Q#66: [Dynamics > Resultant forces](Past Exam Paper – November 2012 Paper 11 Q13)
Mass of 2.0 kg rests on a frictionless surface. It is attached to 1.0 kg mass by a light, thin string which passes over frictionless pulley. 1.0 kg mass is released and accelerates downwards.
What is the speed of 2.0 kg mass as the 1.0 kg mass hits floor, having fallen a distance of 0.50 m?
A 1.8 m s–1 B 2.2 m s–1 C 3.1 m s–1 D 9.8 m s–1
Solution 66:
Answer: A.
The gravitational force acts on the 1.0kg mass, causing an acceleration of 9.81ms-2.
Weight of 1.0kg box = 1 x 9.81 = 9.81N
But as the 1.0kg mass moves, the 2.0kg also undergoes motion. Therefore, considering the whole system, a force of 9.81N is acting on a total mass of (2.0 + 1.0 =) 3.0kg.
The resultant acceleration, a on this total mass can be obtained as follows
3a = 9.81
Acceleration, a = 9.81 / 3 = 3.27ms-2
Hence, each of the 2 masses undergoes a resultant acceleration of 3.27ms-2.
Initial speed of 2.0kg mass, u = 0
Distance travelled, s = 0.5m
v2 = u2 + 2as = 0 + 2 (3.27) (0.5)
Speed, v = √(2 x 3.27 x 0.5) = 1.8ms-1
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