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 ( p = mv ), is always conserved in an isolated system (without external forces).

Given:

– Let v_t and v_c be the velocities of the truck and the car respectively.

– Assume the truck has a mass m_t and the car has a mass m_c .

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

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

**Step 1: Calculate Initial Momenta**

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

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

**Step 2: Use Conservation of Momentum**

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

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

– p_{\text{total, final}} = p_{\text{total, init}} .

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

**Step 3: Change in Momentum**

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

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

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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- Statistics

Beginner

Mathematical

GQ

A pool cue ball, mass 0.7 kg, is traveling at 2 m/s when it collides head on with another ball, mass 0.5 kg, traveling in the opposite direction with a speed of 1.2 m/s. After the collision, the cue ball travels in the opposite direction at 0.3 m/s. What is the velocity of the other ball?

- Momentum

Advanced

Mathematical

GQ

A 2,000 kg car collides with a stationary 1,000 kg car. Afterwards, they slide 6 m before coming to a stop. The coefficient of friction between the tires and the road is 0.7. Find the initial velocity of the 2,000 kg car before the collision?

- 1D Kinematics, Energy, Linear Forces, Momentum

Intermediate

Conceptual

MCQ

A rocket explodes into two fragments, one 25 times heavier than the other. The change in momentum of the lighter fragment is

- Impulse, Momentum

Advanced

Conceptual

MCQ

- Linear Forces, Momentum

Advanced

Mathematical

FRQ

A 20 g piece of clay moving at a speed of 50 m/s strikes a 500 g pendulum bob at rest. The length of a string is 0.8 m. After the collision the clay-bob system starts to oscillate as a simple pendulum.

- Energy, Momentum, Pendulums, Simple Harmonic Motion

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

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

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

a = \frac{\Delta v}{\Delta t} | f = \mu N |

R = \frac{v_i^2 \sin(2\theta)}{g} |

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

F_c = \frac{mv^2}{r} | KE = \frac{1}{2} mv^2 |

a_c = \frac{v^2}{r} | PE = mgh |

KE_i + PE_i = KE_f + PE_f |

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

p = m v | \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 |
---|

F = -k x |

T = 2\pi \sqrt{\frac{l}{g}} |

T = 2\pi \sqrt{\frac{m}{k}} |

Constant | Description |
---|---|

g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |

G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |

\mu_k and \mu_s | Coefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion. |

k | Spring constant, in \text{N/m} |

M_E = 5.972 \times 10^{24} , \text{kg} | Mass of the Earth |

M_M = 7.348 \times 10^{22} , \text{kg} | Mass of the Moon |

M_M = 1.989 \times 10^{30} , \text{kg} | Mass of the Sun |

Variable | SI Unit |
---|---|

s (Displacement) | \text{meters (m)} |

v (Velocity) | \text{meters per second (m/s)} |

a (Acceleration) | \text{meters per second squared (m/s}^2\text{)} |

t (Time) | \text{seconds (s)} |

m (Mass) | \text{kilograms (kg)} |

Variable | Derived SI Unit |
---|---|

F (Force) | \text{newtons (N)} |

E, PE, KE (Energy, Potential Energy, Kinetic Energy) | \text{joules (J)} |

P (Power) | \text{watts (W)} |

p (Momentum) | \text{kilogram meters per second (kgm/s)} |

\omega (Angular Velocity) | \text{radians per second (rad/s)} |

\tau (Torque) | \text{newton meters (Nm)} |

I (Moment of Inertia) | \text{kilogram meter squared (kgm}^2\text{)} |

f (Frequency) | \text{hertz (Hz)} |

General Metric Conversion Chart

Conversion Example

Example of using unit analysis: Convert 5 kilometers to millimeters.

Start with the given measurement:

`\text{5 km}`

Use the conversion factors for kilometers to meters and meters to millimeters:

`\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}`

Perform the multiplication:

`\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}`

Simplify to get the final answer:

`\boxed{5 \times 10^6 \, \text{mm}}`

Prefix | Symbol | Power of Ten | Equivalent |
---|---|---|---|

Pico- | p | 10^{-12} | 0.000000000001 |

Nano- | n | 10^{-9} | 0.000000001 |

Micro- | µ | 10^{-6} | 0.000001 |

Milli- | m | 10^{-3} | 0.001 |

Centi- | c | 10^{-2} | 0.01 |

Deci- | d | 10^{-1} | 0.1 |

(Base unit) | – | 10^{0} | 1 |

Deca- or Deka- | da | 10^{1} | 10 |

Hecto- | h | 10^{2} | 100 |

Kilo- | k | 10^{3} | 1,000 |

Mega- | M | 10^{6} | 1,000,000 |

Giga- | G | 10^{9} | 1,000,000,000 |

Tera- | T | 10^{12} | 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 v_i is written as u ; sometimes \Delta x is written as s .
- 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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