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Two blocks, A and B, of masses 2m and 3m respectively, are attached to the ends of a light string - Edexcel - A-Level Maths Mechanics - Question 3 - 2019 - Paper 1

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Two blocks, A and B, of masses 2m and 3m respectively, are attached to the ends of a light string. Initially A is held at rest on a fixed rough plane. The plane is... show full transcript

Worked Solution & Example Answer:Two blocks, A and B, of masses 2m and 3m respectively, are attached to the ends of a light string - Edexcel - A-Level Maths Mechanics - Question 3 - 2019 - Paper 1

Step 1

Show that T = 12mg/5

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Answer

To establish the tension T in the string, we apply Newton's second law for both blocks A and B.

  1. For block A (mass = 2m):

    • The forces acting on A are the tension T, friction force F, and the component of the weight down the slope.
    • The equation of motion can be expressed as: TF2mgsin(α)=2maT - F - 2mg \sin(\alpha) = 2ma
  2. For block B (mass = 3m):

    • The forces acting on B include its weight and the tension:
    • The equation of motion is: 3mgT=3ma3mg - T = 3ma
  3. Substituting F:

    • Given that the coefficient of friction ( \frac{2}{3} ), friction F is: F=23R=23(2mgcos(α))F = \frac{2}{3} R = \frac{2}{3} (2mg \cos(\alpha))
  4. Relating angles:

    • From the right triangle formed, we know ( \tan(\alpha) = \frac{5}{12} ).
    • We can compute ( \sin(\alpha) ) and ( \cos(\alpha) ): sin(α)=513,cos(α)=1213\sin(\alpha) = \frac{5}{13}, \quad \cos(\alpha) = \frac{12}{13}
  5. Final steps:

    • Expressing R: R=2mgcos(α)=2mg(1213)R = 2mg \cos(\alpha) = 2mg \left(\frac{12}{13}\right)
    • The tension equation thus combines into: T=12mg5T = \frac{12mg}{5}

Step 2

Determine whether A will remain at rest, carefully justifying your answer.

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Answer

After B reaches the ground, we must analyze the forces acting on A:

  1. Forces on A: The only forces acting on block A are the tension T in the string and the friction force F.
  2. Tension and Friction: When B reaches the ground, the tension will be less than it was during the motion. Thus, if the force due to friction exceeds the combined forces acting downward on A, A may remain at rest.
  3. Calculating Forces:
    • The upward force due to friction must be calculated: Ffriction=23(2mgcos(α))F_{\text{friction}} = \frac{2}{3} (2mg \cos(\alpha))
    • The critical comparison is made between T and F. Since T reduces after B hits the ground, there’s a point where the forces may equalize allowing A to stay secure against slipping.
  4. Conclusion: If the resultant friction force is sufficient, A will remain at rest. If not, it will slide down the plane.

Step 3

Suggest two refinements to the model that would make it more realistic.

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Answer

  1. Inclusion of Air Resistance: Adding air resistance to the model could provide a more accurate representation of the forces acting on both blocks, especially at higher velocities.

  2. Weight of the String: Considering the weight of the string itself could affect the tension in the string during the motion and provide a more precise calculation of dynamics as it would contribute additional downward force.

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