Two bodies of mass 1 kg and 3 kg have position vectors i^ + 2j^ + k^and − 3i^ − 2j^ + k^ respectively. The magnitude of position vector of centre of mass of this system will be similar to the magnitude of vector:
Question
Two bodies of mass 1 kg and 3 kg have position vectors i^ + 2j^ + k^and − 3i^ − 2j^ + k^ respectively. The magnitude of position vector of centre of mass of this system will be similar to the magnitude of vector:
Solution
To find the position vector of the center of mass of the system, we use the formula:
R_cm = (m1r1 + m2r2) / (m1 + m2)
where m1 and m2 are the masses of the two bodies, and r1 and r2 are their respective position vectors.
Given that m1 = 1 kg, m2 = 3 kg, r1 = i^ + 2j^ + k^, and r2 = -3i^ - 2j^ + k^, we can substitute these values into the formula:
R_cm = [(1 kg * (i^ + 2j^ + k^)) + (3 kg * (-3i^ - 2j^ + k^))] / (1 kg + 3 kg)
Solving this gives:
R_cm = [i^ + 2j^ + k^ - 9i^ - 6j^ + 3k^] / 4 kg
Simplifying this gives:
R_cm = [-8i^ - 4j^ + 4k^] / 4 kg
Finally, we get:
R_cm = -2i^ - j^ + k^
So, the magnitude of the position vector of the center of mass of the system is similar to the magnitude of the vector -2i^ - j^ + k^.
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