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A rocket-powered hockey puck has a thrust of 4.40 N and a total mass of 1.00 kg . It is released from rest on a frictionless table, 2.10 m from the edge of a 2.10 m drop. The front of the rocket is pointed directly toward the edge. Assuming that the thrust of the rocket present for the entire time of travel, how far does the puck land from the base of the table?
2.81 meters from the base of the table
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | a = \frac{F}{m} | Calculating horizontal acceleration (a) from thrust (F) and mass (m). |
| 2 | s = ut + \frac{1}{2}at^2 | Distance formula for motion with constant acceleration. |
| 3 | 2.10, \text{m} = \frac{1}{2}a t_{\text{table}}^2 | Distance covered on the table is 2.10 m, initial velocity (u) is 0. |
| 4 | t_{\text{table}} = \sqrt{\frac{2 \times 2.10, \text{m}}{a}} | Solving for the time (t_{\text{table}}) taken to travel across the table. |
| 5 | v_{x} = at_{\text{table}} | Calculating the horizontal velocity (v_{x}) at the edge of the table. |
| 6 | v_{x} = 4.2988, m/s | Plug in values and solve. |
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | y = \frac{1}{2}gt_{\text{fall}}^2 | Distance formula for free fall, where y is the fall height and g is the acceleration due to gravity. |
| 2 | t_{\text{fall}} = \sqrt{\frac{2y}{g}} | Solving for the time (t_{\text{fall}}) to fall 2.10 m. |
| 3 | t_{\text{fall}} = .654, s | Plug in values and solve. |
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | d_{\text{air}} = v_{x} t_{\text{fall}} | Horizontal distance (d_{\text{air}}) traveled while falling, using the horizontal velocity at the edge of the table and the fall time. |
| 2 | d_{\text{air}} =4.3 * .65 | Plug in values to get d_{\text{air}} = 2.81, m. Final Answer. |
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Note: This is a bonus question. Skip if you haven’t yet taken calculus.
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A block rests on a flat plane inclined at an angle of 30° with respect to the horizontal. What is the minimum coefficient of friction necessary to keep the block from sliding?
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Does it take longer to move up the distance d or back down the distance d? Or does it take the same amount of time?
Bonus Challenge: Repeat the problem but assume you are not given the initial speed, angle of incline, or µk.
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A baseball rolls off a 0.70 m high desk and strikes the floor 0.25 m away from the base of the desk. How fast was the ball rolling?
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2.81 meters from the base of the table
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| Kinematics | Forces |
|---|---|
| \Delta x = v_i \cdot t + \frac{1}{2} a \cdot t^2 | F = m \cdot a |
| v = v_i + a \cdot t | F_g = \frac{G \cdot m_1 \cdot m_2}{r^2} |
| a = \frac{\Delta v}{\Delta t} | f = \mu \cdot N |
| R = \frac{v_i^2 \cdot \sin(2\theta)}{g} |
| Circular Motion | Energy |
|---|---|
| F_c = \frac{m \cdot v^2}{r} | KE = \frac{1}{2} m \cdot v^2 |
| a_c = \frac{v^2}{r} | PE = m \cdot g \cdot h |
| KE_i + PE_i = KE_f + PE_f |
| Momentum | Torque and Rotations |
|---|---|
| p = m \cdot v | \tau = r \cdot F \cdot \sin(\theta) |
| J = \Delta p | I = \sum m \cdot r^2 |
| p_i = p_f | L = I \cdot \omega |
| Simple Harmonic Motion |
|---|
| F = -k \cdot x |
| T = 2\pi \sqrt{\frac{l}{g}} |
| T = 2\pi \sqrt{\frac{m}{k}} |
| Constant | Description |
|---|---|
| g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |
| G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |
| \mu_k and \mu_s | Coefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion. |
| k | Spring constant, in \text{N/m} |
| Variable | SI Unit |
|---|---|
| s (Displacement) | \text{meters (m)} |
| v (Velocity) | \text{meters per second (m/s)} |
| a (Acceleration) | \text{meters per second squared (m/s}^2\text{)} |
| t (Time) | \text{seconds (s)} |
| m (Mass) | \text{kilograms (kg)} |
| Variable | Derived SI Unit |
|---|---|
| F (Force) | \text{newtons (N)} |
| E, PE, KE (Energy, Potential Energy, Kinetic Energy) | \text{joules (J)} |
| P (Power) | \text{watts (W)} |
| p (Momentum) | \text{kilogram meters per second (kg·m/s)} |
| \omega (Angular Velocity) | \text{radians per second (rad/s)} |
| \tau (Torque) | \text{newton meters (N·m)} |
| I (Moment of Inertia) | \text{kilogram meter squared (kg·m}^2\text{)} |
| f (Frequency) | \text{hertz (Hz)} |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: \text{5 km}
Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}
Perform the multiplication: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}
Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | 10^{-12} | |
Nano- | n | 10^{-9} | |
Micro- | µ | 10^{-6} | |
Milli- | m | 10^{-3} | |
Centi- | c | 10^{-2} | |
Deci- | d | 10^{-1} | |
(Base unit) | – | 10^{0} | |
Deca- or Deka- | da | 10^{1} | |
Hecto- | h | 10^{2} | |
Kilo- | k | 10^{3} | |
Mega- | M | 10^{6} | |
Giga- | G | 10^{9} | |
Tera- | T | 10^{12} |
