JEE Exam > JEE Questions > Three particles of masses 1 kg, 2 kg and 3 kg...
Start Learning for Free
Three particles of masses 1 kg, 2 kg and 3 kg are situated at the comers of an equilateral triangle move at speed 6ms-1, 3ms-1 and 2ms-1 respectively. Each particle maintains a direction towards the particle at the next corner symmetrically. Find velocity of CM of the system at this 2kg instant
- a)3ms-1
- b)5ms-1
- c)6ms-1
- d)Zero
Correct answer is option 'D'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
Three particles of masses 1 kg, 2 kg and 3 kg are situated at the come...
Given:
- Three particles of masses 1 kg, 2 kg and 3 kg are situated at the comers of an equilateral triangle
- They move at speeds 6ms-1, 3ms-1, and 2ms-1 respectively
- Each particle maintains a direction towards the particle at the next corner symmetrically
To find:
Velocity of CM of the system at this instant
Solution:
Let's assume that the equilateral triangle is an XY plane and its center is the origin.
Let's also assume that the 1 kg particle is at (1,0), 2 kg particle is at (-0.5, √3/2), and 3 kg particle is at (-0.5, -√3/2) at time t=0.
At any time t, the positions of the particles can be given by:
- 1 kg particle: (1,0) + (6t,0) = (1+6t, 0)
- 2 kg particle: (-0.5, √3/2) + (3t, -3√3/2) = (-0.5+3t, √3/2-3√3t/2)
- 3 kg particle: (-0.5, -√3/2) + (2t, √3) = (-0.5+2t, -√3/2+√3t)
Now, we can find the position of the center of mass (CM) of the system at time t using the formula:
CM = (m1r1 + m2r2 + m3r3) / (m1 + m2 + m3)
where m1, m2, and m3 are the masses of the particles and r1, r2, and r3 are their positions at time t.
Substituting the values, we get:
CMx = (1*(1+6t) + 2*(-0.5+3t) + 3*(-0.5+2t)) / 6 = t
CMy = (1*0 + 2*(√3/2-3√3t/2) + 3*(-√3/2+√3t)) / 6 = 0
So, the position of CM at time t is (t,0).
Now, we can find the velocity of CM at time t by differentiating its position with respect to time:
vCM = d(CM)/dt = (1,0)
Therefore, the velocity of CM of the system at any instant is (1,0) or simply 1 m/s in the x-direction.
Answer:
d) Zero (Incorrect)
Correct answer is option 'a' 1ms-1
- Three particles of masses 1 kg, 2 kg and 3 kg are situated at the comers of an equilateral triangle
- They move at speeds 6ms-1, 3ms-1, and 2ms-1 respectively
- Each particle maintains a direction towards the particle at the next corner symmetrically
To find:
Velocity of CM of the system at this instant
Solution:
Let's assume that the equilateral triangle is an XY plane and its center is the origin.
Let's also assume that the 1 kg particle is at (1,0), 2 kg particle is at (-0.5, √3/2), and 3 kg particle is at (-0.5, -√3/2) at time t=0.
At any time t, the positions of the particles can be given by:
- 1 kg particle: (1,0) + (6t,0) = (1+6t, 0)
- 2 kg particle: (-0.5, √3/2) + (3t, -3√3/2) = (-0.5+3t, √3/2-3√3t/2)
- 3 kg particle: (-0.5, -√3/2) + (2t, √3) = (-0.5+2t, -√3/2+√3t)
Now, we can find the position of the center of mass (CM) of the system at time t using the formula:
CM = (m1r1 + m2r2 + m3r3) / (m1 + m2 + m3)
where m1, m2, and m3 are the masses of the particles and r1, r2, and r3 are their positions at time t.
Substituting the values, we get:
CMx = (1*(1+6t) + 2*(-0.5+3t) + 3*(-0.5+2t)) / 6 = t
CMy = (1*0 + 2*(√3/2-3√3t/2) + 3*(-√3/2+√3t)) / 6 = 0
So, the position of CM at time t is (t,0).
Now, we can find the velocity of CM at time t by differentiating its position with respect to time:
vCM = d(CM)/dt = (1,0)
Therefore, the velocity of CM of the system at any instant is (1,0) or simply 1 m/s in the x-direction.
Answer:
d) Zero (Incorrect)
Correct answer is option 'a' 1ms-1
Free Test
FREE
| Start Free Test |
Community Answer
Three particles of masses 1 kg, 2 kg and 3 kg are situated at the come...
Given:
Mass of particle 1, m1 = 1 kg
Mass of particle 2, m2 = 2 kg
Mass of particle 3, m3 = 3 kg
Speed of particle 1, v1 = 6 m/s
Speed of particle 2, v2 = 3 m/s
Speed of particle 3, v3 = 2 m/s
To find:
Velocity of the center of mass (CM) of the system at the instant when the speed of particle 2 is 2 m/s.
