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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[p_i = m v_i\] | Initial momentum equals mass times initial velocity \(p_i = (0.25\,\text{kg})(19.0\,\text{m/s})\). |
| 2 | \[p_i = 4.75\; \text{kg\,m/s}\] | Multiplying gives the initial momentum \(4.75\;\text{kg\,m/s}\). |
| 3 | \[p_f = m v_x\] | Final momentum equals mass times final velocity (opposite direction so velocity is negative) \(v_x=-21\,\text{m/s}\). |
| 4 | \[p_f = -5.25\; \text{kg\,m/s}\] | Calculate \(p_f = (0.25)(-21)= -5.25\;\text{kg\,m/s}\). |
| 5 | \[J = p_f – p_i\] | Impulse is the change in momentum \(J=\Delta p\). |
| 6 | \[J = -5.25 – 4.75 = -10\; \text{kg\,m/s}\] | Subtract to find the impulse; the sign shows the direction opposite the initial motion. |
| 7 | \[\boxed{10\; \text{kg\,m/s}}\] | The magnitude is the absolute value, \(|J| = 10\;\text{kg\,m/s}\), matching choice (d). |
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A 0.035 kg bullet moving horizontally at 350 m/s embeds itself into an initially stationary 0.55 kg block. Air resistance is negligible.
A child (\(m = 32 \, \text{kg}\)) in a boat (\(m = 71 \, \text{kg}\)) throws a \(7.1 \, \text{kg}\) package out horizontally with a speed of \(12.2 \, \text{m/s}\). Calculate the velocity of the boat immediately after, assuming it was initially at rest. Ignore water resistance.
A bullet of mass \(0.0500 \, \text{kg}\) traveling at \(50.0 \, \text{m/s}\) is fired horizontally into a wooden block suspended from a long rope. The mass of the wooden block is \(0.300 \, \text{kg}\) and it is initially at rest. The collision is completely inelastic and after impact the bullet + wooden block move together until the center of mass of the system rises a vertical distance \(h\) above its initial position.
A baseball, mass \(0.5 \, \text{kg}\), is traveling to the right at \(32.2 \, \text{m/s}\) when it is hit by a bat and travels the opposite direction at \(72.2 \, \text{m/s}\). The bat hits the ball with a force of \(1,222 \, \text{N}\). What is the ball’s change in momentum and how long was the ball in contact with the bat?
A 1.0-kg object is moving with a velocity of 6.0 m/s to the right. It collides and sticks to a 2.0-kg object moving with a velocity of 3.0 m/s in the same direction. How much kinetic energy was lost in the collision?
A \(15 \, \text{g}\) marble moves to the right at \(3.5 \, \text{m/s}\) and makes an elastic head-on collision with a \(22 \, \text{g}\) marble. The final velocity of the \(22 \, \text{g}\) marble is \(2.0 \, \text{m/s}\) to the right, and the final velocity of the \(15 \, \text{g}\) marble is \(5.4 \, \text{m/s}\) to the left. What was the initial velocity of the \(22 \, \text{g}\) marble?
A bullet at speed [katex] v_0 [/katex] trikes and embeds itself in a block of wood which is suspended by a string, causing the bullet and block to rise to a maximum height h. Which of the following statements is true?
A \(20 \, \text{g}\) piece of clay moving at a speed of \(50 \, \text{m/s}\) strikes a \(500 \, \text{g}\) pendulum bob at rest. The length of a string is \(0.8 \, \text{m}\). After the collision, the clay-bob system starts to oscillate as a simple pendulum.
An egg dropped on the road usually beaks, while one dropped on the grass usually does not break because for the egg dropped on the grass:
A \(4 \, \text{kg}\) mass is traveling at \(10 \, \text{m/s}\) to the right when it collides elastically with a stationary \(7 \, \text{kg}\) mass. The \(7 \, \text{kg}\) mass then travels at \(2 \, \text{m/s}\) at an angle of \(22^\circ\) below the horizontal. What is the velocity of the \(4 \, \text{kg}\) mass?
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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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