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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[m v_i = (m+M) v_x\] | Linear momentum is conserved during the perfectly inelastic collision because no external horizontal forces act. The small block of mass \(m\) with speed \(v_i\) sticks to the block \(M\); both move together with speed \(v_x\). |
| 2 | \[v_x = \frac{m}{m+M}v_i\] | Algebraically solve for the common speed \(v_x\). |
| 3 | \[\boxed{v_x = \frac{m}{m+M}v}\] | Replace \(v_i\) by the given speed \(v\) of the incoming block. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\tfrac{1}{2}(m+M) v_x^2 = \tfrac{1}{2} k A^2\] | The kinetic energy of the joined masses right after impact transforms completely into spring potential energy at maximum compression (amplitude \(A\)). |
| 2 | \[A = v_x\sqrt{\frac{m+M}{k}}\] | Solve the energy equation for \(A\). |
| 3 | \[A = \frac{m v}{m+M}\sqrt{\frac{m+M}{k}}\] | Substitute the expression for \(v_x\) obtained in part (a). |
| 4 | \[\boxed{A = \frac{m v}{\sqrt{k\,(m+M)}}}\] | Simplify the radicals and fractions. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[T = 2\pi \sqrt{\frac{m+M}{k}}\] | The system now behaves as a simple mass–spring oscillator with effective mass \(m+M\) and spring constant \(k\). The standard formula for the period of such an oscillator is used. |
| 2 | \[\boxed{T = 2\pi \sqrt{\dfrac{m+M}{k}}}\] | Final expression for the period. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[E = \tfrac{1}{2}(m+M) v_x^2\] | Total mechanical energy after collision equals the kinetic energy just after impact; this energy stays constant and equals the maximum spring potential energy. |
| 2 | \[E = \tfrac{1}{2}(m+M)\left(\frac{m}{m+M}v\right)^2\] | Insert the value of \(v_x\) from part (a). |
| 3 | \[\boxed{E = \frac{m^2 v^2}{2\,(m+M)}}\] | Simplify the algebraic expression. |
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The launching mechanism of a toy gun consists of a spring with an unknown spring constant, \( k \). When the spring is compressed \( 0.120 \, \text{m} \) vertically, a \( 35.0 \, \text{g} \) projectile is able to be fired to a maximum height of \( 25 \, \text{m} \) above the position of the projectile when the spring is compressed. Assume that the barrel of the gun is frictionless.
A child pushes horizontally on a box of mass m with constant speed v across a rough horizontal floor. The coefficient of friction between the box and the floor is µ. At what rate does the child do work on the box?
A bullet moving with an initial speed of \( v_o \) strikes and embeds itself in a block of wood which is suspended by a string, causing the bullet and block to rise to a maximum height \( h \). Which of the following statements is true of the collision.
A horizontal force of \(110 \, \text{N}\) is applied to a \(12 \, \text{kg}\) object, moving it \(6 \, \text{m}\) on a horizontal surface where the kinetic friction coefficient is \(\mu_k = 0.25\). The object then slides up a \(17^\circ\) inclined plane. Assuming the \(110 \, \text{N}\) force is no longer acting on the incline, and the coefficient of kinetic friction there is \(\mu_k = 0.45\), calculate the distance the object will slide on the incline.
A \( 1.0 \, \text{kg} \) lump of clay is sliding to the right on a frictionless surface with a speed of \( 2 \, \text{m/s} \). It collides head-on and sticks to a \( 0.5 \, \text{kg} \) metal sphere that is sliding to the left with a speed of \( 4 \, \text{m/s} \). What is the kinetic energy of the combined objects after the collision?
A box of mass \( 20 \) \( \text{kg} \) moves to the right on a horizontal frictionless surface with a speed of \( 4.0 \) \( \text{m/s} \). The box collides with and remains attached to one end of a spring of negligible mass whose other end is fixed to a wall. After the collision, the spring compresses a maximum distance of \( 0.50 \) \( \text{m} \), and the box then oscillates back and forth.
A spring stretches \( 8.0 \) \( \text{cm} \) when a \( 13 \) \( \text{N} \) force is applied. How far does it stretch when a \( 26 \) \( \text{N} \) force is applied?
A snowboarder starts from rest and slides down a \(32^\circ\) incline that’s \(75 \, \text{m}\) long.
A constant force of strength \( 20 \) \( \text{N} \) acts on an object of mass \( 3 \) \( \text{kg} \) as it moves a distance of \( 4 \) \( \text{m} \). If this force is applied perpendicular to the \( 4 \) \( \text{m} \) displacement, the work done by the force is equal to:
A block attached to spring demonstrates simple harmonic motion about its equilibrium position with amplitude [katex] A [/katex] and angular frequency [katex] \omega [/katex]. What is the maximum magnitude of the block’s velocity?
\(v_x = \frac{m}{m+M} v\)
\(A = \frac{m v}{\sqrt{k (m+M)}}\)
\(T = 2\pi \sqrt{\frac{m+M}{k}}\)
\(E = \frac{m^2 v^2}{2(m+M)}\)
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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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