Question

Two particles of masses m and M (M > m ) are connected by a cord that passes over a massless, frictionless pulley. The tension T in the string and the acceleration a of the particles is
(a) $T=\frac{2mM}{(M-m)}g;\ a=\frac{Mm}{(M+m)}g$(b) $T=\frac{2mM}{(M+m)}g;\ a=\bigg \lgroup\frac{M-m}{(M+m)}\bigg \rgroup g$(c) $T=\bigg \lgroup \frac{m-M}{(M+m)}\bigg \rgroup g;\ a=\bigg \lgroup\frac{Mm}{(M+m)}\bigg \rgroup g$(d) $T=\bigg \lgroup \frac{mM}{(M+m)}\bigg \rgroup g;\ a=\bigg \lgroup\frac{2Mm}{(M+m)}\bigg \rgroup g$

Answer 1

For mass m, the forces are given as

 T - mg = ma                                            (1 eq)

 Similarly, for mass M, the forces are given as

 Mg - T = Ma                                           (2 eq)

 From equation (1), we have

 T = mg + ma                                          (3 eq)

 Substituting equation (3) in (2) we will get

 Mg - mg - ma = Ma

g(M - m) = a(M + m)

 

Therefore a = (M-m/M+m)/g + mg

Substituting the value of a in equation 1,

T= m× (M-m/M+m)/g + mg

T = (Mm-m²/M+m)g+mg

T = g(Mm-m²/M+m +m)

  = g(Mm-m²+m²+Mm)/M+m

  = g(2Mm/M+m)

Answer 2

you can confirm your answer from above photos it will be easy when you learn free body diagrams.