(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.

Wednesday, 09 April 2025

Register • Login

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

Trending Questions

** Free JAMB 2025 CBT Practice **

In a federal system,the essence of specifying the constitutional relationship between ...

Abraham circumcised his son,Isaac, on the eight days because

In mammals, the exchange of nutrients and metabolic products occurs in the

The problems of conducting census include?

Take Free Practice Test On 2025 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE

(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)

Prev Current Next

STAY UPDATED
FOLLOW US ON :

Facebook

Register • Login