Question

Two masses m and M are connected by a light string passing over a smooth pulley. When set free m moves up by 1.4m in 2 seconds. The ratio of m/M

Answer 1

First refer  to the attachment , then observe the equations which are written below.

Mg - T = Ma

Mg - Ma = T

and ,

T - mg  = ma

T = mg +ma

Mg - Ma = mg + ma

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

or

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

Now here , s = 1/2 at²

or  1.4 =  1/2 × a × 2²

a = 0.7 m/s².

Now we have, a = g (M - m) / (M + m)

0.7 = 9.8 (M - m)/ (M + m)

M + m = 14 M - 14m

15m = 13M

m/ M = 13/15

Hence the ration of m/M is 13/15

Hope it helps :-)

Answer 2

Mg - T = Ma

Mg - Ma = T

and ,

T - mg  = ma

T = mg +ma

Mg - Ma = mg + ma

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

or

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

Now here , s = 1/2 at²

or  1.4 =  1/2 × a × 2²

a = 0.7 m/s².

Now we have, a = g (M - m) / (M + m)

0.7 = 9.8 (M - m)/ (M + m)

M + m = 14 M - 14m

15m = 13M

m/ M = 13/15