According to the principle of conservation of linear momentum, if the external force acting on the system is zero, the linear momentum of the system will remain conserved. It means if the centre of mass of a system is initially at rest, it will remain, at rest in the absence of external force, that is, the displacement of centre of mass will be zero. Two blocks of masses 'm'and '2m' are placed as shown in Fig. There is no friction anywhere. A spring of force constant k is attached to the bigger block. Mass 'm' is kept in touch with the spring but not attached to it. 'A' and 'B' are two supports attached to '2m'. Now m is moved towards left so that spring is compressed by distance 'x' and then the system is released from rest. What is the loss in the energy of the system due to breaking ofB?

According to the principle of conservation of linear momentum, if the

1.	According to the principle of conservation of linear momentum, if the external force acting on the system is zero, the linear momentum of the system will remain conserved. It means if the centre of mass of a system is initially at rest, it will remain, at rest in the absence of external force, that is, the displacement of centre of mass will be zero. Two blocks of masses 'm'and '2m' are placed as shown in Fig. There is no friction anywhere. A spring of force constant k is attached to the bigger block. Mass 'm' is kept in touch with the spring but not attached to it. 'A' and 'B' are two supports attached to '2m'. Now m is moved towards left so that spring is compressed by distance 'x' and then the system is released from rest. What is the loss in the energy of the system due to breaking ofB?
Answer» `3/8kx^(2)` `3/4kx^(2)` `1/8kx^(2)` `5/8kx^(2)` Solution :`KE_(i)=1/2 mv_(1)^(2)+1/2xx2mv_(2)^(2)=1/2kx^(2)` `KE_(F)=1/2m((v_(1))/2)^(2)+1/2xx2m((v_(2))/2)^(2)=1/4[1/2kx^(2)]` LOSS `=KE_(i)-KE_(f)=3/8kx^(2)`