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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."
A spring launches a 4 kg block across a frictionless horizontal surface. The block then ascends a 30° incline with a kinetic friction coefficient of 0.25, stopping after 55 m on the incline. If the spring constant is 800 N/m, find the initial compression of the spring. Disregard friction while in contact with the spring.
A boy of mass \( m \) and a girl of mass \( 2m \) are initially at rest at the center of a frozen pond. They push each other so that she slides to the left at speed \( v \) across the frictionless ice surface and he slides to the right. What is the total work done by the children?
A force F is exerted by a broom handle on the head of a broom, which has a mass m. The handle is at an angle θ to the horizontal. The work done by the force on the head of the broom as it moves a distance d across a horizontal floor is:
A ski lift carries skiers along a 695 meter slope inclined at 34°. To lift a single rider, it is necessary to move 72 kg of mass to the top of the lift. Under maximum load conditions, five riders per minute arrive at the top. If 65 percent of the energy supplied by the motor goes to overcoming friction, what average power must the motor supply?
Ball \(A\) of mass \(m\) is dropped from a building of height \(H\). Ball \(B\) of mass \(1.7 \, \text{m}\) is dropped from a building of height \(1.7H\). Using energy, what the ratio of \(v_A\) to \(v_B\) (final velocity of ball \(A\) to final velocity of ball \(B\)). Air resistance is negligible.
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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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