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| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[m_b v_i = (m_b + M) v_x\] | Linear momentum of the bullet–block system is conserved because the external horizontal forces are negligible. Here \(m_b\) is the bullet mass, \(M\) the block mass, \(v_i\) the bullet’s initial speed, and \(v_x\) the common speed just after impact. |
| 2 | \[v_x = \frac{m_b v_i}{m_b + M}\] | Algebraically solve the previous relation for the unknown \(v_x\). |
| 3 | \[v_x = \frac{0.0500\,\text{kg}\,\times\,50.0\,\text{m/s}}{0.0500\,\text{kg}+0.300\,\text{kg}} = 7.14\,\text{m/s}\] | Substitute the given numerical values: \(m_b = 0.0500\,\text{kg}\), \(M = 0.300\,\text{kg}\), and \(v_i = 50.0\,\text{m/s}\). |
| 4 | \[\boxed{v_x = 7.14\,\text{m/s}}\] | State the final speed of the combined mass immediately after the collision. |
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[\tfrac12 (m_b+M) v_x^2 = (m_b+M) g h\] | After impact, kinetic energy converts into gravitational potential energy at the highest point; mechanical energy is conserved because non-conservative work is negligible. |
| 2 | \[h = \frac{v_x^2}{2g}\] | Solve the energy equation for vertical rise \(h\); the common mass cancels out. |
| 3 | \[h = \frac{(7.14\,\text{m/s})^2}{2\,(9.80\,\text{m/s}^2)} = 2.60\,\text{m}\] | Insert \(v_x = 7.14\,\text{m/s}\) and \(g = 9.80\,\text{m/s}^2\) to compute the height. |
| 4 | \[\boxed{h = 2.60\,\text{m}}\] | Quote the vertical distance the system rises. |
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[h = \frac{(m_b v_i)^2}{2g\,(m_b+M)^2}\] | Combine momentum \(m_b v_i=(m_b+M)v_x\) with energy \(h=v_x^2/2g\) to express \(h\) only in terms of parameters \(m_b\) and \(v_i\). |
| 2 | \[h’_v = 4h\] | Doubling \(v_i\) multiplies the numerator of the above expression by \(2^2=4\), while the denominator is unchanged; thus height becomes four times larger. |
| 3 | \[\frac{h’_m}{h}=4\left(\frac{m_b+M}{2m_b+M}\right)^2\] | Replacing \(m_b\) by \(2m_b\) changes both numerator and denominator; the ratio displayed compares the new height to the original. |
| 4 | \[\frac{h’_m}{h}=3.06\,\text{(for given masses)}\] | With \(m_b=0.0500\,\text{kg}\) and \(M=0.300\,\text{kg}\), doubling mass raises height by a factor \(3.06<4\). |
| 5 | \[\boxed{\text{Doubling }v_i\text{ increases }h\text{ the most}}\] | Because height depends on the square of speed but less strongly on bullet mass, option (i) has the greater effect; option (iii) is therefore false. |
Just ask: "Help me solve this problem."
A golf club exerts an average horizontal force of \(1000 \, \text{N}\) on a \(0.045 \, \text{kg}\) golf ball that is initially at rest on the tee. The club is in contact with the ball for \(1.8 \, \text{milliseconds}\). What is the speed of the golf ball just as it leaves the tee?
A pendulum with a period of \( 1 \) \( \text{s} \) on Earth, where the acceleration due to gravity is \( g \), is taken to another planet, where its period is \( 2 \) \( \text{s} \). The acceleration due to gravity on the other planet is most nearly
Why is more fuel required for a spacecraft to travel from the Earth to the Moon than to return from the Moon to the Earth?
When can the motion of a pendulum be modeled as simple harmonic motion?
The efficiency of a pulley system is 55%. The
pulleys are used to raise a mass of 90.0 kg to a height of
5.60 m. What force is exerted on the rope of the pulley
system if the rope is pulled for 22 m in order to raise
the mass to the required height?
\(7.14\,\text{m/s}\)
\(2.60\,\text{m}\)
\(\text{Doubling bullet’s velocity}\)
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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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