Edexcel M2 2016 October — Question 7 10 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2016
SessionOctober
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeVector form projectile motion
DifficultyStandard +0.8 This M2 projectile question requires students to connect kinetic energy concepts with vector projectile motion, use the constant horizontal velocity principle, apply energy conservation to find λ, then calculate time and position. The multi-step reasoning across mechanics topics and the non-standard energy condition make this harder than typical projectile questions but still within standard M2 scope.
Spec3.02h Motion under gravity: vector form6.02e Calculate KE and PE: using formulae
7. [In this question, the unit vectors \(\mathbf { i }\) and \(\mathbf { j }\) are in a vertical plane, \(\mathbf { i }\) being horizontal and \(\mathbf { j }\) being vertically upwards. Position vectors are given relative to a fixed origin O.] At time \(t = 0\) seconds, the particle \(P\) is projected from \(O\) with velocity ( \(3 \mathbf { i } + \lambda \mathbf { j }\) ) \(\mathrm { ms } ^ { - 1 }\), where \(\lambda\) is a positive constant. The particle moves freely under gravity. As \(P\) passes through the fixed point \(A\) it has velocity \(( 3 \mathbf { i } - 4 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\). The kinetic energy of \(P\) at the instant it passes through \(A\) is half the initial kinetic energy of \(P\). Find the position vector of \(A\), giving the components to 2 significant figures.
(10)
Question 7:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Vertical component of velocity: \(-4 = \lambda - gT\)M1
\(-4 = \lambda - gT\)A1 Accept with their \(\lambda\)
Kinetic energy: \(\frac{1}{2}\times\left(\frac{1}{2}m(3^2+\lambda^2)\right) = \frac{1}{2}m(3^2+4^2)\)M1
A2\(-1\) each error; \(\frac{1}{2}\) on wrong side is one error
Solve for \(\lambda\) and \(T\): \(9+\lambda^2=50\), \(\lambda=\sqrt{41}(=6.40...)\)DM1 Dependent on both preceding M marks
\(T = \frac{4+\lambda}{g} = \frac{4+\sqrt{41}}{g}(=1.06)\)A1
i component of position vector \(= 3T = 3\frac{4+\sqrt{41}}{g} = 3.2\,(\mathbf{i})\) (their \(T\))B1ft \(> 2\) s.f. is B0
j component of position vector \(= \lambda T - \frac{1}{2}gT^2\) for their \(\lambda, T\)M1
\(= 1.3\,(\mathbf{j})\)A1 \(1.1\) from \(T=1.1\) is premature approximation \(\to\) A0
Total: (10)
Alternative for j component: \(v^2=u^2+2as \Rightarrow (-4)^2 = \lambda^2 - 2gs\)M1
\(\Rightarrow s = \frac{25}{2g} = 1.3\)A1 NB the Q asks for 2 s.f. Only penalise \(> 2\) s.f. once
Question 8a:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
CLM: \(12mu = 4mv + kmw\)M1 All 3 terms and dimensionally consistent
\((12u = 4v + kw)\)A1 Correct unsimplified equation
Impact law: \(w - v = \frac{2}{3}\times 3u = 2u\)M1 Must be used the right way round
\((w = v + 2u)\)A1 Correct unsimplified equation
Solve for \(v\): \(12u = 4v + k(v+2u)\)DM1 Dependent on both preceding M marks
\((4+k)v = 12u - 2ku\)
\(v = \left\lvert\frac{2u(6-k)}{k+4}\right\rvert\) (m s\(^{-1}\))
Total: (6)
Question 8b:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Same direction \(\Rightarrow v > 0\)M1
\(\Rightarrow (0 <)\, k < 6\)A1 If stated, minimum value must be 0
Total: (2)
Question 8c:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(k=4 \Rightarrow v = \frac{u}{2}\), \(w = \frac{5u}{2}\)B1
CLM & Impact: \(4m\times\frac{5u}{2} = 4mx + 2my\), \((10u = 4x+2y)\)M1 Both equations
\(y - x = \frac{2}{3}\times\frac{5u}{2}\left(=\frac{5u}{3}\right)\)A1ft In any form, follow their \(w\); both equations correct
Solve for \(x\): \(10u = 4x + 2\left(x + \frac{5u}{3}\right)\)M1
\(6x = \frac{20u}{3}\), \(x = \frac{10u}{9}\)A1
\(\frac{u}{2} < \frac{10u}{9} \Rightarrow A\) will not collide with \(B\) againA1 CSO
Total: (6) [14] 
## Question 7:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Vertical component of velocity: $-4 = \lambda - gT$ | M1 | |
| $-4 = \lambda - gT$ | A1 | Accept with their $\lambda$ |
| Kinetic energy: $\frac{1}{2}\times\left(\frac{1}{2}m(3^2+\lambda^2)\right) = \frac{1}{2}m(3^2+4^2)$ | M1 | |
| | A2 | $-1$ each error; $\frac{1}{2}$ on wrong side is one error |
| Solve for $\lambda$ and $T$: $9+\lambda^2=50$, $\lambda=\sqrt{41}(=6.40...)$ | DM1 | Dependent on both preceding M marks |
| $T = \frac{4+\lambda}{g} = \frac{4+\sqrt{41}}{g}(=1.06)$ | A1 | |
| **i** component of position vector $= 3T = 3\frac{4+\sqrt{41}}{g} = 3.2\,(\mathbf{i})$ (their $T$) | B1ft | $> 2$ s.f. is B0 |
| **j** component of position vector $= \lambda T - \frac{1}{2}gT^2$ for their $\lambda, T$ | M1 | |
| $= 1.3\,(\mathbf{j})$ | A1 | $1.1$ from $T=1.1$ is premature approximation $\to$ A0 |
| **Total: (10)** | | |
| Alternative for **j** component: $v^2=u^2+2as \Rightarrow (-4)^2 = \lambda^2 - 2gs$ | M1 | |
| $\Rightarrow s = \frac{25}{2g} = 1.3$ | A1 | NB the Q asks for 2 s.f. Only penalise $> 2$ s.f. once |

