Step | Derivation/Formula | Reasoning |
---|---|---|
1 | m_1 = 0.7 \, \text{kg}, \quad v_{1i} = 2 \, \text{m/s} | Assign mass and initial velocity to the cue ball. |
2 | m_2 = 0.5 \, \text{kg}, \quad v_{2i} = -1.2 \, \text{m/s} | Assign mass and initial velocity to the second ball. The velocity is negative since it moves in the opposite direction. |
3 | v_{1f} = -0.3 \, \text{m/s} | Cue ball’s final velocity post-collision; given as moving in the opposite direction. |
4 | m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} | Use the law of conservation of momentum since there are no external forces. |
5 | (0.7 \times 2) + (0.5 \times -1.2) = (0.7 \times -0.3) + (0.5 \times v_{2f}) | Substitute known values into the momentum conservation equation. |
6 | 1.4 – 0.6 = -0.21 + 0.5 v_{2f} | Calculate the values on each side using simple arithmetic. |
7 | 0.8 = -0.21 + 0.5 v_{2f} | Simplify to set up the equation to solve for v_{2f}. |
8 | 0.5 v_{2f} = 0.8 + 0.21 | Move terms involving v_{2f} to one side and constants to the other. |
9 | 0.5 v_{2f} = 1.01 | Add the constants to find the new value. |
10 | v_{2f} = \frac{1.01}{0.5} | Divide both sides by 0.5 to isolate v_{2f}. |
11 | v_{2f} = 2.02 \, \text{m/s} | The final velocity of the second ball post-collision. |
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A block with mass m slides at speed v_0 on a smooth surface and hits a stationary block with mass M . They stick together and move at speed v_0/3 . Find M in terms of m .
A block of mass m is moving on a horizontal frictionless surface with a speed v_0 as it approaches a block of mass 2m which is at rest and has an ideal spring attached to one side.
When the two blocks collide, the spring is completely compressed and the two blocks momentarily move at the same speed, and then separate again, each continuing to move.
Block 2 initially is at rest. Block 1 travels towards block 2 and collides with Block 2 as shown above. Find the final velocities of both blocks assuming the collision is elastic.
A 75.0kg log floats downstream with a speed of 1.80 m/s. Eight frogs hop onto the log in a series of perfectly inelastic collisions. If each frog has a mass of 0.30 kg and an upstream speed of 1.3 m/s, what is the change in kinetic energy for this system?
A 4 kg mass is traveling at 10 m/s to the right when it collides inelastically with a stationary 7 kg mass. The 7 kg mass then travels at 2m/s at an angle of 22° below the horizontal. What are the velocity and the angle of the 4 kg mass?
2 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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