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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."
A cardinal (Richmondena cardinalis) of mass \( 3.80 \times 10^{-2} \) \( \text{kg} \) and a baseball of mass \( 0.150 \) \( \text{kg} \) have the same kinetic energy. What is the ratio of the cardinal’s magnitude \( p_c \) of momentum to the magnitude \( p_b \) of the baseball’s momentum?
A pendulum consists of a mass \( M \) hanging at the bottom end of a massless rod of length \( \ell \) which has a frictionless pivot at its top end. A mass \( m \), moving with velocity \( v \), impacts \( M \) and becomes embedded. In terms of the given variables and constants, what is the smallest value of \( v \) sufficient to cause the pendulum (with embedded mass \( m \)) to swing clear over the top of its arc?
A simple pendulum consists of a sphere tied to the end of a string of negligible mass. The sphere is pulled back until the string is horizontal and then released from rest. Assume the gravitational potential energy is zero when the sphere is at its lowest point.
What angle will the string make with the horizontal when the kinetic energy and the potential energy of the sphere-Earth system are equal?
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 box of mass \(m\) is initially at rest at the top of a ramp that is at an angle \(\theta\) with the horizontal. The block is at a height \(h\) and length \(L\) from the bottom of the ramp. The coefficient of kinetic friction between the block and the ramp is \(\mu\). What is the kinetic energy of the box at the bottom of the ramp?
(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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