Figure 11 shows a student launching a toy aeroplane - AQA - GCSE Physics - Question 10 - 2022 - Paper 1

To find the maximum speed of the toy aeroplane, we can use the principle of conservation of energy. The potential energy stored in the spring is converted into kinetic energy when the aeroplane is launched.

The elastic potential energy (E_e) stored in the spring can be calculated using the formula:

$E_e = \frac{1}{2} k x^2$

where:

• $k$ is the spring constant (50 N/m)
• $x$ is the extension of the spring (0.12 m)

Substituting the values:

$E_e = \frac{1}{2} \times 50 \times (0.12)^2 = 0.36\, J$

Next, we equate this to the kinetic energy (E_k) of the aeroplane when it is launched:

$E_k = \frac{1}{2} m v^2$

where:

• $m$ is the mass of the aeroplane (0.020 kg)
• $v$ is the maximum speed of the aeroplane.

Setting the two energies equal:

$0.36 = \frac{1}{2} \times 0.020 \times v^2$

Rearranging for $v^2$:

$v^2 = \frac{0.36 \times 2}{0.020} = 36$

Taking the square root gives:

$v = \sqrt{36} = 6.0\, m/s$

Thus, the maximum speed of the toy aeroplane just after it is launched is 6.0 m/s.