Physics 0625 | Paper 5 Mock 01 | Practical Test

(a) Length of each strip [2]

Measured from the baseline in Fig. 1.1:

• Strip A = 3.0 cm
• Strip B = 4.5 cm
• Strip C = 6.0 cm
• Strip D = 7.5 cm
• Strip E = 9.0 cm

[1] for correct values, [1] for all five recorded. Each strip is longer than the previous one, showing the trolley was speeding up. The dots are further apart on later strips because the trolley covers more distance in each equal time interval.

(b) Time interval per strip [1]

The timer makes 50 dots per second, so consecutive dots are separated by \( \frac{1}{50} = 0.02 \) s. Each strip contains 5 spaces (the distance between 6 consecutive dots), so:

\( t = 5 \times 0.02 = 0.10 \) s

(c) Speed for strips A and E [2]

Speed = distance / time:

• Strip A: \( v_A = \frac{3.0}{0.10} = 30 \) cm/s [1]
• Strip E: \( v_E = \frac{9.0}{0.10} = 90 \) cm/s [1]

Each strip represents the average speed during that 0.10 s interval. The speed more than doubles from A to E, showing significant acceleration.

(d) Bar chart [2]

Calculate all speeds: A = 30, B = 45, C = 60, D = 75, E = 90 cm/s.

[1] Axes labelled: speed / cm/s on y-axis, strip label (A to E) on x-axis. [1] Five bars plotted at the correct heights with equal widths and spacing. The bars increase steadily in height from A to E.

(e) How speed changes [1]

The speed increases steadily as the trolley moves down the ramp. The equal increase of 15 cm/s from one strip to the next indicates uniform (constant) acceleration. This is expected: the component of gravitational force along the ramp provides a constant net force, producing constant acceleration (\( F = ma \)).

(f) Precaution when releasing the trolley [1]

Hold the tape loosely and ensure it runs freely through the timer without twisting, tangling, or catching. Any friction or snagging on the tape would slow the trolley and produce dots that are closer together than they should be, giving a speed that is too low.

(g) Speed difference [1]

\( \Delta v = v_E - v_A = 90 - 30 = 60 \) cm/s

This represents the total increase in speed over the four time intervals from A to E (i.e. over \( 4 \times 0.10 = 0.40 \) s), giving an average acceleration of \( \frac{60}{0.40} = 150 \) cm/s² or 1.5 m/s².

(a) The object distance measured from the diagram is:

\[ u = 5.0 \text{ cm} \] [1]

(The object is at 5.0 cm from the lens centre C, which lies between F (3.0 cm) and 2F (6.0 cm).)

(b), (c), (d) The completed ray diagram showing all three construction rays and the image:

Ray 1 (already partially drawn): travels parallel to the principal axis from the top of the object to the lens, then refracts to pass through F on the far side. [given]

Ray 2: travels from the top of the object straight through the centre C of the lens and continues in the same direction (undeviated). [1]

Ray 3: travels from the top of the object through F on the near side of the lens, then refracts to emerge parallel to the principal axis. [1]

The three rays converge at a point below the principal axis. The image is marked where the rays cross. [1]

(e) The image distance measured from the diagram:

\[ v = 7.5 \text{ cm} \] [1]

(Accept 7.0 to 8.0 cm.)

(f) The image height measured from the diagram:

\[ \text{image height} = 3.0 \text{ cm} \] [1]

(Accept 2.8 to 3.2 cm.)

(g) The magnification:

\[ M = \frac{v}{u} = \frac{7.5}{5.0} = 1.5 \] [1]

(Accept value consistent with measured \( v \).)

This means the image is 1.5 times the size of the object, consistent with the measured image height of 3.0 cm from a 2.0 cm object.

(h) The image is: [2]

• Enlarged (magnification > 1; the image is larger than the object). [1]
• Inverted (upside down) and real (it can be projected onto a screen, and the rays actually converge at the image position). [1]

The object is between F and 2F, which always produces an enlarged, inverted, real image beyond 2F on the other side of the lens.

(i) Verifying the thin lens equation \( 1/f = 1/u + 1/v \):

\[ \frac{1}{u} + \frac{1}{v} = \frac{1}{5.0} + \frac{1}{7.5} = 0.200 + 0.133 = 0.333 \] [1]

\[ \frac{1}{f} = \frac{1}{3.0} = 0.333 \]

The two values agree (both equal 0.333), which verifies the lens equation. [1]

(a) Recording measurements [1]

All three measurements from Table 11.1: mass of evacuated flask = 253.72 g, mass of flask + air = 254.34 g, volume of water to fill flask = 500 cm³.

