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To solve the problem, we must follow several steps to find both the distance and angle at which the two cars skid before coming to a stop after their inelastic collision. The steps include calculating the final velocity of the system using momentum conservation, the resulting kinetic energy, and how friction acts to stop the vehicles.
So, here’s a step-by-step solution detailed in the table format:
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]m_1 = 1000 \text{ kg}, v_{1x} = 20 \text{ m/s}, m_2 = 2000 \text{ kg}, v_{2y} = 15 \text{ m/s}[/katex] | Set initial conditions with masses of the cars and their velocities. The subscript 1 refers to the eastbound car, and 2 refers to the northbound car. |
| 2 | [katex]\vec{p}_{\text{initial}} = m_1 v_{1x} \hat{i} + m_2 v_{2y} \hat{j}[/katex] | Calculate the initial momentum vectorally considering east as [katex]\hat{i}[/katex] direction and north as [katex]\hat{j}[/katex]. |
| 3 | [katex]\vec{p}_{\text{final}} = \vec{p}_{\text{initial}} = (20000 \hat{i} + 30000 \hat{j}) \, \text{kg m/s}[/katex] | In an inelastic collision, total momentum is conserved. Calculate final momentum (which is the same as initial since there are no external forces in the horizontal plane). |
| 4 | [katex]\vec{v}_{\text{final}} = \frac{\vec{p}_{\text{final}}}{m_1 + m_2} = \frac{20000 \hat{i} + 30000 \hat{j}}{3000} = (6.67 \hat{i} + 10 \hat{j}) \, \text{m/s}[/katex] | Calculate the final velocity vector of the system combining both masses (the mass after collision is sum of both masses). |
| 5 | [katex]v_{\text{final}} = \sqrt{6.67^2 + 10^2} \approx 12.08 \, \text{m/s}[/katex] | Calculate the magnitude of final velocity using the Pythagorean theorem. |
| 6 | [katex]\theta = \tan^{-1} \left( \frac{10}{6.67} \right)[/katex] | Calculate the angle of the final velocity vector using tangent inverse. Angle is measured from the east axis (positive x-axis). |
| 7 | [katex]F = \mu_k m g[/katex] | Calculate the friction force acting, which opposes the motion. [katex]\mu_k[/katex] is the coefficient of kinetic friction and [katex]g[/katex] is acceleration due to gravity (approximately [katex]9.8 \, \text{m/s}^2[/katex]). |
| 8 | [katex]\Delta KE = 0.5 (m_1 + m_2) v_{\text{final}}^2[/katex] | Calculate the kinetic energy just after the collision. |
| 9 | [katex]\text{Work done by friction, } W = F \cdot d = \Delta KE[/katex] | Friction does work to bring the cars to a stop, equal to the change in kinetic energy (which is all converted into heat and other forms). |
| 10 | [katex]d = \frac{\Delta KE}{F} = \frac{0.5 \cdot 3000 \cdot 12.08^2}{0.9 \cdot 3000 \cdot 9.8} \approx 8.19 \, \text{m}[/katex] | Calculate the skidding distance. Plugging in all the known values will give the distance over which the cars skid. |
Summary:
– Distance skidded: [katex]8.19[/katex] meters.
– Angle from east direction: [katex]\tan^{-1}(1.5) \approx 56.31°[/katex] north of east.
Just ask: "Help me solve this problem."
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8.19 m, 56.3°
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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] |
General Metric Conversion Chart
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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