Standard +0.3 This is a standard M1 collision problem requiring conservation of momentum applied twice, followed by solving simultaneous equations. The setup is straightforward with clearly defined velocities, and the algebraic manipulation is routine. While it has multiple parts and requires careful sign conventions, it involves no novel insight—just methodical application of a core syllabus technique.
\includegraphics{figure_5_1}
A particle \(P\) of mass \(0.5\) kg is projected with speed \(6\) m s\(^{-1}\) on a smooth horizontal surface towards a stationary particle \(Q\) of mass \(m\) kg (see Fig. 1). After the particles collide, \(P\) has speed \(v\) m s\(^{-1}\) in the original direction of motion, and \(Q\) has speed \(1\) m s\(^{-1}\) more than \(P\). Show that \(v(m + 0.5) = -m + 3\). [3]
\includegraphics{figure_5_2}
\(Q\) and \(P\) are now projected towards each other with speeds \(4\) m s\(^{-1}\) and \(2\) m s\(^{-1}\) respectively (see Fig. 2). Immediately after the collision the speed of \(Q\) is \(v\) m s\(^{-1}\) with its direction of motion unchanged and \(P\) has speed \(1\) m s\(^{-1}\) more than \(Q\). Find another relationship between \(m\) and \(v\) in the form \(v(m + 0.5) = am + b\), where \(a\) and \(b\) are constants. [4]
By solving these two simultaneous equations show that \(m = 0.9\), and hence find \(v\). [4]
\begin{enumerate}[label=(\roman*)]
\item
\includegraphics{figure_5_1}
A particle $P$ of mass $0.5$ kg is projected with speed $6$ m s$^{-1}$ on a smooth horizontal surface towards a stationary particle $Q$ of mass $m$ kg (see Fig. 1). After the particles collide, $P$ has speed $v$ m s$^{-1}$ in the original direction of motion, and $Q$ has speed $1$ m s$^{-1}$ more than $P$. Show that $v(m + 0.5) = -m + 3$. [3]
\item
\includegraphics{figure_5_2}
$Q$ and $P$ are now projected towards each other with speeds $4$ m s$^{-1}$ and $2$ m s$^{-1}$ respectively (see Fig. 2). Immediately after the collision the speed of $Q$ is $v$ m s$^{-1}$ with its direction of motion unchanged and $P$ has speed $1$ m s$^{-1}$ more than $Q$. Find another relationship between $m$ and $v$ in the form $v(m + 0.5) = am + b$, where $a$ and $b$ are constants. [4]
\item By solving these two simultaneous equations show that $m = 0.9$, and hence find $v$. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR M1 2009 Q5 [11]}}