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Nov 2005 p4 q7
3585
Two particles A and B, of masses 0.3 kg and 0.2 kg respectively, are attached to the ends of a light inextensible string which passes over a smooth fixed pulley. Particle B is held on the horizontal floor and particle A hangs in equilibrium. Particle B is released and each particle starts to move vertically with constant acceleration of magnitude a m s-2.
Find the value of a.
Particle A hits the floor 1.2 s after it starts to move, and does not rebound upwards.
Show that A hits the floor with a speed of 2.4 m s-1.
Find the gain in gravitational potential energy by B, from leaving the floor until reaching its greatest height.
Solution
(i) Apply Newton's second law to both particles. For particle A:
\(0.3g - T = 0.3a\)
For particle B:
\(T - 0.2g = 0.2a\)
Adding these equations gives:
\(0.3g - 0.2g = 0.3a + 0.2a\)
Solving for a gives:
\(a = 2\)
(ii) Using the equation of motion:
\(v = u + at\)
\(where initial velocity u = 0, a = 2 m s-2, and t = 1.2 s, we find:\)
\(v = 0 + 2 imes 1.2 = 2.4 ext{ m s}^{-1}\)
(iii) The gain in potential energy while the string is slack is given by: