| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projection from elevated point - angle below horizontal or horizontal |
| Difficulty | Standard +0.3 This is a standard M2 projectiles question with projection from an elevated point at a downward angle. Part (i) requires basic SUVAT equations, part (ii) needs velocity component calculation for direction, and part (iii) involves straightforward energy conservation. All three parts follow routine procedures with no novel problem-solving required, making it slightly easier than average for M2 level. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model6.02e Calculate KE and PE: using formulae |
| Answer | Marks | Guidance |
|---|---|---|
| \(9 = 17 \cos 25° × t\) | M1 | B1 \(y=\tan\theta-4.9x^2/y^2\cos^2\theta\) |
| \(t = 0.584\) (or \(9/17\cos 25°\)) | A1 | M1/A1 \(y=9\tan(-25°)-4.9×9^2/17^2\cos^2 25°\) |
| \(d = 17\sin 25°×0.584 + \frac{1}{2}×9.8×0.584^2\) (d = ht lost (5.87)) | M1 A1 | A1 \(y = -5.87\) 2.13 |
| \(h = 2.13\) | A1 5 |
| Answer | Marks | Guidance |
|---|---|---|
| \(v_u = 17\cos 25°\) (15.4) | B1 | M1/A1 \(dv/dx =\) |
| \(v_v = 17\sin 25° + 9.8×0.584\) | M1 | \(\tan\theta - 9.8v/v^2\cos^2\theta\) |
| or \(v_v^2 = (17\sin 25°)^2 + 2×9.8×5.87\) | ||
| \(v_v = 12.9\) | A1 | A1 \(dv/dx = -0.838\) |
| \(\tan\theta = 12.9/15.4\) | M1 | M1 \(\tan^{-1}(-0.838)\) |
| \(\theta = 40.0°\) below horizontal | A1 5 | or 50.0° to vertical (20.1) |
| Answer | Marks | Guidance |
|---|---|---|
| speed \(= \sqrt{(12.9^2 + 15.4^2)}\) | M1 | NB 0.3 instead of 0.7 gives 11.0 (M0) |
| \(v_{2m}v^2 = v_{2m}×20.1^2 × 0.7\) | M1 A1 | |
| \(v = 16.8 \text{ m s}^{-1}\) | A1 4 |
### Part (i)
$9 = 17 \cos 25° × t$ | M1 | B1 $y=\tan\theta-4.9x^2/y^2\cos^2\theta$
$t = 0.584$ (or $9/17\cos 25°$) | A1 | M1/A1 $y=9\tan(-25°)-4.9×9^2/17^2\cos^2 25°$
$d = 17\sin 25°×0.584 + \frac{1}{2}×9.8×0.584^2$ (d = ht lost (5.87)) | M1 A1 | A1 $y = -5.87$ 2.13
$h = 2.13$ | A1 5 |
### Part (ii)
$v_u = 17\cos 25°$ (15.4) | B1 | M1/A1 $dv/dx =$
$v_v = 17\sin 25° + 9.8×0.584$ | M1 | $\tan\theta - 9.8v/v^2\cos^2\theta$
or $v_v^2 = (17\sin 25°)^2 + 2×9.8×5.87$ | |
$v_v = 12.9$ | A1 | A1 $dv/dx = -0.838$
$\tan\theta = 12.9/15.4$ | M1 | M1 $\tan^{-1}(-0.838)$
$\theta = 40.0°$ below horizontal | A1 5 | or 50.0° to vertical (20.1)
### Part (iii)
speed $= \sqrt{(12.9^2 + 15.4^2)}$ | M1 | NB 0.3 instead of 0.7 gives 11.0 (M0)
$v_{2m}v^2 = v_{2m}×20.1^2 × 0.7$ | M1 A1 |
$v = 16.8 \text{ m s}^{-1}$ | A1 4 | | 147\\
\includegraphics[max width=\textwidth, alt={}, center]{e85c2bf4-21a8-4d9a-93c5-d5679b2a8233-4_440_657_906_744}
A ball is projected with an initial speed of $17 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $25 ^ { \circ }$ below the horizontal from a point on the top of a vertical wall. The point of projection is 8 m above horizontal ground. The ball hits a vertical fence which is at a horizontal distance of 9 m from the wall (see diagram).\\
(i) Calculate the height above the ground of the point where the ball hits the fence.\\
(ii) Calculate the direction of motion of the ball immediately before it hits the fence.\\
(iii) It is given that $30 \%$ of the kinetic energy of the ball is lost when it hits the fence. Calculate the speed of the ball immediately after it hits the fence.
\hfill \mbox{\textit{OCR M2 2009 Q7 [14]}}