| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Vertical motion under gravity |
| Difficulty | Moderate -0.3 This is a straightforward three-part SUVAT question requiring standard application of kinematic equations and momentum calculation. Part (i) uses v²=u²+2as, part (ii) uses v=u+at, and part (iii) applies momentum = mv. All parts follow routine procedures with no problem-solving insight required, making it slightly easier than average but not trivial due to the multi-step nature and need for careful sign conventions. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form6.03a Linear momentum: p = mv6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v^2 = 3.5^2 + 2g \times 5\) | M1 | Uses \(v^2 = 3.5^2 +/- 2g5\); Accept \(-3.5^2\) for \((-3.5)^2\) etc |
| \(v = 10.5 \text{ ms}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(5 = 0.87u - g \times 0.87^2/2\) | M1 | \(+/-5 = 0.87u +/- g\ 0.87^2/2\); May come from \(s = vt - gt^2/2\) |
| \(u = 10.0 \text{ m s}^{-1}\) | A1 | |
| A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Change \(= 0.2 \times 10.5 + 0.2 \times 10\) | M1 | Or \(+/- 0.2(\text{Ans(i)}) +/- \text{Ans(ii)}\) |
| Change \(= 4.1(0) \text{ kg m s}^{-1}\) | A1 | It is OK to get \(-4.1\) from correct work |
# Question 1:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v^2 = 3.5^2 + 2g \times 5$ | M1 | Uses $v^2 = 3.5^2 +/- 2g5$; Accept $-3.5^2$ for $(-3.5)^2$ etc |
| $v = 10.5 \text{ ms}^{-1}$ | A1 | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $5 = 0.87u - g \times 0.87^2/2$ | M1 | $+/-5 = 0.87u +/- g\ 0.87^2/2$; May come from $s = vt - gt^2/2$ |
| $u = 10.0 \text{ m s}^{-1}$ | A1 | |
| | A1 | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Change $= 0.2 \times 10.5 + 0.2 \times 10$ | M1 | Or $+/- 0.2(\text{Ans(i)}) +/- \text{Ans(ii)}$ |
| Change $= 4.1(0) \text{ kg m s}^{-1}$ | A1 | It is OK to get $-4.1$ from correct work |
---1 A particle $P$ is projected vertically downwards with initial speed $3.5 \mathrm {~ms} ^ { - 1 }$ from a point $A$ which is 5 m above horizontal ground.\\
(i) Find the speed of $P$ immediately before it strikes the ground.
After striking the ground, $P$ rebounds and moves vertically upwards and 0.87 s after leaving the ground $P$ passes through $A$.\\
(ii) Calculate the speed of $P$ immediately after it leaves the ground.
It is given that the mass of $P$ is 0.2 kg .\\
(iii) Calculate the change in the momentum of $P$ as a result of its collision with the ground.
\hfill \mbox{\textit{OCR M1 2014 Q1 [7]}}