| 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. |
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From the figure above, determine which characteristic fits this collision best.
In a town’s water system, pressure gauges in still water at street level read \( 150 \) \( \text{kPa} \). If a pipeline connected to the system breaks and shoots water straight up, how high above the street does the water shoot?
A bullet of mass \(0.0500 \, \text{kg}\) traveling at \(50.0 \, \text{m/s}\) is fired horizontally into a wooden block suspended from a long rope. The mass of the wooden block is \(0.300 \, \text{kg}\) and it is initially at rest. The collision is completely inelastic and after impact the bullet + wooden block move together until the center of mass of the system rises a vertical distance \(h\) above its initial position.
The two blocks of masses \( M \) and \( 2M \) travel at the same speed \( v \) but in opposite directions. They collide and stick together. How much mechanical energy is lost to other forms of energy during the collision?
A bowling ball moving with speed \(v\) collides head-on with a stationary tennis ball. The collision is elastic and there is no friction. The bowling ball barely slows down. What is the speed of the tennis ball after the collision?
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 ball of mass \(m\) is released from rest at a distance \(h\) above a frictionless plane inclined at an angle of \(45^\circ\) to the horizontal as shown above. The ball bounces horizontally off the plane at point \(P_1\) with the same speed with which it struck the plane and strikes the plane again at point \(P_2\). In terms of \(g\) and \(h\), determine each of the following quantities:
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?
Block 2 initially is at rest. Block 1 travels towards block 2 and collides with Block 2 as shown above. Find the final velocities of both blocks assuming the collision is elastic.
A \( 1.5 \; \text{kg} \) mass attached to a spring with a force constant of \( 20.0 \; \text{N/m} \) oscillates on a horizontal, frictionless track. At \( t = 0 \), the mass is released from rest at \( x = 10.0 \; \text{cm} \). (That is, the spring is stretched by \( 10.00 \; \text{cm} \).)
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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] |
Metric Prefixes
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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