| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Two particles: different start times, same height |
| Difficulty | Standard +0.3 This is a straightforward two-particle kinematics problem requiring standard SUVAT equations. Part (i) is routine (v²=u²+2as). Part (ii) involves setting up equations for both particles and solving simultaneously, but the working is guided ('show that') and the method is standard for M1. Slightly above average due to the two-particle coordination, but no novel insight required. |
| Spec | 3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Height reached by first particle is given by \(0 = 21^2 - 2 \times 9.8 \times s\) | M1 | Other methods must be complete. Allow \(g = \pm 9.8, \pm 10\) |
| so \(s = 22.5\) so 22.5 m | A1 | Accept with consistent signs |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(t\) seconds after second particle projected its height is \(15t - 4.9t^2\) and the first particle has height \(22.5 - 4.9t^2\) (or \(21t - 4.9t^2\)) | M1 | Allow \(g = \pm 9.8, \pm 10\) |
| A1 | ||
| M1 | Allow \(g = \pm 9.8, \pm 10\) | |
| A1 | Award only if used correctly | |
| either Sub \(t = 1.5\) to show both have same value. State height as 11.475 m | E1 | (or sub \(t = 3.64\) into \(21t - 4.9t^2\) for 1st & \(t = 1.5\) for 2nd) cao. Accept any reasonable accuracy. Don't award if only one correctly used equation obtained. |
| A1 | ||
| or \(15t - 4.9t^2 = 22.5 - 4.9t^2\) giving \(t = 1.5\) and height as 11.475 m | M1 | Both. \(t\) shown. Ht cao (to any reasonable accuracy) |
| A1 | ||
| Sol (1) Total: 6 |
| Answer | Marks | Guidance |
|---|---|---|
| \(t\) seconds after second particle projected its height is \(15t - 4.9t^2\) and the first particle has fallen \(4.9t^2\) | M1 | Allow \(g = \pm 9.8, \pm 10\) |
| A1 | ||
| B1 | ||
| Collide when \(15t - 4.9t^2 + 4.9t^2 = 22.5\) so \(T = 1.5\) \(H = 22.5 - 4.9 \times 1.5^2 = 11.475\) m | M1 | Or other correct method |
| E1 | ||
| A1 | cao. Accept any reasonable accuracy. Don't award if only one correctly used equation obtained. | |
| Sol (2) Total: 6 |
**Part (i)**
| Height reached by first particle is given by $0 = 21^2 - 2 \times 9.8 \times s$ | M1 | Other methods must be complete. Allow $g = \pm 9.8, \pm 10$ |
| so $s = 22.5$ so 22.5 m | A1 | Accept with consistent signs |
| | | **Total: 2** |
**Part (ii) - Sol (1)**
| $t$ seconds after second particle projected its height is $15t - 4.9t^2$ and the first particle has height $22.5 - 4.9t^2$ (or $21t - 4.9t^2$) | M1 | Allow $g = \pm 9.8, \pm 10$ |
| | A1 | |
| | M1 | Allow $g = \pm 9.8, \pm 10$ |
| | A1 | Award only if used correctly |
| **either** Sub $t = 1.5$ to show both have same value. State height as 11.475 m | E1 | (or sub $t = 3.64$ into $21t - 4.9t^2$ for 1st & $t = 1.5$ for 2nd) cao. Accept any reasonable accuracy. Don't award if only one correctly used equation obtained. |
| | A1 | |
| **or** $15t - 4.9t^2 = 22.5 - 4.9t^2$ giving $t = 1.5$ and height as 11.475 m | M1 | Both. $t$ shown. Ht cao (to any reasonable accuracy) |
| | A1 | |
| | | **Sol (1) Total: 6** |
**Part (ii) - Sol (2)**
| $t$ seconds after second particle projected its height is $15t - 4.9t^2$ and the first particle has fallen $4.9t^2$ | M1 | Allow $g = \pm 9.8, \pm 10$ |
| | A1 | |
| | B1 | |
| Collide when $15t - 4.9t^2 + 4.9t^2 = 22.5$ so $T = 1.5$ $H = 22.5 - 4.9 \times 1.5^2 = 11.475$ m | M1 | Or other correct method |
| | E1 | |
| | A1 | cao. Accept any reasonable accuracy. Don't award if only one correctly used equation obtained. |
| | | **Sol (2) Total: 6** |
---4 A particle is projected vertically upwards from a point O at $21 \mathrm {~ms} ^ { - 1 }$.\\
(i) Calculate the greatest height reached by the particle.
When this particle is at its highest point, a second particle is projected vertically upwards from $O$ at $15 \mathrm {~ms} ^ { - 1 }$.\\
(ii) Show that the particles collide 1.5 seconds later and determine the height above O at which the collision takes place.
\hfill \mbox{\textit{OCR MEI M1 2005 Q4 [8]}}