M1 June 2005 Q8
8. [In this question, the unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are horizontal vectors due east and north respectively.]
At time \(t = 0\), a football player kicks a ball from the point \(A\) with position vector \((2\mathbf{i} + \mathbf{j})\) m on a horizontal football field. The motion of the ball is modelled as that of a particle moving horizontally with constant velocity \((5\mathbf{i} + 8\mathbf{j})\) m s\(^{-1}\). Find
The point \(B\) on the field has position vector \((10\mathbf{i} + 7\mathbf{j})\) m.
At time \(t = 0\), another player starts running due north from \(B\) and moves with constant speed \(v\) m s\(^{-1}\). Given that he intercepts the ball,
| Scheme | Marks |
|---|---|
| Speed of ball \(= \sqrt{(5^2 + 8^2)} \approx 9.43\) m s\(^{-1}\) | M1 A1 |
| (2) |
Notes
(a) M1 Valid attempt at speed (square, add and squ. root cpts)
| Scheme | Marks |
|---|---|
| p.v. of ball \(= (2\mathbf{i} + \mathbf{j}) + (5\mathbf{i} + 8\mathbf{j})t\) | M1 A1 |
| (2) |
Notes
(b) M1 needs non-zero p.v. + (attempt at veloc vector) \(\times\, t\). Must be vector
| Scheme | Marks |
|---|---|
| North of \(B\) when \(\mathbf{i}\) components same, i.e. \(2 + 5t = 10\) | M1 |
| \(t = 1.6\) s | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| When \(t = 1.6\), p.v. of ball \(= 10\mathbf{i} + 13.8\mathbf{j}\) (or \(\mathbf{j}\) component \(= 13.8\)) | M1 A1 |
| Distance travelled by 2nd player \(= 13.8 - 7 = 6.8\) | M1 A1 |
| Speed \(= 6.8 \div 1.6 = 4.25\) m s\(^{-1}\) | M1 A1 |
| (6) |
Notes
or
| Scheme | Marks |
|---|---|
| \([(2 + 5t)\mathbf{i} +]\ (1 + 8t)\mathbf{j} = [10\mathbf{i} +]\ (7 + vt)\mathbf{j}\) (pv’s or \(\mathbf{j}\) components same) | M1 A1 |
| Using \(t = 1.6\): \(1 + 12.8 = 7 + 1.6v\) (equn in \(v\) only) | M1 A1 |
| \(v = 4.25\) m s\(^{-1}\) | M1 A1 |
(d) 2nd M1 – allow if finding displacement vector (e.g. if using wrong time)
3rd M1 for getting speed as a scalar (and final answer must be as a scalar). But if they get e.g. ‘\(4.25\mathbf{j}\)’, allow M1 A0
(Corrected from the printed mark scheme: the distance is printed as \(13.8 - 6 = 6.8\); \(B\) has \(\mathbf{j}\) component 7.)
| Scheme | Marks |
|---|---|
| Allow for friction on field (i.e. velocity of ball not constant) or allow for vertical component of motion of ball | B1 |
| (1) | |
| (13 marks) |
Notes
(e) Allow ‘wind’, ‘spin’, ‘time for player to accelerate’, size of ball
Do not allow on their own ‘swerve’, ‘weight of ball’.