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The best way to analyze this situation is to apply the principle of conservation of momentum and understand how momentum changes for individual bodies in a collision. Momentum, defined as the product of mass and velocity ([katex] p = mv [/katex]), is always conserved in an isolated system (without external forces).

Given:

– Let [katex] v_t [/katex] and [katex] v_c [/katex] be the velocities of the truck and the car respectively.

– Assume the truck has a mass [katex] m_t [/katex] and the car has a mass [katex] m_c [/katex].

– Truck’s initial velocity [katex] v_{t, \text{init}} = 15 , \text{km/h} [/katex] (we’ll consider the direction towards the car as positive).

– Car’s initial velocity [katex] v_{c, \text{init}} = -30 , \text{km/h} [/katex] (since it’s head-on and opposite, it’s negative).

**Step 1: Calculate Initial Momenta**

– Momentum of the truck initially: [katex] p_{t, \text{init}} = m_t \times 15 [/katex]

– Momentum of the car initially: [katex] p_{c, \text{init}} = m_c \times (-30) [/katex]

**Step 2: Use Conservation of Momentum**

– Total initial momentum [katex] p_{\text{total, init}} = m_t \times 15 + m_c \times (-30) [/katex].

The collision is an isolated event with no external forces, so total momentum must be preserved:

– [katex] p_{\text{total, final}} = p_{\text{total, init}} [/katex].

After the collision, depending on the details (elastic or inelastic), both truck and car will have new velocities [katex] v_{t, \text{final}} [/katex] and [katex] v_{c, \text{final}} [/katex] but their combined momentum remains [katex] m_t \times v_{t, \text{final}} + m_c \times v_{c, \text{final}} = m_t \times 15 + m_c \times (-30) [/katex].

**Step 3: Change in Momentum**

– Change in momentum for the truck: [katex] \Delta p_t = m_t \times (v_{t, \text{final}} – 15) [/katex]

– Change in momentum for the car: [katex] \Delta p_c = m_c \times (v_{c, \text{final}} + 30) [/katex]

Change in the magnitude of the individual momenta will depend on the masses of the truck and car and their change in velocities but must reflect conservation principles.

**Answer Analysis:**

– (a) Incorrect – A greater mass does not directly mean a greater change in momentum unless the change in velocity is considered.

– (b) Correct – Since momentum is conserved and the system is isolated, both the car and truck experience the equal magnitude of momentum change but in opposite directions.

– (c) Incorrect – Greater initial speed of the car implies greater initial momentum, but not necessarily a greater change in momentum.

– (d) Incorrect – Both vehicles change their momentum, but the total system momentum is conserved.

– (e) Incorrect – Statement (b) is necessarily true considering the law of conservation of momentum in an isolated system.

Thus, the statement that best describes the situation is **(b) They both have the same change in magnitude of momentum because momentum is conserved.**

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Kinematics | Forces |
---|---|

[katex]\Delta x = v_i t + \frac{1}{2} at^2[/katex] | [katex]F = ma[/katex] |

[katex]v = v_i + at[/katex] | [katex]F_g = \frac{G m_1m_2}{r^2}[/katex] |

[katex]a = \frac{\Delta v}{\Delta t}[/katex] | [katex]f = \mu N[/katex] |

[katex]R = \frac{v_i^2 \sin(2\theta)}{g}[/katex] |

Circular Motion | Energy |
---|---|

[katex]F_c = \frac{mv^2}{r}[/katex] | [katex]KE = \frac{1}{2} mv^2[/katex] |

[katex]a_c = \frac{v^2}{r}[/katex] | [katex]PE = mgh[/katex] |

[katex]KE_i + PE_i = KE_f + PE_f[/katex] |

Momentum | Torque and Rotations |
---|---|

[katex]p = m v[/katex] | [katex]\tau = r \cdot F \cdot \sin(\theta)[/katex] |

[katex]J = \Delta p[/katex] | [katex]I = \sum mr^2[/katex] |

[katex]p_i = p_f[/katex] | [katex]L = I \cdot \omega[/katex] |

Simple Harmonic Motion |
---|

[katex]F = -k x[/katex] |

[katex]T = 2\pi \sqrt{\frac{l}{g}}[/katex] |

[katex]T = 2\pi \sqrt{\frac{m}{k}}[/katex] |

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

Conversion Example

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] | 0.000000000001 |

Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |

Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |

Milli- | m | [katex]10^{-3}[/katex] | 0.001 |

Centi- | c | [katex]10^{-2}[/katex] | 0.01 |

Deci- | d | [katex]10^{-1}[/katex] | 0.1 |

(Base unit) | – | [katex]10^{0}[/katex] | 1 |

Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |

Hecto- | h | [katex]10^{2}[/katex] | 100 |

Kilo- | k | [katex]10^{3}[/katex] | 1,000 |

Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |

Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |

Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |

- Some answers may be slightly off by 1% depending on rounding, etc.
- Answers will use different values of gravity. Some answers use 9.81 m/s
^{2}, and other 10 m/s^{2 }for calculations. - Variables are sometimes written differently from class to class. For example, sometime initial velocity [katex] v_i [/katex] is written as [katex] u [/katex]; sometimes [katex] \Delta x [/katex] is written as [katex] s [/katex].
- Bookmark questions that you can’t solve so you can come back to them later.
- Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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