Browsing as Guest. Progress, bookmarks and attempts are disabled.
Log in to track your work.
Nov 2012 p43 q7
3913
Particles A and B have masses 0.32 kg and 0.48 kg respectively. The particles are attached to the ends of a light inextensible string which passes over a small smooth pulley fixed at the edge of a smooth horizontal table. Particle B is held at rest on the table at a distance of 1.4 m from the pulley. A hangs vertically below the pulley at a height of 0.98 m above the floor (see diagram). A, B, the string and the pulley are all in the same vertical plane. B is released and A moves downwards.
(i) Find the acceleration of A and the tension in the string. [5]
A hits the floor and B continues to move towards the pulley. Find the time taken, from the instant that B is released, for
(ii) A to reach the floor, [2]
(iii) B to reach the pulley. [3]
Solution
(i) Applying Newton's second law to particle A:
\(0.32g - T = 0.32a\)
For particle B:
\(T = 0.48a\)
Solving these equations simultaneously:
\(0.32g = 0.32a + 0.48a\)
\(0.32g = 0.8a\)
\(a = \frac{0.32g}{0.8} = 4 \text{ m/s}^2\)
Substitute back to find tension:
\(T = 0.48 \times 4 = 1.92 \text{ N}\)
(ii) Using the equation \(s = \frac{1}{2}at^2\) for A: