Considering a system having two masses ...

Considering a system having two masses `m_(1)` and `m_(2)` in which first mass is pushed towards centre of mass by a distance a, the distance required to be moved for second mass to keep centre of mass at same position is :-

`(m_(1))/(m_(2))a`

`(m_(1)m_(2))/(a)`

`(m_(2))/(m_(1))a`

`((m_(2)m_(1))/(m_(1) + m_(2)))a`

Text Solution

Verified by Experts

The correct Answer is:

`Deltabar(x) = (m_(1).Deltax_(1) + m_(2).Deltax_(2))/(m_(1) + m_(2))`
`rArr 0 = (m_(1)a + m_(2)Deltax_(2))/(m_(1) + m_(2))`
`rArr Deltax_(2) = -(m_(1)a)/(m_(2))`

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Consider a two-particle system with particles having masses m_1 and m_2 . If the first particle is pushed towards the centre of mass through a distance d, by what distance should the second particle be moved, so as to keep the centre of mass at the same position?

`(m_2)/(m_1)d`

`(m_1)/(m_1 + m_2)d`

`(m_1)/(m_1)d`

Consider a two particle system with particles having masses m_1 and m_2 if the first particle is pushed towards the centre of mass through a distance d, by what distance should the second particle is moved, so as to keep the center of mass at the same positon?

`(m_2)/(m_1) d`

`(m_1)/(m_1+m_2)d`

`(m_1)/(m_2)d`

Consider a system of two particles having masses m_(1) and m_(2) . If the particle of mass m_(1) is pushed towards the centre of mass of particles through a distance d , by what distance would the particle of mass m_(2) move so as to keep the mass centre of particles at the original position?

`m_(1)/(m_(1)+m_(2)) d`

`m_(1)/m_(2) d`

`d`

`m_(2)/m_(1) d`