WAEC SSCE - Physics - 2015 (Practical)

Maelezo ya Majibu

(a) Determination of the moment of inertia of a pulley

A mass \(m\) on the hanger unwinds the thread from the pulley (radius \(R\)) and falls through a fixed height \(h = 1.4\ \text{m}\). The apparatus is set up as shown below, with the clamp \(1.5\ \text{m}\) above the floor.

For each value of \(m\) the time of fall \(t\) is recorded twice and averaged, and the derived quantities are computed from

\[ a = \frac{2h}{t^{2}} = \frac{2(1.4)}{t^{2}} = \frac{2.8}{t^{2}},\qquad T = \frac{m}{1000}\,(10 - a),\qquad \alpha = \frac{a}{R},\quad R = 0.08\ \text{m}. \]

xii. Table of readings

S/N	\(m\)/g	\(t_1\)/s	\(t_2\)/s	\(t=\dfrac{t_1+t_2}{2}\)/s	\(t^{2}\)/s\(^2\)	\(a=\dfrac{2.8}{t^{2}}\)/m s\(^{-2}\)	\(T=\dfrac{m}{1000}(10-a)\)/N	\(\alpha=\dfrac{a}{R}\)/rad s\(^{-2}\)
1	50.0	5.00	5.00	5.00	25.000	0.110	0.490	1.380
2	70.0	4.80	4.80	4.80	23.040	0.120	0.690	1.500
3	90.0	4.60	4.60	4.60	21.160	0.130	0.890	1.630
4	110.0	4.40	4.40	4.40	19.360	0.140	1.080	1.750
5	130.0	4.20	4.20	4.20	17.640	0.150	1.280	1.880

where \(h = 1.4\ \text{m} = 140\ \text{cm}\) and \(R = 0.08\ \text{m} = 8\ \text{cm}\).

Worked check of row 1 (\(m = 50.0\) g, \(t = 5.00\) s):

\[ t^{2} = 5.00^{2} = 25.000\ \text{s}^2,\qquad a = \frac{2.8}{25.000} = 0.110\ \text{m s}^{-2}, \] \[ T = \frac{50}{1000}(10 - 0.110) = 0.050 \times 9.890 = 0.490\ \text{N},\qquad \alpha = \frac{0.110}{0.08} = 1.380\ \text{rad s}^{-2}. \]

xiii. Graph of \(\alpha\) against \(T\)

Plotting \(\alpha\) (vertical axis) against \(T\) (horizontal axis) gives a straight line:

xiv. Slope of the graph

Taking two well-separated points on the line of best fit, \((T_1,\alpha_1) = (0.50,\ 1.383)\) and \((T_2,\alpha_2) = (1.30,\ 1.891)\):

\[ s = \frac{\alpha_2 - \alpha_1}{T_2 - T_1} = \frac{1.891 - 1.383}{1.30 - 0.50} = \frac{0.508}{0.800} = 0.635\ \text{rad s}^{-2}\,\text{N}^{-1}. \]

xv. Moment of inertia

Since the driving torque \(TR = I\alpha\), we have \(\alpha = \dfrac{R}{I}\,T\), so the slope \(s = \dfrac{R}{I}\) and

\[ I = \frac{R}{s} = \frac{0.08}{0.635} = 0.126\ \text{kg m}^{2}. \]

xvi. Two precautions

• I avoided parallax error by reading the metre rule and the height marks with the line of sight perpendicular to the scale.
• I ensured the boss head and clamp were firmly tightened so the pulley did not slip or wobble during the fall.

(b)(i) Centripetal force

Centripetal force is the resultant inward force, directed towards the centre of the circular path, that keeps a body moving with constant speed in a circle. Its magnitude is

\[ F = \frac{m v^{2}}{r}. \]

(b)(ii) Speed of a body dropped from a height

An object drops to the ground from a height \(h = 2.0\ \text{m}\). Taking \(g = 10\ \text{m s}^{-2}\) and equating potential energy to kinetic energy:

\[ \tfrac{1}{2} m v^{2} = m g h \;\Rightarrow\; v^{2} = 2 g h = 2 \times 10 \times 2.0 = 40\ \text{m}^2\text{s}^{-2}, \] \[ v = \sqrt{40} = 6.32\ \text{m s}^{-1}. \]

The object strikes the ground with a speed of \(6.32\ \text{m s}^{-1}\).

Maelezo ya Majibu

Refraction through a triangular glass prism

The refracting angle of the prism (the angle at U) is measured as \(A = U = 60^{\circ}\).

A ray is directed at the face UM so that it makes the chosen angle of incidence \(\phi\) with the normal UN. It refracts into the glass, strikes the second face UR, and emerges making an angle of emergence \(e\) with the normal XY. The angle between the incident direction produced and the emergent ray is the angle of deviation \(d\). The complete ray path traced on the drawing paper is shown below.

Table of readings

For each setting of \(\phi\) the corresponding angle of emergence \(e\) and angle of deviation \(d\) are measured with the protractor and recorded.

