| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[m_1 = 4\,\text{kg},\; m_2 = 7\,\text{kg},\; v_i = 10\,\text{m/s}\] | Identify the masses and the initial velocity \(v_i\) of the \(4\,\text{kg}\) mass. The \(7\,\text{kg}\) mass is initially at rest. |
| 2 | \[p_{x,i}=m_1 v_i = 4(10)=40,\; p_{y,i}=0\] | Calculate the initial momentum components. Motion is purely horizontal to the right, so the vertical component is zero. |
| 3 | \[v_{x2}=2\cos22^{\circ},\; v_{y2}=-2\sin22^{\circ}\] | Resolve the \(7\,\text{kg}\) mass’s given final speed (\(2\,\text{m/s}\)) into horizontal and vertical components. The vertical component is negative (below the horizontal). |
| 4 | \[v_{x2}\approx1.854,\; v_{y2}\approx-0.749\] | Numerical evaluation of the trigonometric components. |
| 5 | \[4 v_{x1}+7 v_{x2}=40\] | Apply conservation of momentum in the \(x\)-direction: total initial \(p_x\) equals total final \(p_x\). |
| 6 | \[4 v_{y1}+7 v_{y2}=0\] | Apply conservation of momentum in the \(y\)-direction: initial \(p_y\) is zero, so the final \(p_y\) must also be zero. |
| 7 | \[v_{x1}=\frac{40-7 v_{x2}}{4}\] | Solve the \(x\)-momentum equation for the unknown horizontal component \(v_{x1}\) of the \(4\,\text{kg}\) mass. |
| 8 | \[v_{x1}=\frac{40-7(1.854)}{4}\approx6.755\,\text{m/s}\] | Substitute \(v_{x2}\) and compute \(v_{x1}\). |
| 9 | \[v_{y1}=-\frac{7 v_{y2}}{4}\] | Rearrange the \(y\)-momentum equation to isolate the vertical component \(v_{y1}\) of the \(4\,\text{kg}\) mass. |
| 10 | \[v_{y1}=-\frac{7(-0.749)}{4}\approx1.311\,\text{m/s}\] | Insert \(v_{y2}\) and calculate \(v_{y1}\). The result is positive, meaning the mass moves upward after the collision. |
| 11 | \[v_x=\sqrt{v_{x1}^2+v_{y1}^2}\] | Use the Pythagorean relation to find the magnitude \(v_x\) of the final velocity of the \(4\,\text{kg}\) mass. |
| 12 | \[v_x=\sqrt{(6.755)^2+(1.311)^2}\approx6.88\,\text{m/s}\] | Compute the numerical value of the speed. |
| 13 | \[\theta=\tan^{-1}\!\left(\frac{v_{y1}}{v_{x1}}\right)\] | Determine the direction angle \(\theta\) measured above the horizontal. |
| 14 | \[\theta=\tan^{-1}\!\left(\frac{1.311}{6.755}\right)\approx11^{\circ}\] | Evaluate the inverse tangent to find the angle. |
| 15 | \[\boxed{v_x\approx6.9\,\text{m/s},\;\theta\approx11^{\circ}\,\text{above horizontal}}\] | Present the final boxed answer: the speed and its angle relative to the horizontal. |
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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.
Two boxes are tied together by a string and are sitting at rest on a frictionless surface. Between the two boxes is a massless compressed spring. The string trying the two boxes is then cut and the spring expands, pushing the boxes apart. The box on the left has four times the mass of the box on the right.
A man weighing \( 700 \) \( \text{N} \) and a woman weighing \( 400 \) \( \text{N} \) have the same momentum. What is the ratio of the man’s kinetic energy \( K_m \) to that of the woman \( K_w \)?
A rubber ball with a mass of \(0.25 \, \text{kg}\) and a speed of \(19.0 \, \text{m/s}\) collides perpendicularly with a wall and bounces off with a speed of \(21 \, \text{m/s}\) in the opposite direction. What is the magnitude of the impulse acting on the rubber ball?
A pool cue ball, mass \(0.7 \, \text{kg}\), is traveling at \(2 \, \text{m/s}\) when it collides head-on with another ball, mass \(0.5 \, \text{kg}\), traveling in the opposite direction with a speed of \(1.2 \, \text{m/s}\). After the collision, the cue ball travels in the opposite direction at \(0.3 \, \text{m/s}\). What is the velocity of the other ball?
A block with mass \( m \) slides at speed \( v_0 \) on a smooth surface and hits a stationary block with mass \( M \). They stick together and move at speed \( \frac{v_0}{3} \). Find \( M \) in terms of \( m \).
A space probe far from the Earth is traveling at 14.8 km/s. It has mass 1312 kg. The probe fires its rockets to give a constant thrust of 156 kN for 220 seconds. It accelerates in the same direction as its initial velocity. In this time it burns 150 kg of fuel. Calculate final speed of the space probe in km/s.
Note: This is a bonus question. Skip if you haven’t yet taken calculus.
A rubber ball bounces off of a wall with an initial speed \(v\) and reverses its direction so its speed is \(v\) right after the bounce. As a result of this bounce, which of the following quantities of the ball are conserved?
A baseball, mass \(0.5 \, \text{kg}\), is traveling to the right at \(32.2 \, \text{m/s}\) when it is hit by a bat and travels the opposite direction at \(72.2 \, \text{m/s}\). The bat hits the ball with a force of \(1,222 \, \text{N}\). What is the ball’s change in momentum and how long was the ball in contact with the bat?
A block of mass \(M_1\) travels horizontally with a constant speed \(v_0\) on a plateau of height \(H\) until it comes to a cliff. A toboggan of mass \(M_2\) is positioned on level ground below the cliff. The center of the toboggan is a distance \(D\) from the base of the cliff.
\(6.9\,\text{m/s}\)
\(11^{\circ}\text{ above horizontal}\)
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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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