An example plot of data is shown below, including the curve fit values. The acceleration in this example is $2 \times 19.0 = 38 cm/s^2$ from the top plot of x(t), and $39.6 cm/s^2$ in the bottom plot of v(t). The more reliable value is from x(t), owing to the approximation used to calculate v(t) from x(t). The basic shape of the curves confirms the expected behavior for constant forces.

The acceleration values were derived from the curve fits to x(t) for 5 tilt angles and two masses. The slopes should equal g, which is typically $980 cm/s^2$. As the plots show:

• The acceleration is linearly proportional to $\sin{\theta}$.
• The slope values for the two masses agree to within 0.5% (independent of mass).
• The g values (slopes) are only 0.9% from the accepted value.  

Sled pulled by falling weight

Tying a sled of mass $m_1$ to a falling weight $m_2$ using a string wrapped over a pulley will cause $m_1$ and $m_2$ to experience the same acceleration a. The connecting string applies a tension T to each mass as shown in the diagram. Newton’s second law applied to each mass gives us (ignoring any friction or pulley angular momentum): 

$T = m_1 a$ (x direction)

$-T + m_2 g = m_2 a$ (y direction)

This gives: $$a = g \frac{m_2}{m1 + m2}$$ .

The total energy (kinetic and potential) $E = \frac{1}{2} (m_1 + m_2)v_2 + m_ 2 g (L – x)$, where $v = 0$ at $x = 0$ (bottom of the ramp) and L is the length of the ramp. $E = m_2 g L$ and is constant. From the equations of motion for a constant force, $v = a t$ and $x = \frac{1}{2} a t^2$, which gives $\frac{1}{2}(m_1 + m_2) a^2 t^2 = \frac{1}{2} m_2 g a t^2$ that simplifies to $a = g \frac{m_2}{m_1 + m_2}$, the same result as above.

Physically, the gravitational force only acts on $m_2$, which pulls on both masses due to the string, resulting in reduced acceleration compared to $m_2$ just falling freely. If $m_1 \ll m_2$, then $a \approx g$. If they have the same mass, then $a = \frac{g}{2}$. If m1 >> m2, then g is reduced by $\approx \frac{m_2}{m_1}$.

The Air Track was used to demonstrate this behavior by attaching a pulley and falling weight at the end of the track. 

The sled position vs. time was recorded by slow-motion video using a smart phone (240 fps). The case for sled mass $m_1 = 162g$ and $m_2 = 198g$ is shown in the video below. A pin inserted into one of the track holes was removed to start the acceleration. The pulley and falling weight are out of view. At the end of its motion, the sled broke apart on impact with the pulley (later repaired).

The measurements of x(t) and v(t) for two cases are plotted below. The curve fit to x(t) is a parabola, and the velocity v(t) is linear with time. This is consistent with 1D motion at constant acceleration: $x(t) = \frac{a}{2}t^2 + v_0 t + x_0$, and $v(t) = a t + v_0$. The curve fits show $v_0$ and $x_0$ terms are small in both cases. 

The larger $m_2$ case gives $a = 486 cm/s^2$, which is 90% of the expected value from $a = g \frac{m_2}{m_1 + m_2} = 537 cm/s^2$. For the smaller $m_2$, the measured $a = 99.5 cm/s^2$, while the expected value is $130 cm/s^2$ (76%). The smaller measured values may be due to pulley friction and rotational inertia that are proportionately larger for the smaller pull force. 

Tilted Ramp Cases

When the sled is on an upward ramp and is pulled by a falling weight, we can expect that the acceleration will be reduced since the sled now experiences a downward force along the ramp of $m_1 g \sin{\theta}$, that pulls back on the string. Accounting for this new force, using the coordinates of the tilted ramp (x’, y’):

$T – m_1 g \sin{\theta} = m_1 a$ (x’ direction)

$N – m_1 g \cos{\theta} = 0$ (y’ direction; N = ramp Normal force)

$-T + m_2 g = m_2 a$ (y direction)

The first and third equations gives:

$a = g \frac{m_2 - m_1 \sin{\theta}}{m_1 + m_2}$.

The total energy (kinetic and potential) $E = \frac{1}{2} (m_1 + m2_) v^2 + m_2 g (L – x) + m_1 g x \sin{\theta}$, where v = 0 at x = 0 (bottom of the ramp) and L is the length of the ramp. $E = m_2 g L$ and is constant. From the equations of motion for a constant force, $v = a t$ and $x = \frac{1}{2} a t^2$, which gives  $(m_1 + m_2) a^2 t^2 =  a t^2 (m_2 g - m_1 g \sin{\theta})$, which simplifies to $a = g \frac{m_2 - m_1 \sin{\theta}}{m1 + m2}$, the same result as above.

The Air Track was tilted by placing a spacer of thickness 3.75cm under the right stand, which is a distance along the track of 96.2cm from the left stand. This makes $\theta = \arcsin{\frac{3.75}{96.2}} = 2.2^\circ$.

A video of an accelerating sled (240fps) on a tilted track. A pin inserted into one of the track holes was removed to start the acceleration. The pulley and falling weight on the right are out of view.

The results using $m_2 = 25g$ with the track tilted by $2.2^\circ$ are shown on the right below concluding with $a = 66.7 cm/s^2$.

Based on the equation derived previously, the ratio of the accelerations for the tilted and level tracks should be $\frac{a(2.2^\circ)}{a(0^\circ)} = \frac{m_2 - m_1 \sin{\theta}}{m_2} = 0.746$. The data shows $\frac{a(2.2^\circ)}{a(0^\circ)} = 0.67$, within 10%. The difference may be attributed to friction and rotational energy in the pulley.

Wind Acceleration (Sled with Fan)

Another type of force can be provided by moving air, implemented by placing an electric fan onto a sled as shown in the image and in the video below. Plots of the position vs. time x(t) and the derived velocity vs. time v(t) are also shown. The data indicates that x(t) is well described as a parabola and v(t) is linear, which means the wind force is constant, similar to a falling object. Based on the curve fits, there is a constant acceleration of about $2.2 cm/s^2$. This would be equivalent to an acceleration of $g \sin{\theta}$ if a frictionless ramp were tilted at $\theta = 0.13^\circ$.