Solution:
Step 1: Find the coordinates of the particles
Let the equilateral triangle have sides of length 'a'. Then, the coordinates of the particles are:
Particle 1: (x1, y1) = (a/2, √3a/6)
Particle 2: (x2, y2) = (0, 0)
Particle 3: (x3, y3) = (a, 0)
Step 2: Find the velocity vectors of the particles
The velocity vector of each particle is directed towards the next particle symmetrically. Therefore, we can draw lines connecting the particles and find the direction of the velocity vectors.
Let the angle between the line connecting particles 1 and 2 and the x-axis be θ. Then, the velocity vectors are:
Particle 1: (v1x, v1y) = (v1cos(θ+60°), v1sin(θ+60°))
Particle 2: (v2x, v2y) = (v2cos(θ), v2sin(θ))
Particle 3: (v3x, v3y) = (v3cos(θ-60°), v3sin(θ-60°))
Step 3: Find the velocity of the center of mass
The velocity of the center of mass is given by:
vCM = (m1v1 + m2v2 + m3v3) / (m1 + m2 + m3)
Substituting the values, we get:
vCM = [(1 kg)(6 m/s)(cos(θ+60°), sin(θ+60°)) + (2 kg)(3 m/s)(cos(θ), sin(θ)) + (3 kg)(2 m/s)(cos(θ-60°), sin(θ-60°))] / 6 kg
Simplifying the expression, we get:
vCM = [(3cos(θ) - 3√3sin(θ))i + (3√3cos(θ) + 3sin(θ))j] m/s
Step 4: Find the value of θ when v2 = 2 m/s
From the velocity vector of particle 2, we have:
v2x = v2cos(θ) = 2 m/s
Solving for θ, we get:
θ = cos^-1(2/3)
Step 5: Find the velocity of the center of mass at the given instant
Substituting θ in the expression for vCM, we get:
vCM = [(3cos(cos^-1(2/3)) - 3√3sin(cos^-1(2/3)))i + (3√3cos(cos^-1(2/3
Mass of particle 1, m1 = 1 kg
Mass of particle 2, m2 = 2 kg
Mass of particle 3, m3 = 3 kg
Speed of particle 1, v1 = 6 m/s
Speed of particle 2, v2 = 3 m/s
Speed of particle 3, v3 = 2 m/s
To find:
Velocity of the center of mass (CM) of the system at the instant when the speed of particle 2 is 2 m/s.
Solution:
Step 1: Find the coordinates of the particles
Let the equilateral triangle have sides of length 'a'. Then, the coordinates of the particles are:
Particle 1: (x1, y1) = (a/2, √3a/6)
Particle 2: (x2, y2) = (0, 0)
Particle 3: (x3, y3) = (a, 0)
Step 2: Find the velocity vectors of the particles
The velocity vector of each particle is directed towards the next particle symmetrically. Therefore, we can draw lines connecting the particles and find the direction of the velocity vectors.
Let the angle between the line connecting particles 1 and 2 and the x-axis be θ. Then, the velocity vectors are:
Particle 1: (v1x, v1y) = (v1cos(θ+60°), v1sin(θ+60°))
Particle 2: (v2x, v2y) = (v2cos(θ), v2sin(θ))
Particle 3: (v3x, v3y) = (v3cos(θ-60°), v3sin(θ-60°))
Step 3: Find the velocity of the center of mass
The velocity of the center of mass is given by:
vCM = (m1v1 + m2v2 + m3v3) / (m1 + m2 + m3)
Substituting the values, we get:
vCM = [(1 kg)(6 m/s)(cos(θ+60°), sin(θ+60°)) + (2 kg)(3 m/s)(cos(θ), sin(θ)) + (3 kg)(2 m/s)(cos(θ-60°), sin(θ-60°))] / 6 kg
Simplifying the expression, we get:
vCM = [(3cos(θ) - 3√3sin(θ))i + (3√3cos(θ) + 3sin(θ))j] m/s
Step 4: Find the value of θ when v2 = 2 m/s
From the velocity vector of particle 2, we have:
v2x = v2cos(θ) = 2 m/s
Solving for θ, we get:
θ = cos^-1(2/3)
Step 5: Find the velocity of the center of mass at the given instant
Substituting θ in the expression for vCM, we get:
vCM = [(3cos(cos^-1(2/3)) - 3√3sin(cos^-1(2/3)))i + (3√3cos(cos^-1(2/3
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
Three particles of masses 1 kg, 2 kg and 3 kg are situated at the comers of an equilateral triangle move at speed6ms-1,3ms-1and2ms-1respectively. Each particle maintains a direction towards the particle at the next corner symmetrically. Find velocity of CM of the system at this 2kg instanta)3ms-1 b)5ms-1 c)6ms-1 d)ZeroCorrect answer is option 'D'. Can you explain this answer?
