A small ball B of mass m is suspended with light inelastic string of length L from a block. A of same mass m which can move on smooth horizontal surface as shown in the figure. The ball is displaced by angle θ from lower most position & then released. The displacement of block when ball reaches the lower most position is (a) L sin θ 2 (b) L sin θ (c) L (d) none of these Share with your friends Share 35 Sanjay Singh answered this Dear student Considering both the blocks (A+B) to be our system.Net external force acting on the system along the motion of the block A is zero.So displacement of centre of mass along the surfacesay X axis is zero.Let block A on table moves a distance x towards right by the time the string with block B becomes vertical so displacement of block B with respect to block A that is keeping block A at rest towards left is Lsinθ .So displacement of block B with respect to ground is ∆x1=(x-Lsinθ ) towards right.Formula of displacement of centre of centre of mass in X axisa)∆xcm=m1∆x1+m2∆x2m1+m20=m(x-Lsinθ )+mx2m=2x-Lsinθ22x-Lsinθ=0x=Lsinθ2b)Kinetic energy of block A is maximum when the string bocmes vertical because till then work done by tension of thread on block A is positive.From conservation of momentum along X axis initial momentum along X axis is Pinitial=0Momentum of block( A+B) when string becomes vertical is Pfinal=mv1-mv2 (v1 is velocity of A towards right and v2 is the velocity of B towards left ).Net momentum towards rightmv1-mv2 =0v1=v2=vApplying energy conservation Loss of potential energy block of (A+B) is equal to increase in kinetic energy of block (A+B).Loss of potential energy block of (A+B)=mgL(1-sinθ) increase in kinetic energy of block (A+B)=12mv2×2=mv2mgL(1-sinθ)=mv2v=gL(1-sinθ) displacement of centre of mass in Y axis is∆ycm=m1∆y1+m2∆y2m1+m2=m(0)+m(L-Lcosθ)2m=(L-Lcosθ)2 net displacement of centre is (L-Lcosθ)2Regards 33 View Full Answer