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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."
| Experiment | Initial Velocity of Cart X \( (\text{m/s}) \) | Initial Velocity of Cart Y \( (\text{m/s}) \) | Final Velocity of Cart X \( (\text{m/s}) \) | Final Velocity of Cart Y \( (\text{m/s}) \) |
|---|---|---|---|---|
| \( 1 \) | \( 1 \) | \( 0 \) | \( 0 \) | \( 1 \) |
| \( 2 \) | \( 1 \) | \( -1 \) | \( -1 \) | \( 1 \) |
| \( 3 \) | \( 2 \) | \( 1 \) | \( 1 \) | \( 2 \) |
A student performs several experiments in which two carts collide as they travel along a horizontal surface. Cart X and Cart Y both have a mass of \( 1 \) \( \text{kg} \). Data collected from the three experiments are shown in the table above. During which experiment does the center of mass of the system of two carts have the greatest change in its momentum?
A rubber ball and a lump of clay have equal mass. They are thrown with equal speed against a wall. The ball bounces back with nearly the same speed with which it hit. The clay sticks to the wall. Which one of these objects experiences the greater impulse?
A platform is initially rotating on smooth ice with negligible friction, as shown above. A stationary disk is dropped directly onto the center of the platform. A short time later, the disk and platform rotate together at the same angular velocity, as shown at right in the figure. How does the angular momentum of only the platform change, if at all, after the disk drops? And what is the best justification.
A simple pendulum consists of a sphere tied to the end of a string of negligible mass. The sphere is pulled back until the string is horizontal and then released from rest. Assume the gravitational potential energy is zero when the sphere is at its lowest point.
What angle will the string make with the horizontal when the kinetic energy and the potential energy of the sphere-Earth system are equal?
A \( 25.0 \) \( \text{kg} \) block is placed at the top of an inclined plane set at an angle of \( 35 \) degrees to the horizontal. The block slides down the \( 1.5 \) \( \text{m} \) slope at a constant rate. How much work did friction do on the block?
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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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