Figure 11 shows a toy car in different positions on a racing track
The toy car and racing track can be modelled as a closed system.
Why can the toy car and racing track be considered ‘a closed system’?
[1 mark]
ANSWER
A closed system is a system where there is no external forces acting on it. Therefore, the total energy of the system remains constant.
ii- The car is released from rest at position A and accelerates due to gravity down the track to position B.
mass of toy car = 0.040 kg
vertical height between position A and position B = 90 cm
gravitational field strength = 9.8 N/kg
Calculate the maximum possible speed of the toy car when it reaches position B.
[5 marks]
SOLUTION
Method 1
As the car is released from the position A to B, the GPE at point A would start to convert to KE as the car reaches point B. Therefore, the loss of GPE would be equals to gain in KE.
Method 2
Since the car accelerates down the track from point A to B due to gravity, we may consider using the suvat equation, which is only used when the acceleration of a body is constant.
Using equation of motion from the equation sheet.
iii- Figure 11 is repeated below.
At position C the car’s gravitational potential energy is 0.20 J greater than at position B.
How much kinetic energy does the car need at position B to complete the loop of the track?
Give a reason for your answer.
[2 marks]
ANSWER
We are comparing position B and C, and the car needs more KE at B than 0.20J so that the car could easily complete the circular loop.
Reason: Car needs to be moving at C to complete the loop.
