| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2021 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Projectile motion: trajectory equation |
| Difficulty | Moderate -0.3 This is a straightforward multi-part SUVAT question with standard projectile motion. Parts (a)-(c) involve routine vertical motion calculations with given values. Parts (d)-(e) require recognizing that equal maximum heights mean equal vertical components, then using horizontal motion to find u and α. All steps follow standard textbook methods with no novel insight required, making it slightly easier than average. |
| Spec | 3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Time when \(v = 0\) given by \(0 = 29.4 - 9.8t\), so \(t = 3\) s | E1 [1] | Using suvat equation(s) leading to correct value for \(t\) with \(v = 0\). Allow for verifying that \(t = 3\) gives \(v = 0\) if identified as the maximum point oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Straight line with negative gradient through either \((3, 0)\) or \((0, 29.4)\) | B1 | |
| Both \((3, 0)\) and \((0, 29.4)\) clearly seen; must include negative values of \(v\) for \(t > 3\) | B1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(t = 5\), \(v = 29.4 - 9.8 \times 5\) | M1 | Using suvat equation(s) leading to a value for \(v\) with \(t = 5\). Allow sign errors. If motion from the highest point considered \(u = 0\), \(t = 2\), \(g = +9.8\) then \(v = 19.6\) is fully correct |
| \(v = -19.6\) | A1 | May be implied by 19.6 seen |
| Speed is \(19.6 \text{ m s}^{-1}\) | A1 [3] | FT their negative velocity. Allow M1A1A0 if \(29.4 - 9.8 \times 5 = 19.6\) seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Max height unchanged so \(u_y = 29.4 \text{ m s}^{-1}\) | B1 | Allow if calculated from \(y = 44.1\) m |
| Time to max height unchanged, so 3 s | B1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(u_x \times 3 = 48\) | M1 | Using (their) \(t = 3\) to find \(u_x\) |
| \(u = \sqrt{u_x^2 + u_y^2} = \sqrt{16^2 + 29.4^2} = 33.5\) | M1 | Combining their components to find either one of \(u\) and \(\alpha\) |
| \(\tan\alpha = \frac{u_y}{u_x} = \frac{29.4}{16}\) giving \(\alpha = 61.4°\) | A1 [3] | Both values correct |
## Question 10:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Time when $v = 0$ given by $0 = 29.4 - 9.8t$, so $t = 3$ s | E1 [1] | Using suvat equation(s) leading to correct value for $t$ with $v = 0$. Allow for verifying that $t = 3$ gives $v = 0$ if identified as the maximum point oe |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Straight line with negative gradient through either $(3, 0)$ or $(0, 29.4)$ | B1 | |
| Both $(3, 0)$ and $(0, 29.4)$ clearly seen; must include negative values of $v$ for $t > 3$ | B1 [2] | |
### Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $t = 5$, $v = 29.4 - 9.8 \times 5$ | M1 | Using suvat equation(s) leading to a value for $v$ with $t = 5$. Allow sign errors. If motion from the highest point considered $u = 0$, $t = 2$, $g = +9.8$ then $v = 19.6$ is fully correct |
| $v = -19.6$ | A1 | May be implied by 19.6 seen |
| Speed is $19.6 \text{ m s}^{-1}$ | A1 [3] | FT their negative velocity. Allow M1A1A0 if $29.4 - 9.8 \times 5 = 19.6$ seen |
### Part (d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Max height unchanged so $u_y = 29.4 \text{ m s}^{-1}$ | B1 | Allow if calculated from $y = 44.1$ m |
| Time to max height unchanged, so 3 s | B1 [2] | |
### Part (e):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $u_x \times 3 = 48$ | M1 | Using (their) $t = 3$ to find $u_x$ |
| $u = \sqrt{u_x^2 + u_y^2} = \sqrt{16^2 + 29.4^2} = 33.5$ | M1 | Combining their components to find either one of $u$ and $\alpha$ |
| $\tan\alpha = \frac{u_y}{u_x} = \frac{29.4}{16}$ giving $\alpha = 61.4°$ | A1 [3] | Both values correct |
---10 A ball is thrown upwards with a velocity of $29.4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that the ball reaches its maximum height after 3 s .
\item Sketch a velocity-time graph for the first 5 s of motion.
\item Calculate the speed of the ball 5 s after it is thrown.
A second ball is thrown at $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $\alpha ^ { \circ }$ above the horizontal. It reaches the same maximum height as the first ball.
\item Use this information to write down
\begin{itemize}
\item the vertical component of the second ball's initial velocity,
\item the time taken for the second ball to reach its greatest height.
\end{itemize}
This second ball reaches its greatest height at a point which is 48 m horizontally from the point of projection.
\item Calculate the values of $u$ and $\alpha$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2021 Q10 [11]}}