M1 June 2014 Q7
7.
Three particles \(A\), \(B\) and \(C\) have masses \(3m\), \(2m\) and \(2m\) respectively. Particle \(C\) is attached to particle \(B\). Particles \(A\) and \(B\) are connected by a light inextensible string which passes over a smooth light fixed pulley. The system is held at rest with the string taut and the hanging parts of the string vertical, as shown in Figure 5. The system is released from rest and \(A\) moves upwards.
At the instant when \(A\) is 0.7 m above its original position, \(C\) separates from \(B\) and falls away. In the subsequent motion, \(A\) does not reach the pulley.
| Scheme | Marks |
|---|---|
| \(4mg - T = 4ma\) | M1A1 |
| \(T - 3mg = 3ma\) | M1A1 |
| Condone the use of \(4mg - 3mg = 4ma + 3ma\) in place of one of these equations. | M1A1 |
| Reach given answer \(a = \dfrac{g}{7}\) correctly *** | A1 |
| Form an equation in \(T\): \(T = 3mg + 3\left(mg - \dfrac{T}{4}\right)\), \(T = 3mg + 3m\dfrac{g}{7}\), or \(T = 4mg - 4m\dfrac{g}{7}\) | M1 |
| \(T = \dfrac{24}{7}mg\) or equivalent, \(33.6m\), \(34m\) | A1 |
| (7) |
Notes
First M1 for resolving vertically (up or down) for \(B + C\), with correct no. of terms.
First A1 for a correct equation.
Second M1 for resolving vertically (up or down) for \(A\), with correct no. of terms.
Second A1 for a correct equation.
Third A1 for \(g/7\), obtained correctly. Given answer (1.4 A0)
Third M1 for an equation in \(T\) only
Fourth A1 for \(24mg/7\) oe or \(33.6m\) or \(34m\)
N.B. If they omit \(m\) throughout (which gives \(a = g/7\)), can score max M1A0M1A0A0M1A0 for part (a) BUT CAN SCORE ALL OF THE MARKS in parts (b), (c) and (d).
| Scheme | Marks |
|---|---|
| \(v^2 = u^2 + 2as = 2 \times \dfrac{g}{7} \times 0.7 = 1.96\), \(v = 1.4\) ms\(^{-1}\) | M1A1 |
| (2) |
Notes
M1 for an equation in \(v\) only (usually \(v^2 = u^2 + 2as\))
A1 for 1.4 (ms\(^{-1}\)) allow \(\sqrt{(g/5)}\) oe.
| Scheme | Marks |
|---|---|
| \(3mg - T = 3ma\) | M1A1 |
| \(T - 2mg = 2ma\) | A1 |
| \(a = \dfrac{g}{5}\) | A1 |
| (4) |
Notes
First M1 for resolving vertically (up or down) for \(A\) or \(B\), with correct no. of terms. (N.B. M0 if they use the tension from part (a))
First A1 for a correct equation for \(A\).
Second A1 for a correct equation for \(B\).
N.B. ‘Whole system’ equation: \(3mg - 2mg = 5ma\) earns first 3 marks but any error loses all 3
Third A1 for g/5 oe or 1.96 or 2.0 (ms\(^{-2}\)) (allow a negative answer)
| Scheme | Marks |
|---|---|
| \(0 = 1.96 - 2 \times \dfrac{g}{5} \times s\) | M1 |
| \(s = \dfrac{5 \times 1.96}{2g} = 0.5\) (m) | A1 |
| Total height \(= 0.7 + 0.5 = 1.2\) (m) | A1 ft |
| (3) | |
| (16 marks) |
Notes
M1 for an equation in \(s\) only using their \(v\) from (b) and \(a\) from (c).
either \(0 = 1.4^2 - 2(g/5)s\) or \(1.4^2 = 0 + 2(g/5)s\)
First A1 for \(s = 0.5\) (m) correctly obtained
Second A1 ft for their \(0.5 + 0.7 = 1.2\) (m)
Alternative using conservation of energy
M1 for an equation in \(s\) only, with correct number of terms, using their \(v\) from (b):-
\((3mgs - 2mgs) = \tfrac{1}{2}\,3m\,(1.4)^2 + \tfrac{1}{2}\,2m\,(1.4)^2\)
First A1 for \(s = 0.5\) (m) correctly obtained
Second A1 ft for their \(0.5 + 0.7 = 1.2\) (m)
(Corrected from the printed mark scheme: in Alt d the expression for \(s\) is printed as \(\dfrac{2.5 \times 1.96^2}{g}\).)
Alt d
| Using energy: \(3mgs - 2mgs = \dfrac{1}{2}3m \times 1.4^2 + \dfrac{1}{2}2m \times 1.4^2\) | M1 |
| \(s = \dfrac{2.5 \times 1.4^2}{g} = 0.5\) (m) | A1 |
| Total height \(= 0.7 + 0.5 = 1.2\) (m) | A1 ft |
| (3) |