Theme A · Space, time and motion

A ball is launched from level ground at 20 m s⁻¹ at an angle of 30° above the horizontal. Air resistance is negligible. How long does the ball take to reach its maximum height?

1. A0.51 s
2. B1.0 s
3. C1.8 s
4. D2.0 s

A book rests on a table. Which pair of forces is a Newton's third law pair?

1. Athe weight of the book and the normal force of the table on the book
2. Bthe weight of the book and the force of the book on the table
3. Cthe normal force of the table on the book and the force of the book on the table
4. Dthe weight of the book and the weight of the table

A resultant force acts on a stationary object of mass 0.50 kg. The force–time graph is a triangle rising from zero at t = 0 to a peak of 40 N at t = 0.10 s and falling back to zero at t = 0.20 s. What is the speed of the object at t = 0.20 s?

1. A4.0 m s⁻¹
2. B8.0 m s⁻¹
3. C16 m s⁻¹
4. D20 m s⁻¹

A crane lifts a load of mass 250 kg through a vertical height of 12 m in 8.0 s at constant speed. What is the useful output power of the crane?

1. A375 W
2. B2.5 kW
3. C3.7 kW
4. D29 kW

A disc of moment of inertia I rotates freely about a vertical axis at angular speed ω. A second, stationary disc of moment of inertia 2I is dropped onto it and the two rotate together. What is the new angular speed?

1. Aω/3
2. Bω/2
3. C2ω/3
4. D3ω

A spacecraft travels past Earth at 0.60c. A clock on board measures a time interval of 2.0 s between two events that occur at the same place on the spacecraft. What is the interval between those events as measured from Earth?

1. A1.2 s
2. B1.6 s
3. C2.5 s
4. D3.3 s

A student determines the acceleration of free fall using a simple pendulum. She measures the period T for several lengths L and processes the data as shown. Theory predicts T = 2π√(L/g).

1. (a)

   Show (that) Show that a graph of T² against L should be a straight line through the origin, and state an expression for its gradient.

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   1. 1

      squares the relationship: T² = 4π²L/g

   2. 2

      this is of the form y = mx with no constant term, so the line passes through the origin

   3. 3

      Comparing T² = (4π²/g)L with y = mx gives gradient = 4π²/g.

2. (b)

   Determine Determine the gradient of the line, and hence determine a value for g.

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   1. 1

      gradient = (4.829 − 1.610) / (1.200 − 0.400) = 3.219 / 0.800

   2. 2

      gradient = 4.02 s² m⁻¹

   3. 3

      g = 4π²/4.02 = 9.8 m s⁻²

3. (c)

   Outline The uncertainty in each value of L is ±0.005 m and the uncertainty in each value of T is ±0.002 s. Outline why the uncertainty in T² is not ±0.002 s², and determine the uncertainty in T² for the shortest pendulum.

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   1. 1

      squaring a quantity doubles its fractional (percentage) uncertainty, so the absolute uncertainty is not carried through unchanged

   2. 2

      fractional uncertainty in T = 0.002/1.269 = 0.16%, so fractional uncertainty in T² = 0.32%

   3. 3

      uncertainty in T² = 0.0032 × 1.610 = ±0.005 s²

4. (d)

   Evaluate The student's line of best fit has a small positive intercept on the T² axis rather than passing through the origin. Evaluate two possible reasons for this.

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   1. 1

      a systematic error in the measurement of L — for example measuring to the top of the bob rather than to its centre, so every length is recorded too short

   2. 2

      a systematic error in timing — for example consistently starting the stopwatch late, though this would need to affect longer pendulums proportionately less to produce an intercept rather than a change of gradient

   3. 3

      evaluates which is more likely, noting that a constant offset in L produces exactly a constant intercept and is therefore the better explanation

A ball of mass 0.150 kg is thrown horizontally at 12 m s⁻¹ from the top of a building 25 m high. Air resistance is negligible until part (d).

1. (a)

   Calculate Calculate the time taken for the ball to reach the ground.

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   1. 1

      25 = ½ × 9.81 × t²

   2. 2

      t = 2.26 s

2. (b)

   Determine Determine the speed of the ball as it reaches the ground.

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   1. 1

      vertical component v_y = 9.81 × 2.26 = 22.2 m s⁻¹

   2. 2

      horizontal component is unchanged at 12 m s⁻¹

   3. 3

      speed = √(12² + 22.2²) = 25 m s⁻¹

3. (c)

   Determine Determine the magnitude of the impulse delivered to the ball by gravity during its flight.

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   1. 1

      impulse = change in momentum = m × change in vertical velocity = 0.150 × 22.2

   2. 2

      impulse = 3.3 N s, directed vertically downwards

4. (d)

   Explain In reality air resistance is not negligible. Explain how the horizontal distance travelled and the time of flight each differ from the values calculated above.

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   1. 1

      air resistance acts opposite to the velocity, so it has both a horizontal and a vertical component throughout the flight

   2. 2

      the horizontal component decelerates the ball, so the horizontal velocity is no longer constant and the horizontal distance is smaller

   3. 3

      the vertical component acts upward as the ball falls, reducing the downward acceleration below g

   4. 4

      so the ball takes longer to fall the same 25 m, and the time of flight increases

A cyclist and her bicycle have a combined mass of 78 kg. She rides up a straight road inclined at 4.0° to the horizontal at a constant speed of 5.5 m s⁻¹. The total resistive force opposing her motion is 25 N.

1. (a)

   Define Define power.

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      the rate at which work is done, or the rate at which energy is transferred

2. (b)

   Determine Determine the component of the total weight acting down the slope.

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   1. 1

      component = mg sin θ = 78 × 9.81 × sin 4.0°

   2. 2

      Evaluating mg sin 4.0° gives a downslope weight component of 53 N.

3. (c)

   Calculate Calculate the useful power output of the cyclist.

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   1. 1

      at constant speed the driving force balances the weight component and the resistive force: F = 53 + 25 = 78 N

   2. 2

      P = Fv = 78 × 5.5

   3. 3

      P = 4.3 × 10² W

4. (d)

   Determine The cyclist's body converts chemical energy to mechanical work with an efficiency of 22%. Determine the rate at which she uses chemical energy.

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   1. 1

      input power = 429 / 0.22

   2. 2

      = 1.9 × 10³ W

5. (e)

   Discuss At the top of the hill the cyclist stops pedalling and freewheels down the other side, which has the same gradient. Discuss whether she reaches a constant speed, and what determines its value.

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   1. 1

      on the way down, the component of weight along the slope now acts in the direction of motion and accelerates her

   2. 2

      the resistive force increases with speed, mainly because air resistance rises steeply with speed

   3. 3

      she reaches a constant speed when the resistive force has grown to equal the 53 N component of weight down the slope

   4. 4

      the value of that speed is set by how quickly the resistive force grows with speed — by her frontal area, her posture and the air density, not by her mass alone

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.

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   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

2. (b)

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

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   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

3. (c)

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

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   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

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.

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   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