S/N	\(\phi\,/^{\circ}\)	\(e\,/^{\circ}\)	\(d\,/^{\circ}\)
1	60.0	39.0	39.0
2	55.0	42.5	37.5
3	50.0	47.0	37.0
4	40.0	58.5	38.5
5	35.0	66.0	41.0

Graph of \(d\) against \(e\)

Plotting \(d\) on the vertical axis against \(e\) on the horizontal axis (both axes starting from the origin) and joining the points with a smooth curve gives a shallow U-shaped curve. The lowest point of the curve gives the minimum deviation.

From the graph the turning point (lowest point of the curve) gives:

\[d_{m} = 37^{\circ}, \qquad e_{m} = 48^{\circ}\]

Refractive index

Using the formula with \(U = A = 60^{\circ}\) and \(d_{m} = 37^{\circ}\):

\[n = \frac{\sin\left(\dfrac{d_{m}+U}{2}\right)}{\sin\left(\dfrac{U}{2}\right)} = \frac{\sin\left(\dfrac{37^{\circ}+60^{\circ}}{2}\right)}{\sin\left(\dfrac{60^{\circ}}{2}\right)}\]\[n = \frac{\sin 48.5^{\circ}}{\sin 30^{\circ}} = \frac{0.7490}{0.5000}\]\[\boxed{n = 1.50}\]

Precautions

• The optical pins were fixed vertically (upright) and reasonably spaced apart before the sighting lines were drawn.
• Parallax error was avoided when reading the protractor by placing the eye directly above each mark.

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(b) Total internal reflection

(i) Conditions necessary for total internal reflection:

• Light must travel from an optically denser medium into an optically less dense (rarer) medium.
• The angle of incidence in the denser medium must be greater than the critical angle.

(ii) The refractive index is related to the critical angle \(c\) by:

\[n = \frac{1}{\sin c}\]

With \(c = 39^{\circ}\):

\[n = \frac{1}{\sin 39^{\circ}} = \frac{1}{0.6293}\]\[\boxed{n = 1.59}\]

Maelezo ya Majibu

(a) E.m.f. and internal resistance experiment

The circuit is connected as shown below: the battery (e.m.f. \(E\), internal resistance \(r\)) is joined in series with the key \(K\) and the resistance box \(R\), while the voltmeter is connected across \(R\) to read the terminal potential difference \(V\).

When the key is closed and the box is set to a resistance \(R\), the same current flows through \(R\) and through the internal resistance \(r\), so the terminal p.d. is

\[ V=\frac{ER}{R+r}. \]

Taking reciprocals of both sides,

\[ \frac{1}{V}=\frac{R+r}{ER}=\frac{1}{E}+\frac{r}{E}\cdot\frac{1}{R}. \]

Hence a graph of \(V^{-1}\) against \(R^{-1}\) is a straight line of slope \(s=\dfrac{r}{E}\) and vertical intercept \(C=\dfrac{1}{E}\).

Table of readings

S/N	R / \(\Omega\)	V / V	R\(^{-1}\) / \(\Omega^{-1}\)	V\(^{-1}\) / V\(^{-1}\)
1	1.0	1.80	1.00	0.56
2	2.0	2.10	0.50	0.48
3	3.0	2.30	0.33	0.43
4	4.0	2.40	0.25	0.42
5	5.0	2.50	0.20	0.40
6	6.0	2.60	0.17	0.38

Graph

Plotting \(V^{-1}\) (vertical axis) against \(R^{-1}\) (horizontal axis) gives the straight line below.

Slope and intercept

Reading two points on the line of best fit, \((R^{-1}=1.00,\;V^{-1}=0.57)\) and \((R^{-1}=0.20,\;V^{-1}=0.40)\):

\[ s=\frac{0.57-0.40}{1.00-0.20}=\frac{0.17}{0.80}=0.21\ \Omega. \]

The line cuts the vertical axis (at \(R^{-1}=0\)) at

\[ C=0.36\ \text{V}^{-1}. \]

Evaluating \(C^{-1}\)

Since \(C=\dfrac{1}{E}\), the e.m.f. of the cell is

\[ C^{-1}=E=\frac{1}{0.36}=2.78\ \text{V}. \]

The internal resistance follows from \(s=\dfrac{r}{E}\):

\[ r=sE=0.21\times2.78=0.58\ \Omega. \]

Two precautions

1. I read the voltmeter with my eye directly in line with the pointer and the scale to avoid parallax error, and checked that the pointer read zero (no zero error) before starting.
2. I ensured all connecting wires were pressed tightly onto the terminals and opened the key immediately after each reading so that the battery would not run down and cause the readings to drift.

(b)(i) Potential difference

The potential difference between two points in an electric circuit is the work done in moving one coulomb of positive charge from one point to the other. It is measured in volts (V).

(b)(ii) Why the e.m.f. exceeds the terminal p.d.

The e.m.f. \(E\) is the total energy supplied by the cell to each coulomb of charge it drives round the whole circuit. When the cell delivers a current \(I\) through an external resistance, part of this energy is used up in driving the current through the cell's own internal resistance \(r\); this wasted "lost volt" is \(Ir\). The p.d. available at the terminals is therefore

\[ V=E-Ir, \]

which is less than \(E\). The two are equal only when no current flows (open circuit), where \(Ir=0\).