On this page, we will explore the motion of two particles in the x-y plane. The detailed analysis below will guide you through calculating various important physical quantities for the system.**Problem Statement:**Two particles are moving in the x-y plane. - Particle #1 has a mass \( m_1 = 6.40 \, \text{kg} \) and is located (at any given time) by the position vector \( \vec{r}_1(t) = [0.300 \, \text{m} + (2.00 \, \text{m/s}^3)t^3]\hat{i} + 0.200 \, \text{m}\hat{j} \).- Particle #2 has a mass \( m_2 = 9.00 \, \text{kg} \) and is located (similarly, at any given time) by the position vector \( \vec{r}_2(t) = 0.100 \, \text{m}\hat{i} + [0.300 \, \text{m} + (0.500 \, \text{m/s})t + (1.50 \, \text{m/s}^2)t^2]\hat{j} \).Determine the following at the time \(t = 1.00 \, s\). (Express your answers in vector form.)### Part (a) Location of the center of mass\[ \vec{r}_{cm}(t = 1.00 \, s) = \underline{\hspace{2cm}} \, \text{m} \]### Part (b) Velocity of the center of mass\[ \vec{V}_{cm}(t = 1.00 \, s) = \underline{\hspace{2cm}} \, \text{m/s} \]### Part (c) Acceleration of the center of mass\[ \vec{a}_{cm}(t = 1.00 \, s) = \underline{\hspace{2cm}} \, \text{m/s}^2 \]### Part (d) Momentum of the center of mass\[ \vec{P}_{cm}(t = 1.00 \, s) = \underline{\hspace{2cm}} \, \text{kg} \cdot \text{m/s

Name: On this page, we will explore the motion of two particles in the x-y
Uploaded: 2024-03-20T06:57:37.000Z
Channel: Bartleby
Description: On this page, we will explore the motion of two particles in the x-y plane. The detailed analysis below will guide you through calculating various important physical quantities for the system.**Problem Statement:**Two particles are moving in the x-y

Answered: Image /qna-images/answer/352854fe-bb2f-4b2f-bda4-7a0ace8fa194.jpg

Science

Physics

Two particles are moving in the x-y plane. Particle #1 has a mass m, = 6.40 kg and is located (at any time) by the position vector r, (t) = [0.300 m + (2.00 m/s?)t?jî + 0.200 mj. Particle #2 has a mass m, = 9.00 kg and is located (at any time) by the position vector r,(t) = 0.100 mî + [0.300 m + (0.500 m/s)t + (1.50 m/s?)t?jj. Determine the following at the time t = 1.00 s. (Express your answers in vector form.) (a) location of the center of mass r(t = 1.00 s) = m (b) velocity of the center of mass Vem (t = 1.00 s) = m/s v (c) acceleration of the center of mass a(t = 1.00 s) = m/s?

Two particles are moving in the x-y plane. Particle #1 has a mass m, = 6.40 kg and is located (at any time) by the position vector r, (t) = [0.300 m + (2.00 m/s?)t?jî + 0.200 mj. Particle #2 has a mass m, = 9.00 kg and is located (at any time) by the position vector r,(t) = 0.100 mî + [0.300 m + (0.500 m/s)t + (1.50 m/s?)t?jj. Determine the following at the time t = 1.00 s. (Express your answers in vector form.) (a) location of the center of mass r(t = 1.00 s) = m (b) velocity of the center of mass Vem (t = 1.00 s) = m/s v (c) acceleration of the center of mass a(t = 1.00 s) = m/s?

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Related questions

Q: Problem 18: A particle of mass 0.61 kg begins at rest and is then subject to a force in the…

A: A particle of mass 0.61 kg begins at rest and is then subject to a force in the positive x direction…

Q: (COLLAB Gauss's Law for Mass) Journey through the Center of the Earth. A 1024-kg blue ball is…

A: Given: The mass of the blue ball is 1024 kg. The position is 2.1x106 m. The radius of the planet is…

Q: Three point particles are fixed in position in an xy plane. Two of them, particle A of mass 5 g and…

Q: Calculate the kinetic energies (in J) of the following. (a) a 2,001.0 kg automobile moving at…

Q: At the instant of the figure, a 4.90 kg particle P has a position vector r of magnitude 9.40 m and…

Q: If you have a ball with a mass of 6.00kg, that has an initial velocity of vA=(2.00i + 5.00j)m/s and…

