Two bodies of mass \(1\) kg and \(3\) kg have position vectors \(\hat i+2\hat j+\hat k\) and \(−3\hat i−2 \hat j+\hat k\), respectively. The magnitude of position vector of centre of mass of this system will be similar to the magnitude of vector.

Question

Approximate value of \( \cos 59^\circ \) is:

Answer 1

To find the magnitude of the position vector of the center of mass of the system consisting of two bodies, we use the formula for the center of mass:

\(\vec{R}_{cm} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2}\)

where:

\(\vec{r}_1\) = Position vector of the first mass = \(\hat{i} + 2\hat{j} + \hat{k}\)
\(\vec{r}_2\) = Position vector of the second mass = \(−3\hat{i}−2\hat{j}+\hat{k}\)
\(m_1\) = Mass of the first body = 1 kg
\(m_2\) = Mass of the second body = 3 kg

The total mass \(M = m_1 + m_2 = 1 + 3 = 4\) kg.

\(\vec{R}_{cm} = \frac{1(\hat{i} + 2\hat{j} + \hat{k}) + 3(−3\hat{i}−2\hat{j}+\hat{k})}{4}\)

Calculating the numerator by distributing the scalar multiplication:

\(= \frac{\hat{i} + 2\hat{j} + \hat{k} + (-9\hat{i} - 6\hat{j} + 3\hat{k})}{4}\)

\(= \frac{(\hat{i} - 9\hat{i}) + (2\hat{j} - 6\hat{j}) + (\hat{k} + 3\hat{k})}{4}\)

\(= \frac{-8\hat{i} - 4\hat{j} + 4\hat{k}}{4}\)

Simplifying by dividing each component by 4:

\(= -2\hat{i} - \hat{j} + \hat{k}\)

The magnitude of this position vector is:

\(|\vec{R}_{cm}| = \sqrt{(-2)^2 + (-1)^2 + 1^2}\)

\(= \sqrt{4 + 1 + 1}\)

\(= \sqrt{6}\)

Among the given options, the vector with the same magnitude is \(\hat{i} + 2\hat{j} + \hat{k}\), as its magnitude also equals:

\(\sqrt{1^2 + 2^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}\)

Therefore, the correct answer is \(\hat{i} + 2\hat{j} + \hat{k}\).

The Correct Option is A