A bullet of mass in travelling with a speed v hits a block of mass M initially at rest and gets embedded in it. The combined system is free to move and there is no other force acting on the system. The heat generated in the process will be

A bullet of mass in travelling with a speed v hits a block of mass M i

1.	A bullet of mass in travelling with a speed v hits a block of mass M initially at rest and gets embedded in it. The combined system is free to move and there is no other force acting on the system. The heat generated in the process will be
Answer» zero `(mv^2)/2` `(Mmv^2)/(2(M - m))` `(mMv^2)/(2(M + m))` Solution :If V is the velocity of the combined system (i.e., block + bullet) after the COLLISION, then by the LAW of conseration of momentum `mv + M(0) = (M + m)V " or " V = (mv)/((M + m))` The heat generated in the process `= 1/2 mv^2 - 1/2 (M + m) V^2 = 1/2 mv^2- 1/2 (M + m)((mv)/(M + m))^2` ` = 1/2 mv^2 - 1/2 (m^2v^2)/((M + m)) = 1/2 mv^2 (1 - (m)/((M + m)))` `= 1/2 mv^2 ((M)/(M + m)) = (mMv^2)/(2(M + m))` .