| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | January |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Average power over journey |
| Difficulty | Standard +0.3 This is a standard M2 work-energy-power question with routine multi-part calculations. Parts (i)-(iii) involve straightforward application of formulas (P=W/t, P=Fv, energy conservation), while part (iv) requires showing P=Fv with F=R+ma, then simple reasoning about constant power implying variable acceleration. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| GPE is \(1200 \times 9.8 \times 60 = 705\,600\) | B1 | Need not be evaluated |
| Power is \((705\,600 + 1\,800\,000) \div 120 = 20\,880 \text{ W} = 20\,900 \text{ W}\) (3 s.f.) | M1, A1, A1 | power is WD ÷ time; 120 s; cao |
| Answer | Marks | Guidance |
|---|---|---|
| so \(F = 750\) | M1 | Use of \(P = Fv\) |
| Answer | Marks | Guidance |
|---|---|---|
| Hence resistance is \(750\) N | A1, E1 | cao; Needs some justification |
| We require \(750 \times 200 = 150\,000\) J (= 150 kJ) | M1, F1 | Use of WD = Fd or Pt; FT their F |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 1200 \times 9.8 \times s \sin 5 - 1500x\) | M1, B1, M1 | Use of W-E equation with 'x'; 2 KE terms present; GPE term with resolution |
| GPE term correct | A1 | |
| All correct | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| so \(x = 306.91...\) so 307 m (3 s.f.) | (blank) | (blank) |
| Answer | Marks | Guidance |
|---|---|---|
| and N2L gives \(F - R = 1200a\) | B1, B1 | Shown |
| Answer | Marks | Guidance |
|---|---|---|
| \(P = (R + 1200a)v\) | E1 | Shown |
| If \(a \neq 0\), \(v\) is not constant. But \(P\) and \(R\) are constant so \(a\) cannot be constant. | E1 | (blank) |
## 2(i)
GPE is $1200 \times 9.8 \times 60 = 705\,600$ | B1 | Need not be evaluated
Power is $(705\,600 + 1\,800\,000) \div 120 = 20\,880 \text{ W} = 20\,900 \text{ W}$ (3 s.f.) | M1, A1, A1 | power is WD ÷ time; 120 s; cao
## 2(ii)
Using $P = Fv$. Let resistance be $R$ N:
$13500 = 18F$
so $F = 750$ | M1 | Use of $P = Fv$
As $v$ const, $a = 0$ so $F - R = 0$
Hence resistance is $750$ N | A1, E1 | cao; Needs some justification
We require $750 \times 200 = 150\,000$ J (= 150 kJ) | M1, F1 | Use of WD = Fd or Pt; FT their F
## 2(iii)
$\frac{1}{2} \times 1200 \times (9^2 - 18^2)$
$= 1200 \times 9.8 \times s \sin 5 - 1500x$ | M1, B1, M1 | Use of W-E equation with 'x'; 2 KE terms present; GPE term with resolution
GPE term correct | A1
All correct | A1 | cao
Hence $145800 = 475.04846...x$
so $x = 306.91...$ so 307 m (3 s.f.) | (blank) | (blank)
## 2(iv)
$P = Fv$
and N2L gives $F - R = 1200a$ | B1, B1 | Shown
Substituting gives
$P = (R + 1200a)v$ | E1 | Shown
If $a \neq 0$, $v$ is not constant. But $P$ and $R$ are constant so $a$ cannot be constant. | E1 | (blank)
---2 A car of mass 1200 kg travels along a road for two minutes during which time it rises a vertical distance of 60 m and does $1.8 \times 10 ^ { 6 } \mathrm {~J}$ of work against the resistance to its motion. The speeds of the car at the start and at the end of the two minutes are the same.\\
(i) Calculate the average power developed over the two minutes.
The car now travels along a straight level road at a steady speed of $18 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ while developing constant power of 13.5 kW .\\
(ii) Calculate the resistance to the motion of the car.
How much work is done against the resistance when the car travels 200 m ?
While travelling at $18 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the car starts to go down a slope inclined at $5 ^ { \circ }$ to the horizontal with the power removed and its brakes applied. The total resistance to its motion is now 1500 N .\\
(iii) Use an energy method to determine how far down the slope the car travels before its speed is halved.
Suppose the car is travelling along a straight level road and developing power $P \mathrm {~W}$ while travelling at $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ with acceleration $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$ against a resistance of $R \mathrm {~N}$.\\
(iv) Show that $P = ( R + 1200 a ) v$ and deduce that if $P$ and $R$ are constant then if $a$ is not zero it cannot be constant.
\hfill \mbox{\textit{OCR MEI M2 2010 Q2 [19]}}