| 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 bullet (mass: \(0.05 \, \text{kg}\)) is fired horizontally (\(v = 200 \, \text{m/s}\)) at a block (mass: \(1.3 \, \text{kg}\)) initially at rest on a frictionless surface. The block is attached to a spring (\(k = 2500 \, \text{N/m}\)). The bullet becomes embedded. Calculate:
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?
A \(70 \, \text{kg}\) woman and her \(35 \, \text{kg}\) son are standing at rest on an ice rink. They push against each other for a time of \(0.60 \, \text{s}\), causing them to glide apart. The speed of the woman immediately after they separate is \(0.55 \, \text{m/s}\). Assume that during the push, friction is negligible compared with the forces the people exert on each other.
A space probe far from the Earth is travelling at \( 14.8 \) \( \text{km s}^{-1} \). It has mass \( 1\,312 \) \( \text{kg} \). The probe fires its rockets to give a constant thrust of \( 156 \) \( \text{kN} \) for \( 220. \) \( \text{s} \). It accelerates in the same direction as its initial velocity. In this time it burns \( 150. \) \( \text{kg} \) of fuel.
Calculate the final speed of the space probe in \( \text{km s}^{-1} \).
A rubber ball bounces off of a wall with an initial speed \(v\) and reverses its direction so its speed is \(v\) right after the bounce. As a result of this bounce, which of the following quantities of the ball are conserved?
A rubber ball and a lump of clay have equal mass. They are thrown with equal speed against a wall. The ball bounces back with nearly the same speed with which it hit. The clay sticks to the wall. Which one of these objects experiences the greater impulse?
Two boxes are tied together by a string and are sitting at rest on a frictionless surface. Between the two boxes is a massless compressed spring. The string trying the two boxes is then cut and the spring expands, pushing the boxes apart. The box on the left has four times the mass of the box on the right.
In which of the following is the rate of change of the particle’s momentum zero?
Two students hold a large bed sheet vertically between them. A third student, who happens to be the star pitcher on the school baseball team, throws a raw egg at the center of the sheet. Explain why the egg does not break when it hits the sheet, regardless of its initial speed.
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?
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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