| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Variable power or two power scenarios |
| Difficulty | Standard +0.8 This is a multi-step mechanics problem requiring simultaneous equations from F=ma and P=Fv at two different states, followed by finding terminal velocity. It demands careful algebraic manipulation and understanding of power-resistance relationships, going beyond routine single-concept questions but using standard Further Mechanics techniques. |
| Spec | 6.02l Power and velocity: P = Fv6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (i) | P P |
| Answer | Marks |
|---|---|
| P(cid:32)12040W | B1 |
| Answer | Marks |
|---|---|
| [7] | 1.1a |
| Answer | Marks |
|---|---|
| 1.1 | P |
| Answer | Marks |
|---|---|
| Attempt to solve for P or k BC | Can be with T for M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (ii) | 15000 |
| Answer | Marks |
|---|---|
| v(cid:32)29.8ms-1 | e*M1 |
| Answer | Marks |
|---|---|
| [3] | c |
| Answer | Marks |
|---|---|
| 1.1 | Substituting their k and solving for v |
Question 4:
4 | (i) | P P
or seen
15 20
P
(cid:16)225k (cid:32)1250(0.54) or
15
P
(cid:16)400k (cid:32)1250(0.3)
20
P
(cid:16)225k (cid:32)1250(0.54)
15
P
(cid:16)400k (cid:32)1250(0.3)
20
k (cid:32)0.568
P(cid:32)12040W | B1
M1
A1
A1
M1
A1
A1
[7] | 1.1a
3.3
1.1
1.1
3.4
1.1
i
1.1 | P
Use of tractive force T (cid:32)
v
Attempt to use N2L once
n
e
m
Attempt to solve for P or k BC | Can be with T for M1
N2L is Newton’s second law
0.567567…
12040.540…
4 | (ii) | 15000
kv2 (cid:32)
v
15000 p
v3 (cid:32)
0.567...
v(cid:32)29.8ms-1 | e*M1
dep*M1
A1
[3] | c
2.2a
3.4
1.1 | Substituting their k and solving for v
29.78684…
--- 4(i) ---
4(i)
t
f
a
r
D
--- 4(ii) ---
4(ii)A car of mass 1250 kg experiences a resistance to its motion of magnitude $kv^2$ N, where $k$ is a constant and $v \, \text{m s}^{-1}$ is the car's speed. The car travels in a straight line along a horizontal road with its engine working at a constant rate of $P$ W. At a point $A$ on the road the car's speed is $15 \, \text{m s}^{-1}$ and it has an acceleration of magnitude $0.54 \, \text{m s}^{-2}$. At a point $B$ on the road the car's speed is $20 \, \text{m s}^{-1}$ and it has an acceleration of magnitude $0.3 \, \text{m s}^{-2}$.
\begin{enumerate}[label=(\roman*)]
\item Find the values of $k$ and $P$. [7]
\end{enumerate}
The power is increased to 15 kW.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Calculate the maximum steady speed of the car on a straight horizontal road. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics AS Q4 [10]}}