Question Description
Three particles of masses 1 kg, 2 kg and 3 kg are situated at the comers of an equilateral triangle move at speed6ms-1,3ms-1and2ms-1respectively. Each particle maintains a direction towards the particle at the next corner symmetrically. Find velocity of CM of the system at this 2kg instanta)3ms-1 b)5ms-1 c)6ms-1 d)ZeroCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Three particles of masses 1 kg, 2 kg and 3 kg are situated at the comers of an equilateral triangle move at speed6ms-1,3ms-1and2ms-1respectively. Each particle maintains a direction towards the particle at the next corner symmetrically. Find velocity of CM of the system at this 2kg instanta)3ms-1 b)5ms-1 c)6ms-1 d)ZeroCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Three particles of masses 1 kg, 2 kg and 3 kg are situated at the comers of an equilateral triangle move at speed6ms-1,3ms-1and2ms-1respectively. Each particle maintains a direction towards the particle at the next corner symmetrically. Find velocity of CM of the system at this 2kg instanta)3ms-1 b)5ms-1 c)6ms-1 d)ZeroCorrect answer is option 'D'. Can you explain this answer?.
Three particles of masses 1 kg, 2 kg and 3 kg are situated at the comers of an equilateral triangle move at speed6ms-1,3ms-1and2ms-1respectively. Each particle maintains a direction towards the particle at the next corner symmetrically. Find velocity of CM of the system at this 2kg instanta)3ms-1 b)5ms-1 c)6ms-1 d)ZeroCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Three particles of masses 1 kg, 2 kg and 3 kg are situated at the comers of an equilateral triangle move at speed6ms-1,3ms-1and2ms-1respectively. Each particle maintains a direction towards the particle at the next corner symmetrically. Find velocity of CM of the system at this 2kg instanta)3ms-1 b)5ms-1 c)6ms-1 d)ZeroCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Three particles of masses 1 kg, 2 kg and 3 kg are situated at the comers of an equilateral triangle move at speed6ms-1,3ms-1and2ms-1respectively. Each particle maintains a direction towards the particle at the next corner symmetrically. Find velocity of CM of the system at this 2kg instanta)3ms-1 b)5ms-1 c)6ms-1 d)ZeroCorrect answer is option 'D'. Can you explain this answer?.
Solutions for Three particles of masses 1 kg, 2 kg and 3 kg are situated at the comers of an equilateral triangle move at speed6ms-1,3ms-1and2ms-1respectively. Each particle maintains a direction towards the particle at the next corner symmetrically. Find velocity of CM of the system at this 2kg instanta)3ms-1 b)5ms-1 c)6ms-1 d)ZeroCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of Three particles of masses 1 kg, 2 kg and 3 kg are situated at the comers of an equilateral triangle move at speed6ms-1,3ms-1and2ms-1respectively. Each particle maintains a direction towards the particle at the next corner symmetrically. Find velocity of CM of the system at this 2kg instanta)3ms-1 b)5ms-1 c)6ms-1 d)ZeroCorrect answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Three particles of masses 1 kg, 2 kg and 3 kg are situated at the comers of an equilateral triangle move at speed6ms-1,3ms-1and2ms-1respectively. Each particle maintains a direction towards the particle at the next corner symmetrically. Find velocity of CM of the system at this 2kg instanta)3ms-1 b)5ms-1 c)6ms-1 d)ZeroCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for Three particles of masses 1 kg, 2 kg and 3 kg are situated at the comers of an equilateral triangle move at speed6ms-1,3ms-1and2ms-1respectively. Each particle maintains a direction towards the particle at the next corner symmetrically. Find velocity of CM of the system at this 2kg instanta)3ms-1 b)5ms-1 c)6ms-1 d)ZeroCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of Three particles of masses 1 kg, 2 kg and 3 kg are situated at the comers of an equilateral triangle move at speed6ms-1,3ms-1and2ms-1respectively. Each particle maintains a direction towards the particle at the next corner symmetrically. Find velocity of CM of the system at this 2kg instanta)3ms-1 b)5ms-1 c)6ms-1 d)ZeroCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Three particles of masses 1 kg, 2 kg and 3 kg are situated at the comers of an equilateral triangle move at speed6ms-1,3ms-1and2ms-1respectively. Each particle maintains a direction towards the particle at the next corner symmetrically. Find velocity of CM of the system at this 2kg instanta)3ms-1 b)5ms-1 c)6ms-1 d)ZeroCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup