M1 June 2012 Q7
7.
Two particles \(P\) and \(Q\), of mass 0.3 kg and 0.5 kg respectively, are joined by a light horizontal rod. The system of the particles and the rod is at rest on a horizontal plane. At time \(t = 0\), a constant force \(\mathbf{F}\) of magnitude 4 N is applied to \(Q\) in the direction \(PQ\), as shown in Figure 3. The system moves under the action of this force until \(t = 6\) s. During the motion, the resistance to the motion of \(P\) has constant magnitude 1 N and the resistance to the motion of \(Q\) has constant magnitude 2 N.
Find
At \(t = 6\) s, \(\mathbf{F}\) is removed and the system decelerates to rest. The resistances to motion are unchanged. Find
| Scheme | Marks |
|---|---|
| For system N2L \(4 - 3 = 0.8a\) | M1 A1 |
| \(a = 1.25\) (m s\(^{-2}\)), 1.3 | A1 |
| (3) |
Notes
(In parts (a), (c), (d) and (e) use the value of the mass being used to guide you as to which part of the system is being considered, and mark equation(s) accordingly)
M1 for resolving horizontally to produce an equation in \(a\) ONLY.
First A1 for a correct equation
Second A1 for 1.25
| Scheme | Marks |
|---|---|
| \(v = u + at\ \Rightarrow\ v = 0 + 1.25 \times 6 = 7.5\) (m s\(^{-1}\)) | M1 A1 |
| (2) |
Notes
M1 for a complete method to find the speed
A1 cao 7.5
| Scheme | Marks |
|---|---|
| For \(P\) N2L \(T - 1 = 0.3 \times 1.25\) ft their \(a\) | M1 A1ft |
| \(T = 1.375\) (N) 1.38, 1.4 | A1 |
| (3) |
Notes
M1 for resolving horizontally, for either \(P\) or \(Q\), to produce an equation in \(T\) only.
First A1ft for a correct equation, ft on their \(a\)
Second A1 cao for 1.38 (N) or 1.375 (N)
OR
| For \(Q\) N2L \(4 - 2 - T = 0.5 \times 1.25\) |
| Scheme | Marks |
|---|---|
| For system N2L \(-3 = 0.8a\ \Rightarrow\ a = -3.75\) | M1 A1 |
| \(v^2 = u^2 + 2as\ \Rightarrow\ 0^2 = 7.5^2 - 2 \times 3.75s\) | M1 |
| \(s = 7.5\) (m) | A1 |
| (4) |
Notes
First M1 for resolving horizontally to produce an equation in \(a\) ONLY.
First A1cao for \(-3.75\) (or 3.75)
Second M1 for use of \(v^2 = u^2 + 2as\), with \(v = 0\), \(u =\) their (b) and their \(a\), (or any other complete method which produces an equation in \(s\) only)
M0 if they haven’t calculated a value of \(a\).
Second A1 for 7.5 m
| Scheme | Marks |
|---|---|
| For \(P\) N2L \(T^{\prime} + 1 = 0.3 \times 3.75\) | M1 A1 |
| \(T^{\prime} = 0.125\) (N), 0.13 | A1 |
| (3) | |
| (15 marks) |
Notes
M1 for resolving horizontally, for either \(P\) or \(Q\), to produce an equation in \(T\) only.
M0 if they haven’t calculated a value of \(a\)
First A1cao for a correct equation
Second A1 cao for 0.125 or 0.13 (N) (must be positive)
Alternative for (e)
| For \(Q\) N2L \(2 - T^{\prime} = 0.5 \times 3.75\) | M1 A1 |
| \(T^{\prime} = 0.125\) (N), 0.13 | A1 |
| (3) |