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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?
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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
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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
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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
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