Question #01a56

1 Answer
Jun 5, 2016

Answer:

#=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#