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Question
Consider a system of two identical particles. One of the particles is at rest and the other has an acceleration a. The centre of mass has an Acceleration
Options
zero
\[\frac{1}{2} \vec{a}\]
\[\vec{a}\]
\[2 \vec{a}\]
Solution
\[\frac{1}{2} \vec{a}\]
Acceleration of centre of mass of a two-particle system is given as,
\[\vec{a}_{cm} = \frac{m_1 \vec{a}_1 + m_2 \vec{a_2}}{m_1 + m_2}\] ..............(1)
According to the question,
\[m_1 = m_2 = m\]
\[ a_1 = 0\]
\[ a_2 = a\]
Substituting these values in equation (1), we get:
\[\vec{a}_{cm} = \frac{m \times 0 + m \vec{a}}{2m} = \frac{1}{2} \vec{a}\]
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