JEE Challenger
More from Laws of Motion
PhyMedAdv 2007 • Paper 1

Acceleration of Masses on String Pulled Vertically Upward

Two particles of mass mm each are tied at the ends of a light string of length 2a2a. The whole system is kept on a frictionless horizontal surface with the string held tight so that each mass is at a distance 'aa' from the center P (as shown in the figure).

Diagram 0

Now, the mid-point of the string is pulled vertically upwards with a small but constant force FF. As a result, the particles move towards each other on the surface. The magnitude of acceleration, when the separation between them becomes 2x2x, is

Options

A

F2maa2−x2\frac{F}{2m} \frac{a}{\sqrt{a^2 - x^2}}

B

F2mxa2−x2\frac{F}{2m} \frac{x}{\sqrt{a^2 - x^2}}

Correct
C

F2mxa\frac{F}{2m} \frac{x}{a}

D

F2ma2−x2x\frac{F}{2m} \frac{\sqrt{a^2 - x^2}}{x}

Topics & Concepts

Laws of MotionNewton's Laws

Step-by-Step Solution

To determine the magnitude of the acceleration of each particle, let us analyze the geometry and the forces acting on the system at the instant when the separation between the particles is 2x2x.

  1. Geometric Relations:

    • The total length of the light string is 2a2a, so each half of the string has a length of aa.
    • When the separation between the two particles on the horizontal surface is 2x2x, the distance of each particle from the central vertical line passing through the midpoint PP is xx.
    • Let yy be the height of the midpoint PP above the horizontal surface. By the Pythagorean theorem:
      y=a2−x2y = \sqrt{a^2 - x^2}
    • Let θ\theta be the angle that each half of the string makes with the horizontal surface. From the right-angled triangle formed by the string, the horizontal line, and the vertical line:
      cos⁡θ=xa\cos\theta = \frac{x}{a}
      sin⁡θ=ya=a2−x2a\sin\theta = \frac{y}{a} = \frac{\sqrt{a^2 - x^2}}{a}

  2. Forces on the Mid-point PP:

    • Since the string is light (massless), the net force acting at the midpoint PP must be zero at all times.
    • The force FF acts vertically upwards, and the tension TT in both halves of the string pulls downwards at an angle θ\theta to the horizontal:
      ΣFy=F−2Tsin⁡θ=0\Sigma F_y = F - 2T\sin\theta = 0
    • Solving for the tension TT:
      T=F2sin⁡θ=F2(aa2−x2)T = \frac{F}{2\sin\theta} = \frac{F}{2} \left(\frac{a}{\sqrt{a^2 - x^2}}\right)

  3. Acceleration of each Particle:

    • The force FF is stated to be small, ensuring that the particles remain in contact with the horizontal surface without being lifted off (Tsin⁡θ≤mgT\sin\theta \le mg).
    • The only force acting horizontally on each particle of mass mm is the horizontal component of the tension:
      Fnet, horizontal=Tcos⁡θF_{\text{net, horizontal}} = T\cos\theta
    • Applying Newton's second law for the horizontal motion of each particle:
      m aparticle=Tcos⁡θm\, a_{\text{particle}} = T\cos\theta
      aparticle=Tcos⁡θm=1m(F2sin⁡θ)cos⁡θ=F2mcot⁡θa_{\text{particle}} = \frac{T\cos\theta}{m} = \frac{1}{m} \left(\frac{F}{2\sin\theta}\right) \cos\theta = \frac{F}{2m}\cot\theta
    • Substituting cot⁡θ=xa2−x2\cot\theta = \frac{x}{\sqrt{a^2 - x^2}}:
      aparticle=F2mxa2−x2a_{\text{particle}} = \frac{F}{2m} \frac{x}{\sqrt{a^2 - x^2}}

Therefore, the magnitude of the acceleration of each particle is F2mxa2−x2\frac{F}{2m} \frac{x}{\sqrt{a^2 - x^2}}, which corresponds to Option B.

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