Question

Two particles of mass m_(1) and m_(2)(m_(1) gt m_(2)) are attached to the ends of a light inextensible string passing over a smooth fixed pulley. When the level of m_(1) is higher than that of m_(2) by 49 cm and string is just taut m_(2) is released. 0.5 s after motion started the masses are at the same level. the ratio of m_(1) to m_(2) is (Take g=9.8ms^(-2))

Answer 1

Answer:

m1 come down, m2 goes up

acceleration for both masses=a= m1g-m2g/m1+m2

let m2 be at rest.

so acceleration of m1 wrt m2= 2a.

s=1/2*a*t^2

0.49= 0.5*2a*(0.5)^2

substitute a and solve to get m1/m2