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Two particles of equal mass have coordinates (2m,4m,6m) and (6m,2m,8m). Of these one particle has a velocity v1=2i^ and another particle has velocity v2 =2j^ at time t=0 . The coordinates of centre of mass at time t=1s will be?
Most Upvoted Answer
Two particles of equal mass have coordinates (2m,4m,6m) and (6m,2m,8m)...
Given:
- Two particles of equal mass
- Coordinates of particle 1: (2m,4m,6m)
- Coordinates of particle 2: (6m,2m,8m)
- Velocity of particle 1: v1 = 2i^
- Velocity of particle 2: v2 = 2j^
- Time: t = 0s

To find:
- Coordinates of the center of mass at time t = 1s

Solution:

1. Calculating the initial position of the center of mass:
- The center of mass is given by the formula:
Center of Mass = (m1 * r1 + m2 * r2) / (m1 + m2)
where m1 and m2 are the masses of the particles, and r1 and r2 are their respective positions.
- Since the particles have equal mass, we can simplify the formula to:
Center of Mass = (r1 + r2) / 2

2. Calculating the initial position of the center of mass:
- Substituting the given coordinates of the particles into the formula, we get:
Center of Mass = ((2m, 4m, 6m) + (6m, 2m, 8m)) / 2
= (8m, 6m, 14m) / 2
= (4m, 3m, 7m)

3. Calculating the displacement of the center of mass:
- The displacement of the center of mass can be calculated using the formula:
Displacement = Velocity * Time
- We are given that the velocity of particle 1 is v1 = 2i^ and the velocity of particle 2 is v2 = 2j^.
- At time t = 1s, the displacement of particle 1 will be:
Displacement1 = v1 * t
= (2i^) * 1s
= 2i^
- At time t = 1s, the displacement of particle 2 will be:
Displacement2 = v2 * t
= (2j^) * 1s
= 2j^

4. Calculating the final position of the center of mass:
- Adding the displacements of the center of mass to its initial position, we get:
Final Position = Initial Position + Displacement
= (4m, 3m, 7m) + (2i^) + (2j^)
= (4m + 2m, 3m + 0m, 7m + 0m)
= (6m, 3m, 7m)

Conclusion:
The coordinates of the center of mass at time t = 1s are (6m, 3m, 7m).
Community Answer
Two particles of equal mass have coordinates (2m,4m,6m) and (6m,2m,8m)...
As velocity of both the particles are given... v1 is in x-axis and v2 in y-axis.. and coordinate of 1st particle is (2m, 4m, 6m) i.e. (x, y, z) so multiply 2 with 2m and you get (4m, 4m, 6m) and for the 2nd particle the coordinates are (6m, 2m, 8m) i.e. (x,y,z) and multiply the velocity(2) which is in y-axis.. we get it as (6m, 4m, 8m)

COM of x-axis is= 4+6/2
=5

COM of y-axis=4+4/2
=4
COM of z-axis= 6+8/2
=7
so the final coordinates are (5m, 4m, 7m)
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Two particles of equal mass have coordinates (2m,4m,6m) and (6m,2m,8m). Of these one particle has a velocity v1=2i^ and another particle has velocity v2 =2j^ at time t=0 . The coordinates of centre of mass at time t=1s will be?
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Two particles of equal mass have coordinates (2m,4m,6m) and (6m,2m,8m). Of these one particle has a velocity v1=2i^ and another particle has velocity v2 =2j^ at time t=0 . The coordinates of centre of mass at time t=1s will be? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Two particles of equal mass have coordinates (2m,4m,6m) and (6m,2m,8m). Of these one particle has a velocity v1=2i^ and another particle has velocity v2 =2j^ at time t=0 . The coordinates of centre of mass at time t=1s will be? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two particles of equal mass have coordinates (2m,4m,6m) and (6m,2m,8m). Of these one particle has a velocity v1=2i^ and another particle has velocity v2 =2j^ at time t=0 . The coordinates of centre of mass at time t=1s will be?.
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