Question

answer with correct explanation....
irrelevant answer will b reported...​

Answer 1

let the tension in the block be t

then

mgsin53°-T=ma......1)

T-mgsin37°=ma.......2)

adding both

mg(sin53°-sin 37°)=2ma

mg(4/5-3/5)=2ma

g/5=2a

a=g/10

a=1m/s

this is the acceleration of both blocks...

Now let's find the acceleration of centre of mass of both the blocks, using the vector

for block first

acceleration=acos53°i+asin53°j

acceleration=a(3/5i+4/5j)

for block second

acceleration=acos53°i+asin53°j

acceleration=a(4/5i +3/5 j)

A(cm)= (ma1+ma2)/(m1+m2)

A(cm)=ma(7/5i+7/5j)/2m

put a=1

A(cm)=(7/5i+7/5j)/2

A(cm)={(98/25)^1/2}/2

A(cm)={(3.9)^1/2}/2

3.9=4 (approx)

A(cm)={(4)^1/2}/2

A(cm)=2/2=1

this shows that acceleration of centre of mass of both the blocks is same as acceleration of each block separately......

so option1),will be the correct option

keep asking ur doubts......

Answer 2

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