| 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 \( 1000 \) \( \text{kg} \) 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?
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 \(0.025 \, \text{kg}\) 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 \(10.0 \, \text{m/s}\).
Block 2 initially is at rest. Block 1 travels towards block 2 and collides with Block 2 as shown above. Find the final velocities of both blocks assuming the collision is elastic.
In which of the following is the rate of change of the particle’s momentum zero?
A \(4 \, \text{kg}\) mass is traveling at \(10 \, \text{m/s}\) to the right when it collides inelastically with a stationary \(7 \, \text{kg}\) mass. The \(7 \, \text{kg}\) mass then travels at \(2 \, \text{m/s}\) at an angle of \(22^\circ\) below the horizontal. What are the velocity and the angle of the \(4 \, \text{kg}\) mass?
A cardinal (Richmondena cardinalis) of mass \( 3.80 \times 10^{-2} \) \( \text{kg} \) and a baseball of mass \( 0.150 \) \( \text{kg} \) have the same kinetic energy. What is the ratio of the cardinal’s magnitude \( p_c \) of momentum to the magnitude \( p_b \) of the baseball’s momentum?
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 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 firecracker in a coconut blows the coconut into three pieces. Two pieces of equal mass fly off south and west, perpendicular to each other, at \( 18 \) \( \text{m/s} \). The third piece has \( 2.5 \) times the mass as the other two.
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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