(a)i) Define a torque.
(b)(i) Define free fall.
(ii) A body is thrown vertically upwards from the top of a tower 40.0m high with a velocity of 10.0ms\(^{-1}\). Calculate the time taken for the body to reach the ground. [g.= 10.0ms\(^{-2}\)]
c) A cube of wood of side 8.0cm, floats at the interface between oil and water with 2.0cm of its surface below the interface as shown in the diagram below. Given that the relative densities of oil and water are 0.72 and 1.00 respectively, calculate the mass of the wood
(a)(i) The torque (moment) of a force about a point is the turning effect of the force about that point; it is the product of the force and the perpendicular distance of its line of action from the point (the axis of rotation): \(\tau=F\times d\).
(a)(ii) Factors that determine a torque:
- The magnitude of the applied force.
- The perpendicular distance between the line of action of the force and the pivot (axis).
- The angle at which the force is applied relative to the arm (line of action direction).
(b)(i) Free fall is the motion of a body falling only under the action of gravity, with no other force (such as air resistance) acting on it; its acceleration is g.
(b)(ii) Take upward positive, u = 10.0 m/s, tower height 40.0 m so displacement to the ground = -40.0 m, g = 10.0 m/s\(^2\).
\[s=ut-\tfrac{1}{2}gt^{2}\Rightarrow-40=10t-5t^{2}.\]
\[5t^{2}-10t-40=0\Rightarrow t^{2}-2t-8=0\Rightarrow(t-4)(t+2)=0.\]
Taking the positive root, \(t=4\,\text{s}\).
(c) Cube side 8.0 cm, so face area \(A=8\times8=64\,\text{cm}^2\). It floats with 2.0 cm below the oil-water interface (in water) and \(8-2=6\,\text{cm}\) in the oil. By the principle of flotation, the weight of the wood equals the total upthrust from oil and water.
Mass \(=\rho_{water}V_{water}+\rho_{oil}V_{oil}\) (using densities in g/cm\(^3\): water 1.00, oil 0.72).
\(V_{water}=64\times2=128\,\text{cm}^3\); \(V_{oil}=64\times6=384\,\text{cm}^3\).
\[m=(1.00\times128)+(0.72\times384)=128+276.48=404.48\,\text{g}\approx404.5\,\text{g}.\]