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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 A and B connected with an inextensible string of length...

Two masses AA and BB connected with an inextensible string of length ll lie on a smooth horizontal plane. AA is given a velocity of vms−1vms^{- 1} along the ground perpendicular to line ABAB as shown in the figure. Find the tension in a string during their subsequence motion.

Two masses A and B connected with an inextensible string of length l lie on a smooth horizontal plane. A is given a

Options

  1. A2mv23l\dfrac{2mv^{2}}{3l}
    &
  2. B3mv22l\dfrac{3mv^{2}}{2l}
  3. Cmv22l\dfrac{mv^{2}}{2l}
  4. D4mv23l\dfrac{4mv^{2}}{3l}

Correct answer

A. 2mv23l\dfrac{2mv^{2}}{3l}

&

Step-by-step solution

From conservation of linear momentum

vcm=m×0+2mvm+2m⇒vcm=2v3{v}_{cm} = \dfrac{\text{m} \times 0 + \text{2mv}}{\text{m} + \text{2m}} ⇒ {v}_{cm} = \dfrac{\text{2v}}{3}

position of the centre of mass

ycm=0×2m+m×l2m⇒ycm=l3{y}_{cm} = \dfrac{0 \times \text{2m} + \text{m} \times l}{2m} ⇒ {y}_{cm} = \dfrac{l}{3}

Two masses A and B connected with an inextensible string of length l lie on a smooth horizontal plane. A is given a

blocks will perform circular motion in the center of mass frame

velocity of A in COM frame isvA′=v−2v/3⇒vA′=v/3\text{velocity of A in COM frame is} {v}_A^{′} = \text{v} - 2 \text{v/3} ⇒ {v}_A^{′} = \text{v/3}

Tension will provide the required centripetal force

T=(2m)vA′2(ℓ/3)⇒T=(2m)(v/3)2(ℓ/3)⇒T=2mv23ℓ\text{T} = \dfrac{\left(\text{2m}\right) {v}_A^{′ 2}}{\left(ℓ / 3\right)} ⇒ \text{T} = \dfrac{\left(\text{2m}\right) {\left(\text{v/3}\right)}^{2}}{\left(ℓ / 3\right)} ⇒ \text{T} = \dfrac{{2mv}^{2}}{3 ℓ}
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