| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance with other powers |
| Difficulty | Standard +0.3 This is a standard M2 mechanics question involving power-force-velocity relationships and Newton's second law on an incline. Part (i) is straightforward substitution (P=Fv at constant speed), and part (ii) requires resolving forces down the slope with resistance—routine application of familiar techniques with no novel insight required. |
| Spec | 3.03d Newton's second law: 2D vectors3.03h Newton's third law: action-reaction pairs6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((k25^{\frac{3}{2}}) \times 25 = 15000\) | M1 | Tractive force × speed = power |
| \(k = 4.8\) | A1 | |
| A1 | ||
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(R = 4.8 \times 16^{\frac{3}{2}}\) | B1 | 307.2. N2L, 4 terms to find tractive force (T). Allow \(cv(R)\), R not 600; \((T = 59.866...)\). \(16 \times\) Tractive force |
| \(T - 4.8x16^{\frac{3}{2}} + 700gx1/15 = 700x0.3\) | M1 | |
| \(P = 59.9 \times 16\) | M1 | |
| \(P = 958\) W | A1 | |
| [5] |
**Part (i)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(k25^{\frac{3}{2}}) \times 25 = 15000$ | M1 | Tractive force × speed = power |
| $k = 4.8$ | A1 | |
| | A1 | |
| | **[3]** | |
**Part (ii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $R = 4.8 \times 16^{\frac{3}{2}}$ | B1 | 307.2. N2L, 4 terms to find tractive force (T). Allow $cv(R)$, R not 600; $(T = 59.866...)$. $16 \times$ Tractive force |
| $T - 4.8x16^{\frac{3}{2}} + 700gx1/15 = 700x0.3$ | M1 | |
| $P = 59.9 \times 16$ | M1 | |
| $P = 958$ W | A1 | |
| | **[5]** | |2 The resistance to the motion of a car is $k v ^ { \frac { 3 } { 2 } } \mathrm {~N}$, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the car's speed and $k$ is a constant. The power exerted by the car's engine is 15000 W , and the car has constant speed $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ along a horizontal road.\\
(i) Show that $k = 4.8$.
With the engine operating at a much lower power, the car descends a hill of inclination $\alpha$, where $\sin \alpha = \frac { 1 } { 15 }$. At an instant when the speed of the car is $16 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, its acceleration is $0.3 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(ii) Given that the mass of the car is 700 kg , calculate the power of the engine.\\
\includegraphics[max width=\textwidth, alt={}, center]{941c0c81-a74f-49c0-acb7-1c23266fc2c8-02_579_447_1658_849}
A particle $P$ of mass 0.4 kg is attached to one end of each of two light inextensible strings which are both taut. The other end of the longer string is attached to a fixed point $A$, and the other end of the shorter string is attached to a fixed point $B$, which is vertically below $A$. The string $A P$ makes an angle of $30 ^ { \circ }$ with the vertical and is 0.5 m long. The string $B P$ makes an angle of $60 ^ { \circ }$ with the vertical. $P$ moves with constant angular speed in a horizontal circle with centre vertically below $B$ (see diagram). The tension in the string $A P$ is twice the tension in the string $B P$. Calculate\\
\hfill \mbox{\textit{OCR M2 2011 Q2 [8]}}