Question

Two bodies of masses 2g and 10g have position vectors 3i + 2j - k and i - j +3k respectively find the position vectors and distance of C.M from the origin​

Answer 1

Explanation:

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Answer 2

The distance of the C.M from the origin is √(461/36) units.

To find the position vector of the center of mass (C.M) of the two bodies, we need to use the formula:

Rcm = (m1r1 + m2r2) / (m1 + m2)

where Rcm is the position vector of the C.M, m1 and m2 are the masses of the two bodies, and r1 and r2 are their respective position vectors.

Substituting the given values, we get:

Rcm = (2*(3i + 2j - k) + 10*(i - j + 3k)) / (2 + 10)

= (6i + 4j - 2k + 10i - 10j + 30k) / 12

= 16i/12 + 6j/12 + 28k/12

= (4/3)i + (1/2)j + (7/3)k

Thus, the position vector of the C.M is (4/3)i + (1/2)j + (7/3)k.

To find the distance of the C.M from the origin, we simply need to find the magnitude of the position vector. Using the formula for the magnitude of a vector:

|a| = √(a1² + a2² + a3²)

we get:

|Rcm| = √((4/3)² + (1/2)² + (7/3)²)

= √(64/9 + 1/4 + 49/9)

= √(256/36 + 9/36 + 196/36)

= √(461/36)

Therefore, the distance of the C.M from the origin is √(461/36) units.

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