| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Horizontal projection from height |
| Difficulty | Standard +0.3 This is a standard two-part projectiles question requiring straightforward application of SUVAT equations. Part (i) involves finding time from vertical motion then using it for horizontal distance. Part (ii) requires resolving velocity components and solving a quadratic, which is slightly more involved but still routine for M2 level with no novel problem-solving required. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2} \times 9.8 \times t^2 = 0.2\) | M1 | Using SUVAT to find \(t\), consistent signs for \(g\) and \(0.2\) |
| \(t = 0.2(02)\) | A1 | aef |
| \(s = 14.4 \times t_c\) | M1 | Using their value of \(t\) |
| \(s = 2.91\) m | A1 | |
| [4] | ||
| OR: Use equation of trajectory | M1 | B1 for correct equation of the trajectory seen anywhere but award in part (ii) unless different method seen; consistent signs for \(g\) and \(0.2\) |
| \(-0.2 = x\tan 0 - gx^2\sec^2 0/(2\times 14.4^2)\) | A1 | |
| Solve quadratic for \(x\) | M1 | |
| \(x = 2.91\) | A1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(U\sin 15 \times t - \frac{1}{2} \times 9.8 \times t^2 = -0.2\) | *M1 | Using \(s = ut + \frac{1}{2}at^2\) with \(s = \pm 0.2\) and \(a = \pm g\) |
| \(U\cos 15 \times t = 6\) | A1 | |
| B1 | ||
| Eliminate \(t\) | Dep*M1 | Eliminate \(U\) |
| Attempt to solve to find \(U\) | Dep*M1 | Attempt to solve to find \(t(=0.607)\) |
| \(U = 10.2 \text{ ms}^{-1}\) | A1 | |
| [6] | ||
| OR: \(y = x\tan\theta - gx^2\sec^2\theta/2U^2\) | *B1 | |
| Substitute values for \(y, x, \theta\) | Dep*M1 | |
| \(-0.2 = 6\tan 15 - g.6^2\sec^2 15/2U^2\) | A1 | |
| Attempt to solve for \(U\) | Dep*M2 | |
| \(U = 10.2 \text{ ms}^{-1}\) | A1 | |
| [6] |
## Question 4:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2} \times 9.8 \times t^2 = 0.2$ | M1 | Using SUVAT to find $t$, consistent signs for $g$ and $0.2$ |
| $t = 0.2(02)$ | A1 | aef |
| $s = 14.4 \times t_c$ | M1 | Using their value of $t$ |
| $s = 2.91$ m | A1 | |
| **[4]** | | |
| OR: Use equation of trajectory | M1 | B1 for correct equation of the trajectory seen anywhere but award in part (ii) unless different method seen; consistent signs for $g$ and $0.2$ |
| $-0.2 = x\tan 0 - gx^2\sec^2 0/(2\times 14.4^2)$ | A1 | |
| Solve quadratic for $x$ | M1 | |
| $x = 2.91$ | A1 | |
| **[4]** | | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $U\sin 15 \times t - \frac{1}{2} \times 9.8 \times t^2 = -0.2$ | *M1 | Using $s = ut + \frac{1}{2}at^2$ with $s = \pm 0.2$ and $a = \pm g$ |
| $U\cos 15 \times t = 6$ | A1 | |
| | B1 | |
| Eliminate $t$ | Dep*M1 | Eliminate $U$ |
| Attempt to solve to find $U$ | Dep*M1 | Attempt to solve to find $t(=0.607)$ |
| $U = 10.2 \text{ ms}^{-1}$ | A1 | |
| **[6]** | | |
| OR: $y = x\tan\theta - gx^2\sec^2\theta/2U^2$ | *B1 | |
| Substitute values for $y, x, \theta$ | Dep*M1 | |
| $-0.2 = 6\tan 15 - g.6^2\sec^2 15/2U^2$ | A1 | |
| Attempt to solve for $U$ | Dep*M2 | |
| $U = 10.2 \text{ ms}^{-1}$ | A1 | |
| **[6]** | | |
---4 A boy throws a small ball at a vertical wall. The ball is thrown horizontally, from a point $O$, at a speed of $14.4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and it hits the wall at a point which is 0.2 m below the level of $O$.\\
(i) Find the horizontal distance from $O$ to the wall.
The boy now moves so that he is 6 m from the wall. He throws the ball at an angle of $15 ^ { \circ }$ above the horizontal. The ball again hits the wall at a point which is 0.2 m below the level from which it was thrown.\\
(ii) Find the speed at which the ball was thrown.
\hfill \mbox{\textit{OCR M2 2012 Q4 [10]}}