AP Physics Unit

Unit 5 - Linear Momentum

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Mathematical

GQ

A space probe far from the Earth is traveling at 14.8 km/s. It has mass 1312 kg. The probe fires its rockets to give a constant thrust of 156 kN for 220 seconds. It accelerates in the same direction as its initial velocity. In this time it burns 150 kg of fuel. Calculate final speed of the space probe in km/s.

Note: This is a bonus question. Skip if you haven’t yet taken calculus. 

42.58 km/s

Objective: Calculate the final speed of the space probe after firing its rockets and burning fuel.

Step Formula Derivation Reasoning
1 F = ma Newton’s second law, force equals mass times acceleration.
2 a = \frac{F}{m} Rearrange to solve for acceleration, a.
3 a = \frac{156 \times 10^3\text{ N}}{1312\text{ kg}} Substitute the thrust (F) and initial mass of the probe (m). Note: Mass will change, this is initial acceleration.
4 a_{\text{initial}} = 118.9\text{ m/s}^2 Calculate the initial acceleration. Note: This will change as the probe burns fuel.
5 \Delta v = a \Delta t The change in velocity is the product of acceleration and time.
6 v_{\text{final}} = v_{\text{initial}} + \Delta v The final velocity is the initial velocity plus the change in velocity.

The acceleration will not be constant because the mass of the probe changes as it burns fuel. However, we can integrate the force over the time to find the change in velocity, considering the change in mass.

For the Change in Velocity (Variable Mass)

Step Formula Derivation Reasoning
1 \Delta v = \int_{0}^{t} \frac{F}{m(t)} dt With variable mass, acceleration changes over time.
2 m(t) = m_{\text{initial}} – \frac{dm}{dt}t Mass as a function of time, where \frac{dm}{dt} is the rate of mass loss.
3 \frac{dm}{dt} = \frac{150\text{ kg}}{220\text{ s}} Calculate the rate of mass loss.
4 \Delta v = \int_{0}^{220\text{ s}} \frac{156 \times 10^3\text{ N}}{1312\text{ kg} – \left(\frac{150\text{ kg}}{220\text{ s}}\right)t} dt Substitute m(t), F, and burn time into the integral.

This integral represents the change in momentum over the time period, taking into account the loss of mass from the fuel burn. Let’s calculate the integral to find the change in velocity and then determine the final speed of the space probe.

Final speed of the space probe: \boxed{v_{\text{final}} = 42.58 \text{ km/s}}

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42.58 km/s

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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)}
\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

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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