(b) Mass of air [1]

\( m = 254.34 - 253.72 = 0.62 \) g

This small difference (less than 1 g) shows why a sensitive balance is essential for this experiment.

(c) Density of air [2]

\( \rho = \frac{m}{V} = \frac{0.62}{500} \) [1]

\( \rho = 0.00124 \) g/cm³ [1]

(d) Conversion to kg/m³ [1]

Since 1 g/cm³ = 1000 kg/m³:

\( 0.00124 \text{ g/cm}^3 \times 1000 = 1.24 \text{ kg/m}^3 \)

Alternatively: multiply by \( \frac{1 \text{ kg}}{1000 \text{ g}} \times \frac{(100 \text{ cm})^3}{(1 \text{ m})^3} = \frac{10^6}{10^3} = 1000 \).

(e) Percentage difference [1]

\( \% \text{ difference} = \frac{|1.24 - 1.2|}{1.2} \times 100 = \frac{0.04}{1.2} \times 100 = 3.3\% \)

Accept 2 to 5 % depending on rounding. A small percentage difference indicates the method is reasonably accurate.

(f) Why the flask must be rigid [1]

When the flask is evacuated, the external atmospheric pressure pushes inward. A flexible container would collapse, reducing its internal volume. When air is re-admitted, the volume of air entering would not equal the original internal volume, so the density calculation \( \rho = m/V \) would use the wrong value of \( V \).

(g) Why the pump may not remove all air [1]

No mechanical pump can achieve a perfect vacuum. A small amount of residual air always remains trapped in the flask, in the valve, or in dead spaces in the tubing. This means the measured mass of air (difference between the two weighings) is slightly less than the true mass of air that fills the flask, giving a density that is slightly too low.

(h) Improvement [1]

Any valid suggestion: repeat the experiment several times and calculate the mean density to reduce the effect of random errors; use a larger flask so the mass of air is greater, reducing the percentage uncertainty in the mass measurement; use a more sensitive balance; or use a better vacuum pump to remove more air.

(i) Temperature reading and its significance [1]

Temperature = 22 °C. The density of air depends on temperature: warmer air expands and becomes less dense, while cooler air is denser. Recording the temperature allows the result to be compared with the accepted value at the same temperature, and explains any discrepancy if conditions differ from the standard reference temperature.

(a) Reading the five catch positions from the ruler diagrams in Fig. 5.2: [2]

Each arrow in the diagram points to where the ruler was caught. Read the scale carefully to the nearest millimetre.

(b) [1]

\[ d_{\text{avg}} = \frac{18.5 + 15.2 + 16.8 + 14.5 + 17.0}{5} = \frac{82.0}{5} = 16.4 \text{ cm} \]

(c) Rearranging \( d = \tfrac{1}{2}gt^{2} \) gives \( t = \sqrt{\dfrac{2d}{g}} \). [2]

For the largest distance, \( d = 18.5 \text{ cm} = 0.185 \text{ m} \):

\[ t = \sqrt{\frac{2 \times 0.185}{9.8}} = \sqrt{0.03776} = 0.194 \text{ s} \]

For the smallest distance, \( d = 14.5 \text{ cm} = 0.145 \text{ m} \):

\[ t = \sqrt{\frac{2 \times 0.145}{9.8}} = \sqrt{0.02959} = 0.172 \text{ s} \]

Always convert centimetres to metres before substituting into the equation.

(d) Calculate all five reaction times, then take the mean: [1]

\[ t_{\text{mean}} = \frac{0.194 + 0.176 + 0.185 + 0.172 + 0.186}{5} = \frac{0.913}{5} = 0.183 \text{ s} \]

(e) The same person must catch the ruler every time. Other acceptable answers: the same hand is used, the same ruler, or the fingers start at the same position (level with the zero mark). Keeping these constant ensures only reaction time varies between trials. [1]

(f) The catcher may not yet be familiar with the procedure or may be anxious, leading to a slower or less consistent response. Practice effects mean later trials are usually more reliable. [1]

(g) Take more than five trials and calculate a new mean, or remove the highest and lowest values before averaging. More trials reduce the impact of any single anomalous result on the calculated mean. [1]

(h) The ruler falls under gravity with acceleration \( g = 9.8 \text{ m/s}^{2} \). From \( d = \tfrac{1}{2}gt^{2} \), a longer reaction time \( t \) means the ruler falls for a longer duration before being caught, producing a greater distance \( d \). The relationship is quadratic, so even a small increase in reaction time gives a noticeably larger distance. [1]