A: Given:- The ball of mass m = 6.0 kg The initial velocity of v→A= 2.0i^+5.0j^m/s. The final…

Q: −3.3 × 10-6 C/m2

A: Given, the infinitely long non-conducting sheet of charge density:…

Q: An asteroid with a mass of 75 kg is travelling in deep space, far away from the the gravitational…

Q: While sunbathing on the balcony of your 3rd floor apartment, you notice a gorilla drop a m = 34.7 kg…

A: Step 1: Given Step 2: General solution of Angular momentum of crate when it falls through a height…

Q: A ball of mass m = 0.490 kg is dropped near the surface of the Earth. However, there exists an air…

A: Given data as per the question mass =0.049 kgβ=0.700 N.smg=9.81 ms2

Q: The mass of particle 1 is 11kg. The initial velocity for particle 1 is (-104m/s)i + (216m/s)j. F21…

A: Given informationF21=Fxe-tTi+Fysin(2πtT)j (1)where Fx=16.8 NFy=-106 NT=100sprocess…

Q: A binary star system has two stars, each with the same mass M, that are separated by a distance of…

A: Dear Learner, As per guidelines, I can do only first three parts, Please re-post your pending part…

Q: Three particles qı = 7.90 µC, q2 = -1.10 µC, and q3 = 2.00 µC, have Cartesian coordinates (0.00,…

Q: The diagram shows two identical particles (labeled 1 and 2) that are moving in the x-y plane. The…

A: We need to find the magnitude of the net angular momentum of the particles about the origin.

Q: An asteroid with a mass of 75 kg is travelling in deep space, far away from the the gravitational…

Q: Mass ma=5kg and mg-2kg. Determine accelerations as and as of the two particles. (g=9.81m/s²),…

Q: To test the speed of a bullet, you create a pendulum by attaching a 3.20 kg wooden block to the…

Q: a. The boost rocket, m=12 Mg, generates an altitude dependent thrust of T(y)=By? where B=1.4 and…

A: I have given handwritten solution of the problem. see below steps.

Q: (COLLAB Gauss's Law for Mass) Journey through the Center of the Earth. A 1024.kg blue ball is…

A: The acceleration due to gravity for the given planet, Here G is the universal gravitational…

Q: The radius of Saturn (from the center to just above the atmosphere) is 60300 km (60300 × 10³ m), and…

A: Given Radius of Saturn R= 60300× 103 m = 6.03×107m Mass of Saturn M = 5.7 × 1026 kg

Q: Two blocks are on a frictionless horizontal table. A block of mass m,=1.51kg travelling with a speed…

Q: 9. A particle of mass m is moving along the smooth horizontal floor of a tank which is filled with…

A: Fluids that are viscous do not flow easily. It is possible for cold fluid to become viscous or thick…

Q: kg begins at rest and is then subject to a force in the positive x direction that changes with time…

A: Given:- The mass of the particle is m = 0.61 kg A particle begins at rest and is then subject to a…

Q: P a b m C Z X Q d R A small object of mass m is on a horizontal, frictionless surface, moving along…

Q: Two forces, F, = (2.70î – 3.40j) N and F, = (3.70î – 7.80j) N, act on a particle of mass 2.10 kg…

A: Given The one force is F1=2.70i-3.40j. The second force is F2=3.70i-7.80j. The mass of the particle…

Q: A particle of mass 16kg experiences a time-dependent force for t T P={1 0 F (4t³, 2t) N 0 for a…

A: Given, The force on the particle is given by F=0(4t3,2t)for t<0for 0≤t≤T0for t>T

Q: Problem 14: A particle of mass 1.39 kg begins at rest and is then subject to a force in the positive…

Q: An object of mass 23.0 kg is not moving on a plane with an angle 0, and of µ = 0.23, what is the…

Q: An object of mass 8.6 kg moves along a flat surface and then up an incline with a steepness of…

Q: The two masses m₁ = 4kg and m₂ = 6kg are connected by a massless string passing through a massless…

Q: Two forces, F, = (2.70î – 3.40j) N and F, = (3.70î – 7.80j) N, act on a particle of mass 2.10 kg…

A: F1 = 2.7 i - 3.4 j F2 = 3.7 i - 7.8 j mass, m = 2.1 kg

Q: Two asteroids of equal mass in the asteroid belt between Mars and Jupiter collide with a glancing…

