The toy car and racing track can be modelled as a closed system - AQA - GCSE Physics - Question 8 - 2021 - Paper 1

To find the maximum speed of the toy car at position B, we first calculate the gravitational potential energy (E_p) at position A, which can be given by the formula:

$E_p = mgh$

Where:

• m = mass of the toy car = 0.040 kg
• g = gravitational field strength = 9.8 N/kg
• h = vertical height = 0.90 m

Substituting the values:

$E_p = 0.040 imes 9.8 imes 0.90 = 0.3528 ext{ J}$

At position B, this potential energy is converted into kinetic energy (E_k), which gives us:

E_k = rac{1}{2} mv^2

Equating the two energies:

0.3528 = rac{1}{2} imes 0.040 imes v^2

Rearranging to find v:

v^2 = rac{0.3528}{0.020}

$v^2 = 17.64$

Taking the square root:

$v = 4.2 ext{ m/s}$

The car needs more than 0.20 J of kinetic energy at position B to complete the loop of the track. This is because the car must have enough energy to reach the top of the loop and to overcome any potential energy gained at that height.