| 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 golf club exerts an average horizontal force of \(1000 \, \text{N}\) on a \(0.045 \, \text{kg}\) golf ball that is initially at rest on the tee. The club is in contact with the ball for \(1.8 \, \text{milliseconds}\). What is the speed of the golf ball just as it leaves the tee?
A fisherman is standing in the back of his small fishing boat (the mass of the fisherman is the same as the mass of the boat) and he is a few meters from shore. He is done fishing so he starts walking towards the shore so he can get off the boat. What happens to the boat and the fisherman? Select all that apply and assume there is no friction between the boat and the water.
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.
Car A, mass 1000 kg, is traveling at 40 m/s when it collides with a stationary car B. They stick together and travel at 7 m/s. What is the mass of car B?
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 \( K_w \)?
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 “doomsday” asteroid with a mass of \( 1010 \, \text{kg} \) is hurtling through space. Unless the asteroid’s speed is changed by about \( 0.20 \, \text{cm/s} \), it will collide with Earth and cause tremendous damage. Researchers suggest that a small “space tug” sent to the asteroid’s surface could exert a gentle constant force of \( 2.5 \, \text{N} \). For how long must this force act?
A truck going \(15 \, \text{km/h}\) has a head-on collision with a small car going \(30 \, \text{km/h}\). Which statement best describes the situation?
A \( 1000 \) \( \text{kg} \) car is traveling east at \( 20 \) \( \text{m/s} \) when it collides perfectly inelastically with a northbound \( 2000 \) \( \text{kg} \) car traveling at \( 15 \) \( \text{m/s} \). If the coefficient of kinetic friction is \( 0.9 \), how far, and at what angle do the two cars skid before coming to a stop?
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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