Momentum and CollisionHard
Question
Two masses A and Of mass M and 2M respectively are connected by a compressed ideal spring. The system is placed on a horizontal frictional tables and a velocity
in the z - direction as shown in the figure. The spring is released In the the subsequent motion the line from B to A always points along the î unit vector . At some instant of time mass B has a x - component of velocity as vxî. The velocity
of mass A at that instant is


Options
A.vxî + 
B.- vxî + 
C.- 2vxî + 
D.2vxî + 
Solution
From COLM MvxA+ 2MvxB = 0 ⇒ vxA = - 2vxB
so
A = - 2vxî + v
so
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