Figure 11 shows a toy car in different positions on a racing track - AQA - GCSE Physics - Question 8 - 2021 - Paper 1

To calculate the maximum possible speed, we first need to determine the potential energy at position A and equate it to the kinetic energy at position B.

Using the formula for gravitational potential energy:

$E_p = mgh$

where:

• $m = 0.040 \text{ kg}$ (mass of the toy car)
• $g = 9.8 \text{ N/kg}$ (gravitational field strength)
• $h = 0.90 \text{ m}$ (vertical height)

Calculating the potential energy:

$E_p = 0.040 \times 9.8 \times 0.90 = 0.3528 \text{ J}$

Since the car is released from rest, all this potential energy converts to kinetic energy ($E_k$) at position B:

$E_k = \frac{1}{2} mv^2$

Setting the equations equal:

$0.3528 = \frac{1}{2} \times 0.040 \times v^2$

Rearranging to solve for $v$:

$v^2 = \frac{0.3528 \times 2}{0.040} = 17.64$

Finally, taking the square root gives:

$v = \sqrt{17.64} \approx 4.2 \text{ m/s}$

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 needs to have enough speed to reach the top of the loop. If the car does not have sufficient kinetic energy, it may not be able to ascend the loop successfully.