Two masses m1 and m2 are connected by massless flexible and inextensib

1.	Two masses m1 and m2 are connected by massless flexible and inextensible string passed over massless and frictionless pulley the acceleration of center of mass is
Answer» Answer: Explanation: first find acceleration of each bodies . as M1 > m2 so m1 moves downward while m2 moves upward direction with same acceleration a. for mass m1, downward force - upward force = Net force or, m1g - T = m1a.......(1) for mass m2 upward force - downward force = net force or, T - m2g = m2a .......(2) adding equations (1) and (2), we GET, (m1 - m2)g = (m1 + m2)a or, a = (m1 - m2)g/(m1 + m2) HENCE, magnitude of acceleration of each BODY is a = (m1 - m2)g/(m1 + m2) but directions are just opposite. now, acceleration of centre of mass ; acm = (m1a1 + m2a2)/(m1 + m2) = [m1(m1 - m2)g/(m1 + m2) + m2{-(m1 - m2)g/(m1 + m2)}]/(m1 + m2) = [(m1 - m2)/(m1 + m2)]²g hence , acceleration of centre of mass is [(m1 - m2)/(m1 + m2)]² g Read more on Brainly.in - brainly.in/question/12230741#readmore

Two masses m1 and m2 are connected by massless flexible and inextensible string passed over massless and frictionless pulley the acceleration of center of mass is

Answer»

Answer:

Explanation:

first find acceleration of each bodies .

as M1 > m2 so m1 moves downward while m2 moves upward direction with same acceleration a.

for mass m1,

downward force - upward force = Net force

or, m1g - T = m1a.......(1)

for mass m2

upward force - downward force = net force

or, T - m2g = m2a .......(2)

adding equations (1) and (2),

we GET, (m1 - m2)g = (m1 + m2)a

or, a = (m1 - m2)g/(m1 + m2)

HENCE, magnitude of acceleration of each BODY is a = (m1 - m2)g/(m1 + m2) but directions are just opposite.

now, acceleration of centre of mass ;

acm = (m1a1 + m2a2)/(m1 + m2)

= [m1(m1 - m2)g/(m1 + m2) + m2{-(m1 - m2)g/(m1 + m2)}]/(m1 + m2)

= [(m1 - m2)/(m1 + m2)]²g

hence , acceleration of centre of mass is [(m1 - m2)/(m1 + m2)]² g