0 attempts
0% avg
UBQ Credits
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."
An object undergoes constant acceleration. Starting from rest, the object travels \( 5 \, \text{m} \) in the first second. Then it travels \( 15 \, \text{m} \) in the next second. What total distance will be covered after the 3rd second?
In which of these cases is the rate of change of the particle’s displacement constant?
A super dart of mass 20 g, traveling at 350 m/s, strikes a steel plate at an angle of 30° with the plane of the plate, as shown in the figure. It bounces off the plate at the same angle but at a speed of 320 m/s. What is the magnitude of the impulse that the plate gives to the bullet?
On a harsh winter day, a 1500 kg vehicle takes a circular banked exit ramp (radius R = 150 m; banking angle of 10 degrees) at a speed of 30 mph, since the speed limit is 35 mph. However, the exit ramp is completely iced up (= frictionless). To make matters worse, a wind is blowing parallel to the ramp in a downward direction. The wind exerts a force of 3000 N. Under these conditions, can the driver continue to follow a safe horizontal circle on the exit ramp and stay below the speed limit? To convert mph into m/s use 1 mi = 1607 m and 1 hr is 3600 s.
A baseball is tossed from street level by a student straight up at a speed of \(25.3 \text{ m/s}\). After reaching maximum height, it is caught by another student on the roof of a building, \(17.4 \text{ m}\) above the street. How long did this take?
8.19 m, 56.3°
By continuing you (1) agree to our Terms of Sale and Terms of Use and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.
| 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] |
The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
Understand you mistakes quicker.
Phy automatically provides feedback so you can improve your responses.
10 Free Credits To Get You Started
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.
