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# Part A: Calculate velocity of the bullet + wooden block just after impact
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]m_1 = 0.0500 \, \text{kg}[/katex] | Mass of the bullet. |
| 2 | [katex]u_1 = 50.0 \, \text{m/s}[/katex] | Initial velocity of the bullet. |
| 3 | [katex]m_2 = 0.300 \, \text{kg}[/katex] | Mass of the wooden block. |
| 4 | [katex]u_2 = 0 \, \text{m/s}[/katex] | Initial velocity of the wooden block (at rest). |
| 5 | [katex]v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}[/katex] | Formula for the final velocity after a completely inelastic collision, where [katex] m_1 [/katex] and [katex] m_2 [/katex] are masses of the bullet and block, [katex] u_1 [/katex] and [katex] u_2 [/katex] are their initial velocities respectively. |
| 6 | [katex]v = \frac{(0.0500 \, \text{kg}) (50.0 \, \text{m/s}) + (0.300 \, \text{kg}) (0 \, \text{m/s})}{0.0500 \, \text{kg} + 0.300 \, \text{kg}}[/katex] | Substitute the values of [katex] m_1 [/katex], [katex] m_2 [/katex], [katex] u_1 [/katex], and [katex] u_2 [/katex] into the formula. |
| 7 | [katex]\mathbf{v = 7.14 \, \text{m/s}}[/katex] | Compute to find the final velocity of the bullet + wooden block system just after the collision. |
# Part B: Calculate the vertical distance [katex] h [/katex] reached by the bullet and wooden block
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]v = 7.14 \, \text{m/s}[/katex] | Initial velocity of the system after impact (from part A). |
| 2 | [katex]v^2 = u^2 + 2gh[/katex] | Use the kinematic equation relating initial velocity ([katex]u[/katex]), final velocity ([katex]v[/katex]), acceleration due to gravity ([katex]g[/katex]), and height ([katex]h[/katex]). Since the system comes to a stop at the highest point, final velocity [katex]v[/katex] is 0. |
| 3 | [katex]0 = (7.14 \, \text{m/s})^2 + 2(-9.81 \, \text{m/s}^2)h[/katex] | At the highest point the final velocity is 0, and acceleration due to gravity (a negative value, since it acts downwards) impacts the rising object. |
| 4 | [katex]h = \frac{(7.14 \, \text{m/s})^2}{2 \times 9.81 \, \text{m/s}^2}[/katex] | Rearrange to solve for [katex]h[/katex]. |
| 5 | [katex]\mathbf{h = 2.59 \, \text{m}}[/katex] | Compute the vertical height. |
# Part C: Compare effects of doubling the bullet’s mass vs. velocity on [katex] h [/katex]
| Comparison | Analysis | Conclusion |
|---|---|---|
| Doubling bullet’s velocity | [katex] h_f \propto v^2 [/katex], doubling [katex] v [/katex] would quadruple [katex] h [/katex] (since [katex] h_f = \frac{4v^2}{2g} [/katex]) | Increase in velocity has a significant effect, leading to an increase by a factor of four for [katex] h [/katex]. |
| Doubling bullet’s mass | [katex] h_f [/katex] determined more by [katex] v [/katex] (system velocity) than by mass directly. Increasing mass lowers system velocity, hence slightly lowering [katex] h [/katex]. | Little to no increase in [katex] h [/katex]; could actually decrease due to lower velocity after collision. |
| None | Analysis shows that velocity change impacts height significantly compared to mass change. | (ii) Doubling the bullet’s velocity has a greater effect on height than doubling the bullet’s mass. |
Just ask: "Help me solve this problem."
An apple is released from rest 500 m above the ground. Due to the combined forces of air resistance and gravity, it has a speed of 40 m/s when it reaches the ground. What percentage of the initial mechanical energy of the apple-Earth system was dissipated due to air resistance? Take the potential energy of the apple-Earth system to be zero when the apple reaches the ground.
A 4 kg mass is traveling at 10 m/s to the right when it collides elastically with a stationary 7 kg mass. The 7 kg mass then travels at 2 m/s at an angle of 22° below the horizontal. What is the velocity of the 4 kg mass?
A \( 0.0350 \) \( \text{kg} \) bullet moving horizontally at \( 425 \) \( \text{m/s} \) embeds itself into an initially stationary \( 0.550 \) \( \text{kg} \) block.
A net force of \( 8.0 \) \( \text{N} \) accelerates a \( 4.0 \) \( \text{kg} \) body from rest to a speed of \( 5.0 \) \( \text{m s}^{-1} \). Which of the following is equal to the work done by the force?
A \( 4700 \, \text{kg} \) truck carrying a \( 900 \, \text{kg} \) crate is traveling at \( 25 \, \text{m/s} \) to the right along a straight, level highway, as shown above. The truck driver then applies the brakes, and as it slows down, the truck travels \( 55 \, \text{m} \) in the next \( 3.0 \, \text{s} \). The crate does not slide on the back of the truck.
(a) 7.1 m/s
(b) 2.6 m
(c) ii
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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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