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June 2014 p42 q7
3899
A light inextensible string of length 5.28 m has particles A and B, of masses 0.25 kg and 0.75 kg respectively, attached to its ends. Another particle P, of mass 0.5 kg, is attached to the mid-point of the string. Two small smooth pulleys P1 and P2 are fixed at opposite ends of a rough horizontal table of length 4 m and height 1 m. The string passes over P1 and P2 with particle A held at rest vertically below P1, the string taut and B hanging freely below P2. Particle P is in contact with the table halfway between P1 and P2 (see diagram). The coefficient of friction between P and the table is 0.4. Particle A is released and the system starts to move with constant acceleration of magnitude a m s-2. The tension in the part AP of the string is TA N and the tension in the part PB of the string is TB N.
Find TA and TB in terms of a.
Show by considering the motion of P that a = 2.
Find the speed of the particles immediately before B reaches the floor.
Find the deceleration of P immediately after B reaches the floor.
Solution
(i) Applying Newton's second law to particle A:
\(T_A - 2.5 = 0.25a\)
\(T_A = 2.5 + 0.25a\)
For particle B:
\(7.5 - T_B = 0.75a\)
\(T_B = 7.5 - 0.75a\)
(ii) For particle P, the frictional force \(F = 0.4 \times 5 = 2\) N. Using Newton's second law:
\(T_B - T_A - F = 0.5a\)
Substituting the expressions for \(T_A\) and \(T_B\):
\(7.5 - 0.75a - (2.5 + 0.25a) - 2 = 0.5a\)
\(7.5 - 2.5 - 2 = 0.75a + 0.25a + 0.5a\)
\(3 = 1.5a\)
\(a = 2\)
(iii) Using \(v^2 = 2as\) with \(s = 1 - \frac{1}{2}(5.28 - 4) = 0.36\) m:
\(v^2 = 2 \times 2 \times 0.36\)
\(v = 1.2\) m/s
(iv) Immediately after B reaches the floor, apply Newton's second law to P: