| Derivation | Reasoning |
|---|---|
| \[ m v_{x} – 2m v = 0 \] | Since the boy and girl start at rest, the total momentum is zero. Taking right as positive and left as negative, the boy’s momentum \( m v_{x} \) cancels the girl’s momentum \( -2m v \). |
| \[ v_{x} = 2v \] | Solving the momentum conservation equation shows that the boy must have a speed of \(2v\) to balance the girl’s momentum. |
| \[ KE_{\text{boy}} = \frac{1}{2} m (2v)^2 = 2m v^2 \] | Calculating the kinetic energy of the boy using \( KE = \frac{1}{2} m v^2 \) and substituting his speed \(2v\). |
| \[ KE_{\text{girl}} = \frac{1}{2} (2m)v^2 = m v^2 \] | Calculating the kinetic energy of the girl with her mass \(2m\) and speed \(v\). |
| \[ KE_{\text{total}} = 2m v^2 + m v^2 = 3m v^2 \] | The total work done by the children is equal to the sum of their kinetic energies since they started from rest. |
| \[ \boxed{3m v^2} \] | This final expression for the total work done matches option (d). |
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A \(0.5 \, \text{kg}\) cart, on a frictionless \(2 \, \text{m}\) long table, is being pulled by a \(0.1 \, \text{kg}\) mass connected by a string and hanging over a pulley. The system is released from rest. After the hanging mass falls \(0.5 \, \text{m}\), calculate the speed of the cart on the table. Use ONLY forces and energy.
In a controlled experiment, engineers test a firecracker. The firecracker has mass \( m \) and is placed at rest on a horizontal surface. When the firecracker is lit, it explodes and breaks apart into two pieces. In the first trial, one piece with mass \( \frac{m}{2} \) moves to the left with speed \( v_L \) and the other piece moves to the right with speed \( v_R \). A second trial is performed with an identical firecracker, and one piece with mass \( \frac{3m}{4} \) moves to the left, again with speed \( v_L \). What will the speed of the other piece be in this second trial?
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 \(\theta\) 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 sphere of mass \( M \) and radius \( r \), and rotational inertia \( I \) is released from the top of an inclined plane of height \( h \). The surface has considerable friction. Using only the variables mentioned, derive an expression for the sphere’s center of mass velocity.
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 car traveling to the right with a speed \( v \) brakes to a stop in a distance \( d \). What is the work done on the car by the frictional force \( F \)? (Assume that the frictional force is constant)
A spring with a spring constant of \( 50. \) \( \text{N/m} \) is hanging from a stand. A second spring with a spring constant of \( 100. \) \( \text{N/m} \) is hanging from the first spring. How far do they stretch if a \( 0.50 \) \( \text{kg} \) mass is hung from the bottom spring?
Two particles of equal mass \( m_0 \) are moving with equal speeds \( v_0 \) along paths inclined at \( 60^\circ \) to the \( x \)-axis, as shown above. They collide and stick together in a perfectly inelastic collision. Their velocity after the collision has magnitude:
A spring launches a \(4 \, \text{kg}\) block across a frictionless horizontal surface. The block then ascends a \(30^\circ\) incline with a kinetic friction coefficient of \(\mu_k = 0.25\), stopping after \(55 \, \text{m}\) on the incline. If the spring constant is \(800 \, \text{N/m}\), find the initial compression of the spring. Disregard friction while in contact with the spring.
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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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