Velocity just after the collision
Step | Formula Derivation | Reasoning |
---|---|---|
1 | m_1v_1 + m_2v_2 = (m_1 + m_2)v' | Conservation of momentum, where m_1 and m_2 are the masses of the bullet and block, v_1 and v_2 are their initial velocities, and v' is their final velocity. |
2 | v' = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} | Solve for v'. Given: m_1 = 0.0350 , \text{kg}, v_1 = 425 , \text{m/s}, m_2 = 0.550 , \text{kg}, v_2 = 0 , \text{m/s}. |
3 | v' = 25.43 , \text{m/s} | The velocity just after the collision |
Step | Formula Derivation | Reasoning |
---|---|---|
1 | v^2 = v'^2 + 2ad | Kinematic equation for motion under constant acceleration, where v is the final velocity, v' is the initial velocity, a is acceleration, and d is the distance. |
2 | a = -\mu_k g | Acceleration due to kinetic friction, where \mu_k is the coefficient of kinetic friction and g is the acceleration due to gravity (approximately 9.81 , \text{m/s}^2). Given: \mu_k = 0.40. |
3 | v = \sqrt{v'^2 + 2ad} | Solve for v. Given: v' = 25.43 , \text{m/s}, d = 10.0 , \text{m}. |
4 | v = 23.84 , \text{m/s} | Velocity after sliding 10.0 meters |
Distance traveled by the combined system (2 blocks and the bullet)
Step | Formula Derivation | Reasoning |
---|---|---|
1 | m_1v_1 + m_2v_2 = (m_1 + m_2)v' | Conservation of momentum for the collision between the bullet-block system and the second block, where m_1 and v_1 are the mass and velocity of the bullet-block system, m_2 and v_2 are the mass and velocity of the second block, and v' is the final velocity of the combined system. |
2 | v' = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} | Solve for v'. Given: m_1 = 0.585 , \text{kg}, v_1 = 23.84 , \text{m/s}, m_2 = 2.50 , \text{kg}, v_2 = 0 , \text{m/s}. |
3 | 0 = v'^2 + 2ad | Kinematic equation for motion under constant acceleration when the final velocity is 0. |
4 | d = \frac{-v'^2}{2a} | Solve for d. The acceleration a remains -\mu_k g as before. |
5 | v' = 4.52 , \text{m/s} | Final velocity of the combined system after the second collision |
6 | d = 2.60 , \text{m} | Distance traveled by the combined system before stopping. |
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Ball A of mass m is dropped from a building of height H. Ball B of mass 1.7m is dropped from a building of height 3.5H. Using energy, what the ratio of vA to vB (final velocity of ball A to final velocity of ball B). Air resistance is negligible.
A uniform solid cylinder of mass M and radius R is initially at rest on a frictionless horizontal surface. A massless string is attached to the cylinder and is wrapped around it. The string is then pulled with a constant force F , causing the cylinder to rotate about its center of mass. After the cylinder has rotated through an angle \theta , what is the kinetic energy of the cylinder in terms of F and \theta ?
A skier with a mass of 58 kg glides up a snowy incline that forms an angle of 28 degrees with the horizontal. The skier initially moves at a speed of 7.2 m/s. After traveling a distance of 2.3 meters up the slope, the skier’s speed reduces to 3.8 m/s.
A 2kg object slides east at 4 m/s and collides with a stationary 3 kg object. After the collision, the 2 kg object is traveling at an unknown velocity at 15° north of east and the 3 kg object is traveling at 38° south of east. What is each object’s final velocity?
A 100 kg person is riding a 10 kg bicycle up a 25° hill. The hill is long and the coefficient of static friction is 0.9. The person rides 10 m up the hill then takes a rest at the top. If she then starts from rest from the top of the hill and rolls down a distance of 7 m before squeezing hard on the brakes locking the wheels. How much work is done by friction to bring the bicycle to a full stop, knowing that the coefficient of kinetic friction is 0.65?
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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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