| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile passing through given point |
| Difficulty | Standard +0.3 This is a standard two-part projectile question requiring application of SUVAT equations in 2D. Part (i) involves finding time to reach the wall then calculating vertical velocity component. Part (ii) requires solving a quadratic equation to find initial speed. Both parts use routine mechanics techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v_x = 12\cos20\) | *B1 | \(11.27631\ldots\) |
| \(8 = 12t\cos20\) | B1 | Using suvat to find expression in \(t\) only. \((t = 0.70945\ldots)\) |
| \(v_y = 12\sin20 - gcv(t)\) | *M1 | Attempt at use of \(v = u + at\) |
| A1 | \(-2.84838\ldots\) | |
| \(\tan\theta = v_y/v_x\) | Dep**M1 | Use trig to find a relevant angle |
| \(14.2°\) below horizontal | A1 | \(14.1763\ldots\) (\(75.8°\) downward vertical) |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(8 = Vt\cos20\) | B1 | |
| \(1.5 = Vt\sin20 - gt^2/2\) | *M1 | Attempt at use of \(s = ut + \frac{1}{2}at^2\) |
| A1 | ||
| Eliminate \(t\) | dep*M1 | OR Eliminate \(V\) and solve for \(t\) |
| Attempt to solve a quadratic for \(V\) | dep*M1 | AND Sub value for \(t\) and solve for \(V\) |
| \(V = 15.9\) | A1 | \(V = 15.8606\ldots\) |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = x\tan\theta - \dfrac{gx^2\sec^2\theta}{2u^2}\) | *B1 | Use equation of trajectory |
| Substitute values for \(y\), \(x\), \(\theta\) | dep*M1 | |
| \(1.5 = 8\tan20 - g8^2\sec^2 20/2V^2\) | A1 | |
| Attempt to solve a quadratic for \(V\) | dep*M2 | SC M1 for solving for \(V^2\) |
| \(V = 15.9\) | A1 | \(V = 15.8606\ldots\) |
| [6] |
## Question 8:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v_x = 12\cos20$ | *B1 | $11.27631\ldots$ |
| $8 = 12t\cos20$ | B1 | Using suvat to find expression in $t$ only. $(t = 0.70945\ldots)$ |
| $v_y = 12\sin20 - gcv(t)$ | *M1 | Attempt at use of $v = u + at$ |
| | A1 | $-2.84838\ldots$ |
| $\tan\theta = v_y/v_x$ | Dep**M1 | Use trig to find a relevant angle |
| $14.2°$ below horizontal | A1 | $14.1763\ldots$ ($75.8°$ downward vertical) |
| **[6]** | | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $8 = Vt\cos20$ | B1 | |
| $1.5 = Vt\sin20 - gt^2/2$ | *M1 | Attempt at use of $s = ut + \frac{1}{2}at^2$ |
| | A1 | |
| Eliminate $t$ | dep*M1 | OR Eliminate $V$ and solve for $t$ |
| Attempt to solve a quadratic for $V$ | dep*M1 | AND Sub value for $t$ and solve for $V$ |
| $V = 15.9$ | A1 | $V = 15.8606\ldots$ |
| **[6]** | | |
**OR**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = x\tan\theta - \dfrac{gx^2\sec^2\theta}{2u^2}$ | *B1 | Use equation of trajectory |
| Substitute values for $y$, $x$, $\theta$ | dep*M1 | |
| $1.5 = 8\tan20 - g8^2\sec^2 20/2V^2$ | A1 | |
| Attempt to solve a quadratic for $V$ | dep*M2 | SC M1 for solving for $V^2$ |
| $V = 15.9$ | A1 | $V = 15.8606\ldots$ |
| **[6]** | | |8 A child is trying to throw a small stone to hit a target painted on a vertical wall. The child and the wall are on horizontal ground. The child is standing a horizontal distance of 8 m from the base of the wall. The child throws the stone from a height of 1 m with speed $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $20 ^ { \circ }$ above the horizontal.\\
(i) Find the direction of motion of the stone when it hits the wall.
The child now throws the stone with a speed of $\mathrm { Vm } \mathrm { s } ^ { - 1 }$ from the same initial position and still at an angle of $20 ^ { \circ }$ above the horizontal. This time the stone hits the target which is 2.5 m above the ground.\\
(ii) Find $V$.
\hfill \mbox{\textit{OCR M2 2014 Q8 [12]}}