(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.

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(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.

Explanation

(i)

T_{1} = (\begin{matrix} - T_{1} \cos 40 ° \\ T_{1} \sin 40 ° \end{matrix}) = (\begin{matrix} - 0.7660 T_{1} \\ 0.6428 T_{2} \end{matrix})

T_{2} = (\begin{matrix} T_{2} \cos 30 ° \\ T_{2} \cos 30 ° \end{matrix}) = (\begin{matrix} 0.8660 T_{2} \\ 0.5000 T_{2} \end{matrix})

W = (\begin{matrix} 0 \\ - 65 \end{matrix})

(ii) P is at rest, the vector equation is

T_{1} + T_{2} + W = 0

(iii)

(\begin{matrix} - 0.7660 T_{1} \\ 0.6428 T_{1} \end{matrix}) + (\begin{matrix} 0.8660 T_{2} \\ 0.5000 T_{2} \end{matrix}) + (\begin{matrix} 0 \\ - 65 \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix})

0.7660 T_{1} = 0.8660 T_{2} . . . (1)

0.6428 T_{1} + 0.5000 T_{2} - 65 = 0 . . . (2)

From (1),

T_{1} = \frac{0.8660 T_{2}}{0.7660} . . . (3)

Put (3) in (2) :

0.6428 (\frac{0.8660 T_{2}}{0.7660}) + 0.5000 T_{2} = 65

0.7627 T_{2} + 0.5000 T_{2} = 65

1.2267 T_{2} = 65 ⟹ T_{2} = \frac{65}{1.2267}

T_{2} = 52.988 N ≊ 53.0 N

(to 1 d.p)
From (3),

T_{1} = \frac{0.866 \times 52.988}{0.766} = 59.905 N

≊ 59.9 N

(to 1 d.p)

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(a) An object P of mass 6.5kg is suspended by two light inextensible strings, AP and ...

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