The velocities of two particles A and B of same mass are VA-a- and VB-b- where a and b are constants- The acceleration of particle A is 2a-4b- and acceleration of particle B is a-b- in m-s2- The centre of mass of two particles will move in

Given, V_A=aî and V_B=bĵ a_A=2aî+4bĵ a_B=aî bĵ V_com=m_Av_A+m_Bv_Bm_A+m_B ⇒ V_com=aî+bĵ2 a_com=m_Aa_A+m_Ba_Bm_A+m_B a_com=a_A+a_B2=32(aî+bĵ) Clearly, a_ ...

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Standard XII

Physics

Newton's Laws of Motion

The velocitie...

Question

The velocities of two particles $A$ and $B$ of same mass are $V_{A} = a^i$ and $V_{B} = b^j$ , where $a$ and $b$ are constants. The acceleration of particle $A$ is $(2 a^i + 4 b^j)$ and acceleration of particle $B$ is $(a^i - b^j)$ (in ${m/s}^{2})$ . The centre of mass of two particles will move in

Straight line

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Solution

The correct option is A Straight line
Given,

$V_{A} = a^i$ and $V_{B} = b^j$

$a_{A} = 2 a^i + 4 b^j$

$a_{B} = a^i - b^j$

${\to V}_{c o m} = \frac{m_{A} {\to v}_{A} + m_{B} {\to v}_{B}}{m_{A} + m_{B}}$

$\Rightarrow V_{c o m} = \frac{a^i + b^j}{2}$

${\to a}_{c o m} = \frac{m_{A} {\to a}_{A} + m_{B} {\to a}_{B}}{m_{A} + m_{B}}$

$a_{c o m} = \frac{{\to a}_{A} + {\to a}_{B}}{2} = \frac{3}{2} (a^i + b^j)$

Clearly, ${\to a}_{c o m} = 3 {\to V}_{c o m}$

So, ${\to a}_{c o m} | | {\to V}_{c o m}$

Hence, the trajectory of $COM$ is a straight line.

Hence, option $(A)$ is the correct answer.

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The velocities of two particles A and B of same mass are VA=a^i and VB=b^j, where a and b are constants. The acceleration of particle A is (2a^i+4b^j) and acceleration of particle B is (a^i−b^j) (in m/s2). The centre of mass of two particles will move in

The velocities of two particles $A$ and $B$ of same mass are $V_{A} = a^i$ and $V_{B} = b^j$ , where $a$ and $b$ are constants. The acceleration of particle $A$ is $(2 a^i + 4 b^j)$ and acceleration of particle $B$ is $(a^i - b^j)$ (in ${m/s}^{2})$ . The centre of mass of two particles will move in