4 An object is projected vertically upwards with speed \(7 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Calculate
- the speed of the object when it is 2.1 m above the point of projection,
- the greatest height above the point of projection reached by the object,
- the time after projection when the object is travelling downwards with speed \(5.7 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{57725055-7bce-4ad0-bb1c-59d07d56e2bd-3_227_897_635_664}
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\caption{Fig. 1}
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A particle \(P\) of mass 0.5 kg is projected with speed \(6 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) on a smooth horizontal surface towards a stationary particle \(Q\) of mass \(m \mathrm {~kg}\) (see Fig. 1). After the particles collide, \(P\) has speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in its original direction of motion, and \(Q\) has speed \(1 \mathrm {~ms} ^ { - 1 }\) more than \(P\). Show that \(v ( m + 0.5 ) = - m + 3\). - \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{57725055-7bce-4ad0-bb1c-59d07d56e2bd-3_229_901_1265_662}
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\caption{Fig. 2}
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\(Q\) and \(P\) are now projected towards each other with speeds \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and \(2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) respectively (see Fig. 2). Immediately after the collision the speed of \(Q\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) with its direction of motion unchanged and \(P\) has speed \(1 \mathrm {~ms} ^ { - 1 }\) more than \(Q\). Find another relationship between \(m\) and \(v\) in the form \(v ( m + 0.5 ) = a m + b\), where \(a\) and \(b\) are constants. - By solving these two simultaneous equations show that \(m = 0.9\), and hence find \(v\).