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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ m_{\text{bullet}}\, v_i = \left(m_{\text{bullet}} + m_{\text{block}}\right)\, v_x \] | Apply conservation of momentum for this inelastic collision where the bullet embeds in the block. |
| 2 | \[ 0.0350 \times 425 = \left(0.0350 + 0.550\right)\, v_x \] | Substitute the given values: the bullet mass is \(0.0350\) kg, its velocity is \(425\) m/s, and the block mass is \(0.550\) kg. |
| 3 | \[ v_x = \frac{0.0350 \times 425}{0.0350 + 0.550} = \frac{14.875}{0.585} \] | Compute the bullet’s momentum \(0.0350 \times 425 = 14.875\) and the total mass \(0.0350 + 0.550 = 0.585\) kg to solve for \(v_x\). |
| 4 | \[ \boxed{v_x \approx 25.4 \; \text{m/s}} \] | The velocity of the bullet and block together right after the collision is approximately \(25.4\) m/s. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ a = -\mu_k\, g \] | Friction produces a deceleration given by the product of the kinetic friction coefficient \(\mu_k\) and gravitational acceleration \(g\). The negative sign indicates deceleration. |
| 2 | \[ a = -0.40 \times 9.80 = -3.92 \; \text{m/s}^2 \] | Substitute \(\mu_k = 0.40\) and \(g = 9.80 \; \text{m/s}^2\) to calculate the acceleration. |
| 3 | \[ v_x^2 = (25.4)^2 + 2\,(-3.92)\,(10.0) \] | Use the kinematic equation where the initial velocity is the \(25.4\) m/s from part (a) and the displacement \(\Delta x\) is \(10.0\) m. |
| 4 | \[ v_x^2 \approx 645.16 – 78.4 = 566.76 \] | Simplify the expression by computing \((25.4)^2 \approx 645.16\) and \(2 \times 3.92 \times 10.0 = 78.4\). |
| 5 | \[ v_x \approx \sqrt{566.76} \approx 23.8 \; \text{m/s} \] \quad \text{or} \quad \boxed{v_x \approx 23.8 \; \text{m/s}} \] |
Taking the square root yields the final velocity after sliding \(10.0\) m: approximately \(23.8\) m/s. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ m_1\, v_i = \left(m_1 + m_2\right)\, v_x \] | Apply conservation of momentum for the second collision where the bullet-embedded block (\(m_1 = 0.585\) kg) collides inelastically with a stationary block (\(m_2 = 2.50\) kg). |
| 2 | \[ v_x = \frac{0.585 \times 23.8}{0.585 + 2.50} \] | Substitute \(v_i = 23.8\) m/s from part (b) and add the masses \(0.585\) kg and \(2.50\) kg for the collision. |
| 3 | \[ v_x \approx \frac{13.923}{3.085} \approx 4.51 \; \text{m/s} \] | Calculate the post-collision velocity; the numerator \(0.585 \times 23.8 \approx 13.923\) and the total mass is \(3.085\) kg. |
| 4 | \[ 0 = (4.51)^2 + 2\,(-3.92)\,d \] | Use the kinematic equation to find the distance \(d\) traveled before coming to a stop, with \(a = -3.92\) m/s² due to friction. |
| 5 | \[ d = \frac{(4.51)^2}{2 \times 3.92} \] | Solve for \(d\) by rearranging the kinematics equation. |
| 6 | \[ \boxed{d \approx 2.60 \; \text{m}} \] | Evaluating the expression gives a stopping distance of approximately \(2.60\) m after the collision. |
Just ask: "Help me solve this problem."
List at least 2 everyday forces that are not conservative, and explain why they aren’t.
A 6 kg cube rests against a compressed spring with a force constant of 1,800 N/m, initially compressed by 0.3 m. Upon release, the cube slides on a horizontal surface with a kinetic friction coefficient of 0.12 for 3 m, then ascends a 12° slope, stopping after 4.5 m. Determine the coefficient of kinetic friction on the slope.
An average adult elephant \( (5000 \, \text{kg}) \) is strapped to a spring, which is then pulled \( 2 \, \text{meters} \) away from its equilibrium position and released. The elephant starts oscillating back and forth with a period of \( 10 \) seconds.
Two identical blocks are connected to the opposite ends of a compressed spring. The blocks initially slide together on a frictionless surface with velocity \( v \) to the right. The spring is then released by remote control. At some later instant, the left block is moving at \( \frac{v}{2} \) to the left, and the other block is moving to the right. What is the speed of the center of mass of the system at that instant?
Two balls are dropped from the roof of a building. One ball has twice as massive as the other and air resistance is negligible. Just before hitting the ground, the more massive ball has ball ____ the kinetic energy of the less massive ball.
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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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