## WAEC Physics

## General WAEC physics instructions:

The standard of the two alternative papers (A & B) compared favorably with each other and with those of the previous years. The questions were well framed in simple language while marking schemes were quite adequate and clear.

The papers tested the candidates' ability to:

- Set up simple experiments
- Carry out the experimental procedure
- Collect and analyze data
- Make deductions from the analysis
- Show understanding of the theories behind the experiments

The performance of the candidates was quite similar to those in previous years. The candidates recorded a mean score of 25 and a standard deviation of 10.25 as against a mean score of 25 and standard deviation of 11.14 in 2007.

## WAEC Physics Paper 1, May/June Answers

### Section ASection BPrecautions

### Physics Paper 1 Question 2Section ASection BPrecautions

### Physics Paper 1 Question 3Section ASection BPrecautionsPhysics Paper 1 Question 4Section ASection BPrecautions

- This question was popular among candidates but some didn't understand the difference between projectile motion from a height and from level ground.
- Some candidates used the wrong equations or substituted values incorrectly.

## Question 2

A body of mass **0.5 kg** is thrown vertically upwards with a speed of **25 ms⁻¹**. Calculate the potential energy of the body at the maximum height reached.

### Comment:

This question tested candidates understanding on the concept of conservation of energy. Majority of the candidates were able to solve this problem. However, few candidates could not solve this question correctly because they lack the knowledge that at maximum height the potential energy P.E is equal to the kinetic energy K.E.

### Expected Answer:

At max. Ht, P.E = initial K.E

= ½ mu²

= ½ × 0.5 × (25)²

= 156.25 J

**OR**

V² = u² - 2gh

At maximum height, v = 0

0 = 25² - 2 × 10h

h = ^{25²}/_{2 × 10}

h = 31.25m

P.E = mgh

= 0.5 × 10 × 31.25

= 156.25J

### Question 12

**(a)** State the principle of conservation of energy.

**(b)** State three effects of heat on a substance.

**(c)** Explain why when the bulb of thermometer is dipped into ice-cold water the mercury level first rises before falling.

**(d)** Water of volume 200 cm^{3} in an aluminum container at 20°C is cooled to – 10°C when put in a freezer.

Calculate the quantity of heat energy extracted from the water.

If heat is extracted at a rate of 120 Js^{-1} by the freezer, calculate the time taken to produce this ice.

State a reason why the calculated time will be less than the actual time.

### Comments:

**P)** This was well answered by most candidates; however, few candidates omitted the phrase "in a closed system" or "isolated system" in their statement.

**(b)** The three effects of heat on a substance were generally well understood by candidates, with satisfactory performance.

Explanation of how the mercury level changes when dipped in ice-cold water proved difficult for most candidates, as they could not provide satisfactory explanation. The numerical section was poorly tackled by most candidates. The formula \( m = v \) eluded most of them. Additionally, applying the formula \( Q = mc \) appropriately was another challenge due to their poor understanding of the concepts of specific heat capacity and latent heat.

### Expected Answers:

**(a)** In an isolated/closed system, the total amount of energy is (always) constant, although energy may be transformed from one form to another.

**(b)** The effects of heat on a substance:

- Change in temperature
- Change in size (expansion/contraction)
- Change of phase/state

**(c)** The bulb comes in contact with the cold water before the mercury, causing the glass to cool first, contract, and squeeze the mercury up the stem until thermal equilibrium is reached. The mercury, being a better conductor of heat, cools faster than the glass, resulting in the fall in mercury level.

**(d)(i)** \( m = \rho V = \frac{1000 \times 200}{1,000,000} = 0.2 \, \text{kg} \)

\( Q = M_wC_w\Delta\theta_w + M_wL_f + M_iC_i\Delta\theta_i \)

\( = 0.2 \times 4200 \times (20-0) + 0.2 \times 3.36 \times 10^5 + 0.2 \times 2100 \times (0-(-10)) \)

\( = 16800 + 67200 + 4200 = 88200 \, \text{J} \)

**(ii)** \( Q = Pt \) or \( P = \frac{Q}{t} = \frac{88200}{120} = 735 \, \text{s} \) or 12.25 minutes

**(iii)** Why the calculated time is less than the actual time:

- No consideration is given to heat absorbed by the aluminum container
- Heat lost / absorbed from the surroundings
- Effect of impurities on the freezing point of the liquid.

## Physics Paper 1, MAy/June. 2011

### Question 4

### (a) Diagram

You are provided with two retort stands, two metre rules, pieces of thread and other necessary apparatus.

- Set up the apparatus as illustrated above ensuring that the strings are permanently 10 cm from either end of the rule.
- Measure and record the length L = 80 cm of the two strings.
- Hold both ends of the rule and displace the rule slightly, then release so that it oscillates about a vertical axis through its centre.
- Determine and record the time t for 10 complete oscillations.
- Determine the period T of oscillations.
- Evaluate log T and log L.
- Repeat the procedure for four other values of L = 70, 60, 50, and 40 cm.
- Tabulate your readings.
- Plot a graph with log T on the vertical axis and log L on the horizontal axis.
- Determine the slope, s, and the intercept, c, on the vertical axis.
- State two precautions taken to ensure accurate results.

### (b)

(i) Define simple harmonic motion.

(ii) Determine the value of L corresponding to t = 12s from the graph in (a) above.

### Observation

The observation and its tabulation were well handled by most responding candidates. However, the evaluation of Log L and Log T to the nearest decimal places was poorly tackled.

The plotting of the graph of log T against Log L was poorly handled. Evaluation of slope and determination of intercept on the vertical axis was fairly attempted. Candidates were able to state the precautions correctly in acceptable language.

The part b was fairly handled by most responding candidates as they were able to define simple harmonic motion but could not satisfactorily deduce the value of L from the graph when t = 12s probably because of the Logarithm involved.

In part (a) candidates are expected to:

- Measure and record five values of L to at least 1 d.p. and in trend. Trend: As L increases, t decreases
- Evaluate five values each of T, log L, and log I to at least 3 s.f.
- Record data in a composite table.
- Plot five points, correctly using reasonable scales and distinguished axes.
- Draw a line of best fit.
- Determine the intercept on the vertical axes and the slope of the graph using a large right-angled triangle.
- State any two of the following precautions in acceptable language:
- Ensured supports of pendula were rigged
- Avoided parallax error on meter rule/stopwatch
- Zero error was noted and corrected on meter rule/stopwatch/clock
- Ensured smooth and regular oscillations in a horizontal plane
- Repeated readings shown on the table

### Part b Expected Answers:

Simple harmonic motion is the motion of a body whose acceleration is always directed towards a fixed point and is proportional to the displacement from the fixed point.

T = 1.2 secs, Log 1.2 = 0.079

0.079 shown on the graph with corresponding log L read. L correctly determined.