In the system shown in figure, mass m is released from rest from position A. Suppose potential energy of m at point A with respect to point B is E. Volume of m is negligible and all surfaces are smooth. When mass m reaches at point B

`{:(,"column I","columnII",),((A),"Kinetic Energy of m",(P),E//3),((B),"Kinetic Energy of 2m ",(Q),2E//3),((C),"Momentum of m",(R),sqrt(4/3 mE)),((D),"Momentum of 2m",(S),sqrt(2/3 mE)),(,,(T),"None"):}`

Answer

A – Q; B – P; C – R; D – R

Explanation

Given: mgR = E
When m reaches point B:
Applying momentum conservation in horizontal: `" " 0 = mv_1 + 2m (-v_2) implies v_1 = 2v_2`
Applying energy conservation:
`-mgR + 0 + (1/2m v_1^2 - 0) + (1/2 2 m v_2^2 - 0) = 0 , 1/2 m (2v_2)^2 + 1/2 2 mv_2^2 = mgR`.
`implies 3 mv_2^2 = mgR implies v_2 = sqrt((gR)/(3)) , v_1 = 2v_2 = 2sqrt((gR)/(3))`
`K.E. "of" m = 1/2 m v_1^2 = 1/2 m(2sqrt((gR)/(3)))^(2) = 2/3 m gR = (2E)/3`
`K.E "of" 2m = 1/2 2m v_2^2 = 1/2 2m (sqrt((gR)/(3)))^(2) = (m gR)/3 = E/3`
Momentum of `m =m v_1 = m2 sqrt((gR)/3) = sqrt(4/3 m E)`, Momentum of `2m = 2V_2 = 2m sqrt((gR)/(3)) = sqrt(4/3 ME)`