| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Minor (Further Mechanics Minor) |
| Year | 2020 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Find steady/maximum speed given power |
| Difficulty | Challenging +1.2 This is a multi-part mechanics question requiring vector magnitude calculation, application of F=ma with power considerations, and use of P=Fv. Part (a) is routine vector algebra. Parts (b-c) require understanding that zero power means resistance does all the work, and connecting power/force/acceleration equations. The question is longer than average and requires careful interpretation of physical constraints, but uses standard A-level mechanics techniques without requiring novel insight or particularly complex mathematical manipulation. |
| Spec | 1.10d Vector operations: addition and scalar multiplication6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks |
|---|---|
| 3(a) | R =( cβ1 ) i+ ( 2 c ) j,R= ( cβ1 )2 +4c |
| = c2 +2c+1= ( c+1 )2 =c+1* | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 2.2a | πΌπΌ ππππππ π½π½ |
| Answer | Marks |
|---|---|
| 3(b) | 8(c+1)=900(6βv) |
| Answer | Marks |
|---|---|
| ( 0<) c<674 | M1* |
| Answer | Marks |
|---|---|
| [3] | 3.3 |
| Answer | Marks |
|---|---|
| 1.1 | Applies impulse = change in |
| Answer | Marks |
|---|---|
| Must be strict inequality | Where v is the speed of the |
| Answer | Marks |
|---|---|
| 3(c) | 18000=5D |
| Answer | Marks |
|---|---|
| c=449 | B1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | Applies N2L β correct number of |
| Answer | Marks |
|---|---|
| condone sign errors | Where D is the tractive |
| Answer | Marks | Guidance |
|---|---|---|
| 3(d) | Model resistance as varying or model the power as not | |
| constant. | B1 | |
| [1] | 3.5c | Any alternative correct reason e.g. air |
Question 3:
--- 3(a) ---
3(a) | R =( cβ1 ) i+ ( 2 c ) j,R= ( cβ1 )2 +4c
= c2 +2c+1= ( c+1 )2 =c+1* | M1
A1
[2] | 1.1
2.2a | πΌπΌ ππππππ π½π½
Modulus (or modulus squared) of R
stated correctly
AG
--- 3(b) ---
3(b) | 8(c+1)=900(6βv)
8
6β (c+1)>0
900
( 0<) c<674 | M1*
M1dep*
A1
[3] | 3.3
3.4
1.1 | Applies impulse = change in
momentum (allow sign errors) or
application of N2L.
Uses the fact that v > 0 or considers
the limiting case when v = 0
Must be strict inequality | Where v is the speed of the
car at B
--- 3(c) ---
3(c) | 18000=5D
18000
β(c+1)=900Γ3.5
5
c=449 | B1
M1
A1
[3] | 1.1
3.3
1.1 | Applies N2L β correct number of
terms (allow D for tractive force) β
condone sign errors | Where D is the tractive
force
--- 3(d) ---
3(d) | Model resistance as varying or model the power as not
constant. | B1
[1] | 3.5c | Any alternative correct reason e.g. air
resistance incorporated into the
model.3 In this question $\mathbf { i }$ and $\mathbf { j }$ are perpendicular unit vectors and $c$ is a positive real number.\\
The resultant of two forces $c \mathbf { i N }$ and $- \mathbf { i } + 2 \sqrt { c } \mathbf { j N }$ is denoted by $R \mathrm {~N}$.
\begin{enumerate}[label=(\alph*)]
\item Show that the magnitude of $R$ is $c + 1$.
A car of mass 900 kg travels along a straight horizontal road with constant resistance to motion of magnitude $( c + 1 ) \mathrm { N }$. The car passes through point A on the road with speed $6 \mathrm {~ms} ^ { - 1 }$, and 8 seconds later passes through a point B on the same road.
The power developed by the car while travelling from A to B is zero. Furthermore, while travelling between A and B, the car's direction of motion is unchanged.
\item Determine the range of possible values of $c$.
The car later passes through a point C on the road. While travelling between B and C the power developed by the car is modelled as constant and equal to 18 kW . The car passes through C with speed $5 \mathrm {~ms} ^ { - 1 }$ and acceleration $3.5 \mathrm {~ms} ^ { - 2 }$.
\item Determine the value of $c$.
\item Suggest how one of the modelling assumptions made in this question could be improved.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Minor 2020 Q3 [9]}}