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# Part A: Calculate velocity of the bullet + wooden block just after impact
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]m_1 = 0.0500 \, \text{kg}[/katex] | Mass of the bullet. |
| 2 | [katex]u_1 = 50.0 \, \text{m/s}[/katex] | Initial velocity of the bullet. |
| 3 | [katex]m_2 = 0.300 \, \text{kg}[/katex] | Mass of the wooden block. |
| 4 | [katex]u_2 = 0 \, \text{m/s}[/katex] | Initial velocity of the wooden block (at rest). |
| 5 | [katex]v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}[/katex] | Formula for the final velocity after a completely inelastic collision, where [katex] m_1 [/katex] and [katex] m_2 [/katex] are masses of the bullet and block, [katex] u_1 [/katex] and [katex] u_2 [/katex] are their initial velocities respectively. |
| 6 | [katex]v = \frac{(0.0500 \, \text{kg}) (50.0 \, \text{m/s}) + (0.300 \, \text{kg}) (0 \, \text{m/s})}{0.0500 \, \text{kg} + 0.300 \, \text{kg}}[/katex] | Substitute the values of [katex] m_1 [/katex], [katex] m_2 [/katex], [katex] u_1 [/katex], and [katex] u_2 [/katex] into the formula. |
| 7 | [katex]\mathbf{v = 7.14 \, \text{m/s}}[/katex] | Compute to find the final velocity of the bullet + wooden block system just after the collision. |
# Part B: Calculate the vertical distance [katex] h [/katex] reached by the bullet and wooden block
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]v = 7.14 \, \text{m/s}[/katex] | Initial velocity of the system after impact (from part A). |
| 2 | [katex]v^2 = u^2 + 2gh[/katex] | Use the kinematic equation relating initial velocity ([katex]u[/katex]), final velocity ([katex]v[/katex]), acceleration due to gravity ([katex]g[/katex]), and height ([katex]h[/katex]). Since the system comes to a stop at the highest point, final velocity [katex]v[/katex] is 0. |
| 3 | [katex]0 = (7.14 \, \text{m/s})^2 + 2(-9.81 \, \text{m/s}^2)h[/katex] | At the highest point the final velocity is 0, and acceleration due to gravity (a negative value, since it acts downwards) impacts the rising object. |
| 4 | [katex]h = \frac{(7.14 \, \text{m/s})^2}{2 \times 9.81 \, \text{m/s}^2}[/katex] | Rearrange to solve for [katex]h[/katex]. |
| 5 | [katex]\mathbf{h = 2.59 \, \text{m}}[/katex] | Compute the vertical height. |
# Part C: Compare effects of doubling the bullet’s mass vs. velocity on [katex] h [/katex]
| Comparison | Analysis | Conclusion |
|---|---|---|
| Doubling bullet’s velocity | [katex] h_f \propto v^2 [/katex], doubling [katex] v [/katex] would quadruple [katex] h [/katex] (since [katex] h_f = \frac{4v^2}{2g} [/katex]) | Increase in velocity has a significant effect, leading to an increase by a factor of four for [katex] h [/katex]. |
| Doubling bullet’s mass | [katex] h_f [/katex] determined more by [katex] v [/katex] (system velocity) than by mass directly. Increasing mass lowers system velocity, hence slightly lowering [katex] h [/katex]. | Little to no increase in [katex] h [/katex]; could actually decrease due to lower velocity after collision. |
| None | Analysis shows that velocity change impacts height significantly compared to mass change. | (ii) Doubling the bullet’s velocity has a greater effect on height than doubling the bullet’s mass. |
Just ask: "Help me solve this problem."
A 90 kg individual is cycling up a hill inclined at 30 degrees on a 12 kg bicycle. The hill is quite steep, and the coefficient of static friction is 0.85. The cyclist ascends 12 meters up the hill and then pauses at the summit. If they then start descending from the peak at rest and travel 9 meters before firmly applying the brakes, causing the wheels to lock.
An apple is released from rest 500 m above the ground. Due to the combined forces of air resistance and gravity, it has a speed of 40 m/s when it reaches the ground. What percentage of the initial mechanical energy of the apple-Earth system was dissipated due to air resistance? Take the potential energy of the apple-Earth system to be zero when the apple reaches the ground.
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?
A box of mass \(m\) is initially at rest at the top of a ramp that is at an angle \(\theta\) with the horizontal. The block is at a height \(h\) and length \(L\) from the bottom of the ramp. The coefficient of kinetic friction between the block and the ramp is \(\mu\). What is the kinetic energy of the box at the bottom of the ramp?
A karate master is about to split a piece of wood with her hand. Select all she must do in order to deliver the maximum force to split the wood.
(a) 7.1 m/s
(b) 2.6 m
(c) ii
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque and Rotations |
|---|---|
| \(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |
| \(J = \Delta p\) | \(I = \sum mr^2\) |
| \(p_i = p_f\) | \(L = I \cdot \omega\) |
| Simple Harmonic Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| Constant | Description |
|---|---|
| [katex]g[/katex] | Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface |
| [katex]G[/katex] | Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex] |
| [katex]\mu_k[/katex] and [katex]\mu_s[/katex] | Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion. |
| [katex]k[/katex] | Spring constant, in [katex]\text{N/m}[/katex] |
| [katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex] | Mass of the Earth |
| [katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex] | Mass of the Moon |
| [katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex] | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| [katex]s[/katex] (Displacement) | [katex]\text{meters (m)}[/katex] |
| [katex]v[/katex] (Velocity) | [katex]\text{meters per second (m/s)}[/katex] |
| [katex]a[/katex] (Acceleration) | [katex]\text{meters per second squared (m/s}^2\text{)}[/katex] |
| [katex]t[/katex] (Time) | [katex]\text{seconds (s)}[/katex] |
| [katex]m[/katex] (Mass) | [katex]\text{kilograms (kg)}[/katex] |
| Variable | Derived SI Unit |
|---|---|
| [katex]F[/katex] (Force) | [katex]\text{newtons (N)}[/katex] |
| [katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy) | [katex]\text{joules (J)}[/katex] |
| [katex]P[/katex] (Power) | [katex]\text{watts (W)}[/katex] |
| [katex]p[/katex] (Momentum) | [katex]\text{kilogram meters per second (kgm/s)}[/katex] |
| [katex]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/s)}[/katex] |
| [katex]\tau[/katex] (Torque) | [katex]\text{newton meters (Nm)}[/katex] |
| [katex]I[/katex] (Moment of Inertia) | [katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex] |
| [katex]f[/katex] (Frequency) | [katex]\text{hertz (Hz)}[/katex] |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: [katex]\text{5 km}[/katex]
Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]
Perform the multiplication: [katex]\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}[/katex]
Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | [katex]10^{-12}[/katex] | |
Nano- | n | [katex]10^{-9}[/katex] | |
Micro- | µ | [katex]10^{-6}[/katex] | |
Milli- | m | [katex]10^{-3}[/katex] | |
Centi- | c | [katex]10^{-2}[/katex] | |
Deci- | d | [katex]10^{-1}[/katex] | |
(Base unit) | – | [katex]10^{0}[/katex] | |
Deca- or Deka- | da | [katex]10^{1}[/katex] | |
Hecto- | h | [katex]10^{2}[/katex] | |
Kilo- | k | [katex]10^{3}[/katex] | |
Mega- | M | [katex]10^{6}[/katex] | |
Giga- | G | [katex]10^{9}[/katex] | |
Tera- | T | [katex]10^{12}[/katex] |
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