| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[v_i = +32.2\,\text{m/s},\; v_x = -72.2\,\text{m/s}\] | Choose the positive x–direction to the right; the ball reverses, so its final velocity is negative. |
| 2 | \[\Delta v = v_x – v_i = (-72.2) – (32.2) = -104.4\,\text{m/s}\] | Find the change in velocity using the definition \(\Delta v = v_x – v_i\). |
| 3 | \[\Delta p = m\,\Delta v = (0.5\,\text{kg})(-104.4\,\text{m/s})\] | Momentum change is mass times change in velocity. |
| 4 | \[\boxed{\Delta p = -52.2\,\text{kg·m/s}}\] | The negative sign indicates the ball’s momentum is now to the left; magnitude is \(52.2\,\text{kg·m/s}\). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[J = F_{\text{avg}}\,\Delta t = \Delta p\] | Impulse–momentum theorem: impulse equals change in momentum. |
| 2 | \[\Delta t = \frac{\Delta p}{F_{\text{avg}}} = \frac{-52.2}{-1222}\] | Rearrange for \(\Delta t\) and insert \(\Delta p\) and the average force (direction signs cancel). |
| 3 | \[\boxed{\Delta t \approx 4.27\times10^{-2}\,\text{s}}\] | Compute the time of contact; result is about 0.0427 s. |
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An astronaut initially at rest in space throws a wrench, and recoils in the opposite direction. Select all that is true.
A man weighing \( 700 \) \( \text{N} \) and a woman weighing \( 400 \) \( \text{N} \) have the same momentum. What is the ratio of the man’s kinetic energy \( K_m \) to that of the woman \( K_w \)?
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.
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 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 small boat coasts at constant speed under a bridge. A heavy sack of sand is dropped from the bridge onto the boat. The speed of the boat
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 mass \( m_1 \) traveling with an initial velocity \( v \) has an elastic collision with a mass \( m_2 \) that is initially at rest.
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 rubber ball with a mass of \(0.25 \, \text{kg}\) and a speed of \(19.0 \, \text{m/s}\) collides perpendicularly with a wall and bounces off with a speed of \(21 \, \text{m/s}\) in the opposite direction. What is the magnitude of the impulse acting on the rubber ball?
\(-52.2\,\text{kg·m/s}\)
\(4.27\times10^{-2}\,\text{s}\)
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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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