(a) In his first attempt, a long jumper took off from the springboard with a speed of 8 ms\(^{-1}\) at 30° to the horizontal. He makes a second attempt with the same speed at 45° to the horizontal. Given that the expression for the horizontal range of a projectile is \(\frac{v^2 sin \theta}{g}\) where all the symbols have their usual meanings, show that he gains a distance of 0.8576 m in his second attempt.
(b)(i) State Hooke’s law of elasticity.
(ii) Describe an experiment to verify Hooke’s law.
(iii) State two precautions you would take if you were to perform this experiment in the laboratory.
(c) A spiral spring of natural length 20.00 cm has a scale pan hanging freely in its lower end. When an object of mass 40 g is placed in the pan, its length becomes 21.80 cm. When another object of mass 60g Is placed in the pan, the length becomes 22.05cm. Calculate the mass of the scale pan. [g = 10 ms\(^{-2}\)]
Explanation
(a) Range = \(\frac{U^2 sin 2\theta}{g}\)
for first attempy, Range = \(\frac{8^2 \sin 60}{10} = \frac{64 x\times 0.8660}{10}\)
= 5.5424m
for second attemt; Range = \(\frac{8^2 \sin 90}{10} = \frac{64 \times 1}{10}\)
= 6.4m
gain 6.40 - 5.54 = 0.8576m
(b)(i) Hooke's law states that provided the limit of proportionality is not exceeded, the force is proportional to the extension in an elastic material.
(ii) Description: Suspend the spiral spring from a rigid support with a pointer attached at the free end. Suspend a stretching weight from the free end, and note the position of the pointer. Suspend a known mass, note the new position and determine the extension. Repeat the experiment using various known masses. Repeat the above procedure while unloading and determine mean extension. Plot a graph of load against mean extension. A straight line graph through the origin verifies Hooke's law.
Precautions: (i) Loading beyond elastic limit should be avoided
(ii) Pointer to be rigidly attached to the free end and made perpendicular to the spring.
(iii) Avoid parallax error while taking readings.
(c) mg = Ke, m = total mass If m = mass of scale pan then;
(m + 40)g = 1.8k-----------(i)
(m + 60) = 2.05k --------- (ii)
Dividing (i) by (ii) \(\frac{m + 40}{m + 60} = \frac{1.8}{2.05}\)
2.05(m + 40) = 1.8 (m + 60)
M = \(\frac{26}{0.25}\) = 104gm