JAMB - Physics (2002 - No. 5)
If the distance between two suspended masses 10kg each is tripled, the gravitational force of attraction between them is reduced by
one half
one third
one quarter
one ninth
Selitys
Let the distance apart the two masses = m.
From the equation F & M1 m2/ϒ2
F1 α 10 x 10/ϒ2 = F α 100/ ϒ2
F2 α 10 X 10/(3ϒ)2 => F2 α 100/9ϒ2
therefore F2/F1 = 100/9ϒ2/100/ϒ2 = 100/9ϒ2 ϒ2/100 = 1/9
Therefore F 2 : F 1 = 1 : 9
From the equation F & M1 m2/ϒ2
F1 α 10 x 10/ϒ2 = F α 100/ ϒ2
F2 α 10 X 10/(3ϒ)2 => F2 α 100/9ϒ2
therefore F2/F1 = 100/9ϒ2/100/ϒ2 = 100/9ϒ2 ϒ2/100 = 1/9
Therefore F 2 : F 1 = 1 : 9