0 attempts
0% avg
UBQ Credits
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."
The launching mechanism of a toy gun consists of a spring with an unknown spring constant, \( k \). When the spring is compressed \( 0.120 \, \text{m} \) vertically, a \( 35.0 \, \text{g} \) projectile is able to be fired to a maximum height of \( 25 \, \text{m} \) above the position of the projectile when the spring is compressed. Assume that the barrel of the gun is frictionless.
A 0.035 kg bullet moving horizontally at 350 m/s embeds itself into an initially stationary 0.55 kg block. Air resistance is negligible.
A pool cue ball, mass 0.7 kg, is traveling at 2 m/s when it collides head on with another ball, mass 0.5 kg, traveling in the opposite direction with a speed of 1.2 m/s. After the collision, the cue ball travels in the opposite direction at 0.3 m/s. What is the velocity of the other ball?
Find the escape speed from a planet of mass 6.89 x 1025 kg and radius 6.2 x 106 m.
If a small motor does 520 J of work to move a toy car 260 meters in a time of 37 seconds.
Block 1 (1.5 kg): 1.0 m/s to the right
Block 2 (0.75 kg): 4.0 m/s to the right
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
Try our free calculator to see what you need to get a 5 on the upcoming AP Physics 1 exam.
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.
