M1 June 2016 Q7
7. Two forces \(\mathbf{F}_1\) and \(\mathbf{F}_2\) act on a particle \(P\).
The force \(\mathbf{F}_1\) is given by \(\mathbf{F}_1 = (-\mathbf{i} + 2\mathbf{j})\) N and \(\mathbf{F}_2\) acts in the direction of the vector \((\mathbf{i} + \mathbf{j})\).
Given that the resultant of \(\mathbf{F}_1\) and \(\mathbf{F}_2\) acts in the direction of the vector \((\mathbf{i} + 3\mathbf{j})\),
The acceleration of \(P\) is \((3\mathbf{i} + 9\mathbf{j})\) m s\(^{-2}\). At time \(t = 0\), the velocity of \(P\) is \((3\mathbf{i} - 22\mathbf{j})\) m s\(^{-1}\)
| Scheme | Marks |
|---|---|
| \(\mathbf{F}_2 = k\mathbf{i} + k\mathbf{j}\) | B1 |
| \((-1 + a)\mathbf{i} + (2 + b)\mathbf{j}\) | M1 |
| \(\dfrac{-1 + a}{2 + b} = \dfrac{1}{3}\) | DM1 A1 |
| \(a = b = k = 2.5;\ \ \mathbf{F}_2 = 2.5\mathbf{i} + 2.5\mathbf{j}\) | DM1 A1; A1 |
| (7) |
Notes
B1 for \(\mathbf{F}_2 = k\mathbf{i} + k\mathbf{j}\) (\(k \neq 1\)) seen or implied in working, including for an incorrect final answer, with the wrong \(k\) value.
First M1 for adding the 2 forces (for this M mark we only need \(\mathbf{F}_2 = a\mathbf{i} + b\mathbf{j}\)), with \(\mathbf{i}\)’s and \(\mathbf{j}\)’s collected (which can be implied by later working) but allow a slip.
(M0 if \(a\) and \(b\) both assumed to be 1)
Second M1, dependent on first M1, for ratio of their cpts = 1/3 or 3/1 (Must be correct way up for the M mark)
First A1 for a correct equation which may involve two unknowns
Third M1, dependent on first and second M1, for solving for \(k\) oe
Second A1 for a correct \(k\) value
Third A1 for \(2.5\mathbf{i} + 2.5\mathbf{j}\)
ALTERNATIVE: Using two simultaneous equations
| \(\mathbf{F}_2 = k\mathbf{i} + k\mathbf{j}\) | B1 |
| \((-1 + a)\mathbf{i} + (2 + b)\mathbf{j} = p(\mathbf{i} + 3\mathbf{j})\) | M1 for LHS |
| \(-1 + a = p\) | |
| \(2 + b = 3p\) | DM1 A1 |
| \(a = b = k = 2.5;\ \ \mathbf{F}_2 = 2.5\mathbf{i} + 2.5\mathbf{j}\) | DM1 A1; A1 |
B1 for \(\mathbf{F}_2 = k\mathbf{i} + k\mathbf{j}\) (\(k \neq 1\)) seen or implied in working.
First M1 for adding the 2 forces (for this M mark we only need \(\mathbf{F}_2 = a\mathbf{i} + b\mathbf{j}\)), with \(\mathbf{i}\)’s and \(\mathbf{j}\)’s collected (LHS of equation) (M0 if \(a\) and \(b\) both assumed to be 1) but allow a slip
Second M1, dependent on first M1, for equating coeffs to produce two equations in 2 or 3 unknowns. Must have \(p\) and \(3p\) (M0 if \(p\) is assumed to be 1 or \(k\))
First A1 for two correct equations
Third M1, dependent on first and second M1, for solving for \(k\) oe
Second A1 for a correct \(k\) value
Third A1 for \(2.5\mathbf{i} + 2.5\mathbf{j}\)
ALTERNATIVE: Using magnitudes and directions
 | |
| \(\mathbf{F}_2 = k\mathbf{i} + k\mathbf{j}\), seen or implied | B1 |
| Correct vector triangle | M1 |
| \(\dfrac{k\sqrt{2}}{\sin 45^\circ} = \dfrac{\sqrt{5}}{\sin(90^\circ - \alpha)}\), \(\alpha = \arctan 2\) | DM1 A1 |
| \(2k = 5\) | |
| \(k = 2.5;\ \ \mathbf{F}_2 = 2.5\mathbf{i} + 2.5\mathbf{j}\) | DM1 A1; A1 |
B1 for \(\mathbf{F}_2 = k\mathbf{i} + k\mathbf{j}\) seen or implied in working.
First M1 for a correct vector triangle (for this M mark we only need \(\mathbf{F}_2 = a\mathbf{i} + b\mathbf{j}\)). (M0 if \(a\) and \(b\) both assumed to be 1 and/or longest side is assumed to be \(\sqrt{10}\))
Second M1, dependent on first M1, for using sine rule on vector triangle
First A1 for a correct equation. 45\(^\circ\) may not appear exactly.
Third M1, dependent on first and second M1, for solving for \(k\) oe
Second A1 for a correct \(k\) value
Third A1 for \(2.5\mathbf{i} + 2.5\mathbf{j}\)
| Scheme | Marks |
|---|---|
| \(\mathbf{v} = 3\mathbf{i} - 22\mathbf{j} + 3(3\mathbf{i} + 9\mathbf{j})\) | M1 |
| \(= 12\mathbf{i} + 5\mathbf{j}\) | A1 |
| \(|\mathbf{v}| = \sqrt{12^2 + 5^2} = 13\) m s\(^{-1}\) | M1 A1 cso |
| (4) | |
| (11 marks) |
Notes
First M1 for use of \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) with \(t = 3\)
First A1 for \(12\mathbf{i} + 5\mathbf{j}\) seen or implied. However, if a wrong \(\mathbf{v}\) is seen A0
Second M1 for finding magnitude of their \(\mathbf{v}\)
Second A1 for 13