TA Practice Problems Classical Mechanics 1 Chapter 8: Conservation of Energy

Fundamental Definitions and Orbital Mechanics

• Conservative Forces: When a physical force is not path-dependent (meaning the work done by the force depends only on the initial and final positions, not the specific route taken), it is defined as a conservative force.

• Escape Velocity and Mass: The escape velocity ($V_{esc}$) of a rocket or any projectile does not depend on its overall mass (the combined mass of the rocket and its fuel). This is considered a false statement in physics because escape velocity is determined by the mass and radius of the celestial body being escaped from.

• Relationship Between Escape and Circular Velocity: The relationship between escape velocity ($V_{esc}$) and circular orbit velocity ($V_{circ}$) is defined by the following mathematical expression:   - $V_{esc} = V_{circ} imes ext{coefficient}$   - The correct relation is $V_{esc} = V_{circ} imes ext{sqrt}(2)$, which can be written as $V_{esc} = V_{circ} imes rac{ ext{sqrt}(2) imes 2}{2}$ or simply expressed as $V_{esc} = V_{circ} imes ext{sqrt}(2)$.

Conservation of Energy: Sifat and Salman Case Studies

• Scenario 1: The Mountain Chase   - Character Instance: Sifat is on the top of a rocky mountain standing alone. Upon seeing Salman, he begins a chase.   - Parameters: The track is frictionless; Sifat's mass ($m$) is $75\,kg$.   - Energy at Point A (Top of Mountain): Total energy is calculated based on height and potential energy. At Point A, the total energy is calculated to be $7350\,J$ (approximated as $7300\,J$, $7400\,J$, or $7200\,J$ in multiple choice contexts).   - Potential Energy at Point B: For a lower elevation (e.g., $2\,m$), potential energy ($U$) equals $m \times g \times h$. For Sifat, this is $75\,kg \times 9.8\,m/s^2 \times 2.0\,m = 1470\,J$.   - Speed at Point B: Derived from the conservation of mechanical energy ($K_i + U_i = K_f + U_f$). Given the elevation changes, the speed at point B is calculated as $12.6\,m/s$.

• Scenario 2: The Roof Fight   - Character Instance: Sifat and Salman are on a roof $100\,m$ above ground level. Salman throws a snowball at Sifat.   - Parameters: Snowball mass ($m$) is $0.50\,kg$; initial velocity ($v_0$) is $20\,m/s$.   - Total Mechanical Energy at the Roof:     - $E_{total} = KE + PE$     - $E_{total} = (\frac{1}{2} \times 0.50\,kg \times (20\,m/s)^2) + (0.50\,kg \times 9.8\,m/s^2 \times 100\,m)$     - $E_{total} = 100\,J + 490\,J = 590\,J$.   - Impact Speed: The speed at which the snowball hits a car on the ground is determined by setting final kinetic energy equal to initial total mechanical energy (neglecting air resistance):     - $590\,J = \frac{1}{2} \times 0.50\,kg \times v^2$     - $v = \text{sqrt}(\frac{590\,J \times 2}{0.50\,kg}) = \text{sqrt}(2360) \approx 48.6\,m/s$.

Comparative Motion: Anik and Alvi at Splish Splash

• Max Trax Ride Competition: Anik and Alvi are competing on a water ride. Anik is heavier than Alvi.

• Speed Comparison: At the bottom of the slide, both Anik and Alvi will have the same speed. In a frictionless environment (or where friction is proportional to mass), speed depends only on height ($v = \text{sqrt}(2gh)$), meaning mass cancels out of the equation.

• Race Outcome: In an idealized physics problem, they reach simultaneously, but in realistic conditions involving air resistance, the heavier person (Anik) typically maintains speed better against drag.

