(a) An object P of mass 6.5kg is suspended by two light inextensible strings, AP and BP. The strings make angles 50° and 60° respectively with the downward vertical.
(i) Express the forces acting on P in component form; (ii) If P is at rest, write down the vector equation connecting all the forces; (iii) Calculate, correct to one decimal place, the tensions in the strings.
(b) A particle of mass 5 kg moves with initial velocity \(\frac{1}{2} m/s\) and final velocity \(\frac{3}{4} m/s \). Find the magnitude of its change in momentum.
(i) Express the forces acting on P in component form; (ii) If P is at rest, write down the vector equation connecting all the forces; (iii) Calculate, correct to one decimal place, the tensions in the strings.
(b) A particle of mass 5 kg moves with initial velocity \(\frac{1}{2} m/s\) and final velocity \(\frac{3}{4} m/s \). Find the magnitude of its change in momentum.
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Correct Answer: Option n
Explanation:
(i) \(T_{1} = \begin{pmatrix} -T_{1} \cos 40° \\ T_{1} \sin 40° \end{pmatrix} = \begin{pmatrix} -0.7660T_{1} \\ 0.6428T_{2} \end{pmatrix}\)
\(T_{2} = \begin{pmatrix} T_{2} \cos 30° \\ T_{2} \cos 30° \end{pmatrix} = \begin{pmatrix} 0.8660T_{2} \\ 0.5000T_{2} \end{pmatrix}\)
\(W = \begin{pmatrix} 0 \\ -65 \end{pmatrix}\)
(ii) P is at rest, the vector equation is
\(T_{1} + T_{2} + W = 0\)
(iii) \(\begin{pmatrix} -0.7660T_{1} \\ 0.6428T_{1} \end{pmatrix} + \begin{pmatrix} 0.8660T_{2} \\ 0.5000T_{2} \end{pmatrix} + \begin{pmatrix} 0 \\ -65 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\)
\(0.7660T_{1} = 0.8660T_{2} ... (1)\)
\(0.6428T_{1} + 0.5000T_{2} - 65 = 0 ... (2)\)
From (1), \(T_{1} = \frac{0.8660T_{2}}{0.7660} ... (3)\)
Put (3) in (2) :
\(0.6428(\frac{0.8660T_{2}}{0.7660}) + 0.5000T_{2} = 65 \)
\(0.7627T_{2} + 0.5000T_{2} = 65\)
\(1.2267T_{2} = 65 \implies T_{2} = \frac{65}{1.2267}\)
\(T_{2} = 52.988N \approxeq 53.0N\) (to 1 d.p)
From (3), \(T_{1} = \frac{0.866 \times 52.988}{0.766} = 59.905N\)
\(\approxeq 59.9N\) (to 1 d.p)
(i) \(T_{1} = \begin{pmatrix} -T_{1} \cos 40° \\ T_{1} \sin 40° \end{pmatrix} = \begin{pmatrix} -0.7660T_{1} \\ 0.6428T_{2} \end{pmatrix}\)
\(T_{2} = \begin{pmatrix} T_{2} \cos 30° \\ T_{2} \cos 30° \end{pmatrix} = \begin{pmatrix} 0.8660T_{2} \\ 0.5000T_{2} \end{pmatrix}\)
\(W = \begin{pmatrix} 0 \\ -65 \end{pmatrix}\)
(ii) P is at rest, the vector equation is
\(T_{1} + T_{2} + W = 0\)
(iii) \(\begin{pmatrix} -0.7660T_{1} \\ 0.6428T_{1} \end{pmatrix} + \begin{pmatrix} 0.8660T_{2} \\ 0.5000T_{2} \end{pmatrix} + \begin{pmatrix} 0 \\ -65 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\)
\(0.7660T_{1} = 0.8660T_{2} ... (1)\)
\(0.6428T_{1} + 0.5000T_{2} - 65 = 0 ... (2)\)
From (1), \(T_{1} = \frac{0.8660T_{2}}{0.7660} ... (3)\)
Put (3) in (2) :
\(0.6428(\frac{0.8660T_{2}}{0.7660}) + 0.5000T_{2} = 65 \)
\(0.7627T_{2} + 0.5000T_{2} = 65\)
\(1.2267T_{2} = 65 \implies T_{2} = \frac{65}{1.2267}\)
\(T_{2} = 52.988N \approxeq 53.0N\) (to 1 d.p)
From (3), \(T_{1} = \frac{0.866 \times 52.988}{0.766} = 59.905N\)
\(\approxeq 59.9N\) (to 1 d.p)