Q: if A⃗ =2i+3j+6kA→=2i+3j+6k and A⃗ =3i−4j+2kA→=3i−4j+2k, then the angle between the two vectors is

Q: 10. An 8.00-kg point mass and a 15.0-kg point mass are held in place 50.0 cm apart. A particle of…

A: The two masses will have gravitational fore between them. Gravitational force is given as :-Where G…

Q: In a car crash, a Volvo of mass 1.85 x 104 kg collides with a stationary truck of mass 2.20 x 104 kg…

A: 1. Source at rest - When the sound source (whistle) is stationary relative to the observer, the…

Q: Two forces, 1 = (−6.55î + 4.75ĵ) N and 2 = (−4.05î + 6.40ĵ) N, act on a particle of mass 2.20 kg…

Q: Four electrons are located at the vertices of a square, one at each vertex. The length of a side of…

A: Four electrons at the vertices of the square. length of the side of the square is d=1.0nm=1.0×10-9m…

Q: The motion of the mass as shown for 0 <t< 2.1 seconds is defined by s = 10sin[cos (2t)lj + 15…

Q: Freight car A with a gross weight of 159000 lb is moving along the horizontal track in a switching…

Topic Video

Question

100%

On this page, we will explore the motion of two particles in the x-y plane. The detailed analysis below will guide you through calculating various important physical quantities for the system.

**Problem Statement:**

Two particles are moving in the x-y plane.

- Particle #1 has a mass \( m_1 = 6.40 \, \text{kg} \) and is located (at any given time) by the position vector \( \vec{r}_1(t) = [0.300 \, \text{m} + (2.00 \, \text{m/s}^3)t^3]\hat{i} + 0.200 \, \text{m}\hat{j} \).
- Particle #2 has a mass \( m_2 = 9.00 \, \text{kg} \) and is located (similarly, at any given time) by the position vector \( \vec{r}_2(t) = 0.100 \, \text{m}\hat{i} + [0.300 \, \text{m} + (0.500 \, \text{m/s})t + (1.50 \, \text{m/s}^2)t^2]\hat{j} \).

Determine the following at the time \(t = 1.00 \, s\). (Express your answers in vector form.)

### Part (a) Location of the center of mass
\[ \vec{r}_{cm}(t = 1.00 \, s) = \underline{\hspace{2cm}} \, \text{m} \]

### Part (b) Velocity of the center of mass
\[ \vec{V}_{cm}(t = 1.00 \, s) = \underline{\hspace{2cm}} \, \text{m/s} \]

### Part (c) Acceleration of the center of mass
\[ \vec{a}_{cm}(t = 1.00 \, s) = \underline{\hspace{2cm}} \, \text{m/s}^2 \]

### Part (d) Momentum of the center of mass
\[ \vec{P}_{cm}(t = 1.00 \, s) = \underline{\hspace{2cm}} \, \text{kg} \cdot \text{m/s

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

Step 2

Step 3

Step 4

Step 5

Trending now

This is a popular solution!

Step by step

Solved in 5 steps with 4 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions

