| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2005 |
| Session | June |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Acceleration from power and speed |
| Difficulty | Standard +0.3 This is a standard M2 work-energy-power question with multiple routine parts. Parts (i)-(iii) involve direct application of P=Fv and basic power formulas. Part (iv) uses work-energy principle in a straightforward way. Part (v) requires combining gravitational PE with work against resistance, but follows standard methods. All parts are textbook-style applications with no novel insight required, making it slightly easier than average. |
| Spec | 6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02i Conservation of energy: mechanical energy principle6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{mgh}{t} = \frac{850 \times 9.8 \times 60}{20} = 24990\) | M1 | Use of \(\frac{mgh}{t}\) |
| so approx 25 kW | E1 | Shown |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Driving force – resistance = 0 | B1 | May be implied |
| \(25000 = 800v\) | M1 | Use of \(P = Fv\) |
| so \(v = 31.25\) and speed is \(31.25 \text{ m s}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Force is \(\frac{25000}{10} = 2500\) N | B1 | |
| N2L: \(2500 - 800 = 850a\) | M1 | Use of N2L with all terms |
| \(a = 2\) so \(2 \text{ m s}^{-2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(0.5 \times 850 \times 20^2 = 0.5 \times 850 \times 15^2\) | M1 | W-E equation with KE and power term |
| \(+25000 \times 6.90\) | B1 | One KE term correct |
| \(-800x\) | B1 | Use of \(Pt\). Accept wrong sign |
| B1 | WD against resistance. Accept wrong sign | |
| \(x = 122.6562...\) so \(123\) m (3 s.f.) | A1 | All correct |
| A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(0.5 \times 850 \times v^2 = 0.5 \times 850 \times 20^2\) | M1 | W-E equation inc KE, GPE and WD |
| \(-850 \times 9.8 \times \frac{105}{20}\) | M1 | GPE term with attempt at resolution |
| A1 | Correct. Accept expression. Condone wrong sign | |
| \(-800 \times 105\) | B1 | WD term. Neglect sign |
| \(v^2 = 99.452...\) so \(9.97 \text{ m s}^{-1}\) | A1 | cao |
| or N2L + ve up plane: \(-(800 + 850g \times 0.05) = 850a\) | M1 | N2L. All terms present. Allow sign errors |
| \(a = -1.43117...\) | A1 | Accept \(\pm\) |
| \(v^2 = 20^2 + 2 \times (-1.43117...) \times 105\) | M1 | Appropriate \(uvast\). Neglect signs |
| \(v^2 = 99.452...\) so \(9.97 \text{ m s}^{-1}\) | A1 | All correct including consistent signs. Need not follow sign of \(a\) above. cao |
# Question 2:
## Part (i):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{mgh}{t} = \frac{850 \times 9.8 \times 60}{20} = 24990$ | M1 | Use of $\frac{mgh}{t}$ |
| so approx 25 kW | E1 | Shown |
## Part (ii):
| Working | Mark | Guidance |
|---------|------|----------|
| Driving force – resistance = 0 | B1 | May be implied |
| $25000 = 800v$ | M1 | Use of $P = Fv$ |
| so $v = 31.25$ and speed is $31.25 \text{ m s}^{-1}$ | A1 | |
## Part (iii):
| Working | Mark | Guidance |
|---------|------|----------|
| Force is $\frac{25000}{10} = 2500$ N | B1 | |
| N2L: $2500 - 800 = 850a$ | M1 | Use of N2L with all terms |
| $a = 2$ so $2 \text{ m s}^{-2}$ | A1 | |
## Part (iv):
| Working | Mark | Guidance |
|---------|------|----------|
| $0.5 \times 850 \times 20^2 = 0.5 \times 850 \times 15^2$ | M1 | W-E equation with KE and power term |
| $+25000 \times 6.90$ | B1 | One KE term correct |
| $-800x$ | B1 | Use of $Pt$. Accept wrong sign |
| | B1 | WD against resistance. Accept wrong sign |
| $x = 122.6562...$ so $123$ m (3 s.f.) | A1 | All correct |
| | A1 | cao |
## Part (v):
| Working | Mark | Guidance |
|---------|------|----------|
| $0.5 \times 850 \times v^2 = 0.5 \times 850 \times 20^2$ | M1 | W-E equation inc KE, GPE and WD |
| $-850 \times 9.8 \times \frac{105}{20}$ | M1 | GPE term with attempt at resolution |
| | A1 | Correct. Accept expression. Condone wrong sign |
| $-800 \times 105$ | B1 | WD term. Neglect sign |
| $v^2 = 99.452...$ so $9.97 \text{ m s}^{-1}$ | A1 | cao |
| **or** N2L + ve up plane: $-(800 + 850g \times 0.05) = 850a$ | M1 | N2L. All terms present. Allow sign errors |
| $a = -1.43117...$ | A1 | Accept $\pm$ |
| $v^2 = 20^2 + 2 \times (-1.43117...) \times 105$ | M1 | Appropriate $uvast$. Neglect signs |
| $v^2 = 99.452...$ so $9.97 \text{ m s}^{-1}$ | A1 | All correct including consistent signs. Need not follow sign of $a$ above. cao |
---2 A car of mass 850 kg is travelling along a road that is straight but not level.\\
On one section of the road the car travels at constant speed and gains a vertical height of 60 m in 20 seconds. Non-gravitational resistances to its motion (e.g. air resistance) are negligible.\\
(i) Show that the average power produced by the car is about 25 kW .
On a horizontal section of the road, the car develops a constant power of exactly 25 kW and there is a constant resistance of 800 N to its motion.\\
(ii) Calculate the maximum possible steady speed of the car.\\
(iii) Find the driving force and acceleration of the car when its speed is $10 \mathrm {~ms} ^ { - 1 }$.
When travelling along the horizontal section of road, the car accelerates from $15 \mathrm {~ms} ^ { - 1 }$ to $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in 6.90 seconds with the same constant power and constant resistance.\\
(iv) By considering work and energy, find how far the car travels while it is accelerating.
When the car is travelling at $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ up a constant slope inclined at $\arcsin ( 0.05 )$ to the horizontal, the driving force is removed. Subsequently, the resistance to the motion of the car remains constant at 800 N .\\
(v) What is the speed of the car when it has travelled a further 105 m up the slope?
\hfill \mbox{\textit{OCR MEI M2 2005 Q2 [19]}}