| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Speed at specific time or position |
| Difficulty | Standard +0.3 Part (i) is a standard projectile motion calculation using v² = u² + 2as. Part (ii) requires finding the time when the velocity angle is -10°, then calculating distance from origin, which involves more steps but uses routine M2 techniques. Slightly above average due to the angle condition in part (ii), but still a typical textbook exercise. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| \(v_v = 42\sin30°\) (=21) | B1 | |
| \(0 = 21^2 - 2x9.8xh\) | M1 | |
| \(h = 22.5\) | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(v_h = 42\cos30°\) (=36.4) | B1 | |
| \(v_x = \frac{1}{2}v_h \times \tan10°\) | M1 | |
| \(v_c = ±6.41 = 21\sqrt{3}\tan10°\) | A1 | or \(42\cos30°\tan10°\) |
| M1 | ** | must be −6.41(also see "or" x 2) |
| A1 | ** | |
| M1 | ** | \(\sqrt{\text{their } t}\) |
| \(t = 2.80\) | A1 | ** |
| \(y=42\sin30°x2.8 - 4.9x2.8^2\) | M1 | ** |
| \(y = 20.4\) | A1 | ** |
| \(x = 42\cos30° \times 2.80\) | M1 | |
| \(x = 102\) | A1 | ** |
| \(\sqrt{(x^2 + y^2)}\) | M1 | |
| \(d = 104\) | A1 | 11 |
| or | \(6.41^2 = 21t^2 + 2 \times -9.8s\) | M1 |
| \(s = 20.4\) | A1 | ** |
| \(20.4 = 21t + \frac{1}{2} \cdot -9.8t^2\) | M1 | ** |
| \(t = 2.80\) | A1 | ** |
| M1 | ** | dist from top (s = 2.096) |
| A1 | ** | |
| M1 | ** | 2 separate times (2.143, 0.654) |
| \(22.5 \& 2.1 = \frac{1}{2}9.8t^2\) | M1 | ** |
| \(t = 2.80\) | A1 | ** |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = x/\sqrt{3} - x^2/270\) | aef | B1 |
| \(dy/dx = 1/\sqrt{3} - x/135\) | M1 | for differentiating |
| A1 | aef | |
| \(dy/dx = - \tan10°\) | M1 | must be −tan10° |
| \(1/\sqrt{3} - x/135 = - \tan10°\) | A1 | |
| solve for \(x\) | M1 | |
| \(x = 102\) | A1 | ** |
| \(y = x/\sqrt{3} - x^2/270\) | M1 | |
| \(y = 20.4\) | A1 | ** |
| \(\sqrt{(x^2 + y^2)}\) | M1 | |
| \(d = 104\) | A1 | (11) |
**(i)**
$v_v = 42\sin30°$ (=21) | B1 |
$0 = 21^2 - 2x9.8xh$ | M1 |
$h = 22.5$ | A1 | 3 |
**(ii)**
$v_h = 42\cos30°$ (=36.4) | B1 |
$v_x = \frac{1}{2}v_h \times \tan10°$ | M1 |
$v_c = ±6.41 = 21\sqrt{3}\tan10°$ | A1 | or $42\cos30°\tan10°$ |
| M1 | ** | must be −6.41(also see "or" x 2) |
| A1 | ** |
| M1 | ** | $\sqrt{\text{their } t}$
$t = 2.80$ | A1 | ** |
$y=42\sin30°x2.8 - 4.9x2.8^2$ | M1 | ** |
$y = 20.4$ | A1 | ** |
$x = 42\cos30° \times 2.80$ | M1 |
$x = 102$ | A1 | ** | $\sqrt{\text{their } t}$ |
$\sqrt{(x^2 + y^2)}$ | M1 |
$d = 104$ | A1 | 11 |
or | $6.41^2 = 21t^2 + 2 \times -9.8s$ | M1 | ** | vert dist first then time
$s = 20.4$ | A1 | ** |
$20.4 = 21t + \frac{1}{2} \cdot -9.8t^2$ | M1 | ** |
$t = 2.80$ | A1 | ** |
| M1 | ** | dist from top (s = 2.096)
| A1 | ** |
| M1 | ** | 2 separate times (2.143, 0.654)
$22.5 \& 2.1 = \frac{1}{2}9.8t^2$ | M1 | ** | 2.143 + 0.654
$t = 2.80$ | A1 | ** | 14 |
**(ii) alternatively**
$y = x/\sqrt{3} - x^2/270$ | aef | B1 | $y=x\tan30°-9.8x^2/2.42^2.\cos^230°$
$dy/dx = 1/\sqrt{3} - x/135$ | M1 | for differentiating
| A1 | aef |
$dy/dx = - \tan10°$ | M1 | must be −tan10°
$1/\sqrt{3} - x/135 = - \tan10°$ | A1 |
solve for $x$ | M1 |
$x = 102$ | A1 | ** | $\sqrt{\text{on their } dy/dx}$ |
$y = x/\sqrt{3} - x^2/270$ | M1 |
$y = 20.4$ | A1 | ** | $\sqrt{\text{their } x}$ |
$\sqrt{(x^2 + y^2)}$ | M1 |
$d = 104$ | A1 | (11) |8 A missile is projected with initial speed $42 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $30 ^ { \circ }$ above the horizontal. Ignoring air resistance, calculate\\
(i) the maximum height of the missile above the level of the point of projection,\\
(ii) the distance of the missile from the point of projection at the instant when it is moving downwards at an angle of $10 ^ { \circ }$ to the horizontal.
\hfill \mbox{\textit{OCR M2 2007 Q8 [14]}}