Question #01a56

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Answer 1 · 2016-06-05T02:08:20

$= \frac{3}{2} m g$ $=3/2mg$

Given

$m \to Mass of each body$ $m->"Mass of each body"$
$l \to Length of inextensible string$ $l->" Length of inextensible string"$
$v \to Horizontal velocity = \sqrt{2 g l}$ $v->" Horizontal velocity " = sqrt(2gl$

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

Let

$v_{1} \to Velocity of **united mass** after inelastic collision$ $v_1 ->"Velocity of **united mass** after inelastic collision"$

$2 m \times v_{1} = m \times v$ $2mxxv_1=mxxv$

$v_{1} = \frac{1}{2} v$ $v_1=1/2v$

$Tension of string beore collision T_{b} = \frac{m v^{2}}{l}$ $"Tension of string beore collision " T_b=(mv^2)/l$

$Tension of string jus after the collision T_{a} = \frac{m v_{1}^{2}}{l}$ $"Tension of string jus after the collision " T_a=(mv_1^2)/l$

Increase in Tension

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

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