| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[m_b v_i = (m_b + M) v_x\] | Linear momentum of the bullet–block system is conserved because the external horizontal forces are negligible. Here \(m_b\) is the bullet mass, \(M\) the block mass, \(v_i\) the bullet’s initial speed, and \(v_x\) the common speed just after impact. |
| 2 | \[v_x = \frac{m_b v_i}{m_b + M}\] | Algebraically solve the previous relation for the unknown \(v_x\). |
| 3 | \[v_x = \frac{0.0500\,\text{kg}\,\times\,50.0\,\text{m/s}}{0.0500\,\text{kg}+0.300\,\text{kg}} = 7.14\,\text{m/s}\] | Substitute the given numerical values: \(m_b = 0.0500\,\text{kg}\), \(M = 0.300\,\text{kg}\), and \(v_i = 50.0\,\text{m/s}\). |
| 4 | \[\boxed{v_x = 7.14\,\text{m/s}}\] | State the final speed of the combined mass immediately after the collision. |
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[\tfrac12 (m_b+M) v_x^2 = (m_b+M) g h\] | After impact, kinetic energy converts into gravitational potential energy at the highest point; mechanical energy is conserved because non-conservative work is negligible. |
| 2 | \[h = \frac{v_x^2}{2g}\] | Solve the energy equation for vertical rise \(h\); the common mass cancels out. |
| 3 | \[h = \frac{(7.14\,\text{m/s})^2}{2\,(9.80\,\text{m/s}^2)} = 2.60\,\text{m}\] | Insert \(v_x = 7.14\,\text{m/s}\) and \(g = 9.80\,\text{m/s}^2\) to compute the height. |
| 4 | \[\boxed{h = 2.60\,\text{m}}\] | Quote the vertical distance the system rises. |
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[h = \frac{(m_b v_i)^2}{2g\,(m_b+M)^2}\] | Combine momentum \(m_b v_i=(m_b+M)v_x\) with energy \(h=v_x^2/2g\) to express \(h\) only in terms of parameters \(m_b\) and \(v_i\). |
| 2 | \[h’_v = 4h\] | Doubling \(v_i\) multiplies the numerator of the above expression by \(2^2=4\), while the denominator is unchanged; thus height becomes four times larger. |
| 3 | \[\frac{h’_m}{h}=4\left(\frac{m_b+M}{2m_b+M}\right)^2\] | Replacing \(m_b\) by \(2m_b\) changes both numerator and denominator; the ratio displayed compares the new height to the original. |
| 4 | \[\frac{h’_m}{h}=3.06\,\text{(for given masses)}\] | With \(m_b=0.0500\,\text{kg}\) and \(M=0.300\,\text{kg}\), doubling mass raises height by a factor \(3.06<4\). |
| 5 | \[\boxed{\text{Doubling }v_i\text{ increases }h\text{ the most}}\] | Because height depends on the square of speed but less strongly on bullet mass, option (i) has the greater effect; option (iii) is therefore false. |
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Using only work and energy, find the velocity of the masses after they have traveled \(0.8 \, \text{m}\). Refer to the image above.
A ski lift carries skiers along a \(695 \, \text{m}\) slope inclined at \(34^\circ\). To lift a single rider, it is necessary to move \(72 \, \text{kg}\) of mass to the top of the lift. Under maximum load conditions, five riders per minute arrive at the top. If \(65\%\) of the energy supplied by the motor goes to overcoming friction, what average power must the motor supply?
A stone is falling at a constant velocity vertically down a tube filled with oil. Which of the following statements about the energy changes of the stone during its motion are correct?
I. The gain in kinetic energy is less than the loss in gravitational potential energy.
II. The sum of kinetic and gravitational potential energy of the stone is constant.
III. The work done by the force of gravity has the same magnitude as the work done by friction.
Two identical blocks are connected to the opposite ends of a compressed spring. The blocks initially slide together on a frictionless surface with velocity \( v \) to the right. The spring is then released by remote control. At some later instant, the left block is moving at \( \frac{v}{2} \) to the left, and the other block is moving to the right. What is the speed of the center of mass of the system at that instant?
A \(2 \, \text{kg}\) model rocket is launched with a thrust force of \(275 \, \text{N}\) and reaches a height of \(90 \, \text{m}\), at which point the thrust cuts out, but the rocket continues moving at \(150 \, \text{m/s}\). What is the average air resistance force acting on the rocket during its ascent?
Two identical arrows, one with \( 2 \) times the speed of the other, are fired into a bale of hay. Assuming the hay exerts a constant “frictional” force on the arrows, the faster arrow will penetrate how much farther than the slower arrow?
Two balls are thrown off a building with the same speed, one straight up and one at a 45° angle. Which statement is true if air resistance can be ignored?
| Speed | \( 10 \, \mathrm{m/s} \) | \( 20 \, \mathrm{m/s} \) | \( 30 \, \mathrm{m/s} \) |
| Braking Distance | \( 6.1 \, \mathrm{m} \) | \( 23.9 \, \mathrm{m} \) | \( 53.5 \, \mathrm{m} \) |
A car of mass \( 1500 \, \mathrm{kg} \) is traveling at one of the speeds listed when the brakes are first applied. Using the data above, what is the magnitude of the average braking force required to stop the car?
Astronaut Jennifer’s lifeline to her spaceship comes loose and she finds herself stranded, “floating” \( 100 \) \( \text{m} \) from the mothership. She suddenly throws her \( 2.00 \) \( \text{kg} \) wrench at \( 20 \) \( \text{m/s} \) in a direction away from the ship. If she and her spacesuit have a combined mass of \( 200 \) \( \text{kg} \), how long does it take her to coast back to her spaceship?
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:
\(7.14\,\text{m/s}\)
\(2.60\,\text{m}\)
\(\text{Doubling bullet’s velocity}\)
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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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