Edexcel M1 2004 June — Question 3 9 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2004
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions
TypeCollision with friction after impact
DifficultyModerate -0.3 This is a standard two-part M1 mechanics question combining conservation of momentum with equations of motion. Part (a) requires straightforward application of momentum conservation with given relationships between velocities. Part (b) uses work-energy theorem or SUVAT equations with known values. While it requires multiple steps and careful sign conventions, it follows a predictable template with no novel problem-solving required, making it slightly easier than average.
Spec6.02i Conservation of energy: mechanical energy principle6.03b Conservation of momentum: 1D two particles

A particle \(P\) of mass \(2\) kg is moving with speed \(u\) m s\(^{-1}\) in a straight line on a smooth horizontal plane. The particle \(P\) collides directly with a particle \(Q\) of mass \(4\) kg which is at rest on the same horizontal plane. Immediately after the collision, \(P\) and \(Q\) are moving in opposite directions and the speed of \(P\) is one-third the speed of \(Q\).
  1. Show that the speed of \(P\) immediately after the collision is \(\frac{1}{5}u\) m s\(^{-1}\). [4]
After the collision \(P\) continues to move in the same straight line and is brought to rest by a constant resistive force of magnitude \(10\) N. The distance between the point of collision and the point where \(P\) comes to rest is \(1.6\) m.
  1. Calculate the value of \(u\). [5]

Question 3:
AnswerMarks
3u
2 k g 4 k g CLM: 2u = –2v + 4w M1 A1
v w Using w = 3v (⇒ 2u = –2v + 12v) and solve M1
⇒ v = 1u (*) A1 cso
5
(4)
(b) 10 = 2a ⇒ a = 5 m s–2 B1
0 = 1 u2 – 2 x 5 x 1.6 M1 A1√
25
→ u = 20 m s–1 M1 A1
(5)
(a) 1st M1 for valid CLM equn
2nd M1 for correct equn for ‘v’ and ‘w’ and solving for v or w.
Final A1 is cso (dropping u and reinserting loses last A1)
(b) Allow B1 for a = ± 5
M1 for using ‘v2 = u2 + 2as’ with v = 0 and with a value for a
A1 f.t. on their a (provided this is not g), but signs must be correct
SC For using u instead of u/5 ( u = 4), allow M1 A0 M0.
Energy: ½ x 2 x (u/5)2 = 10 x 1.6 M1 A1 A1
→ u = 20 dep M1 A1
Question 3:
3 | u
2 k g 4 k g CLM: 2u = –2v + 4w M1 A1
↓
v w Using w = 3v (⇒ 2u = –2v + 12v) and solve M1
⇒ v = 1u (*) A1 cso
5
(4)
(b) 10 = 2a ⇒ a = 5 m s–2 B1
0 = 1 u2 – 2 x 5 x 1.6 M1 A1√
25
↓
→ u = 20 m s–1 M1 A1
(5)
(a) 1st M1 for valid CLM equn
2nd M1 for correct equn for ‘v’ and ‘w’ and solving for v or w.
Final A1 is cso (dropping u and reinserting loses last A1)
(b) Allow B1 for a = ± 5
M1 for using ‘v2 = u2 + 2as’ with v = 0 and with a value for a
A1 f.t. on their a (provided this is not g), but signs must be correct
SC For using u instead of u/5 ( u = 4), allow M1 A0 M0.
Energy: ½ x 2 x (u/5)2 = 10 x 1.6 M1 A1 A1
→ u = 20 dep M1 A1
A particle $P$ of mass $2$ kg is moving with speed $u$ m s$^{-1}$ in a straight line on a smooth horizontal plane. The particle $P$ collides directly with a particle $Q$ of mass $4$ kg which is at rest on the same horizontal plane. Immediately after the collision, $P$ and $Q$ are moving in opposite directions and the speed of $P$ is one-third the speed of $Q$.

\begin{enumerate}[label=(\alph*)]
\item Show that the speed of $P$ immediately after the collision is $\frac{1}{5}u$ m s$^{-1}$. [4]
\end{enumerate}

After the collision $P$ continues to move in the same straight line and is brought to rest by a constant resistive force of magnitude $10$ N. The distance between the point of collision and the point where $P$ comes to rest is $1.6$ m.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Calculate the value of $u$. [5]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2004 Q3 [9]}}