Work, Energy, and Momentum Unit Review Notes

Conservation of Energy and Power in Motion: Roller Coaster Dynamics

In the study of mechanical systems, the principle of conservation of energy states that in a closed system with no non-conservative forces like friction, the total mechanical energy remains constant. This is exemplified by a roller coaster scenario involving a $4500\,kg$ car moving through various elevations on a frictionless track.

Energy Analysis of a 4500 kg Roller Coaster

Consider a roller coaster with a mass ($m$) of $4500\,kg$. The track consists of various points at different heights: the top of the first hill at $30\,m$, a subsequent point at $25\,m$, and Point D at $12\,m$.

At the top of the first hill ($h_{1} = 30\,m$), the coaster is traveling at a speed ($v_{1}$) of $5\,m/s$. To find the velocity at Point D ($h_{D} = 12\,m$), we apply the Law of Conservation of Energy, where the total energy at the first hill equals the total energy at Point D:

$PE_{1} + KE_{1} = PE_{D} + KE_{D}$

$mgh_{1} + \frac{1}{2}mv_{1}^2 = mgh_{D} + \frac{1}{2}mv_{D}^2$

Substituting the known values (where $g = 9.8\,m/s^2$):

$(4500\,kg)(9.8\,m/s^2)(30\,m) + \frac{1}{2}(4500\,kg)(5\,m/s)^2 = (4500\,kg)(9.8\,m/s^2)(12\,m) + \frac{1}{2}(4500\,kg)(v_{D})^2$

Solving for $v_{D}$ yields a velocity of $19.44\,m/s$ at Point D.

Power Requirements for Steep Ascents

Power ($P$) is defined as the rate at which work is done or energy is transferred. For the same roller coaster car to move from rest ($v_{i} = 0\,m/s$) at a loading platform located at a height of $h_{i} = 5\,m$ to the top of the first hill ($h_{f} = 30\,m$), reaching a speed of $v_{f} = 5\,m/s$ in exactly $49\,s$, we must calculate the total change in mechanical energy ($\Delta E$).

$\Delta E = (PE_{f} + KE_{f}) - (PE_{i} + KE_{i})$

$\Delta E = [ (4500\,kg)(9.8\,m/s^2)(30\,m) + \frac{1}{2}(4500\,kg)(5\,m/s)^2 ] - [ (4500\,kg)(9.8\,m/s^2)(5\,m) + 0 ]$

$\Delta E = 1,323,000\,J + 56,250\,J - 220,500\,J = 1,158,750\,J$

Power is calculated as energy divided by time:

$P = \frac{1,158,750\,J}{49\,s} = 23,647.96\,W$

To express this in horsepower ($hp$), we use the conversion factor $1\,hp = 746\,W$:

$P = \frac{23,647.96\,W}{746\,W/hp} = 31.7\,hp$

Elastic Potential Energy and Spring Mechanics

Springs store energy as elastic potential energy when compressed or stretched. This energy can be converted into kinetic energy to launch objects, as seen in a $1500\,kg$ roller coaster propelled by a spring.

Launching a 1500 kg Roller Coaster

When the coaster is launched from a spring with an initial compression ($x$) of $3.5\,m$, it reaches a launch speed of $24\,m/s$. Assuming all elastic potential energy is converted to kinetic energy:

$PE_{spring} = KE$

$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$

$k(3.5\,m)^2 = (1500\,kg)(24\,m/s)^2$

$k = \frac{1500 \times 576}{12.25} = 70,530.6\,N/m$

Hooke's Law and Manual Compression

Hooke's Law describes the relationship between the force ($F$) exerted on a spring and its displacement ($x$):

$F = kx$

If a manual force of $445\,N$ is applied to the same spring ($k = 70,530.6\,N/m$), the resulting compression would be:

$x = \frac{F}{k} = \frac{445\,N}{70,530.6\,N/m} = 0.00631\,m$

Converted to centimeters, the displacement is $0.63\,cm$.

Impulse and Momentum: Rocket Propulsion and Impact

Impulse ($J$) is the change in momentum ($\Delta p$) and is equal to the average force ($F$) multiplied by the time interval ($\Delta t$) over which the force acts. Momentum ($p$) is defined as the product of mass ($m$) and velocity ($v$).

Case Study: Saturn V Rocket First Stage

The first stage of a Saturn V rocket utilizes 5 Rocketdyne engines. Each engine delivers a force of $6.77 \times 10^{6}\,N$, for a total force ($F_{total}$) of:

$F_{total} = 5 \times (6.77 \times 10^{6}\,N) = 33.85 \times 10^{6}\,N$

The duration of the burn ($\Delta t$) is $165\,s$. The impulse provided by stage 1 is:

$J = F_{total} \times \Delta t = (33.85 \times 10^{6}\,N)(165\,s) = 5.58525 \times 10^{9}\,kg \cdot m/s$

Rounded to significant figures, the impulse is approximately $5.59 \times 10^{9}\,kg \cdot m/s$. Given the rocket mass of $2,800,000\,kg$ ($2.8 \times 10^{6}\,kg$), the change in velocity ($\Delta v$) starting from rest is:

$J = m \Delta v$

$\Delta v = \frac{5.59 \times 10^{9}\,kg \cdot m/s}{2.8 \times 10^{6}\,kg} = 1994.7\,m/s$

Soccer Kick Mechanics

The impulse-momentum theorem also applies to collisions. A $0.25\,kg$ soccer ball rolling at $8.0\,m/s$ toward a player is kicked back in the opposite direction at $16\,m/s$. We must assign direction; let the initial direction be negative ($-8.0\,m/s$) and the final direction positive ($+16\,m/s$).

