| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Engine power on road constant/variable speed |
| Difficulty | Standard +0.3 This is a standard M2 power-resistance problem requiring the equation P=Fv to find driving force, then F=ma with component of weight down the slope. Part (ii) uses constant acceleration equations with average velocity. All techniques are routine for M2 students with no novel insight required, making it slightly easier than average. |
| Spec | 6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(25000/10\) | B1 | |
| \(1500g\sin5\) | B1 | 1281.1 |
| M1 | Attempt at N2L with 4 terms | |
| \(2500 - 750 - 1500g\sin5 = 1500a\) | A1 | \(cv(1500g\sin5)\); \(cv(2500)\) not 25000 |
| \(a = 0.313\) | A1 | Allow 0.31 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| WD against resistance \(= 750d\) | B1 | \(750h/\sin5\) |
| WD by engine \(= 25000 \times 28\ (= 700000)\) | B1 | |
| Change in PE \(= 1500g \times d\sin5\) | B1 | \(1500g \times h\) |
| Change in KE \(= \pm\frac{1}{2} \times 1500 \times (20^2 - 10^2)\) | B1 | Use of correct formula for KE |
| M1 | Use of conservation of energy, at least 3 used including WD by engine | |
| \(25000\times28 = \frac{1}{2}\times1500\times(20^2-10^2) + 750d + 1500g\times d\sin5\) | A1 | |
| \(d = 234\) | A1 |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $25000/10$ | B1 | |
| $1500g\sin5$ | B1 | 1281.1 |
| | M1 | Attempt at N2L with 4 terms |
| $2500 - 750 - 1500g\sin5 = 1500a$ | A1 | $cv(1500g\sin5)$; $cv(2500)$ not 25000 |
| $a = 0.313$ | A1 | Allow 0.31 |
---
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| WD against resistance $= 750d$ | B1 | $750h/\sin5$ |
| WD by engine $= 25000 \times 28\ (= 700000)$ | B1 | |
| Change in PE $= 1500g \times d\sin5$ | B1 | $1500g \times h$ |
| Change in KE $= \pm\frac{1}{2} \times 1500 \times (20^2 - 10^2)$ | B1 | Use of correct formula for KE |
| | M1 | Use of conservation of energy, at least 3 used including WD by engine |
| $25000\times28 = \frac{1}{2}\times1500\times(20^2-10^2) + 750d + 1500g\times d\sin5$ | A1 | |
| $d = 234$ | A1 | |
---5 A car of mass 1500 kg travels up a line of greatest slope of a straight road inclined at $5 ^ { \circ }$ to the horizontal. The power of the car's engine is constant and equal to 25 kW and the resistance to the motion of the car is constant and equal to 750 N . The car passes through point $A$ with speed $10 \mathrm {~ms} ^ { - 1 }$.\\
(i) Find the acceleration of the car at $A$.
The car later passes through a point $B$ with speed $20 \mathrm {~ms} ^ { - 1 }$. The car takes 28s to travel from $A$ to $B$.\\
(ii) Find the distance $A B$.
\hfill \mbox{\textit{OCR M2 2012 Q5 [12]}}