| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2002 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Velocity from two position vectors |
| Difficulty | Moderate -0.8 This is a straightforward M1 mechanics question testing basic vector kinematics. Parts (a)-(e) involve simple displacement/velocity calculations using the formula v = Δr/Δt and position vectors r = r₀ + vt. The only slightly challenging aspect is finding the unit vector for Q's velocity direction, but this is standard bookwork. All steps are routine applications of formulas with no problem-solving insight required. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(v_P = \frac{(50\mathbf{i} - 25\mathbf{j}) - (20\mathbf{i} + 35\mathbf{j})}{\frac{1}{2}} = 60\mathbf{i} - 120\mathbf{j}\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(\mathbf{p} = 20\mathbf{i} + 35\mathbf{j} + (60\mathbf{i} - 120\mathbf{j})t\) | M1 A1 ft | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(v_Q = \frac{120}{5}(4\mathbf{i} - 3\mathbf{j}) = 96\mathbf{i} - 72\mathbf{j}\) | M1 | |
| \(\mathbf{q} = 96t\mathbf{i} - 72t\mathbf{j}\) | M1 A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(t = 2\): \(\mathbf{p} = 140\mathbf{i} - 205\mathbf{j}\), \(\mathbf{q} = 192\mathbf{i} - 144\mathbf{j}\) | M1 | |
| Use \(\overrightarrow{PQ} = \mathbf{q} - \mathbf{p}\) or \(\mathbf{p} - \mathbf{q}\) \(= 52\mathbf{i} + 61\mathbf{j}\) | M1 | |
| \(PQ = \sqrt{52^2 + 61^2} \approx 80\) km | M1 A1 | (4) |
## Question 7:
### Part (a):
| Answer/Working | Marks | Notes |
|---|---|---|
| $v_P = \frac{(50\mathbf{i} - 25\mathbf{j}) - (20\mathbf{i} + 35\mathbf{j})}{\frac{1}{2}} = 60\mathbf{i} - 120\mathbf{j}$ | M1 A1 | |
### Part (b):
| Answer/Working | Marks | Notes |
|---|---|---|
| $\mathbf{p} = 20\mathbf{i} + 35\mathbf{j} + (60\mathbf{i} - 120\mathbf{j})t$ | M1 A1 ft | **(2)** |
### Part (c):
| Answer/Working | Marks | Notes |
|---|---|---|
| $v_Q = \frac{120}{5}(4\mathbf{i} - 3\mathbf{j}) = 96\mathbf{i} - 72\mathbf{j}$ | M1 | |
| $\mathbf{q} = 96t\mathbf{i} - 72t\mathbf{j}$ | M1 A1 | **(3)** |
### Part (d):
| Answer/Working | Marks | Notes |
|---|---|---|
| $t = 2$: $\mathbf{p} = 140\mathbf{i} - 205\mathbf{j}$, $\mathbf{q} = 192\mathbf{i} - 144\mathbf{j}$ | M1 | |
| Use $\overrightarrow{PQ} = \mathbf{q} - \mathbf{p}$ or $\mathbf{p} - \mathbf{q}$ $= 52\mathbf{i} + 61\mathbf{j}$ | M1 | |
| $PQ = \sqrt{52^2 + 61^2} \approx 80$ km | M1 A1 | **(4)** |
---7. Two helicopters $P$ and $Q$ are moving in the same horizontal plane. They are modelled as particles moving in straight lines with constant speeds. At noon $P$ is at the point with position vector $( 20 \mathbf { i } + 35 \mathbf { j } ) \mathrm { km }$ with respect to a fixed origin $O$. At time $t$ hours after noon the position vector of $P$ is $\mathbf { p } \mathrm { km }$. When $t = \frac { 1 } { 2 }$ the position vector of $P$ is $( 50 \mathbf { i } - 25 \mathbf { j } ) \mathrm { km }$. Find
\begin{enumerate}[label=(\alph*)]
\item the velocity of $P$ in the form $( a \mathbf { i } + b \mathbf { j } ) \mathrm { km } \mathrm { h } ^ { - 1 }$,
\item an expression for $\mathbf { p }$ in terms of $t$.
At noon $Q$ is at $O$ and at time $t$ hours after noon the position vector of $Q$ is $\mathbf { q } \mathrm { km }$. The velocity of $Q$ has magnitude $120 \mathrm {~km} \mathrm {~h} ^ { - 1 }$ in the direction of $4 \mathbf { i } - 3 \mathbf { j }$. Find\\
(d) an expression for $\mathbf { q }$ in terms of $t$,\\
(e) the distance, to the nearest km , between $P$ and $Q$ when $t = 2$.
\section*{8.}
\section*{Figure 4}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{14703bfa-abd8-4a8d-bc18-20d66eea409e-6_695_1153_322_562}
\end{center}
Two particles $A$ and $B$, of mass $m \mathrm {~kg}$ and 3 kg respectively, are connected by a light inextensible string. The particle $A$ is held resting on a smooth fixed plane inclined at $30 ^ { \circ }$ to the horizontal. The string passes over a smooth pulley $P$ fixed at the top of the plane. The portion $A P$ of the string lies along a line of greatest slope of the plane and $B$ hangs freely from the pulley, as shown in Fig. 4. The system is released from rest with $B$ at a height of 0.25 m above horizontal ground. Immediately after release, $B$ descends with an acceleration of $\frac { 2 } { 5 } g$. Given that $A$ does not reach $P$, calculate\\
(a) the tension in the string while $B$ is descending,\\
(b) the value of $m$.
The particle $B$ strikes the ground and does not rebound. Find
\item the magnitude of the impulse exerted by $B$ on the ground,
\item the time between the instant when $B$ strikes the ground and the instant when $A$ reaches its highest point.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2002 Q7 [11]}}