Problem 2: An object of mass M is hanging from a spring (with I_o =0 and a spring constant k). There is a constant force from the wind, F_wind, blowing to the right, so the mass is initially at rest, but hanging at an angle Theta from the vertical because of the presence of the wind. Suddenly, spring breaks, and the mass drops under the influence of F_wind (note F_air and F_wind are the same constant vector!) and its weight. If the object (in equilibrium before the spring breaks) is a height H above the ground, how far laterally (in the direction of the wind) does the object hit the ground, L (given theta, k, M, g, H)?
An unfortunate astronaut loses his grip during a spacewalk and finds himself floating away from the space station, carrying only a rope and a bag of tools. First he tries to throw a rope to his fellow astronaut, but the rope is too short. In a last ditch effort, the astronaut throws his bag of tools in the direction of his motion, away from the space station. The astronaut has a mass of ma = 102 kg and the bag of tools has a mass of mp = 19.0 kg. If the astronaut is moving away from the space station at Vj = 2.10 m/s initially, what is the minimum final speed vp.f of the bag of tools with respect to the space station that will keep the astronaut from drifting away forever? Ub.f = m/s
, are: Three point particles are fixed in position in an xy plane. Two of them, particle A of mass 5 g and particle B of mass 11 g, are shown in the figure with a separation of dAB = 0.519 m at angle O = 30°. Particle C, with mass 9 g, is not shown. The net gravitational force acting on particle A due to particles B and Cis 2.71 x 1014 N at an angle of -163.8°. from the positive x axis. What are (a) the x coordinate and (b) the y coordinate of particle C? В dAB A
The proton is traveling at a constant speed of 5.5 × 105 m/s, and the radius of the kissing circle is 0.02 m. The mass of a proton is 1.7 × 10-27 kg. When the proton is at location A, what are the magnitude and direction (d|p→|/d⁢t)(p^) of the parallel component of dp→/d⁢t?
A package of mass 7 kg sits at the equator of an airless asteroid of mass 4.0 1020 kg and radius 5.9 105 m. We want to launch the package in such a way that it will never come back, and when it is very far from the asteroid it will be traveling with speed 188 m/s. We have a large and powerful spring whose stiffness is 2.1 105 N/m. How much must we compress the spring?
A football player, kicked a 0.550 [kg] ball off a rooftop that is 90.0 [m] off the ground, at 20.0 [m/s] oriented 30.0° degree above his horizon. His friend, Josh, caught the ball from a window at a distant building that is 40.0 [m] off the ground. Josh then released the ball from this height and let the ball fall to the ground. What is the speed of the ball just before it reaches Josh? ____ [m/s] What is the speed of the ball just before it reaches the ground? ____ [m/s]
Three point particles are fixed in position in an xy plane. Two of them, particle A of mass 5 g and particle B of mass 11 g, are shown in the figure with a separation of dAB = 0.579 m at angle = 30°. Particle C, with mass 8 g, is not shown. The net gravitational force acting on particle A due to particles B and C is 2.02 x 10-¹4 N at an angle of -163.8��. from the positive x axis. What are (a) the x coordinate and (b) the y coordinate of particle C? B dAB 0 A
Freight car A with a gross weight of 152000 Ib is moving along the horizontal track in a switching yard at 2.0 mi/hr. Freight car B with a gross weight of 162000 Ib and moving at 2.6 mi/hr overtakes car A and is coupled to it. Determine (a) the common velocity V of the two cars as they move together after being coupled and (b) the loss of energy |AE| due to the impact. 2.6 mi/hr 2.0 mi/hr A. Answers: (a) V = mi/hr ( b) |ΔΕΙ- i ft-lb
At the instant of the figure, a 1.00 kg particle P has a position vector 7 of magnitude 5.40 m and angle 01 = 41.0° and a velocity vector v of magnitude 4.50 m/s and angle 62 = 32.0°. Force F ,of magnitude 6.20 N and angle 03 = 32.0° acts on P. All three vectors lie in the xy plane. About the origin, what are the magnitude of (a) the angular momentum of the particle and (b) the torque acting on the particle? F (a) Number i Units (b) Number i Units
k A 13.00 kg particle starts from the origin at time zero. Its velocity as a function of time is given by = 10t² + 1tj where is in meters per second and t is in seconds. (Use the following as necessary: t.) V (a) Find its position as a function of time. ŕ= esc 10/³1 + ²/2 2²; 3 (b) Describe its motion qualitatively. The particle starts from rest at the origin, starts moving in the y direction, and gains speed faster and faster while turning to move more and more nearly parallel to the x axis. Score: 0.06 out of 0.06 Comment: (c) Find its acceleration as a function of time. a = 20ti + lj m/s² (d) Find the net force exerted on the particle as a function of time. F= T= (e) Find the net torque about the origin exerted on the particle as a function of time. N m X X (f) Find the angular momentum of the particle as a function of time. Ľ Need Help? Read It F1 N x (g) Find the kinetic energy of the particle as a function of time. K= x kg m²/s X (h) Find the power injected into the particle as a…
A particle of mass 100gm is moving to a uniform circular motion (with radius 50cm) on a table, which is connected to a hanging mass of 1Kg through a rope of negligible mass. What is the speed of the particle?
The robot moves the particle A (mass M) in the vertical plane using polar coordinate formulas r(t) = 1,2-0,6sin(2πt) [m] θ(t) = 0,5-1,5cos(2πt) [rad] in accordance with Determine the acceleration of the object in the θ direction [m/s2] at time t0=1.9s Use [m/s2] units GIVE ANSWER TO THREE DECIMALS