Edexcel M1 — Question 5 12 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTravel graphs
TypeForces and Newton's second law with kinematics
DifficultyStandard +0.3 This is a standard M1 momentum and impulse question requiring conservation of momentum, impulse-momentum theorem, and Newton's third law. While it involves multiple parts and algebraic manipulation with masses and velocities, the techniques are routine textbook applications with no novel problem-solving insight required. The 12 marks reflect length rather than conceptual difficulty, making it slightly easier than average.
Spec6.03b Conservation of momentum: 1D two particles6.03e Impulse: by a force6.03f Impulse-momentum: relation

Two trucks \(P\) and \(Q\), of masses 18 000 kg and 16 000 kg respectively, collide while moving towards each other in a straight line. Immediately before the collision, both trucks are travelling at the same speed, \(u\) ms\(^{-1}\). Immediately after the collision, \(P\) is moving at half its original speed, its direction of motion having been reversed.
  1. Find, in terms of \(u\), the speed of \(Q\) immediately after the collision. \hfill [5 marks]
  2. State, with a reason, whether the direction of \(Q\)'s motion has been reversed. \hfill [1 mark]
  3. Find, in terms of \(u\), the magnitude of the impulse exerted by \(P\) on \(Q\) in the collision, stating the units of your answer. \hfill [3 marks]
The force exerted by each truck on the other in the impact has magnitude \(108000u\) N.
  1. Find the time for which the trucks are in contact. \hfill [3 marks]

AnswerMarks Guidance
(a) Momentum: \(18u - 16v = -18(u/2) + 16v\)M1 A1 A1
\(2u = -9u + 16v\) → \(11u = 16v\) → \(v = \frac{11u}{16}\)M1 A1
(b) Velocity of Q was negative, now positive, so direction reversedB1
(c) Impulse \(= 16000(u + \frac{11u}{16}) = 27000u\) NsM1 A1 B1
(d) \(108000u = 27000u\) → \(t = 0.25\) sM1 A1 A1 Total: 12 marks
**(a)** Momentum: $18u - 16v = -18(u/2) + 16v$ | M1 A1 A1 |

$2u = -9u + 16v$ → $11u = 16v$ → $v = \frac{11u}{16}$ | M1 A1 |

**(b)** Velocity of Q was negative, now positive, so direction reversed | B1 |

**(c)** Impulse $= 16000(u + \frac{11u}{16}) = 27000u$ Ns | M1 A1 B1 |

**(d)** $108000u = 27000u$ → $t = 0.25$ s | M1 A1 A1 | **Total: 12 marks**
Two trucks $P$ and $Q$, of masses 18 000 kg and 16 000 kg respectively, collide while moving towards each other in a straight line. Immediately before the collision, both trucks are travelling at the same speed, $u$ ms$^{-1}$. Immediately after the collision, $P$ is moving at half its original speed, its direction of motion having been reversed.

\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $u$, the speed of $Q$ immediately after the collision. \hfill [5 marks]
\item State, with a reason, whether the direction of $Q$'s motion has been reversed. \hfill [1 mark]
\item Find, in terms of $u$, the magnitude of the impulse exerted by $P$ on $Q$ in the collision, stating the units of your answer. \hfill [3 marks]
\end{enumerate}

The force exerted by each truck on the other in the impact has magnitude $108000u$ N.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Find the time for which the trucks are in contact. \hfill [3 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1  Q5 [12]}}