Question

Two masses m and m2(m, <m2) are positioned as shown in figure, m, being on the ground
and m2 at a height 'h' above the ground. When m2 is released, the speed at which it hits the
ground will be

A) Root2ghm/m2
B) Root2gh(m-m2)/(m-m2)
C) Root2gh(m+m2)/(m-m2)
D) Root2gh(m2-m)/(m+m2)​

Answer 1

Explanation:

Here, we are given that the mass m

1

and m

2

pass over the fixed pulley. So the forces corresponding to the mass m

1

and m

2

are,

m

1

g−T=m

1

a.....(1)

T−m

2

g=m

2

a.....(2)

Adding equations (1) and (2), we get

(m

1

−m

2

)g=(m

1

+m

2

)a

⇒a=(

m

1

+m

2

m

1

−m

2

)g

The downward acceleration =(

m

1

+m

2

m

1

−m

2

)g

Hope it's correct and help full