| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\sum \vec p_i=\sum \vec p_f\] | Conservation of momentum (always valid for the two-body system during the collision, assuming negligible external impulse). |
| 2 | For the \(7\,\text{kg}\) mass after collision (angle \(22^\circ\) below \(+x\)): \[v_{2x}=2\cos 22^\circ,\qquad v_{2y}=-2\sin 22^\circ\] |
Decompose the given velocity into components. “Below the horizontal” means the \(y\)-component is negative. |
| 3 | Momentum in \(x\): \[m_1u_{1x}+m_2u_{2x}=m_1v_{1x}+m_2v_{2x}\] \[4(10)+7(0)=4v_{1x}+7\bigl(2\cos 22^\circ\bigr)\] |
Apply conservation of momentum in the horizontal direction. |
| 4 | \[ 40=4v_{1x}+14\cos 22^\circ \] \[ 14\cos 22^\circ \approx 12.98 \] |
Compute the \(x\)-momentum contribution of the \(7\,\text{kg}\) mass using the correct trig value. |
| 5 | \[ 4v_{1x}=40-12.98=27.02 \] \[ v_{1x}=\frac{27.02}{4}\approx 6.755\ \text{m/s} \] |
Solve for the \(4\,\text{kg}\) mass’s horizontal component. |
| 6 | Momentum in \(y\): \[m_1u_{1y}+m_2u_{2y}=m_1v_{1y}+m_2v_{2y}\] \[0=4v_{1y}+7\bigl(-2\sin 22^\circ\bigr)\] |
Initial vertical momentum is zero. Since the \(7\,\text{kg}\) mass goes downward (negative \(y\)), the \(4\,\text{kg}\) mass must have positive \(v_{1y}\) to keep total \(y\)-momentum zero. |
| 7 | \[ 0=4v_{1y}-14\sin 22^\circ \] \[ 14\sin 22^\circ \approx 5.244 \] |
Compute the vertical momentum magnitude associated with the \(7\,\text{kg}\) mass (downward). |
| 8 | \[ 4v_{1y}=5.244 \] \[ v_{1y}=\frac{5.244}{4}\approx 1.311\ \text{m/s} \] |
Solve for the \(4\,\text{kg}\) mass’s vertical component (positive = upward). |
| 9 | Speed and direction of the \(4\,\text{kg}\) mass: \[ v_1=\sqrt{v_{1x}^2+v_{1y}^2},\qquad \theta=\arctan\!\left(\frac{v_{1y}}{v_{1x}}\right) \] |
Combine components to get the velocity magnitude and the angle relative to the \(+x\) axis. |
| 10 | \[ v_1=\sqrt{(6.755)^2+(1.311)^2} =\sqrt{45.63+1.72} =\sqrt{47.35}\approx 6.88\ \text{m/s} \] |
Compute the magnitude using the corrected components. |
| 11 | \[ \theta=\arctan\!\left(\frac{1.311}{6.755}\right)\approx 11.0^\circ \] |
The \(4\,\text{kg}\) mass travels about \(11^\circ\) above the horizontal (since \(v_{1y}>0\)). |
| 12 | Elastic-collision check (kinetic energy must also be conserved): \[ K_i=\tfrac12(4)(10^2)=200\ \text{J} \] \[ K_f=\tfrac12(4)v_1^2+\tfrac12(7)(2^2) =2(47.35)+14\approx 108.7\ \text{J} \] |
An elastic collision requires \(K_i=K_f\). Using the given \(2\,\text{m/s}\) for the \(7\,\text{kg}\) mass gives \(K_f\neq K_i\), so the stated data are not consistent with an elastic collision (they describe a non-elastic outcome if momentum is conserved). |
| 13 | \[ \boxed{\vec v_1 \approx (6.755\,\hat i+1.311\,\hat j)\ \text{m/s}} \] \[ \boxed{v_1\approx 6.88\ \text{m/s at }11.0^\circ\text{ above horizontal}} \] |
Final velocity of the \(4\,\text{kg}\) mass from momentum conservation with the given \(7\,\text{kg}\) motion; note this contradicts the “elastic” requirement because kinetic energy is not conserved. |
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A karate master is about to split a piece of wood with her hand. Select all she must do in order to deliver the maximum force to split the wood.
A fisherman is standing in the back of his small fishing boat (the mass of the fisherman is the same as the mass of the boat) and he is a few meters from shore. He is done fishing so he starts walking towards the shore so he can get off the boat. What happens to the boat and the fisherman? Select all that apply and assume there is no friction between the boat and the water.
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.
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?
An object at rest suddenly explodes into two fragments (\(m_1\) and \(m_2\)) by an explosion. Fragment \(m_1\) acquires \(3\) times the kinetic energy of the other. What is the ratio of \(m_1\) to \(m_2\)?
A block of mass [katex] m [/katex] is moving on a horizontal frictionless surface with a speed [katex] v_0 [/katex] as it approaches a block of mass [katex] 2m [/katex] which is at rest and has an ideal spring attached to one side.
When the two blocks collide, the spring is completely compressed and the two blocks momentarily move at the same speed, and then separate again, each continuing to move.
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.
You are lying in bed and want to shut your bedroom door. You have a bouncy “superball” and a blob of clay, both with the same mass \( m \). Which one would be more effective to throw at your door to close it?
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?
The two blocks of masses \( M \) and \( 2M \) travel at the same speed \( v \) but in opposite directions. They collide and stick together. How much mechanical energy is lost to other forms of energy during the collision?
6.81 m/s at 8.8° above the 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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