0 attempts
0% avg
UBQ Credits
# 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."
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?
How does the time t1 of a block m reaching the bottom of slide 1 compare with t2, the time taken block of mass 2m to reach the end of slide 2 that’s curved? The blocks are released from the same height.
A 75.0kg log floats downstream with a speed of 1.80 m/s. Eight frogs hop onto the log in a series of perfectly inelastic collisions. If each frog has a mass of 0.30 kg and an upstream speed of 1.3 m/s, what is the change in kinetic energy for this system?
A person is making homemade ice cream. She exerts a force of magnitude 23 N on the free end of the crank handle on the ice-cream maker, and this end moves on a circular path of radius 0.25 m. The force is always applied parallel to the motion of the handle. If the handle is turned once every 1.7 s, what is the average power being expended?
A mechanic pushes a [katex]2500 \, \text{kg}[/katex] car from rest to a final speed [katex]v[/katex] by doing [katex]5.0 \times 10^3 \, \text{J}[/katex] of work on the car. Frictional effect between the car and the ground are negligible. What is the final speed of the car?
(a) 7.1 m/s
(b) 2.6 m
(c) ii
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.
