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| 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. |
Just ask: "Help me solve this problem."
A small block of mass \( M \) is released from rest at the top of the curved frictionless ramp shown above. The block slides down the ramp and is moving with a speed \( 3.5v_0 \) when it collides with a larger block of mass \( 1.5M \) at rest at the bottom of the incline. The larger block moves to the right at a speed \( 2v_0 \) immediately after the collision.
Express your answers to the following questions in terms of the given quantities and fundamental constants.
A bowling ball moving with speed v collides head-on with a stationary tennis ball. The collision is elastic and there is no friction. The bowling ball barely slows down. What is the speed of the tennis ball after the collision?
Two particles of equal mass \( m_0 \) are moving with equal speeds \( v_0 \) along paths inclined at \( 60^\circ \) to the \( x \)-axis, as shown above. They collide and stick together in a perfectly inelastic collision. Their velocity after the collision has magnitude:
An astronaut initially at rest in space throws a wrench, and recoils in the opposite direction. Select all that is true.
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 \( \text{km s}^{-1} \).
\(-52.2\,\text{kg·m/s}\)
\(4.27\times10^{-2}\,\text{s}\)
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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] |
General Metric Conversion Chart
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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