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Space, time and motion · Question 10

Space, time and motion · Original GioPhysics question with a detailed, mark-by-mark answer guide.

Demand
discriminating
Marks
14
Topics
2
Answer
Complete
Solid cylinder released from rest on a sloperadius 0.15 msolid cylinder, 2.0 kg, released from rest1.2 mbottom of the slope
Figure 5Side view. A straight hatched slope runs down from the upper left to a horizontal surface at the lower right. A cylinder is drawn end-on resting on the slope near the top, labelled solid cylinder, 2.0 kg, released from rest, with a line from its centre to the rim labelled radius 0.15 m. A dashed horizontal line runs to the right from the level of the cylinder's centre, and a dimension line marks 1.2 m between that level and the horizontal surface at the bottom of the slope.
structured14 marks

Part 1. A uniform solid cylinder of mass 2.0 kg and radius 0.15 m rolls without slipping down a slope, starting from rest at a height of 1.2 m above the bottom. The moment of inertia of a uniform solid cylinder about its axis is ½MR². Part 2. A spacecraft passes Earth at a constant speed of 0.80c. The spacecraft has a proper length of 90 m.

  1. (a)

    Show (that) Show that the translational speed of the cylinder at the bottom of the slope is about 4.0 m s⁻¹.

    4 marks
  2. (b)

    Determine Determine the fraction of the cylinder's total kinetic energy at the bottom that is rotational.

    3 marks
  3. (c)

    Calculate Calculate the length of the spacecraft as measured by an observer on Earth.

    3 marks
  4. (d)

    Explain An observer on the spacecraft claims that it is the Earth that is 0.80c and that Earth's distances are contracted, not the spacecraft's. Explain why both observers are correct, and outline what would have to change for one of them to be wrong.

    4 marks
Ready to self-mark?Reveal the detailed answer guide
Answer overviewKey answer: Mgh = ½Mv² + ½Iω², with I = ½MR² applies the rolling condition ω = v/R, so ½Iω² = ¼Mv² Mgh = ¾Mv², so v = √(4gh/3)
01

(a)

4 marks

Show (that) Show that the translational speed of the cylinder at the bottom of the slope is about 4.0 m s⁻¹.

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

  1. 1

    Mgh = ½Mv² + ½Iω², with I = ½MR²

  2. 2

    applies the rolling condition ω = v/R, so ½Iω² = ¼Mv²

  3. 3

    Mgh = ¾Mv², so v = √(4gh/3)

  4. 4

    v = √(4 × 9.81 × 1.2 / 3) = 3.96 ≈ 4.0 m s⁻¹

02

(b)

3 marks

Determine Determine the fraction of the cylinder's total kinetic energy at the bottom that is rotational.

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

  1. 1

    rotational KE = ¼Mv², translational KE = ½Mv²

  2. 2

    Adding the translational and rotational terms gives total kinetic energy = ¾Mv².

  3. 3

    fraction rotational = (¼)/(¾) = 1/3

03

(c)

3 marks

Calculate Calculate the length of the spacecraft as measured by an observer on Earth.

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

  1. 1

    γ = 1/√(1 − 0.80²) = 1/0.60 = 1.67

  2. 2

    L = L₀/γ = 90/1.67

  3. 3

    L = 54 m

04

(d)

4 marks

Explain An observer on the spacecraft claims that it is the Earth that is 0.80c and that Earth's distances are contracted, not the spacecraft's. Explain why both observers are correct, and outline what would have to change for one of them to be wrong.

How to approach it

State the outcome first, then link cause to effect with the relevant physical principle. Each link in the reasoning should be explicit enough to earn its own marking point.

  1. 1

    both frames are inertial, and the principle of relativity states that the laws of physics are the same in all inertial frames

  2. 2

    there is no experiment either observer can do to establish that they are the one "really" moving, so neither frame is privileged

  3. 3

    each measures the other's length as contracted because they disagree about which events are simultaneous, and a length measurement requires locating both ends at the same time

  4. 4

    the symmetry would be broken only if one observer accelerated — an accelerating frame is not inertial, and that observer would feel the acceleration and know it

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