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
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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)...
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![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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd0afef86-41bd-49f7-9deb-bb70dd2f5172%2F352854fe-bb2f-4b2f-bda4-7a0ace8fa194%2F0y0jykg.png&w=3840&q=75)
Transcribed Image Text: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
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