Login

AP Physics

Unit 5 - Linear Momentum

17136 views

Advanced

Mathematical

GQ

You're a Pro Member

Supercharge UBQ

Go Pro

0 attempts

0% avg

UBQ Credits

Verfied Answer
Verfied Explanation 0 likes
Written by Phy
0

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

Step Formula Derivation Reasoning
1 [katex]F = ma[/katex] Newton’s second law, force equals mass times acceleration.
2 [katex]a = \frac{F}{m}[/katex] Rearrange to solve for acceleration, [katex]a[/katex].
3 [katex]a = \frac{156 \times 10^3\text{ N}}{1312\text{ kg}}[/katex] Substitute the thrust ([katex]F[/katex]) and initial mass of the probe ([katex]m[/katex]). Note: Mass will change, this is initial acceleration.
4 [katex]a_{\text{initial}} = 118.9\text{ m/s}^2[/katex] Calculate the initial acceleration. Note: This will change as the probe burns fuel.
5 [katex]\Delta v = a \Delta t[/katex] The change in velocity is the product of acceleration and time.
6 [katex]v_{\text{final}} = v_{\text{initial}} + \Delta v[/katex] 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 [katex]\Delta v = \int_{0}^{t} \frac{F}{m(t)} dt[/katex] With variable mass, acceleration changes over time.
2 [katex]m(t) = m_{\text{initial}} – \frac{dm}{dt}t[/katex] Mass as a function of time, where [katex]\frac{dm}{dt}[/katex] is the rate of mass loss.
3 [katex]\frac{dm}{dt} = \frac{150\text{ kg}}{220\text{ s}}[/katex] Calculate the rate of mass loss.
4 [katex]\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[/katex] Substitute [katex]m(t)[/katex], [katex]F[/katex], 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: [katex]\boxed{v_{\text{final}} = 42.58 \text{ km/s}}[/katex]

Need Help? Ask Phy To Explain

Just ask: "Help me solve this problem."

Just Drag and Drop!
Quick Actions ?
×

Topics in this question

Join 1-to-1 Elite Tutoring

  • Statistics

See how Others Did on this question | Coming Soon

Intermediate
Conceptual
MCQ

A truck of mass 3500 kg hits the back of a small car of mass 1400 kg. Which car exerted more force on the other and why?

  • Linear Forces
Solve
Beginner
Mathematical
GQ

A spring in a pogo-stick is compressed \( 12 \) \( \text{cm} \) when a \( 40. \) \( \text{kg} \) girl stands on it. What is the spring constant for the pogo-stick spring?

  • Linear Forces, Springs
Solve
Advanced
Mathematical
GQ

A \(6 \, \text{kg}\) cube rests against a compressed spring with a force constant of \(1{,}800 \, \text{N/m}\), initially compressed by \(0.3 \, \text{m}\). Upon release, the cube slides on a horizontal surface with a kinetic friction coefficient of \(\mu_k = 0.12\) for \(3 \, \text{m}\), then ascends a \(12^\circ\) slope, stopping after \(4.5 \, \text{m}\). Determine the coefficient of kinetic friction on the slope.

  • Energy, Inclines, Linear Forces
Solve
Intermediate
Proportional Analysis
FRQ

Imagine a hypothetical planet that has two moons. Moon \(\#1\) is in a circular orbit of radius \(R\) and has a mass \(M\).

  • Circular Motion, Gravitation, Orbits
Solve
Advanced
Mathematical
GQ

Four blocks of masses \( 20 \, \text{kg}, \, 30 \, \text{kg}, \, 40 \, \text{kg}, \, \text{and} \, 50 \, \text{kg} \) are stacked on top of one another in an elevator in order of decreasing mass with the lightest mass on the top of the stack. The elevator moves downward with an acceleration of \( 3.2 \, \text{m/s}^2 \). Find the contact force between the \( 30 \, \text{kg} \) and \( 40 \, \text{kg} \) masses.

  • Elevators, Multi-Body Systems
Solve

42.58 km/s

Nerd Notes

Discover the world's best Physics resources

Continue with

By continuing you (1) agree to our Terms of Use and Terms of Sale and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.

Error Report

Sign in before submitting feedback.

Sign In

Sign In to View Your Questions

Sign In

Share This Question

Enjoying UBQ? Share the 🔗 with friends!

Link Copied!
nerd-notes.com
Download Equation Sheets ⬇︎
KinematicsForces
\(\Delta x = v_i t + \frac{1}{2} at^2\)\(F = ma\)
\(v = v_i + at\)\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\) 
Circular MotionEnergy
\(F_c = \frac{mv^2}{r}\)\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)\(KE_i + PE_i = KE_f + PE_f\)
 \(W = Fd \cos\theta\)
MomentumTorque 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 MotionFluids
\(F = -kx\)\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)\(F_b = \rho V g\)
\(a = -\omega^2 x\)\(A_1v_1 = A_2v_2\)
ConstantDescription
[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
VariableSI 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]
VariableDerived 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

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

  1. Start with the given measurement: [katex]\text{5 km}[/katex]

  2. 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]

  3. 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]

  4. 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]

Phy Pro

The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.

$11.99

per month

Subscribe

Billed Monthly. Cancel Anytime.

Trial  –>  Phy Pro

  • Unlimited Messages and Images
  • Unlimited UBQ Credits
  • 157% Better than GPT
  • 30 → 300 Word Input
  • 3 --> 15 MB Image Size Limit
  • 1 --> 3 Images per Message
  • All Smart Actions
  • No Ads

Physics is Hard, But It Does NOT Have to Be

We crafted THE Ultimate A.P Physics 1 course so you can learn faster and score higher. 

  • Rapid Video Lessons
  • AI Graded Practice Sets
  • Self Paced, Year-Round Enrollment
Trusted by 10k+ Students
Learn More
I'm Doing Alright For Now

📚 Predict Your AP Physics Exam Score

Try our free calculator to see what you need to get a 5 on the upcoming AP Physics 1 exam.

Calculate

You can close this ad in 5 seconds.

Ads show frequently. Upgrade to Phy Pro to remove ads.

You can close this ad in 7 seconds.

Ads display every few minutes. Upgrade to Phy Pro to remove ads.

You can close this ad in 5 seconds.

Ads show frequently. Upgrade to Phy Pro to remove ads.

How to UBQ

A quick explanation

1. UBQ Credits

Credits are used to grade your FRQs and GQs. Pro users get unlimited credits. 

2. Auto Track Stats

Submitting counts as 1 attempt. 

Viewing answers or explanations count as a failed attempts.

3. Point System

Phy gives partial credit if needed

MCQs and GQs are are 1 point each. FRQs will state points for each part.

Phy helps customize your learning experience.

4. Question Explantions

Phy customizes problem explanations based on what you struggle with. Just hit the explanation button to see.

Phy gives feedback.

Understand you mistakes quicker.

4. Feedback

Phy automatically provides feedback so you can improve your responses. 

10 Free Credits To Get You Started 

Start Solving

By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data. 

Feeling uneasy about your next physics test? We help boost grades in 3 lesson or less.

Join Here

NEW! PHY AI accurately solves all questions

Try For Free

🔥 Get up to 30% off Elite Physics Tutoring

Learn About Memberships

🧠  NEW! Learn Physics From Scratch Self Paced Course

Enrollment Details

🎯 Need exam style practice questions?

Use UBQ For Free
We use cookies to improve your experience. By continuing to browse on Nerd Notes, you accept the use of cookies as outlined in our privacy policy.
I Accept