Question
(a) A truck R of mass \(9400\,\mathrm{kg}\) moves with constant acceleration in a straight line down a slope, as illustrated in Fig. 3.1.
At point A the speed of the truck is \(13\,\mathrm{m\,s^{-1}}\) and at point B the speed of the truck is \(22\,\mathrm{m\,s^{-1}}\). A and B are a distance of \(180\,\mathrm{m}\) apart.
(i) Calculate the acceleration of the truck between A and B. (2 marks)
acceleration = ____________________ \(\mathrm{m\,s^{-2}}\)
(ii) Determine the gain in kinetic energy of the truck between A and B. (3 marks)
gain in kinetic energy = ____________________ \(\mathrm{J}\)
(b) A short time after passing point B truck R moves in a straight line on horizontal ground. The driver of the truck applies the brakes. Fig. 3.2 shows the variation with time of the momentum of the truck.
(i) Define force. (1 mark)
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(ii) Show that the average resultant force \(F\) acting on truck R between time \(t=0\) and \(t=15\,\mathrm{s}\) is \(-1.2\times10^4\,\mathrm{N}\). (1 mark)
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(iii) An identical truck S has the same initial momentum as truck R. Truck S experiences a constant force equal to the force \(F\) in (b)(ii).
State and explain whether truck S will take more, less or the same amount of time to come to rest as truck R. (3 marks)
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Syllabus Topic Codes (A Level Physics P2 syllabus):
• 5.2: Gravitational potential energy and kinetic energy — part (a)(ii)
• 3.1: Momentum and Newton’s laws of motion — parts (b)(i), (b)(ii) and (b)(iii)
▶️ Answer/Explanation
(a)(i) Acceleration [2 marks]
Use the equation
\(v^2=u^2+2as\)
Rearranging,
\(a=\dfrac{v^2-u^2}{2s}\)
\(a=\dfrac{22^2-13^2}{2\times180}\)
\(a=0.875\,\mathrm{m\,s^{-2}}\)
Answer: \( \boxed{0.88\,\mathrm{m\,s^{-2}}} \)
(a)(ii) Gain in kinetic energy [3 marks]
The change in kinetic energy is
\(\Delta E_{\mathrm{k}}=\dfrac{1}{2}m(v^2-u^2)\)
\(\Delta E_{\mathrm{k}}=\dfrac{1}{2}\times9400\times(22^2-13^2)\)
\(\Delta E_{\mathrm{k}}=1.487\times10^6\,\mathrm{J}\)
Answer: \( \boxed{1.5\times10^6\,\mathrm{J}} \)
(b)(i) Definition of force [1 mark]
Force is the rate of change of momentum.
(b)(ii) Average resultant force [1 mark]
The average resultant force is given by
\(F=\dfrac{\Delta p}{\Delta t}\)
From the graph, the momentum changes from approximately \(21\times10^4\,\mathrm{kg\,m\,s^{-1}}\) at \(t=0\) to \(2.5\times10^4\,\mathrm{kg\,m\,s^{-1}}\) at \(t=15\,\mathrm{s}\).
\(F=\dfrac{2.5\times10^4-21\times10^4}{15}\)
\(F=-1.23\times10^4\,\mathrm{N}\)
To the required precision,
Answer: \( \boxed{-1.2\times10^4\,\mathrm{N}} \)
(b)(iii) Comparing the stopping times [3 marks]
Both trucks have the same initial momentum and both must have the same change in momentum to come to rest.
Truck R has an average braking force of approximately
\(\dfrac{21\times10^4}{23}\approx0.91\times10^4\,\mathrm{N}\)
This is less than the constant force of \(1.2\times10^4\,\mathrm{N}\) acting on truck S.
Since the change in momentum is the same and truck S experiences the greater force, truck S takes less time to undergo the required change in momentum.
Answer: \( \boxed{\text{Truck S takes less time to come to rest.}} \)
Equivalently, for truck S, the stopping time is approximately
\(t=\dfrac{-21\times10^4}{-1.2\times10^4}\approx17.5\,\mathrm{s}\)
whereas truck R takes approximately \(23\,\mathrm{s}\).