$\Delta p = m(v_{f} - v_{i}) = 0.25\,kg(16\,m/s - (-8.0\,m/s)) = 0.25\,kg(24\,m/s) = 6\,kg \cdot m/s$

If the kick lasts for $\Delta t = 3.0 \times 10^{-2}\,s$, the average force exerted is:

$F = \frac{\Delta p}{\Delta t} = \frac{6\,kg \cdot m/s}{3.0 \times 10^{-2}\,s} = 200\,N$

Law of Conservation of Momentum in Collisions

In an isolated system, the total momentum before an interaction is equal to the total momentum after the interaction.

Explosive Interactions: Hockey Players on Ice

Two sisters ($m_{1} = 60\,kg$, $m_{2} = 45\,kg$) are at rest ($p_{initial} = 0$) on frictionless ice. When they push off each other, it is an "explosion" where total momentum must remain zero.

$0 = m_{1}v_{1f} + m_{2}v_{2f}$

If the smaller sister ($45\,kg$) moves at $7\,m/s$, and we define her direction as positive:

$0 = (60\,kg)(v_{1f}) + (45\,kg)(7\,m/s)$

$60(v_{1f}) = -315$

$v_{1f} = -5.25\,m/s$

Inelastic Collisions: Slugs and Football

1. Slug Collision: Two slugs collide on frictionless trails. Slug A ($280\,g = 0.28\,kg$) travels North at $7\,m/s$. Slug B ($180\,g = 0.18\,kg$) travels South at $10\,m/s$. After the collision, Slug A moves South at $5\,m/s$. Define North as positive.

    $m_{A}v_{Ai} + m_{B}v_{Bi} = m_{A}v_{Af} + m_{B}v_{Bf}$

    $(0.28\,kg)(7\,m/s) + (0.18\,kg)(-10\,m/s) = (0.28\,kg)(-5\,m/s) + (0.18\,kg)(v_{Bf})$

    $1.96 - 1.8 = -1.4 + 0.18(v_{Bf})$

    $0.16 = -1.4 + 0.18(v_{Bf})$

    $1.56 = 0.18(v_{Bf}) \implies v_{Bf} = 8.67\,m/s$ North.

    Note: The handwritten transcript logic evaluates this as $7\,m/s$ North based on specific rounding or unit assumptions ($2.8\,kg$ and $1.8\,kg$ were used in handwritten notes though the prompt states $280\,g$ and $180\,g$).

2. Football Tackle: A defender ($115\,kg$) running at $4.5\,m/s$ tackles a stationary receiver ($v_{2i} = 0$). They stick together and move at $2.6\,m/s$ after the hit.

    $m_{1}v_{1i} + m_{2}v_{2i} = (m_{1} + m_{2})v_{f}$

    $(115\,kg)(4.5\,m/s) + 0 = (115\,kg + m_{2})(2.6\,m/s)$

    $517.5 = 299 + 2.6(m_{2})$

    $218.5 = 2.6(m_{2}) \implies m_{2} = 84\,kg$

Gun Recoil

A $2.5\,kg$ gun fires a $15\,g$ ($0.015\,kg$) bullet at $900\,m/s$. Momentum conservation from rest:

$0 = m_{gun}v_{gun} + m_{bullet}v_{bullet}$

$0 = (2.5\,kg)(v_{gun}) + (0.015\,kg)(900\,m/s)$

$2.5(v_{gun}) = -13.5 \implies v_{gun} = -5.4\,m/s$

Work-Energy Theorem and Friction

Friction performs negative work on a system, reducing kinetic energy until an object comes to a stop ($W = \Delta KE$).

Stopping Distance of a Car

An $800\,kg$ car traveling at $30\,m/s$ skids to a stop ($v_{f} = 0$). Friction provides a constant force of $3,000\,N$. The work done by friction equals the change in kinetic energy:

$F_{f} \Delta x = \frac{1}{2}mv_{i}^2$

$(3,000\,N)(\Delta x) = \frac{1}{2}(800\,kg)(30\,m/s)^2$

$3,000(\Delta x) = 400(900) = 360,000$

$\Delta x = 120\,m$

Calculating Initial Speed from Skid Marks

A $1700\,kg$ car leaves skid marks $25\,m$ long. The coefficient of friction ($\mu_{k}$) is $0.89$. The frictional force is $F_{f} = \mu_{k} m g$. The work done by friction is:

$W = \mu_{k} m g \Delta x = \frac{1}{2}mv_{i}^2$

Dividing by mass ($m$) on both sides:

$\mu_{k} g \Delta x = \frac{1}{2}v_{i}^2$

$(0.89)(9.8\,m/s^2)(25\,m) = 0.5 v_{i}^2$

$218.05 = 0.5 v_{i}^2$

$v_{i}^2 = 436.1 \implies v_{i} = 20.88\,m/s$