2023 Physics WAEC (WASSCE) Essay Past Questions (Paper 2)

Question 1

The force, F, acting on the wings of an aircraft moving through the air of velocity, v, and density, ρ, is given by the equation F = kv^x ρ^y A^z , where k is a dimensionless constant and A is the surface area of the wings of the aircraft. Use dimensional analysis to determine the values of x, y, and z.

Solution

F = k [v]^x [ρ]^y [A]^z

• F = Force (N or kg ms^-2 ) = \(\scriptsize MLT^{-2}\)
• k = constant
• v = velocity (ms^-1 ) = \(\scriptsize LT^{-1}\)
• ρ = density (kg/m³ ) = \(\scriptsize ML^{-3}\)
• A = surface area (m² ) = \(\scriptsize L^{2}\)

∴  \(\scriptsize  \left(MLT^{-2}\right) = k \left(LT^{-1}\right)^x \left(ML^{-3}\right)^y \left(L^{2}\right)^z\)

Since k is dimensionless,

⇒ \(\scriptsize  \left(MLT^{-2}\right) = \left(LT^{-1}\right)^x \left(ML^{-3}\right)^y \left(L^{2}\right)^z\)

⇒ \(\scriptsize  \left(MLT^{-2}\right) = L^x\: \times \: T^{(-1)x} \: \times \:  M^y\: \times \: L^{(-3)^y} \: \times \: L^{(2)z}\)

⇒ \(\scriptsize  \left(MLT^{-2}\right) = L^x\: \times \: T^{-x} \: \times \:  M^y\: \times \: L^{-3y} \: \times \: L^{2z}\)

⇒ \(\scriptsize  \left(MLT^{-2}\right) = L^{x\:-\:3y\:+\: 2z}\: \times \: T^{-x} \: \times \:  M^y \)

⇒ \(\scriptsize  M^1L^1T^{-2} = L^{x\:-\:3y\:+\: 2z}\: \times \: T^{-x} \: \times \:  M^y \)

For M:

y = 1

For L:

x – 3y +  2z = 1 ……(i)

For T:

-x = -2

∴ x = 2

To find z, substitute x and y into equation (i)

(2) – 3(1) +  2z = 1

2 – 3 +  2z = 1

-1 +  2z = 1

2z = 1 + 1

2z = 2

z = \(\frac{2}{2}\)

z = 1

Hence, x = 2, y = 1 and z = 1.

Question 2

a. Define strain energy.

Answer:

Strain energy is the potential energy stored within an elastic material as a result of its deformation when an external force is applied. It is the ratio of extension produced in a solid/wire to the original length.

b. Write an expression for the energy stored, E, in a stretched wire of original length, l, cross-sectional area, A, extension, e, and Young’s modulus, Y, of the material of the wire.

Solution

Strain energy , E = \(\frac{1}{2} \scriptsize Fe\)

where F = Force, e = extension

⇒ \(\scriptsize Y = \normalsize \frac{Stress}{Strain}\)

⇒ \(\scriptsize Strain = \normalsize \frac{Extension}{Original \: Length} = \frac{e}{L}\)

⇒ \(\scriptsize Stress, \: \sigma\) = \(\frac{Force}{Area} \\ = \frac{F}{A}\)

∴ Y = \(\normalsize \large \frac{\frac{F}{A}}{\frac{e}{L}}\)

Y = \(\frac{FL}{Ae}\)

∴ F = \(\frac{YAe}{L}\)

Substituting F into the strain energy equation, we have:

E = \(\normalsize \frac{1}{2} \scriptsize \: \times\: \normalsize \frac{YAe}{L} \scriptsize\: \times \:  e\)

E = \(\normalsize \frac{1}{2} \scriptsize \: \times\: \normalsize \frac{YAe^2}{L} \)

E = \(\frac{YAe^2}{2L} \)