At one instant, the centre of mass of a system of two particles is located on the x-axis at `x=3.0m` and has a velocity of `(6.0m//s)hatj`. One of the particles is at the origin, the other particle has a mass of `0.10kg` and is at rest on the `x-`axis at `x=12.0m`. (a) What is the mass of the particle at the origin? (b) Calculate the total momentum of this system. (c) What is the velocity of the particle at the origin?

At one instant, the centre of mass of a system of two particles is loc

1.	At one instant, the centre of mass of a system of two particles is located on the x-axis at `x=3.0m` and has a velocity of `(6.0m//s)hatj`. One of the particles is at the origin, the other particle has a mass of `0.10kg` and is at rest on the `x-`axis at `x=12.0m`. (a) What is the mass of the particle at the origin? (b) Calculate the total momentum of this system. (c) What is the velocity of the particle at the origin?
Answer» Correct Answer - A::B::C::D (a) `x_(CM)=(m_1x_1+m_2x_2)/(m_1+m_2)` `implies 3=(m_1(0)+(0.10)(12))/(m_1+0.1)` Solving this equation we get, `=m_1+0.3kg` (b) `P_(CM)=m_(CM)v_(CM)` `=(0.1+0.3)(6hatj)` `=(2.4hatj)kg*m//s` (c) `P_(CM)=P_1+P_2` `:. (2.4hatj)=(0.3)v_1+(0.1)(0)` `:. v_1=(8hatj)m//s`

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