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Solved conceptually: There are no external forces in the system that can change the momentum of the system. Therefore linear momentum is conversed, even after the blocks separate, and velocity of the center of mass remains the same.
Proved mathematically:
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(2m v = m\left(v_x – \frac{v}{2}\right)\) | The total initial momentum is \(2m v\) since both identical blocks (mass \(m\)) move right with speed \(v\). At the later instant the left block moves \(-\frac{v}{2}\) while the right block moves with unknown speed \(v_x\); momentum is conserved. |
| 2 | \(2v = v_x – \frac{v}{2}\) | Dividing both sides by \(m\) cancels the mass, yielding a relation solely in terms of \(v\) and \(v_x\). |
| 3 | \(v_x = 2v + \frac{v}{2} = \frac{5v}{2}\) | Solve for \(v_x\) by adding \(\frac{v}{2}\) to \(2v\); this yields the speed of the right block. |
| 4 | \(v_{\text{cm}} = \frac{-\frac{v}{2} + \frac{5v}{2}}{2} = \frac{\frac{4v}{2}}{2} = v\) | The center-of-mass velocity is the total momentum divided by total mass or equivalently the average of the block speeds. With \(-\frac{v}{2}\) and \(\frac{5v}{2}\), the average is \(v\). This confirms that the \(\text{COM}\) continues moving right at speed \(v\). |
| 5 | \(\text{Total momentum} = 2m v\) | Since no external forces act in the horizontal direction, the center-of-mass velocity remains unchanged from its initial value \(v\).
The other options [(a) \(\frac{5v}{2}\), (b) \(\frac{3v}{2}\), (d) \(\frac{v}{2}\)] incorrectly suggest that internal forces (from the spring) alter the \(\text{COM}\) speed, but internal forces do not affect the overall momentum of the system. |
Just ask: "Help me solve this problem."
A \( 1.0 \, \text{kg} \) lump of clay is sliding to the right on a frictionless surface with a speed of \( 2 \, \text{m/s} \). It collides head-on and sticks to a \( 0.5 \, \text{kg} \) metal sphere that is sliding to the left with a speed of \( 4 \, \text{m/s} \). What is the kinetic energy of the combined objects after the collision?
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?
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 1.0-kg object is moving with a velocity of 6.0 m/s to the right. It collides and sticks to a 2.0-kg object moving with a velocity of 3.0 m/s in the same direction. How much kinetic energy was lost in the collision?
A boy of mass \( m \) and a girl of mass \( 2m \) are initially at rest at the center of a frozen pond. They push each other so that she slides to the left at speed \( v \) across the frictionless ice surface and he slides to the right. What is the total work done by the children?
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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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