Question

Question #01a56

Answer 1

`=3/2mg`

Explanation:

Given

• `m->"Mass of each body"`

• `l->" Length of inextensible string"`

• `v->" Horizontal velocity " = sqrt(2gl`

After inelastic collision the two body will be united to a mass 2m

Let

• `v_1 ->"Velocity of **united mass** after inelastic collision"`

Then by conservation of momentum

`2mxxv_1=mxxv`

`v_1=1/2v`

`"Tension of string beore collision " T_b=(mv^2)/l`

`"Tension of string jus after the collision " T_a=(mv_1^2)/l`

Increase in Tension

`Delta T=T_a-T_b=m/l(v^2-v_1^2)=m/l(v^2-v^2/4)=m/l(3v^2)/4`

`:.Delta T=(3m)/(4l)xx(sqrt(2gl))^2=3/2mg`