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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

NTA Abhyas 2020 · Physics · Center of Mass, Momentum and Collision

Two masses m _ 1 and m _ 2 are connected by a string of length l ....

Two masses m1{m}_{1} and m2{m}_{2} are connected by a string of length ll. They are held in a horizontal plane at a height HH above two heavy plates AA and BB made of different materials placed on the floor. Initially, the distance between the two masses is a<la < l. When the masses are released they fall freely and collide with the plates simultaneously. The coefficient of restitution for the collision of the first ball-plate pair is 0.80. 8 and for the second ball-plate pair is 0.40. 4. The time after the collision when the string becomes taut is

Two masses m _ 1 and m _ 2 are connected by a string of length l . They are held in a horizontal plane at a height H

Options

  1. At=52l2−a22gHt = \dfrac{5}{2}\sqrt{\dfrac{l^{2} - a^{2}}{2gH}}
  2. Bt=54l2−a22gHt = \dfrac{5}{4}\sqrt{\dfrac{l^{2} - a^{2}}{2gH}}
  3. Ct=52la2gHt = \dfrac{5}{2}\sqrt{\dfrac{la}{2gH}}
  4. Dt=54la2gHt = \dfrac{5}{4}\sqrt{\dfrac{la}{2gH}}

Correct answer

A. t=52l2−a22gHt = \dfrac{5}{2}\sqrt{\dfrac{l^{2} - a^{2}}{2gH}}

Step-by-step solution

The string will become taut& when

h1−h2=l2−a2{h}_{1} - {h}_{2} = \sqrt{l^{2} - a^{2}}

h1=0.8vt−12gt2{h}_{1} = 0. 8vt - \dfrac{1}{2}gt^{2}

h2=0.4vt−12gt2{h}_{2} = 0. 4vt - \dfrac{1}{2}gt^{2}

h1−h2=0.4vt{h}_{1} - {h}_{2} = 0. 4vt

0.4vt=l2−a20. 4vt = \sqrt{l^{2} - a^{2}}, where v=2gHv = \sqrt{2gH}

Two masses m _ 1 and m _ 2 are connected by a string of length l . They are held in a horizontal plane at a height H

t=52l2−a22gHt = \dfrac{5}{2}\sqrt{\dfrac{l^{2} - a^{2}}{2gH}}

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