| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\sum \vec p_i=\sum \vec p_f\] | Conservation of momentum (always valid for the two-body system during the collision, assuming negligible external impulse). |
| 2 | For the \(7\,\text{kg}\) mass after collision (angle \(22^\circ\) below \(+x\)): \[v_{2x}=2\cos 22^\circ,\qquad v_{2y}=-2\sin 22^\circ\] |
Decompose the given velocity into components. “Below the horizontal” means the \(y\)-component is negative. |
| 3 | Momentum in \(x\): \[m_1u_{1x}+m_2u_{2x}=m_1v_{1x}+m_2v_{2x}\] \[4(10)+7(0)=4v_{1x}+7\bigl(2\cos 22^\circ\bigr)\] |
Apply conservation of momentum in the horizontal direction. |
| 4 | \[ 40=4v_{1x}+14\cos 22^\circ \] \[ 14\cos 22^\circ \approx 12.98 \] |
Compute the \(x\)-momentum contribution of the \(7\,\text{kg}\) mass using the correct trig value. |
| 5 | \[ 4v_{1x}=40-12.98=27.02 \] \[ v_{1x}=\frac{27.02}{4}\approx 6.755\ \text{m/s} \] |
Solve for the \(4\,\text{kg}\) mass’s horizontal component. |
| 6 | Momentum in \(y\): \[m_1u_{1y}+m_2u_{2y}=m_1v_{1y}+m_2v_{2y}\] \[0=4v_{1y}+7\bigl(-2\sin 22^\circ\bigr)\] |
Initial vertical momentum is zero. Since the \(7\,\text{kg}\) mass goes downward (negative \(y\)), the \(4\,\text{kg}\) mass must have positive \(v_{1y}\) to keep total \(y\)-momentum zero. |
| 7 | \[ 0=4v_{1y}-14\sin 22^\circ \] \[ 14\sin 22^\circ \approx 5.244 \] |
Compute the vertical momentum magnitude associated with the \(7\,\text{kg}\) mass (downward). |
| 8 | \[ 4v_{1y}=5.244 \] \[ v_{1y}=\frac{5.244}{4}\approx 1.311\ \text{m/s} \] |
Solve for the \(4\,\text{kg}\) mass’s vertical component (positive = upward). |
| 9 | Speed and direction of the \(4\,\text{kg}\) mass: \[ v_1=\sqrt{v_{1x}^2+v_{1y}^2},\qquad \theta=\arctan\!\left(\frac{v_{1y}}{v_{1x}}\right) \] |
Combine components to get the velocity magnitude and the angle relative to the \(+x\) axis. |
| 10 | \[ v_1=\sqrt{(6.755)^2+(1.311)^2} =\sqrt{45.63+1.72} =\sqrt{47.35}\approx 6.88\ \text{m/s} \] |
Compute the magnitude using the corrected components. |
| 11 | \[ \theta=\arctan\!\left(\frac{1.311}{6.755}\right)\approx 11.0^\circ \] |
The \(4\,\text{kg}\) mass travels about \(11^\circ\) above the horizontal (since \(v_{1y}>0\)). |
| 12 | Elastic-collision check (kinetic energy must also be conserved): \[ K_i=\tfrac12(4)(10^2)=200\ \text{J} \] \[ K_f=\tfrac12(4)v_1^2+\tfrac12(7)(2^2) =2(47.35)+14\approx 108.7\ \text{J} \] |
An elastic collision requires \(K_i=K_f\). Using the given \(2\,\text{m/s}\) for the \(7\,\text{kg}\) mass gives \(K_f\neq K_i\), so the stated data are not consistent with an elastic collision (they describe a non-elastic outcome if momentum is conserved). |
| 13 | \[ \boxed{\vec v_1 \approx (6.755\,\hat i+1.311\,\hat j)\ \text{m/s}} \] \[ \boxed{v_1\approx 6.88\ \text{m/s at }11.0^\circ\text{ above horizontal}} \] |
Final velocity of the \(4\,\text{kg}\) mass from momentum conservation with the given \(7\,\text{kg}\) motion; note this contradicts the “elastic” requirement because kinetic energy is not conserved. |
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A \(3800 \, \text{kg}\) open railroad car coasts along with a constant speed of \(8.60 \, \text{m/s}\) along a level track. Snow begins to fall vertically and fills the car at a rate of \(3.50 \, \text{kg/min}\). Ignoring friction with the tracks, what is the speed of the car after \(90 \, \text{min}\)?
A 1.0-kg object is moving with a velocity of 6.0 m/s to the right. It collides and sticks to a 2.0-kg object moving with a velocity of 3.0 m/s in the same direction. How much kinetic energy was lost in the collision?
A karate master is about to split a piece of wood with her hand. Select all she must do in order to deliver the maximum force to split the wood.
You are lying in bed and want to shut your bedroom door. You have a bouncy “superball” and a blob of clay, both with the same mass \( m \). Which one would be more effective to throw at your door to close it?
A \(0.025 \, \text{kg}\) golf ball moving at \(18.0 \, \text{m/s}\) crashes through the window of a house in \(5.0 \times 10^{-4} \, \text{s}\). After the crash, the ball continues in the same direction with a speed of \(10.0 \, \text{m/s}\).
Astronaut Jennifer’s lifeline to her spaceship comes loose and she finds herself stranded, “floating” \( 100 \) \( \text{m} \) from the mothership. She suddenly throws her \( 2.00 \) \( \text{kg} \) wrench at \( 20 \) \( \text{m/s} \) in a direction away from the ship. If she and her spacesuit have a combined mass of \( 200 \) \( \text{kg} \), how long does it take her to coast back to her spaceship?
A \(0.10 \, \text{kg}\) ball, traveling horizontally at \(25 \, \text{m/s}\), strikes a wall and rebounds at \(19 \, \text{m/s}\). What is the magnitude of the change in the momentum of the ball during the rebound?
A 75.0kg log floats downstream with a speed of 1.80 m/s. Eight frogs hop onto the log in a series of perfectly inelastic collisions. If each frog has a mass of 0.30 kg and an upstream speed of 1.3 m/s, what is the change in kinetic energy for this system?
A platform is initially rotating on smooth ice with negligible friction, as shown above. A stationary disk is dropped directly onto the center of the platform. A short time later, the disk and platform rotate together at the same angular velocity, as shown at right in the figure. How does the angular momentum of only the platform change, if at all, after the disk drops? And what is the best justification.
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 \( m \).
6.81 m/s at 8.8° above the 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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