---

## Question 8a:

| Answer/Working | Marks | Guidance |
|---|---|---|
| CLM: $12mu = 4mv + kmw$ | M1 | All 3 terms and dimensionally consistent |
| $(12u = 4v + kw)$ | A1 | Correct unsimplified equation |
| Impact law: $w - v = \frac{2}{3}\times 3u = 2u$ | M1 | Must be used the right way round |
| $(w = v + 2u)$ | A1 | Correct unsimplified equation |
| Solve for $v$: $12u = 4v + k(v+2u)$ | DM1 | Dependent on both preceding M marks |
| $(4+k)v = 12u - 2ku$ | | |
| $|v| = \left\lvert\frac{2u(6-k)}{k+4}\right\rvert$ (m s$^{-1}$) | A1 | Or equivalent. **Must be positive** |
| **Total: (6)** | | |

## Question 8b:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Same direction $\Rightarrow v > 0$ | M1 | |
| $\Rightarrow (0 <)\, k < 6$ | A1 | If stated, minimum value must be 0 |
| **Total: (2)** | | |

## Question 8c:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $k=4 \Rightarrow v = \frac{u}{2}$, $w = \frac{5u}{2}$ | B1 | |
| CLM & Impact: $4m\times\frac{5u}{2} = 4mx + 2my$, $(10u = 4x+2y)$ | M1 | Both equations |
| $y - x = \frac{2}{3}\times\frac{5u}{2}\left(=\frac{5u}{3}\right)$ | A1ft | In any form, follow their $w$; both equations correct |
| Solve for $x$: $10u = 4x + 2\left(x + \frac{5u}{3}\right)$ | M1 | |
| $6x = \frac{20u}{3}$, $x = \frac{10u}{9}$ | A1 | |
| $\frac{u}{2} < \frac{10u}{9} \Rightarrow A$ will not collide with $B$ again | A1 | CSO |
| **Total: (6) [14]** | | |
7. [In this question, the unit vectors $\mathbf { i }$ and $\mathbf { j }$ are in a vertical plane, $\mathbf { i }$ being horizontal and $\mathbf { j }$ being vertically upwards. Position vectors are given relative to a fixed origin O.]

At time $t = 0$ seconds, the particle $P$ is projected from $O$ with velocity ( $3 \mathbf { i } + \lambda \mathbf { j }$ ) $\mathrm { ms } ^ { - 1 }$, where $\lambda$ is a positive constant. The particle moves freely under gravity. As $P$ passes through the fixed point $A$ it has velocity $( 3 \mathbf { i } - 4 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$. The kinetic energy of $P$ at the instant it passes through $A$ is half the initial kinetic energy of $P$.

Find the position vector of $A$, giving the components to 2 significant figures.\\
(10)\\

\hfill \mbox{\textit{Edexcel M2 2016 Q7 [10]}}
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