| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Two particles: relative motion |
| Difficulty | Moderate -0.3 This is a straightforward SUVAT question with vertical projection. Parts (i)-(iii) involve routine application of kinematic equations and simple algebra. Parts (iv)-(v) require solving a quadratic and interpreting the result, but the setup is standard. Slightly easier than average due to the guided structure and familiar context, though the multi-part nature prevents it from being significantly below average. |
| Spec | 3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Heights are \(7t - \frac{1}{2}gt^2\) and \(10.5t - \frac{1}{2}gt^2\) | B1 [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Expression is \(3.5t\) | B1 [1] | From correct (i) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0 = 7 - 9.8t\) | M1 | For using \(v = u - gt\) with \(v = 0\) |
| \(t = 5/7\) or \(0.714\) | A1 | |
| Difference is \(2.5\) m | A1ft [3] | ft value of \(t\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(t = 1\) | B1ft | For using ans(ii) \(= 3.5\) correctly |
| Greater than \(5/7\) (may be implied) or \(7 - g \times 1\) is \(-ve\) | M1 | For comparing this \(t\) with the time to greatest height or considering the sign of \(v_A\) for this \(t\) |
| Direction is downwards | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(h_A = 7 \times 1 - \frac{1}{2} \times 9.8 \times 1^2\) | M1 | For using \(h = ut - \frac{1}{2}gt^2\) with relevant \(t\) |
| Height is \(2.1\) m | A1 [2] |
# Question 5:
## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Heights are $7t - \frac{1}{2}gt^2$ and $10.5t - \frac{1}{2}gt^2$ | B1 [1] | |
## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Expression is $3.5t$ | B1 [1] | From correct (i) |
## Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $0 = 7 - 9.8t$ | M1 | For using $v = u - gt$ with $v = 0$ |
| $t = 5/7$ or $0.714$ | A1 | |
| Difference is $2.5$ m | A1ft [3] | ft value of $t$ |
## Part (iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| $t = 1$ | B1ft | For using ans(ii) $= 3.5$ correctly |
| Greater than $5/7$ (may be implied) or $7 - g \times 1$ is $-ve$ | M1 | For comparing this $t$ with the time to greatest height or considering the sign of $v_A$ for this $t$ |
| Direction is downwards | A1 [3] | |
## Part (v):
| Answer | Mark | Guidance |
|--------|------|----------|
| $h_A = 7 \times 1 - \frac{1}{2} \times 9.8 \times 1^2$ | M1 | For using $h = ut - \frac{1}{2}gt^2$ with relevant $t$ |
| Height is $2.1$ m | A1 [2] | |
---5 Two particles $A$ and $B$ are projected vertically upwards from horizontal ground at the same instant. The speeds of projection of $A$ and $B$ are $7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $10.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ respectively.\\
(i) Write down expressions for the heights above the ground of $A$ and $B$ at time $t$ seconds after projection.\\
(ii) Hence find a simplified expression for the difference in the heights of $A$ and $B$ at time $t$ seconds after projection.\\
(iii) Find the difference in the heights of $A$ and $B$ when $A$ is at its maximum height.
At the instant when $B$ is 3.5 m above $A$, find\\
(iv) whether $A$ is moving upwards or downwards,\\
(v) the height of $A$ above the ground.
\hfill \mbox{\textit{OCR M1 2005 Q5 [10]}}