Question

37) Two masses m1 and m2 (m1 > m2 ) are connected by massless flexible and inextensible string passed over massless and frictionless pulley. The acceleration of centre of mass is.

(a) (m1-m2m1+m2)2 g(b) m1-m2m1+m2 g(c) m1+m2m1-m2 g(d) Zero

Answer 1

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

Answer 2

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

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