Question
An object of constant mass moves in a straight line. The variation with time \(t\) of the momentum \(p\) of the object is shown in Fig. 2.1.
(a) Define momentum. (1 mark)
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(b) Calculate the change in momentum of the object from time \(t=0\) to \(t=12\,\mathrm{s}\). (1 mark)
change in momentum = ______________________________ \(\mathrm{kg\,m\,s^{-1}}\)
(c) Calculate the magnitude of the resultant force acting on the object. (2 marks)
force = ______________________________ \(\mathrm{N}\)
(d) Describe the variation of the speed of the object from time \(t=0\) to \(t=8.0\,\mathrm{s}\). (1 mark)
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(e) By reference to Fig. 2.1, explain why the resultant force acting on the object during the first \(8.0\,\mathrm{s}\) of its motion cannot be due to air resistance. (2 marks)
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(f) At time \(t=0\) the displacement of the object is zero.
On Fig. 2.2, sketch the variation of \(d\) with time \(t\) from \(t=0\) to \(t=12\,\mathrm{s}\).
Numerical values of \(d\) are not required. (3 marks)
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Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 3.2: Non-uniform motion — parts (d) and (e)
• 2.1: Equations of motion — part (f)
▶️ Answer/Explanation
(a) Definition of momentum [1 mark]
Momentum is the product of the mass of an object and its velocity.
\(p=mv\)
Answer: \( \boxed{\text{momentum}=\text{mass}\times\text{velocity}} \)
(b) Change in momentum [1 mark]
From Fig. 2.1:
Initial momentum at \(t=0\): \(p_i=+2.8\,\mathrm{kg\,m\,s^{-1}}\)
Final momentum at \(t=12\,\mathrm{s}\): \(p_f=-1.4\,\mathrm{kg\,m\,s^{-1}}\)
Therefore,
\(\Delta p=p_f-p_i\)
\(\Delta p=(-1.4)-(+2.8)\)
\(\Delta p=-4.2\,\mathrm{kg\,m\,s^{-1}}\)
Answer: \( \boxed{-4.2\,\mathrm{kg\,m\,s^{-1}}} \)
(c) Resultant force [2 marks]
Force is the rate of change of momentum:
\(F=\dfrac{\Delta p}{\Delta t}\)
The magnitude of the force is
\(F=\dfrac{4.2}{12}\)
\(F=0.35\,\mathrm{N}\)
Answer: \( \boxed{0.35\,\mathrm{N}} \)
(d) Variation of speed [1 mark]
Since the mass is constant, the magnitude of momentum is proportional to speed.
The magnitude of momentum decreases linearly from its initial value to zero at \(t=8.0\,\mathrm{s}\).
Answer: \( \boxed{\text{The speed decreases uniformly to zero.}} \)
(e) Air resistance [2 marks]
The gradient of the momentum-time graph is constant, so the resultant force is constant during the first \(8.0\,\mathrm{s}\).
Air resistance would vary as the speed changes and therefore would not remain constant. Also, at \(t=8.0\,\mathrm{s}\), the speed is zero but the resultant force is still non-zero.
Answer: \( \boxed{\text{The force is constant even though the speed changes, so it cannot be due to air resistance.}} \)
(f) Displacement-time graph [3 marks]
The gradient of a displacement-time graph represents velocity.
From \(t=0\) to \(t=8.0\,\mathrm{s}\), the object has positive velocity which decreases uniformly to zero. Therefore, \(d\) increases from the origin with a decreasing positive gradient.
At \(t=8.0\,\mathrm{s}\), the velocity is zero, so the displacement-time graph has a horizontal tangent.
After \(t=8.0\,\mathrm{s}\), the momentum and velocity are negative. The gradient therefore becomes negative and its magnitude increases as the speed increases. The displacement remains positive at \(t=12\,\mathrm{s}\).
Required sketch: \( \boxed{\text{A curve from the origin with decreasing positive gradient, horizontal at }8.0\,\mathrm{s},\text{ then negative gradient of increasing magnitude.}} \)