Question
(a) Define acceleration. (1 mark)
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(b) An Olympic diver stands on a platform above a pool of water, as shown in Fig. 2.1.
When the diver is on the platform his centre of gravity is a vertical height of \(9.0\,\mathrm{m}\) above the surface of the water. The diver jumps from the platform with a velocity of \(5.9\,\mathrm{m\,s^{-1}}\) at an angle of \(60^\circ\) to the horizontal.
Air resistance is negligible.
When the diver hits the surface of the water, his centre of gravity is a vertical height of \(1.2\,\mathrm{m}\) above the surface of the water.
Calculate the speed of the diver at the instant he hits the surface of the water. (3 marks)
speed = ________________________________________________ \( \mathrm{m\,s^{-1}} \)
(c) The diver in (b) enters the water and decelerates.
(i) Describe and explain the variation of the viscous drag force acting on the diver in the water as he moves downwards. (2 marks)
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(ii) The diver has a volume of \(7.5\times10^{-2}\,\mathrm{m^3}\). The density of the water is \(1.0\times10^3\,\mathrm{kg\,m^{-3}}\).
Show that the upthrust acting on the diver when he is entirely underwater is \(740\,\mathrm{N}\). (1 mark)
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(iii) At a particular instant when the diver is entirely underwater his horizontal velocity is zero. The viscous drag force acting on him at this instant is \(950\,\mathrm{N}\) vertically upwards. The diver has mass \(78\,\mathrm{kg}\).
Determine the magnitude and direction of the acceleration of the diver. (4 marks)
acceleration = ________________________________________________ \( \mathrm{m\,s^{-2}} \)
direction = ________________________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 3.1: Momentum and Newton’s laws of motion — part (c)(iii)
• 3.2: Non-uniform motion — part (c)(i)
• 4.3: Density and pressure — part (c)(ii)
▶️ Answer/Explanation
(a) Definition of acceleration [1 mark]
Acceleration is the rate of change of velocity.
Answer: \( \boxed{\text{rate of change of velocity}} \)
(b) Speed of the diver when he reaches the water [3 marks]
The diver falls through a vertical distance of
\( \Delta h=9.0-1.2=7.8\,\mathrm{m} \)
Using the equation of motion for the vertical component, or equivalently conservation of mechanical energy:
\( v^2=u^2+2a\Delta h \)
The initial speed is \(5.9\,\mathrm{m\,s^{-1}}\), and the vertical acceleration is \(9.81\,\mathrm{m\,s^{-2}}\).
\( v^2=(5.9)^2+2(9.81)(7.8) \)
\( v^2=188 \)
\( v=\sqrt{188}=13.7\,\mathrm{m\,s^{-1}} \)
To two significant figures:
\( v=14\,\mathrm{m\,s^{-1}} \)
Answer: \( \boxed{14\,\mathrm{m\,s^{-1}}} \)
(c)(i) Variation of viscous drag force [2 marks]
As the diver moves downwards through the water, he decelerates, so his speed decreases.
The viscous drag force depends on the speed of the diver. Therefore, as the speed decreases, the viscous drag force also decreases.
Answer: The speed decreases, so the viscous drag force decreases.
(c)(ii) Upthrust on the diver [1 mark]
The upthrust is equal to the weight of water displaced:
\( F=\rho gV \)
Substituting \( \rho=1.0\times10^3\,\mathrm{kg\,m^{-3}} \), \(g=9.81\,\mathrm{m\,s^{-2}}\) and \(V=7.5\times10^{-2}\,\mathrm{m^3}\):
\( F=(1.0\times10^3)(9.81)(7.5\times10^{-2}) \)
\( F=735.75\,\mathrm{N} \)
Therefore, to an appropriate number of significant figures:
\( F=740\,\mathrm{N} \)
Answer: \( \boxed{740\,\mathrm{N}} \)
(c)(iii) Acceleration of the diver [4 marks]
The forces acting vertically on the diver are:
• Upthrust \(=740\,\mathrm{N}\) upwards
• Viscous drag \(=950\,\mathrm{N}\) upwards
• Weight \(=mg\) downwards
The weight of the diver is:
\( W=mg=(78)(9.81) \)
\( W=765.18\,\mathrm{N} \)
Taking upwards as positive, the resultant force is:
\( F=740+950-765.18 \)
\( F=924.82\,\mathrm{N} \)
Using Newton’s second law:
\( F=ma \)
\( a=\frac{F}{m} \)
\( a=\frac{924.82}{78} \)
\( a=11.9\,\mathrm{m\,s^{-2}} \)
To two significant figures:
\( a=12\,\mathrm{m\,s^{-2}} \)
Since the resultant force is upwards, the acceleration is vertically upwards.
Answer: \( \boxed{12\,\mathrm{m\,s^{-2}}} \), vertically upwards