Step | Formula Derivation | Reasoning |
---|---|---|
1 | F_f = \mu N | The force of friction, F_f, is the product of the coefficient of friction, \mu, and the normal force, N. |
2 | N = (m_1 + m_2)g | Normal force is the combined weight of the two cars post-collision. |
3 | F_f = \mu (m_1 + m_2)g | Combining formulas for friction force. |
4 | W = F_f d | Work done by friction, W, equals the force of friction times the distance slid, d. |
5 | \frac{1}{2}(m_1 + m_2)v^2 = \mu (m_1 + m_2)g d | The work done by friction equals the initial kinetic energy of the system post-collision. |
6 | v^2 = 2 \mu g d | Simplifying the equation for v, the combined speed post-collision. |
7 | v = \sqrt{2 \mu g d} | Solving for v. |
8 | m_1 v_1 = (m_1 + m_2)v | Conservation of momentum to relate the initial velocity of the 2,000 kg car to the combined velocity post-collision. |
9 | v_1 = \frac{(m_1 + m_2)v}{m_1} | Isolating v_1 for the initial velocity of the 2,000 kg car. |
10 | v = \sqrt{2 \cdot 0.7 \cdot 9.81 \cdot 6} | Plugging in values for \mu = 0.7, g = 9.81,m/s^2, and d = 6,m. |
11 | v \approx 9.077,m/s | Calculating the combined speed post-collision. |
12 | v_1 = \frac{2000,kg}{(2000,kg + 1000,kg)} \cdot 9.077,m/s | Use conservation of momentum. Plug in masses and the combined speed to find the initial velocity of the 2,000 kg car. |
13 | v_1 \approx 13.616,m/s | Calculating the initial velocity of the 2,000 kg car. |
Phy can also check your working. Just snap a picture!
A car decelerates from 25 \, m/s to 5 \, m/s at 10 \, m/s^2. How far does the car travel during this deceleration?
Three blocks of masses m3 = 1.0, m2 = 2.0, and m1 = 4.0 kilograms are connected by massless strings, one of which passes over a frictionless pulley of negligible mass, as shown above.
A 70 kg woman and her 35 kg son are standing at rest on an ice rink, as shown above. They push against each other for a time of 0.60 s, causing them to glide apart. The speed of the woman immediately after they separate is 0.55 m/s.
Assume that during the push, friction is negligible compared with the forces the people exert on each other.
An object travels along a path shown above, with changing velocity as indicated by vectors A and B. Which vector best represents the net acceleration of the object from time t_A to t_B?
Jill does twice as much work as Jack does and in half the time. Jill’s power output is
13.616 m/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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