M3 January 2006 Q1
1.
A particle \(P\) of mass 0.8 kg is attached to one end of a light inelastic string, of natural length 1.2 m and modulus of elasticity 24 N. The other end of the string is attached to a fixed point \(A\). A horizontal force of magnitude \(F\) newtons is applied to \(P\). The particle \(P\) in in equilibrium with the string making an angle 60\(^\circ\) with the downward vertical, as shown in Figure 1.
Calculate
(a) the value of \(F\), (3)
(b) the extension of the string, (3)
(c) the elasticity stored in the string. (2)
| Scheme | Marks |
|---|---|
| \(\rightarrow\ \ F = T\sin 60^\circ \qquad \uparrow\ \ T\cos 60^\circ = 0.8g\) both | M1 |
| [or (perpendicular to the string) \(\ F\cos 60^\circ = 0.8g\cos 30^\circ\) ] | (M2) |
| \(F = 0.8g\tan 60^\circ \approx 14\) (N) accept 13.6 | M1 A1 |
| (3) |
Notes
The scheme prints a direction symbol before the alternative equation; it resolves perpendicular to the string.
| Scheme | Marks |
|---|---|
| \(T = \dfrac{0.8g}{\sin 30^\circ}\ (= 15.68)\) allow in (a) | M1 |
| HL \(15.68 = \dfrac{24 \times x}{1.2} \Rightarrow x \approx 0.78\) (m) accept 0.784 | M1 A1 |
| (3) |
Notes
(Corrected from the printed mark scheme: the unit is printed as (cm).)
| Scheme | Marks |
|---|---|
| \(E = \dfrac{24 \times x^2}{2 \times 1.2} \approx 6.1\) (J) accept 6.15 | M1 A1ft |
| (2) | |
| (8 marks) |