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To solve the problem of finding the final velocities of both blocks after an elastic collision, we need to use both the conservation of momentum and conservation of kinetic energy principles. The masses and initial velocities will be plugged into these equations to determine the final velocities.
Let:
– [katex] m_1 = 1.5 \, \text{kg} [/katex] (mass of Block 1)
– [katex] m_2 = 0.75 \, \text{kg} [/katex] (mass of Block 2)
– [katex] u_1 = 3 \, \text{m/s} [/katex] (initial velocity of Block 1)
– [katex] u_2 = 0 \, \text{m/s} [/katex] (initial velocity of Block 2, as it is at rest)
– [katex] v_1 [/katex] (final velocity of Block 1)
– [katex] v_2 [/katex] (final velocity of Block 2)
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 [/katex] | Conservation of momentum equation. In elastic collisions, momentum is conserved before and after the collision. |
| 2 | [katex] 1.5 \times 3 + 0.75 \times 0 = 1.5 v_1 + 0.75 v_2 [/katex] [katex] 4.5 = 1.5 v_1 + 0.75 v_2 [/katex] |
Substitute known values |
| 3 | [katex] \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 [/katex] | Conservation of kinetic energy equation, also conserved in elastic collisions. |
| 4 | [katex] \frac{1}{2} \times 1.5 \times 3^2 + \frac{1}{2} \times 0.75 \times 0^2 = \frac{1}{2} \times 1.5 \times v_1^2 + \frac{1}{2} \times 0.75 \times v_2^2 [/katex] [katex] 6.75 = 0.75 v_1^2 + 0.375 v_2^2 [/katex] |
Substitute known values |
| 5 | Solve two equations (from previous steps) simultaneously:
Equation 1: [katex] 4.5 = 1.5 v_1 + 0.75 v_2 [/katex] Equation 2: [katex] 6.75 = 0.75 v_1^2 + 0.375 v_2^2 [/katex] |
Use algebraic methods (substitution, elimination) to solve for [katex] v_1 [/katex] and [katex] v_2 [/katex] from equations from step 2 and step 4. |
| 6 | After solving: [katex] v_1 = 1 \, \text{m/s} [/katex] [katex] v_2 = 4 \, \text{m/s} [/katex] |
Final solution |
This solution strategy provides the final velocities of each block after the collision, assuming perfectly elastic conditions where both momentum and kinetic energy are conserved.
Just ask: "Help me solve this problem."
In the figure above, the marble rolls down the track and around a loop-the-loop of radius \( R \). The marble has mass \( m \) and radius \( r \). What minimum height \( h_{min} \) must the track have for the marble to make it around the loop-the-loop without falling off? Express your answer in terms of the variables \( R \) and \( r \).
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Find the escape speed from a planet of mass 6.89 x 1025 kg and radius 6.2 x 106 m.
A projectile of mass 0.750 kg is shot straight up with an initial speed of 18.0 m/s.
In which of the following is the rate of change of the particle’s momentum zero?
Block 1 (1.5 kg): 1.0 m/s to the right
Block 2 (0.75 kg): 4.0 m/s to the right
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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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