Edexcel M2 2006 June — Question 1 6 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2006
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable acceleration (1D)
TypeVelocity from acceleration by integration
DifficultyModerate -0.8 This is a straightforward M2 kinematics question requiring integration of acceleration to find velocity, applying initial conditions, then solving v=0. It's a standard textbook exercise with clear steps and no conceptual challenges, making it easier than average but not trivial due to the algebraic manipulation required.
Spec1.08b Integrate x^n: where n != -1 and sums3.02f Non-uniform acceleration: using differentiation and integration

A particle \(P\) moves on the \(x\)-axis. At time \(t\) seconds, its acceleration is \((5 - 2t)\) m s\(^{-2}\), measured in the direction of \(x\) increasing. When \(t = 0\), its velocity is 6 m s\(^{-1}\) measured in the direction of \(x\) increasing. Find the time when \(P\) is instantaneously at rest in the subsequent motion. [6]

AnswerMarks
\(a = 5 - 2t \Rightarrow v = 5t - t^2 + 6\)M1 A1, A1
\(v = 0 \Rightarrow t^2 - 5t - 6 = 0\)indep M1
\((t - 6)(t + 1) = 0\)dep M1
\(t = 6\text{ s}\)A1
Total: 6 marks
Question 2(a):
AnswerMarks
\(\frac{P}{24} = 600 \text{ or } \frac{1000P}{24} = 600 \Rightarrow P = 14.4\text{ kW}\)M1 A1
Total: 2 marks
Question 2(b):
AnswerMarks
\(\frac{30000}{20} - 1200 \times 9.8 \times \sin \alpha - 600 = 1200a\)M1 A2, 1.0
\(\Rightarrow a = 0.4\text{ m s}^{-2}\)A1
Total: 4 marks
Question 3(a):
AnswerMarks
\(I = \pm 0.5(16\mathbf{i} + 20\mathbf{j} - (-30\mathbf{i}))\)M1
\(= \pm(23\mathbf{i} + 10\mathbf{j})\)indep M1
\(\text{magn} = \sqrt{23^2 + 10^2} \approx 25.1\text{ Ns}\)indep M1 A1
Total: 4 marks
Question 3(b):
AnswerMarks
\(v = 16\mathbf{i} + (20 - 10t)\mathbf{j}\)M1
\(t = 3 \Rightarrow v = 16\mathbf{i} - 10\mathbf{j}\)indep M1
\(v = \sqrt{16^2 + 10^2} \approx 18.9\text{ m s}^{-1}\)indep M1 A1
Total: 4 marks
Question 4(a):
AnswerMarks
Total mass = \(12m\) (used)M1
(i) M(AB): \(m.3a/2 + m.3a/2 + m.3a + 6m.3a + 2m.3a = 12m.x\)indep M1 A1
\(\Rightarrow x = \frac{5}{2}a\)A1
(ii) M(AD): \(m.a + m.a + m.2a + 6m.2a = 12m.y\)indep M1 A1
\(\Rightarrow y = \frac{4}{3}a\)A1
Total: 7 marks
Question 4(b):
AnswerMarks
\(\tan \alpha = \frac{2a - 4a/3}{5a/2}\)M1 A1 f.t.
\(\Rightarrow \alpha \approx 14.9°\)A1 cao
Total: 3 marks
Question 5(a):
AnswerMarks Guidance
\(x_A = 28t\)\(x_B = 35 \cos \alpha t\) B1 B1
Meet \(\Rightarrow 28t = 35 \cos \alpha t \Rightarrow \cos \alpha = 28/35 = 4/5\) *M1 A1
Total: 4 marks
Question 5(b):
AnswerMarks Guidance
\(v_A = 73.5 - \frac{1}{2}gt^2\)\(v_B = 21t - \frac{1}{2}gt^2\) B1 B1
Meet \(\Rightarrow 73.5 = 21t \Rightarrow t = 3.5\text{ s}\)M1 A1
Total: 4 marks
Question 6(a):
AnswerMarks
M(A): \(S.3a = 4mg.2a \cos \alpha + mg.4a \cos \alpha\)M1 A1
\(= \frac{48}{5}mga \Rightarrow S = \frac{16}{5}mg\) *A1
Total: 3 marks
Question 6(b):
AnswerMarks
R(↑): \(R + S \cos\alpha = 5mg\)M1 A1
R(→): \(F = S \sin\alpha\)M1 A1
\(F \leq \mu R \Rightarrow \mu \geq \frac{48}{61}\) *dep on both previous M's M1 A1
Total: 6 marks
Question 6(c):
AnswerMarks
Direction of S is perpendicular to plank or No friction at the pegB1
Total: 1 mark
Question 7(a):
AnswerMarks
\(R = 4g \cos \alpha = 16g/5 \Rightarrow F = 2/7 \times 16g/5\)M1 A1
Work done \(= F \times 2.5 = 22.4\text{ J}\) or \(22\text{ J}\)indep M1 A1
Total: 4 marks
Question 7(b):
AnswerMarks
\(\frac{1}{2} \times 4 \times u^2 = 22.4 + 4g \times 2.5 \times 3/5\)M1 A2, 1.0 f.t.
\(\Rightarrow u \approx 6.37\text{ m s}^{-1}\) or \(6.4\text{ m s}^{-1}\)A1 cao
Total: 4 marks
Question 7(c):
AnswerMarks
\(\frac{1}{2} \times 4 \times v^2 = \frac{1}{2} \times 4 \times u^2 - 44.8\)M1 A2, 1.0 f.t.
[OR \(\frac{1}{2} \times 4 \times v^2 = 0 + 4g \times 2.5 \times 3/5 - 22.4\)]
\(\Rightarrow v \approx 4.27\text{ m s}^{-1}\) or \(4.3\text{ m s}^{-1}\)A1
Total: 4 marks
Question 8(a):
AnswerMarks
\(mu = 4mw - mv\)M1 A1
\(eu = w + v\)M1 A1
\(\Rightarrow w = \left(\frac{1 + e}{5}\right)u, v = \left(\frac{4c - 1}{5}\right)u\)indep M1 A1 A1
Total: 7 marks
Question 8(b):
AnswerMarks
\(w' = \left(\frac{4 + 4e}{25}\right)u\)B1 f.t.
Second collision \(\Rightarrow w' > v\)
\(\Rightarrow \frac{4 + 4e}{25} > \frac{4e - 1}{5}\)M1
\(\Rightarrow e < 9/16\)dep M1 A1
Also \(v > 0 \Rightarrow e > 1/4\) Hence result (*)B1
Total: 5 marks
Question 8(c):
AnswerMarks
KE lost \(= \frac{1}{2}mu^2 - \left[\frac{1}{2}.4m\left(\frac{(u/5)(1 + e)}{}\right)^2 + \frac{1}{2}m\left(\frac{(u/5)(4e - 1)}{}\right)^2\right]\)M1 A1 f.t.
\(= \frac{3}{10}mu^2\)A1 cao
Total: 3 marks
$a = 5 - 2t \Rightarrow v = 5t - t^2 + 6$ | M1 A1, A1 |

