Block A has a mass `3kg` and is sliding on a rough horizontal surface with a velocity `u_A=2m//s` when it makes a direct collision with block B, which has a mass of `2kg` and is originally at rest. The collision is perfectly elastic. Determine the velocity of each block just after collision and the distance between the blocks when they stop sliding. The coefficient of kinetic friction between the blocks and the plane is `mu_k=0.3` ( Take `g=10m//s^2`)

Answer

A, B, C, D

Explanation

`v_3^()'=((m_3-m_2)/(m_3+m_2))v_3+((2m_2)/(m_2+m_3))v_2`
`=((3-2)/(3+2))(2)+0=0.4m//s`
`v_2^()'=0+((2xx3)/(2+3))(2)=2.4m//s`
Retardation of each block,
`a=(mu_kmg)/(m)=mu_kg=3m//s^2`
Before coming to rest
`s_3=(v_3^('2))/(2a)=((0.4)^2)/(2xx3)=0.03m`
`s_2=(v_2^('2))/(2a)=((2.4)^2)/(2xx3)=0.96m`
`:. d=s_2-s_3=0.93m`