Work, Friction, and Power Calculations

• Kinetic Friction on a Pulled Box:   - Parameters: Box mass $m = 60\,kg$; Distance $d = 5\,m$; Pulling force $F = 700\,N$; Angle $\theta = 45^\circ$; Net work $W_{net} = 2300\,J$.   - Calculating Coefficient of Friction (\mu_k): The net work is the sum of the work done by the applied force and the work done by friction ($W_{net} = W_{app} + W_{fric}$).   - $W_{app} = F \times d \times \cos(45^\circ) = 700\,N \times 5\,m \times 0.707 = 2474.5\,J$.   - $W_{fric} = W_{net} - W_{app} = 2300\,J - 2474.5\,J = -174.5\,J$.   - Friction force $f_k = \frac{|W_{fric}|}{d} = \frac{174.5\,J}{5\,m} = 34.9\,N$.   - Normal force $N = mg - F \sin(45^\circ) = (60\,kg \times 9.8\,m/s^2) - (700\,N \times 0.707) = 588\,N - 494.9\,N = 93.1\,N$.   - $\mu_k = \frac{f_k}{N} = \frac{34.9}{93.1} \approx 0.38$.

• Naima's Stair Run (Power and Energy):   - Parameters: Mass $m = 58\,kg$; Vertical height $h = 5\,m$; Time $t = 5\,s$.   - Energy Required: Energy is equal to the work done against gravity ($W = mgh$).     - $E = 58\,kg \times 9.8\,m/s^2 \times 5\,m = 2842\,J$ (approximated as $2800\,J$ or $2900\,J$).   - Power in Watts: Power ($P$) is the rate of doing work.     - $P = \frac{W}{t} = \frac{2842\,J}{5\,s} = 568.4\,W$ (approximated as $570\,W$ or $560\,W$).

Multi-Stage Mechanical Systems: The Truck and Spring

• System Description: A truck features a frictionless curved quadrant (AB) with radius $R = 2.0\,m$, followed by a horizontal span (BC) of length $3.0\,m$ with friction ($\mu_k = 0.19$), and a frictionless spring section (CD).

• Initial State: A block of mass $m = 1\,kg$ is released from rest at point A.

• Calculations:   - Velocity at point B: Using conservation of energy ($mgh = \frac{1}{2}mv^2$), where $h = R = 2.0\,m$.     - $v_B = \text{sqrt}(2 \times 9.8\,m/s^2 \times 2.0\,m) = \text{sqrt}(39.2) \approx 6.26\,m/s$.   - Work Done by Friction (B to C): $W_{fric} = -\mu_k \times m \times g \times d_{BC}$.     - $W_{fric} = -0.19 \times 1\,kg \times 9.8\,m/s^2 \times 3.0\,m = -5.586\,J$.   - Speed at Point C: Using the work-energy theorem ($\frac{1}{2}m v_B^2 + W_{fric} = \frac{1}{2}m v_C^2$).     - $19.6\,J - 5.586\,J = 14.014\,J$     - $v_C = \text{sqrt}(\frac{2 \times 14.014\,J}{1\,kg}) \approx 5.29\,m/s$.   - Stiffness Constant (k): The remaining kinetic energy at C is converted into spring potential energy ($\frac{1}{2} k \Delta x^2$) with compression $\Delta x = 0.25\,m$.     - $14.014\,J = \frac{1}{2} \times k \times (0.25\,m)^2$     - $k = \frac{14.014\,J \times 2}{0.0625} = 448.45\,N/m$.

Hanging Rope Dynamics: Sifat at the Lake

• Initial Motion: Sifat runs at $5\,m/s$ and grabs a hanging rope. He swings out over a lake and releases when his velocity reaches zero.

• Parameters: Mass $m = 56\,kg$; Initial velocity $v_i = 5\,m/s$.

• Energy Analysis:   - Initial Mechanical Energy: Determined at the moment he touches the rope.     - $E_i = KE_i = \frac{1}{2} \times 56\,kg \times (5\,m/s)^2 = 700\,J$.   - Final Mechanical Energy: At the release point ($v = 0$), energy is entirely potential ($E_f = mgh$). Due to conservation, $E_f = E_i = 700\,J$.   - Release Height: $h = \frac{E_f}{mg} = \frac{700\,J}{56\,kg \times 9.8\,m/s^2} \approx 1.275\,m$ relative to the lowest point of the swing.

