| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ KE_{\text{initial}} = \frac{1}{2} m_b v_0^2 \] | Initial kinetic energy of the bullet of mass \( m_b \) moving with speed \( v_0 \). |
| 2 | \[ p_{\text{initial}} = m_b v_0 \] | Initial linear momentum of the bullet (the block is at rest). |
| 3 | \[ p_{\text{after}} = (m_b + m_w) \, v \] | After impact the bullet and wooden block (mass \( m_w \)) move together with speed \( v \); their momenta add because they form a single body. |
| 4 | \[ m_b v_0 = (m_b + m_w) \, v \] | During the short collision interval external impulses (tension, gravity) are negligible, so linear momentum is conserved. |
| 5 | \[ v = \frac{m_b}{m_b + m_w} \, v_0 \] | Solving the conservation-of-momentum equation for the common speed immediately after the collision. |
| 6 | \[ KE_{\text{after}} = \frac{1}{2}(m_b + m_w) v^2 = \frac{1}{2}\, \frac{m_b^2}{m_b + m_w} \, v_0^2 \] | Kinetic energy of the bullet–block system just after the perfectly inelastic collision (note \( KE_{\text{after}} < KE_{\text{initial}} \)). |
| 7 | \[ PE_{\text{max}} = (m_b + m_w) g h \] | Gravitational potential energy when the system rises to its maximum height \( h \) above the collision point; \( g \) is the acceleration due to gravity. |
| 8 | \[ \frac{1}{2}(m_b + m_w) v^2 = (m_b + m_w) g h \] | After the collision, while the system swings upward, mechanical energy is conserved (tension does no work and air resistance is neglected). Thus all post-collision kinetic energy converts to gravitational potential energy. |
| 9 | \[ \frac{KE_{\text{after}}}{KE_{\text{initial}}} = \frac{m_b}{m_b + m_w} \lt 1 \] | Only a fraction \( \tfrac{m_b}{m_b + m_w} \) of the bullet’s original kinetic energy reappears as potential energy; the rest is dissipated as heat, sound, and deformation during the inelastic impact. |
| 10 | Evaluation of statements | (a) False – kinetic energy is not conserved in an inelastic collision. (b) True – \( KE_{\text{after}} = PE_{\text{max}} \) (Step 8). (c) False – bullet alone does not conserve momentum; only the system does. (d) False – at the highest point velocity is zero, so momentum is zero. (e) False – Step 9 shows initial KE exceeds final PE. Therefore the correct statement is (b). |
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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 mass is attached to the end of a spring and set into simple harmonic motion with an amplitude \( A \) on a horizontal frictionless surface. Determine the following in terms of only the variable \( A \).
A typical \( 68 \text{-kg} \) person generates a steady mechanical power output of \( 120 \text{ W} \) at the pedals of a bicycle. Approximately how many Calories are “burned” (total metabolic energy expended) when the person rides a bicycle for \( 15 \text{ minutes} \)? A typical energy efficiency for the human body is \( 25\% \), which takes into account the release of thermal energy. Note (\( 1 \text{ Cal} = 4186 \text{ J} \)).
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.
A \(6 \, \text{kg}\) cube rests against a compressed spring with a force constant of \(1{,}800 \, \text{N/m}\), initially compressed by \(0.3 \, \text{m}\). Upon release, the cube slides on a horizontal surface with a kinetic friction coefficient of \(\mu_k = 0.12\) for \(3 \, \text{m}\), then ascends a \(12^\circ\) slope, stopping after \(4.5 \, \text{m}\). Determine the coefficient of kinetic friction on the slope.
A \( 0.0350 \) \( \text{kg} \) bullet moving horizontally at \( 425 \) \( \text{m/s} \) embeds itself into an initially stationary \( 0.550 \) \( \text{kg} \) block.
On a frictionless horizontal air table, puck A (with mass \( 0.249 \) \( \text{kg} \)) is moving toward puck B (with mass \( 0.375 \) \( \text{kg} \)), which is initially at rest. After the collision, puck A has velocity \( 0.115 \) \( \text{m/s} \) to the left, and puck B has velocity \( 0.645 \) \( \text{m/s} \) to the right.
A box having a mass of \( 1.5 \) \( \text{kg} \) is accelerated across a table at \( 1.5 \) \( \text{m/s}^2 \). The coefficient of kinetic friction on the box is \( 0.3 \).
You kick a ball straight up. Compare the sign of the work done by gravity on the ball while it goes up with the sign of the work done by gravity while it goes down.
A student is designing an experiment to find the spring constant \( k \) of a spring using only a set of known masses and a stopwatch. Which procedure would work?
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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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