Edexcel M1 2005 January — Question 6 13 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2005
SessionJanuary
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions
TypeRebound from wall or barrier
DifficultyModerate -0.3 This is a standard M1 mechanics question testing basic kinematics (SUVAT equations) and Newton's second law with straightforward application of impulse-momentum theorem. Parts (a)-(c) are routine calculations requiring only direct formula application. Part (d) requires more steps but follows a predictable structure: find rebound speed from impulse, then use equations of motion with constant deceleration. The multi-step nature and 6 marks elevate it slightly, but it remains a textbook exercise with no novel problem-solving required, making it slightly easier than average overall.
Spec3.02d Constant acceleration: SUVAT formulae6.02a Work done: concept and definition6.03e Impulse: by a force6.03f Impulse-momentum: relation

A stone \(S\) is sliding on ice. The stone is moving along a straight horizontal line \(ABC\), where \(AB = 24\) m and \(AC = 30\) m. The stone is subject to a constant resistance to motion of magnitude 0.3 N. At \(A\) the speed of \(S\) is 20 m s\(^{-1}\), and at \(B\) the speed of \(S\) is 16 m s\(^{-1}\). Calculate
  1. the deceleration of \(S\), [2]
  2. the speed of \(S\) at \(C\). [3]
  3. Show that the mass of \(S\) is 0.1 kg. [2]
At \(C\), the stone \(S\) hits a vertical wall, rebounds from the wall and then slides back along the line \(CA\). The magnitude of the impulse of the wall on \(S\) is 2.4 Ns and the stone continues to move against a constant resistance of 0.3 N.
  1. Calculate the time between the instant that \(S\) rebounds from the wall and the instant that \(S\) comes to rest. [6]

Question 6:
AnswerMarks
6(a) 162 = 202 – 2 x a x 24 ⇒ a = 3 m s–2 M1 A1
(2)
(b) v2 = 202 – 2 x 3 x 30 M1 A1√
v = √220 or 14.8 m s–1 A1
(3)
(c) 0.3 = m x 3 ⇒ m = 0.1 kg (*) M1 A1
(2)
(d) 0.1(w + √220) = 2.4 M1 A1√
w = 9.17 A1
0 = 9,17 – 3 x t M1 A1√
t ≈ 3.06 s A1
(6)
Question 6:
6 | (a) 162 = 202 – 2 x a x 24 ⇒ a = 3 m s–2 M1 A1
(2)
(b) v2 = 202 – 2 x 3 x 30 M1 A1√
v = √220 or 14.8 m s–1 A1
(3)
(c) 0.3 = m x 3 ⇒ m = 0.1 kg (*) M1 A1
(2)
(d) 0.1(w + √220) = 2.4 M1 A1√
w = 9.17 A1
↓
0 = 9,17 – 3 x t M1 A1√
t ≈ 3.06 s A1
(6)
A stone $S$ is sliding on ice. The stone is moving along a straight horizontal line $ABC$, where $AB = 24$ m and $AC = 30$ m. The stone is subject to a constant resistance to motion of magnitude 0.3 N. At $A$ the speed of $S$ is 20 m s$^{-1}$, and at $B$ the speed of $S$ is 16 m s$^{-1}$. Calculate

\begin{enumerate}[label=(\alph*)]
\item the deceleration of $S$, [2]
\item the speed of $S$ at $C$. [3]
\item Show that the mass of $S$ is 0.1 kg. [2]
\end{enumerate}

At $C$, the stone $S$ hits a vertical wall, rebounds from the wall and then slides back along the line $CA$. The magnitude of the impulse of the wall on $S$ is 2.4 Ns and the stone continues to move against a constant resistance of 0.3 N.

\begin{enumerate}[label=(\alph*)]\setcounter{enumi}{3}
\item Calculate the time between the instant that $S$ rebounds from the wall and the instant that $S$ comes to rest. [6]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2005 Q6 [13]}}