• Geometry of the Swing:   - The height can be expressed in terms of rope length ($l$) and angle ($\theta$) as:     - $h = l - l \cos(\theta) = l(1 - \cos(\theta))$   - Release Angle: Found by equating the calculated height with the geometric height: $\cos(\theta) = 1 - \frac{h}{l}$.

• Tension Calculations:   - At Release Point: Since velocity is zero, tension ($T$) only counteracts the component of gravity:     - $T = mg \cos(\theta)$.   - At Catch Point (Bottom): Sifat is moving at $5\,m/s$ in a circular path. The tension must provide the centripetal force and support his weight:     - $T = mg + \frac{mv^2}{l}$.

Two-Bucket Linked System

• System Configuration: Two buckets connected by a light-weight rope. A $12\,kg$ bucket is positioned $2\,m$ above the floor. One mass is $m$ and the other is likely defined by the specific diagram as $M$.

• Energy Expressions:   - Initial Mechanical Energy: $E_i = m_1 g h_1 + m_2 g h_2$. If released from rest, initial $KE = 0$.   - Final Mechanical Energy: $E_f = PE_f + KE_f = PE_f + \frac{1}{2}(m_1 + m_2)v^2$.

• Impact Speed: Calculated using $\Delta PE = \Delta KE$. As the $12\,kg$ mass descends $2\,m$, the other mass rises $2\,m$. The change in potential energy is converted into kinetic energy for both masses.

Popy's Roller Coaster Physics

• Parameters: Total mass (car + Popy) = $500\,kg$. Point A is at height $h_A = 70\,m$ with velocity $v_A = 10\,m/s$.

• Calculations for Different Points:   - Total Mechanical Energy at A:     - $E_A = (\frac{1}{2} \times 500 \times 10^2) + (500 \times 9.8 \times 70) = 25,000 + 343,000 = 368,000\,J$.   - Speed at point B/C: Conserving energy ($E_A = E_B = E_C$). Speed at any point is $v = \text{sqrt}(\frac{2(E_{total} - mgh)}{m})$.

• Loop Dynamics (Point E):   - Normal Force at E: At the top of a loop of radius $r$, the net force provides centripetal acceleration:     - $N + mg = \frac{mv_E^2}{r}$ therefore $N = \frac{mv_E^2}{r} - mg$.

• Friction Track (Exit D):   - Deceleration: After point D, the car enters a track with $\mu_k = 0.2$.   - Stopping Distance: Found by setting the kinetic energy at D equal to the work done by friction ($KE_D = \mu_k m g d$).     - $d = \frac{v_D^2}{2 \mu_k g}$.

Orbital Mechanics: Alvi and the ISS

• Constants:   - Gravitational Constant ($G$): $6.67 \times 10^{-11}\,Nm^2kg^{-2}$   - Mass of Earth ($M$): $5.9 \times 10^{24}\,kg$   - Radius of Earth ($R$): $6.3 \times 10^6\,m$

• Mission Parameters: Alvi ($250\,kg$) is at distance $x$ below the ISS. The ISS is at altitude $h = 200\,km$ ($200,000\,m$). Alvi has a jet pack velocity of $4\,m/s$.

• Gravitational Potential Energy (GPE):   - Formula: $U_G = -\frac{G M m}{r}$.   - At point A (radial distance $R + h - x$): $U_A = -\frac{G M m}{R + h - x}$.   - At point B (ISS altitude): $U_B = -\frac{G M m}{R + h}$.

• Energy Conservation for Safe Reach: To reach the ISS, the total energy at A (Kinetic + Potential) must equal the total energy at B.   - $\frac{1}{2}mv_A^2 - \frac{G M m}{R_A} = \frac{1}{2}mv_B^2 - \frac{G M m}{R_B}$.   - If reaching the ISS with near-zero relative velocity, the required distance $x$ or required escape/transfer velocity can be calculated. The value for $x$ is typically very small relative to the Earth's radius.