| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Variable resistance or force differential equation |
| Difficulty | Standard +0.3 This is a standard M2 power-resistance problem requiring routine application of P=Fv, F=ma, and resolving forces on an incline. Part (i) is a 'show that' using direct substitution, part (ii) applies Newton's second law, and part (iii) requires equilibrium of forces on an incline with the given resistance formula. All steps are textbook-standard with no novel insight required, making it slightly easier than average. |
| Spec | 3.03d Newton's second law: 2D vectors3.03v Motion on rough surface: including inclined planes6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| \(D = 3000/5^2 = 120\) | M1 | |
| A1 | 2 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(120 - 75 = 100a\) | M1 | |
| \(a = 0.45 \text{ ms}^{-2}\) | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(100x9.8x1/98\) | B1 | weight component |
| \(3000/v^2=3v^2+100x9.8x1/98\) | M1 | |
| \(3000 = 3v^4 + 10v^2\) | A1 | aef |
| solving quad in \(v^2\) | M1 | (\(v^2 = 30\)) |
| \(v = 5.48 \text{ ms}^{-1}\) | A1 | 5 |
**(i)**
$D = 3000/5^2 = 120$ | M1 |
| A1 | 2 | AG |
**(ii)**
$120 - 75 = 100a$ | M1 |
$a = 0.45 \text{ ms}^{-2}$ | A1 | 2 |
**(iii)**
$100x9.8x1/98$ | B1 | weight component
$3000/v^2=3v^2+100x9.8x1/98$ | M1 |
$3000 = 3v^4 + 10v^2$ | A1 | aef
solving quad in $v^2$ | M1 | ($v^2 = 30$)
$v = 5.48 \text{ ms}^{-1}$ | A1 | 5 | accept $\sqrt{30}$ | 9 |5 A model train has mass 100 kg . When the train is moving with speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ the resistance to its motion is $3 v ^ { 2 } \mathrm {~N}$ and the power output of the train is $\frac { 3000 } { v } \mathrm {~W}$.\\
(i) Show that the driving force acting on the train is 120 N at an instant when the train is moving with speed $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(ii) Find the acceleration of the train at an instant when it is moving horizontally with speed $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
The train moves with constant speed up a straight hill inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = \frac { 1 } { 98 }$.\\
(iii) Calculate the speed of the train.
\hfill \mbox{\textit{OCR M2 2007 Q5 [9]}}