M3 June 2015 Q6
6.
Two points \(A\) and \(B\) are 6 m apart on a smooth horizontal floor. A particle \(P\) of mass 0.5 kg is attached to one end of a light elastic spring, of natural length 2.5 m and modulus of elasticity 20 N. The other end of the spring is attached to \(A\). A second light elastic spring, of natural length 1.5 m and modulus of elasticity 18 N, has one end attached to \(P\) and the other end attached to \(B\), as shown in Figure 3. Initially \(P\) rests in equilibrium at the point \(O\), where \(AOB\) is a straight line.
The particle \(P\) now receives an impulse of magnitude 6 N s acting in the direction \(OB\) and \(P\) starts to move towards \(B\).
| Scheme | Marks |
|---|---|
| \(T_A = \dfrac{20x}{2.5}\ \ (= 8x) \qquad T_B = \dfrac{18(2 - x)}{1.5}\ \ \left(= 12(2 - x)\right)\) | |
| \(\dfrac{20x}{2.5} = \dfrac{18(2 - x)}{1.5}\) | M1A1 |
| \(x = \dfrac{12}{10} = 1.2\) | A1 |
| \(AO = 3.7\) m | A1ft |
| (4) |
Notes
M1 Using Hooke's law to find both tensions and equating them. The extension in \(BP\) can be used instead of the extension in \(AP\). ALT: Use both extensions and use \(e_a + e_b = 2\) later
A1 Correct equation
A1 Correct value found for either extension
A1ft Correct length for \(AO\); follow through their extension
| Scheme | Marks |
|---|---|
| \(\dfrac{18(0.8 - y)}{1.5} - \dfrac{20(1.2 + y)}{2.5} = 0.5\ddot{y}\) | M1A1A1 |
| \(-40y = \ddot{y}\ \ \therefore\) SHM (or \(\ddot{y} = (-20/m)\,y\)) | A1cso |
| (4) |
Notes
M1 Forming an equation of motion at a general point. Difference of 2 tensions, both including. the variable. Use of \(a\) instead of \(\ddot{x}\) can score M1A1A0A0 max (ie an A error)
A1 A1 A1A1 fully correct; A1A0 one error May have \(m\) instead of 0.5 Extensions measured from \(O\)
A1cso A correct simplified equation. Any equivalent form, including having \(m\) instead of 0.5. There must be a concluding statement.
| Scheme | Marks |
|---|---|
| (Max) speed \(= \dfrac{6}{0.5} = 12\) m s\(^{-1}\) | B1 |
| \(\omega = \sqrt{40} = 2\sqrt{10}\) | B1ft |
| \(12 = a \times 2\sqrt{10}\) | M1 |
| \(a = \dfrac{6}{\sqrt{10}}\) or \(\dfrac{3\sqrt{10}}{5}\) m (accept 1.897... ie 1.9, 1.90 or better) | A1ft |
| (4) |
Notes
B1 Correct speed following impulse Can be awarded if seen in (b) or (d)
B1ft Correct value of \(\omega\); must be numerical. FT from (b) Can be awarded if seen in (b) or (d)
M1 Using \(v_{\max} = a\omega\) (their values). By energy – equation must have all terms
A1ft Correct value of \(a\) any equivalent form including decimals. Follow through their \(\omega\)
| Scheme | Marks |
|---|---|
| \(1.2 = a\sin\omega t\) | M1(their \(a, \omega\)) |
| \(t = \dfrac{1}{2\sqrt{10}}\sin^{-1}\left(\dfrac{1.2\sqrt{10}}{6}\right)\) | M1(must use radians) |
| \(t = 0.1082\ldots\) s (Accept 0.11 or better) | A1cso |
| (3) | |
| (15 marks) |
Notes
M1 Using \(y = a\sin\omega t\) with their \(a\) and \(\omega\) If \(y = a\cos\omega t\) is used there must be some indication of moving from the time obtained to the required time.
M1 Solving their equation to find a time. Must use radians
A1cso Correct time, min 2 sf. \(\omega\) and \(a\) must have been obtained from correct work.
Alternative: by reference circle
M1 Finding the required angle in radians.
M1 Using the period \(\left(\dfrac{2\pi}{\omega}\right)\) and their angle to find the required time.
A1 Correct time.
(Corrected from the printed mark scheme: this alternative is headed “Question 6(c)” in the mark scheme, but it is a method for part (d).)