Q.32. Two identical bodies A and B of equal masses have initial velocities and respectively. The b

Q: What is the correct answer for this exam Physics Q32?

The correct answer is C. The velocity of the center of mass is $\vec{v}_{cm} = \frac{m_A \vec{v}_A + m_B \vec{v}_B}{m_A + m_B}$. Since $m_A = m_B = m$, $\vec{v}_{cm} = \frac{m(4\hat{i}) + m(4\hat{j})}{2m} = \frac{4\hat{i} + 4\hat{j}}{2} = 2\hat{i} + 2\hat{j}$. The acceleration of the center of mass is $\vec{a}_{cm} = \frac{m_A \vec{a}_A + m_B \vec{a}_B}{m_A + m_B}$. Since $m_A = m_B = m$ and $\vec{a}_B = 0$, $\vec{a}_{cm}

Q: Q.32. Two identical bodies A and B of equal masses have initial velocities and respectively. The body A has acceleration while the acceleration of the other body B is zero. The centre of mass of…

The correct answer is C. The velocity of the center of mass is $\vec{v}_{cm} = \frac{m_A \vec{v}_A + m_B \vec{v}_B}{m_A + m_B}$. Since $m_A = m_B = m$, $\vec{v}_{cm} = \frac{m(4\hat{i}) + m(4\hat{j})}{2m} = \frac{4\hat{i} + 4\hat{j}}{2} = 2\hat{i} + 2\hat{j}$. The acceleration of the center of mass is $\vec{a}_{cm} = \frac{m_A \vec{a}_A + m_B \vec{a}_B}{m_A + m_B}$. Since $m_A = m_B = m$ and $\vec{a}_B = 0$, $\vec{a}_{cm}

Q: What are the options for this this exam Physics question?

A. circular · B. parabolic · C. straight line · D. elliptical

Question

Q.32. Two identical bodies A and B of equal masses have initial velocities  and  respectively. The body A has acceleration  while the acceleration of the other body B is zero. The centre of mass of the two bodies moves in _______ path.

FirstInTest · Accepted Answer

The velocity of the center of mass is $\vec{v}_{cm} = \frac{m_A \vec{v}_A + m_B \vec{v}_B}{m_A + m_B}$. Since $m_A = m_B = m$, $\vec{v}_{cm} = \frac{m(4\hat{i}) + m(4\hat{j})}{2m} = \frac{4\hat{i} + 4\hat{j}}{2} = 2\hat{i} + 2\hat{j}$. The acceleration of the center of mass is $\vec{a}_{cm} = \frac{m_A \vec{a}_A + m_B \vec{a}_B}{m_A + m_B}$. Since $m_A = m_B = m$ and $\vec{a}_B = 0$, $\vec{a}_{cm} = \frac{m(6\hat{i}+6\hat{j}) + m(0)}{2m} = \frac{6\hat{i} + 6\hat{j}}{2} = 3\hat{i} + 3\hat{j}$. Since $\vec{v}_{cm}$ and $\vec{a}_{cm}$ are in the same direction (both have equal $\hat{i}$ and $\hat{j}$ components, so they are parallel), the path of the center of mass is a straight line.

Answer

circular

Answer

parabolic

Answer

straight line

Answer

elliptical

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