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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 ( 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.

– (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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KinematicsForces
\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_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 MotionEnergy
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
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
ConstantDescription
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients 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.
kSpring 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
VariableSI 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)}
VariableDerived 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)}
\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

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

1. Start with the given measurement: \text{5 km}

2. 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}}

3. 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}

4. Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

10^{-12}

Nano-

n

10^{-9}

Micro-

µ

10^{-6}

Milli-

m

10^{-3}

Centi-

c

10^{-2}

Deci-

d

10^{-1}

(Base unit)

10^{0}

Deca- or Deka-

da

10^{1}

Hecto-

h

10^{2}

Kilo-

k

10^{3}

Mega-

M

10^{6}

Giga-

G

10^{9}

Tera-

T

10^{12}

1. Some answers may be slightly off by 1% depending on rounding, etc.
2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
3. 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 .
4. Bookmark questions that you can’t solve so you can come back to them later.
5. 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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