$v = 0 \Rightarrow t^2 - 5t - 6 = 0$ | indep M1 |

$(t - 6)(t + 1) = 0$ | dep M1 |

$t = 6\text{ s}$ | A1 |

**Total: 6 marks**

## Question 2(a):

$\frac{P}{24} = 600 \text{ or } \frac{1000P}{24} = 600 \Rightarrow P = 14.4\text{ kW}$ | M1 A1 |

**Total: 2 marks**

## Question 2(b):

$\frac{30000}{20} - 1200 \times 9.8 \times \sin \alpha - 600 = 1200a$ | M1 A2, 1.0 |

$\Rightarrow a = 0.4\text{ m s}^{-2}$ | A1 |

**Total: 4 marks**

## Question 3(a):

$I = \pm 0.5(16\mathbf{i} + 20\mathbf{j} - (-30\mathbf{i}))$ | M1 |

$= \pm(23\mathbf{i} + 10\mathbf{j})$ | indep M1 |

$\text{magn} = \sqrt{23^2 + 10^2} \approx 25.1\text{ Ns}$ | indep M1 A1 |

**Total: 4 marks**

## Question 3(b):

$v = 16\mathbf{i} + (20 - 10t)\mathbf{j}$ | M1 |

$t = 3 \Rightarrow v = 16\mathbf{i} - 10\mathbf{j}$ | indep M1 |

$v = \sqrt{16^2 + 10^2} \approx 18.9\text{ m s}^{-1}$ | indep M1 A1 |

**Total: 4 marks**

## Question 4(a):

Total mass = $12m$ (used) | M1 |

(i) M(AB): $m.3a/2 + m.3a/2 + m.3a + 6m.3a + 2m.3a = 12m.x$ | indep M1 A1 |

$\Rightarrow x = \frac{5}{2}a$ | A1 |

(ii) M(AD): $m.a + m.a + m.2a + 6m.2a = 12m.y$ | indep M1 A1 |

$\Rightarrow y = \frac{4}{3}a$ | A1 |

**Total: 7 marks**

## Question 4(b):

$\tan \alpha = \frac{2a - 4a/3}{5a/2}$ | M1 A1 f.t. |

$\Rightarrow \alpha \approx 14.9°$ | A1 cao |

**Total: 3 marks**

## Question 5(a):

$x_A = 28t$ | $x_B = 35 \cos \alpha t$ | B1 B1 |

Meet $\Rightarrow 28t = 35 \cos \alpha t \Rightarrow \cos \alpha = 28/35 = 4/5$ * | M1 A1 |

