Edexcel M2 2004 January — Question 5 12 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2004
SessionJanuary
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeBasic trajectory calculations
DifficultyModerate -0.3 This is a straightforward M2 projectiles question requiring standard application of SUVAT equations and the range formula. Part (a) uses the range formula with given values, part (b) applies time of flight formula, and part (c) uses energy conservation or velocity components. All techniques are routine for M2 students with no novel problem-solving required, making it slightly easier than average.
Spec3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae

5. A particle \(P\) is projected with velocity \(( 2 u \mathbf { i } + 3 u \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\) from a point \(O\) on a horizontal plane, where \(\mathbf { i }\) and \(\mathbf { j }\) are horizontal and vertical unit vectors respectively. The particle \(P\) strikes the plane at the point \(A\) which is 735 m from \(O\).
  1. Show that \(u = 24.5\).
  2. Find the time of flight from \(O\) to \(A\). The particle \(P\) passes through a point \(B\) with speed \(65 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
  3. Find the height of \(B\) above the horizontal plane.

AnswerMarks Guidance
(a) \(2ut = 735\)M1 A1
\(0 = 3ut - \frac{1}{2} gt^2\)M1 A1
eliminating \(t\)dep. M1
\(u = 24.5\) *A1 (6)
(b) \(t = \frac{735}{49} = 15\)M1 A1 (2)
(c) Initially: \(v^2 = (2u)^2 + (3u)^2\) (7803.25)M1
\(\frac{1}{2} mv^2 - \frac{1}{2} m 65^2 = mgh\)M1 A1
\(h = 180 \text{ m (183 m)}\)A1 (4)
OR\(w_x^2 = 65^2 - (2u)^2\) (1824) M1
\(w_y^2 = (3u)^2 - 2gh\)M1 A1
\(h = 180 \text{ m (183 m)}\)A1 (4)
(12 marks)
**(a)** $2ut = 735$ | M1 A1 |

$0 = 3ut - \frac{1}{2} gt^2$ | M1 A1 |

eliminating $t$ | dep. M1 |

$u = 24.5$ * | A1 | (6)

**(b)** $t = \frac{735}{49} = 15$ | M1 A1 | (2)

**(c)** Initially: $v^2 = (2u)^2 + (3u)^2$ (7803.25) | M1 |

$\frac{1}{2} mv^2 - \frac{1}{2} m 65^2 = mgh$ | M1 A1 |

$h = 180 \text{ m (183 m)}$ | A1 | (4)

| OR | $w_x^2 = 65^2 - (2u)^2$ (1824) | M1 |

$w_y^2 = (3u)^2 - 2gh$ | M1 A1 |

$h = 180 \text{ m (183 m)}$ | A1 | (4)

| | | **(12 marks)** |

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5. A particle $P$ is projected with velocity $( 2 u \mathbf { i } + 3 u \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$ from a point $O$ on a horizontal plane, where $\mathbf { i }$ and $\mathbf { j }$ are horizontal and vertical unit vectors respectively. The particle $P$ strikes the plane at the point $A$ which is 735 m from $O$.
\begin{enumerate}[label=(\alph*)]
\item Show that $u = 24.5$.
\item Find the time of flight from $O$ to $A$.

The particle $P$ passes through a point $B$ with speed $65 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\item Find the height of $B$ above the horizontal plane.
\end{enumerate}

\hfill \mbox{\textit{Edexcel M2 2004 Q5 [12]}}