Question
(a) Table 1.1 lists some physical quantities. Identify with ticks (✓) which quantities are vectors and which are scalars.
Table 1.1
| quantity | scalar | vector |
| acceleration | ||
| displacement | ||
| gravitational potential energy | ||
| speed | ||
| temperature |
(2 marks)
(b) A constant resultant force \(F\) acts on a car of mass \(m\). The car moves from rest with constant acceleration \(a\) along a horizontal ground. When the car has displacement \(s\), the speed of the car is \(v\).
(i) Using the concept of work done on the car, show that the kinetic energy \(E_{\mathrm{K}}\) of the car is given by the equation
\(E_{\mathrm{K}}=\dfrac{1}{2}mv^2\) (3 marks)
(ii) The mass of the car is \(920\,\mathrm{kg}\). At time \(t=0\), the car is at rest. At time \(t=5.8\,\mathrm{s}\), its velocity is \(17\,\mathrm{m\,s^{-1}}\).
Calculate the kinetic energy of the car at time \(t=5.8\,\mathrm{s}\).
kinetic energy = ______________________________ \(\mathrm{J}\) (1 mark)
(iii) Between time \(t=0\) and time \(t=5.8\,\mathrm{s}\), the work done against resistive forces is \(4.7\times10^4\,\mathrm{J}\).
Determine the average output power of the car during this time.
power = ______________________________ \(\mathrm{W}\) (3 marks)
(iv) At time \(t=5.8\,\mathrm{s}\), the speed of the car becomes constant.
State and explain whether the output power of the car is greater than, less than or the same as the output power just before \(t=5.8\,\mathrm{s}\).
________________________________________________________________________________
________________________________________________________________________________ (1 mark)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 5.1: Energy conservation — parts (b)(i), (b)(iii) and (b)(iv)
• 5.2: Gravitational potential energy and kinetic energy — part (b)(ii)
▶️ Answer/Explanation
(a) Scalars and vectors [2 marks]
| quantity | scalar | vector |
| acceleration | ✓ | |
| displacement | ✓ | |
| gravitational potential energy | ✓ | |
| speed | ✓ | |
| temperature | ✓ |
Answer: acceleration and displacement are vectors. Gravitational potential energy, speed and temperature are scalars.
(b)(i) Derivation of kinetic energy [3 marks]
The work done by the resultant force is
\(W=Fs\)
Using Newton’s second law, \(F=ma\), so
\(W=mas\)
The car starts from rest, so \(u=0\). Using the equation of motion
\(v^2=u^2+2as\)
\(v^2=2as\)
Therefore,
\(as=\dfrac{v^2}{2}\)
Substituting into the work equation,
\(W=m\dfrac{v^2}{2}\)
The work done on the car becomes its kinetic energy, so
Answer: \( \boxed{E_{\mathrm{K}}=\dfrac{1}{2}mv^2} \)
(b)(ii) Kinetic energy at \(t=5.8\,\mathrm{s}\) [1 mark]
Using
\(E_{\mathrm{K}}=\dfrac{1}{2}mv^2\)
\(E_{\mathrm{K}}=\dfrac{1}{2}\times920\times17^2\)
\(E_{\mathrm{K}}=133\,060\,\mathrm{J}\)
\(E_{\mathrm{K}}\approx1.3\times10^5\,\mathrm{J}\)
Answer: \( \boxed{1.3\times10^5\,\mathrm{J}} \)
(b)(iii) Average output power [3 marks]
The output work is the work used to increase the kinetic energy plus the work done against resistive forces.
Therefore,
\(W=4.7\times10^4+1.3\times10^5\)
\(W=1.77\times10^5\,\mathrm{J}\)
Average power is
\(P=\dfrac{W}{t}\)
\(P=\dfrac{4.7\times10^4+1.3\times10^5}{5.8}\)
\(P=3.1\times10^4\,\mathrm{W}\)
Answer: \( \boxed{3.1\times10^4\,\mathrm{W}} \)
(b)(iv) Output power when speed becomes constant [1 mark]
When the speed becomes constant, the kinetic energy of the car no longer increases.
Therefore, no further work is required to accelerate the car. The output power is then only required to overcome the resistive forces.
Just before \(t=5.8\,\mathrm{s}\), some of the output power was also being used to increase the kinetic energy.
Answer: \( \boxed{\text{The output power is less.}} \)