**Total: 4 marks**

## Question 5(b):

$v_A = 73.5 - \frac{1}{2}gt^2$ | $v_B = 21t - \frac{1}{2}gt^2$ | B1 B1 |

Meet $\Rightarrow 73.5 = 21t \Rightarrow t = 3.5\text{ s}$ | M1 A1 |

**Total: 4 marks**

## Question 6(a):

M(A): $S.3a = 4mg.2a \cos \alpha + mg.4a \cos \alpha$ | M1 A1 |

$= \frac{48}{5}mga \Rightarrow S = \frac{16}{5}mg$ * | A1 |

**Total: 3 marks**

## Question 6(b):

R(↑): $R + S \cos\alpha = 5mg$ | M1 A1 |

R(→): $F = S \sin\alpha$ | M1 A1 |

$F \leq \mu R \Rightarrow \mu \geq \frac{48}{61}$ * | dep on both previous M's M1 A1 |

**Total: 6 marks**

## Question 6(c):

Direction of S is perpendicular to plank or No friction at the peg | B1 |

**Total: 1 mark**

## Question 7(a):

$R = 4g \cos \alpha = 16g/5 \Rightarrow F = 2/7 \times 16g/5$ | M1 A1 |

Work done $= F \times 2.5 = 22.4\text{ J}$ or $22\text{ J}$ | indep M1 A1 |

**Total: 4 marks**

## Question 7(b):

$\frac{1}{2} \times 4 \times u^2 = 22.4 + 4g \times 2.5 \times 3/5$ | M1 A2, 1.0 f.t. |

$\Rightarrow u \approx 6.37\text{ m s}^{-1}$ or $6.4\text{ m s}^{-1}$ | A1 cao |

**Total: 4 marks**

## Question 7(c):

$\frac{1}{2} \times 4 \times v^2 = \frac{1}{2} \times 4 \times u^2 - 44.8$ | M1 A2, 1.0 f.t. |

[OR $\frac{1}{2} \times 4 \times v^2 = 0 + 4g \times 2.5 \times 3/5 - 22.4$] |

$\Rightarrow v \approx 4.27\text{ m s}^{-1}$ or $4.3\text{ m s}^{-1}$ | A1 |

**Total: 4 marks**

## Question 8(a):

$mu = 4mw - mv$ | M1 A1 |

$eu = w + v$ | M1 A1 |

$\Rightarrow w = \left(\frac{1 + e}{5}\right)u, v = \left(\frac{4c - 1}{5}\right)u$ | indep M1 A1 A1 |

**Total: 7 marks**

## Question 8(b):

$w' = \left(\frac{4 + 4e}{25}\right)u$ | B1 f.t. |

Second collision $\Rightarrow w' > v$ |

$\Rightarrow \frac{4 + 4e}{25} > \frac{4e - 1}{5}$ | M1 |

$\Rightarrow e < 9/16$ | dep M1 A1 |

Also $v > 0 \Rightarrow e > 1/4$ Hence result (*) | B1 |

**Total: 5 marks**

## Question 8(c):

KE lost $= \frac{1}{2}mu^2 - \left[\frac{1}{2}.4m\left(\frac{(u/5)(1 + e)}{}\right)^2 + \frac{1}{2}m\left(\frac{(u/5)(4e - 1)}{}\right)^2\right]$ | M1 A1 f.t. |

$= \frac{3}{10}mu^2$ | A1 cao |

**Total: 3 marks**
A particle $P$ moves on the $x$-axis. At time $t$ seconds, its acceleration is $(5 - 2t)$ m s$^{-2}$, measured in the direction of $x$ increasing. When $t = 0$, its velocity is 6 m s$^{-1}$ measured in the direction of $x$ increasing. Find the time when $P$ is instantaneously at rest in the subsequent motion.
[6]

\hfill \mbox{\textit{Edexcel M2 2006 Q1 [6]}}