Objective: Calculate the final speed of the space probe after firing its rockets and burning fuel.
Step | Formula Derivation | Reasoning |
---|---|---|
1 | F = ma | Newton’s second law, force equals mass times acceleration. |
2 | a = \frac{F}{m} | Rearrange to solve for acceleration, a. |
3 | a = \frac{156 \times 10^3\text{ N}}{1312\text{ kg}} | Substitute the thrust (F) and initial mass of the probe (m). Note: Mass will change, this is initial acceleration. |
4 | a_{\text{initial}} = 118.9\text{ m/s}^2 | Calculate the initial acceleration. Note: This will change as the probe burns fuel. |
5 | \Delta v = a \Delta t | The change in velocity is the product of acceleration and time. |
6 | v_{\text{final}} = v_{\text{initial}} + \Delta v | The final velocity is the initial velocity plus the change in velocity. |
The acceleration will not be constant because the mass of the probe changes as it burns fuel. However, we can integrate the force over the time to find the change in velocity, considering the change in mass.
For the Change in Velocity (Variable Mass)
Step | Formula Derivation | Reasoning |
---|---|---|
1 | \Delta v = \int_{0}^{t} \frac{F}{m(t)} dt | With variable mass, acceleration changes over time. |
2 | m(t) = m_{\text{initial}} – \frac{dm}{dt}t | Mass as a function of time, where \frac{dm}{dt} is the rate of mass loss. |
3 | \frac{dm}{dt} = \frac{150\text{ kg}}{220\text{ s}} | Calculate the rate of mass loss. |
4 | \Delta v = \int_{0}^{220\text{ s}} \frac{156 \times 10^3\text{ N}}{1312\text{ kg} – \left(\frac{150\text{ kg}}{220\text{ s}}\right)t} dt | Substitute m(t), F, and burn time into the integral. |
This integral represents the change in momentum over the time period, taking into account the loss of mass from the fuel burn. Let’s calculate the integral to find the change in velocity and then determine the final speed of the space probe.
Final speed of the space probe: \boxed{v_{\text{final}} = 42.58 \text{ km/s}}
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If you want to double the momentum of a gas molecule, by what factor must you increase its kinetic energy?
For linear motion the term “inertia” refers to the same physical concept of
42.58 km/s
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Kinematics | Forces |
---|---|
\Delta x = v_i t + \frac{1}{2} at^2 | F = ma |
v = v_i + at | F_g = \frac{G m_1m_2}{r^2} |
a = \frac{\Delta v}{\Delta t} | f = \mu N |
R = \frac{v_i^2 \sin(2\theta)}{g} |
Circular Motion | Energy |
---|---|
F_c = \frac{mv^2}{r} | KE = \frac{1}{2} mv^2 |
a_c = \frac{v^2}{r} | PE = mgh |
KE_i + PE_i = KE_f + PE_f |
Momentum | Torque and Rotations |
---|---|
p = m v | \tau = r \cdot F \cdot \sin(\theta) |
J = \Delta p | I = \sum mr^2 |
p_i = p_f | L = I \cdot \omega |
Simple Harmonic Motion |
---|
F = -k 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} |
M_E = 5.972 \times 10^{24} , \text{kg} | Mass of the Earth |
M_M = 7.348 \times 10^{22} , \text{kg} | Mass of the Moon |
M_M = 1.989 \times 10^{30} , \text{kg} | Mass of the Sun |
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 (kgm/s)} |
\omega (Angular Velocity) | \text{radians per second (rad/s)} |
\tau (Torque) | \text{newton meters (Nm)} |
I (Moment of Inertia) | \text{kilogram meter squared (kgm}^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} | 0.000000000001 |
Nano- | n | 10^{-9} | 0.000000001 |
Micro- | µ | 10^{-6} | 0.000001 |
Milli- | m | 10^{-3} | 0.001 |
Centi- | c | 10^{-2} | 0.01 |
Deci- | d | 10^{-1} | 0.1 |
(Base unit) | – | 10^{0} | 1 |
Deca- or Deka- | da | 10^{1} | 10 |
Hecto- | h | 10^{2} | 100 |
Kilo- | k | 10^{3} | 1,000 |
Mega- | M | 10^{6} | 1,000,000 |
Giga- | G | 10^{9} | 1,000,000,000 |
Tera- | T | 10^{12} | 1,000,000,000,000 |
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