| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[m_1 = 4\,\text{kg},\; m_2 = 7\,\text{kg},\; v_i = 10\,\text{m/s}\] | Identify the masses and the initial velocity \(v_i\) of the \(4\,\text{kg}\) mass. The \(7\,\text{kg}\) mass is initially at rest. |
| 2 | \[p_{x,i}=m_1 v_i = 4(10)=40,\; p_{y,i}=0\] | Calculate the initial momentum components. Motion is purely horizontal to the right, so the vertical component is zero. |
| 3 | \[v_{x2}=2\cos22^{\circ},\; v_{y2}=-2\sin22^{\circ}\] | Resolve the \(7\,\text{kg}\) mass’s given final speed (\(2\,\text{m/s}\)) into horizontal and vertical components. The vertical component is negative (below the horizontal). |
| 4 | \[v_{x2}\approx1.854,\; v_{y2}\approx-0.749\] | Numerical evaluation of the trigonometric components. |
| 5 | \[4 v_{x1}+7 v_{x2}=40\] | Apply conservation of momentum in the \(x\)-direction: total initial \(p_x\) equals total final \(p_x\). |
| 6 | \[4 v_{y1}+7 v_{y2}=0\] | Apply conservation of momentum in the \(y\)-direction: initial \(p_y\) is zero, so the final \(p_y\) must also be zero. |
| 7 | \[v_{x1}=\frac{40-7 v_{x2}}{4}\] | Solve the \(x\)-momentum equation for the unknown horizontal component \(v_{x1}\) of the \(4\,\text{kg}\) mass. |
| 8 | \[v_{x1}=\frac{40-7(1.854)}{4}\approx6.755\,\text{m/s}\] | Substitute \(v_{x2}\) and compute \(v_{x1}\). |
| 9 | \[v_{y1}=-\frac{7 v_{y2}}{4}\] | Rearrange the \(y\)-momentum equation to isolate the vertical component \(v_{y1}\) of the \(4\,\text{kg}\) mass. |
| 10 | \[v_{y1}=-\frac{7(-0.749)}{4}\approx1.311\,\text{m/s}\] | Insert \(v_{y2}\) and calculate \(v_{y1}\). The result is positive, meaning the mass moves upward after the collision. |
| 11 | \[v_x=\sqrt{v_{x1}^2+v_{y1}^2}\] | Use the Pythagorean relation to find the magnitude \(v_x\) of the final velocity of the \(4\,\text{kg}\) mass. |
| 12 | \[v_x=\sqrt{(6.755)^2+(1.311)^2}\approx6.88\,\text{m/s}\] | Compute the numerical value of the speed. |
| 13 | \[\theta=\tan^{-1}\!\left(\frac{v_{y1}}{v_{x1}}\right)\] | Determine the direction angle \(\theta\) measured above the horizontal. |
| 14 | \[\theta=\tan^{-1}\!\left(\frac{1.311}{6.755}\right)\approx11^{\circ}\] | Evaluate the inverse tangent to find the angle. |
| 15 | \[\boxed{v_x\approx6.9\,\text{m/s},\;\theta\approx11^{\circ}\,\text{above horizontal}}\] | Present the final boxed answer: the speed and its angle relative to the horizontal. |
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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?
Two blocks connected to a compressed spring move right at speed \( v \). After releasing the spring, the left block moves left at speed \( v_2 \), the right block moves right. What is the center speed of the blocks then?
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?
On a frictionless horizontal air table, puck A (with mass \( 0.249 \) \( \text{kg} \)) is moving toward puck B (with mass \( 0.375 \) \( \text{kg} \)), which is initially at rest. After the collision, puck A has velocity \( 0.115 \) \( \text{m/s} \) to the left, and puck B has velocity \( 0.645 \) \( \text{m/s} \) to the right.
A space probe far from the Earth is traveling at 14.8 km/s. It has mass 1312 kg. The probe fires its rockets to give a constant thrust of 156 kN for 220 seconds. It accelerates in the same direction as its initial velocity. In this time it burns 150 kg of fuel. Calculate final speed of the space probe in km/s.
Note: This is a bonus question. Skip if you haven’t yet taken calculus.
Block 2 initially is at rest. Block 1 travels towards block 2 and collides with Block 2 as shown above. Find the final velocities of both blocks assuming the collision is elastic.
A \(1200 \, \text{kg}\) car moving at \(15.6 \, \text{m/s}\) suddenly collides with a stationary car of mass \(1500 \, \text{kg}\). If the two vehicles lock together, what is their combined velocity immediately after the collision?
A platform is initially rotating on smooth ice with negligible friction, as shown above. A stationary disk is dropped directly onto the center of the platform. A short time later, the disk and platform rotate together at the same angular velocity, as shown at right in the figure. How does the angular momentum of only the platform change, if at all, after the disk drops? And what is the best justification.
A moderate force will break an egg. However, an egg dropped on the road usually breaks, while one dropped on the grass usually does not break because for the egg dropped on the grass:
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?
\(6.9\,\text{m/s}\)
\(11^{\circ}\text{ above horizontal}\)
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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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