| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ m_b \, v_i = \Bigl(m_b + m_{\text{block}}\Bigr) \, v_x \] | Apply conservation of momentum for the inelastic collision where the bullet embeds in the block. |
| 2 | \[ 0.05 \times 200 = (0.05 + 1.3) \, v_x \] | Substitute the given values: bullet mass \(m_b=0.05\,\text{kg}\), bullet initial speed \(v_i=200\,\text{m/s}\), and block mass \(1.3\,\text{kg}\). |
| 3 | \[ 10 = 1.35 \, v_x \] | Simplify the multiplication and sum of masses. |
| 4 | \[ v_x = \frac{10}{1.35} \] | Solve for the block’s speed immediately after the collision. |
| 5 | \[ \boxed{ v_x \approx 7.41\,\text{m/s} } \] | This is the final speed of the block (with bullet embedded) immediately after impact. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ \frac{1}{2} (m_b+m_{\text{block}}) \, v_x^2 = \frac{1}{2} k \, (\Delta x)^2 \] | At maximum compression of the spring the block’s kinetic energy is completely converted into spring potential energy. |
| 2 | \[ (m_b+m_{\text{block}}) \, v_x^2 = k \, (\Delta x)^2 \] | Simplify by canceling the common factor \( \frac{1}{2} \) on both sides. |
| 3 | \[ (\Delta x)^2 = \frac{(m_b+m_{\text{block}}) \, v_x^2}{k} \] | Rearrange to solve for the square of the displacement (amplitude) \( \Delta x \). |
| 4 | \[ \Delta x = \sqrt{\frac{1.35 \times (7.41)^2}{2500}} \] | Substitute \(m_b+m_{\text{block}}=1.35\,\text{kg}\), \(v_x\approx7.41\,\text{m/s}\), and \(k=2500\,\text{N/m}\). |
| 5 | \[ \Delta x \approx \sqrt{\frac{1.35 \times 54.93}{2500}} \] | Since \((7.41)^2 \approx 54.93\), the numerator calculates to approximately \(74.15\). |
| 6 | \[ \Delta x \approx \sqrt{0.02966} \] | Divide the numerator \(74.15\) by \(2500\) to obtain the value inside the square root. |
| 7 | \[ \boxed{ \Delta x \approx 0.172\,\text{m} } \] | This is the amplitude of the resulting oscillation of the block-spring system. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ \omega = \sqrt{\frac{k}{m_b+m_{\text{block}}}} \] | For a mass-spring system executing simple harmonic motion, the angular frequency \( \omega \) is determined by this formula. |
| 2 | \[ \omega = \sqrt{\frac{2500}{1.35}} \] | Substitute \(k=2500\,\text{N/m}\) and \(m_b+m_{\text{block}}=1.35\,\text{kg}\) into the formula. |
| 3 | \[ \omega \approx 43.03\,\text{rad/s} \] | Calculate the square root to approximate the angular frequency. |
| 4 | \[ f = \frac{\omega}{2\pi} \] | The relationship between angular frequency \( \omega \) and frequency \( f \) is given by \( f = \omega/(2\pi) \). |
| 5 | \[ f \approx \frac{43.03}{2\pi} \] | Substitute the computed value of \( \omega \) into the frequency expression. |
| 6 | \[ \boxed{ f \approx 6.85\,\text{Hz} } \] | This is the frequency of the oscillatory motion of the block. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ x(t) = \Delta x \, \sin(\omega t) \] | The standard equation for simple harmonic motion where the displacement is zero at \(t=0\) and the velocity is maximum. |
| 2 | \[ x(t) = 0.172 \, \sin(43.03\,t) \] | Substitute the amplitude \(\Delta x \approx 0.172\,\text{m}\) and angular frequency \(\omega \approx 43.03\,\text{rad/s}\) into the general equation. |
| 3 | \[ \boxed{ x(t) = 0.172 \, \sin(43.03\,t) } \] | This is the complete equation of motion for the block on the spring, with \(x(0)=0\). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ T = \frac{2\pi}{\omega} \] | The period \(T\) of a simple harmonic oscillator is given by this formula. |
| 2 | \[ T = \frac{2\pi}{43.03} \] | Substitute the angular frequency \(\omega \approx 43.03\,\text{rad/s}\) into the period formula. |
| 3 | \[ \boxed{ T \approx 0.146\,\text{s} } \] | This is the period of the oscillation of the block-spring system. |
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Car A, mass 1000 kg, is traveling at 40 m/s when it collides with a stationary car B. They stick together and travel at 7 m/s. What is the mass of car B?
A block attached to spring demonstrates simple harmonic motion about its equilibrium position with amplitude \( A \) and angular frequency \( \omega \). What is the maximum magnitude of the block’s velocity?
Ball \(A\) of mass \(m\) is dropped from a building of height \(H\). Ball \(B\) of mass \(1.7 \, \text{m}\) is dropped from a building of height \(1.7H\). Using energy, what the ratio of \(v_A\) to \(v_B\) (final velocity of ball \(A\) to final velocity of ball \(B\)). Air resistance is negligible.
A block is attached to a horizontal spring and is initially at rest at the equilibrium position \( x = 0 \), as shown in Figure \( 1 \). The block is then moved to position \( x = -A \), as shown in Figure \( 2 \), and released from rest, undergoing simple harmonic motion. At the instant the block reaches position \( x = +A \), another identical block is dropped onto and sticks to the block, as shown in Figure \( 3 \). The two–block–spring system then continues to undergo simple harmonic motion. Which of the following correctly compares the total mechanical energy \( E_{\text{tot},2} \) of the two–block–spring system after the collision to the total mechanical energy \( E_{\text{tot},1} \) of the one–block–spring system before the collision?
For an object oscillating in SHM, what is the relationship between its displacement, velocity, and acceleration graphs as a function of time?
From the figure above, determine which characteristic fits this collision best.
A vehicle is moving at a speed of 12.3 m/s on a decline when the brakes of all four wheels are fully applied, causing them to lock. The slope of the decline forms an angle of 18.0 degrees with the horizontal plane. Given that the coefficient of kinetic friction between the tires and the road surface is 0.650.
A \( 7.3 \) \( \text{kg} \) mass is placed on a spring with a spring constant of \( 34 \) \( \text{N/cm} \). How much does this stretch the spring?
What is the defining characteristic of the restoring force that causes an object to undergo Simple Harmonic Motion (SHM)?
A firecracker in a coconut blows the coconut into three pieces. Two pieces of equal mass fly off south and west, perpendicular to each other, at \( 18 \) \( \text{m/s} \). The third piece has \( 2.5 \) times the mass as the other two.
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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] |
Metric Prefixes
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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