JAMB - Physics (2005 - No. 45)
A force of 200N acts between two objects at a certain distance apart. The value of the force when the distance is halved is
400N
200N
100N
800N
Selitys
Let the masses of the two objects be M1 and M2
and their distance apart = r
Therefore the force acting F1 = \( \frac{GM1M2}{r^2}\)
for a new distance of r/2, we have a force of
F2 = \( \frac{GM2M2}{r/2}\)
\(\text{Thus } F1 \times r^2 = GM_1M_1 \text{and}\)
F2 \(\times (r/2)^2 = GM_1M_1\)
\(\text{Since } GM1M1 = GM_1M_1\)
\(\implies F_1 \times r^2 = F2 \times (r/2)^2\)
\(\text{Therefore } 200 \times r^2 = F2 \times r^2/4\)
F2 = \(\frac{4 \times 200 \times r^2}{r^2} = 800N\)
and their distance apart = r
Therefore the force acting F1 = \( \frac{GM1M2}{r^2}\)
for a new distance of r/2, we have a force of
F2 = \( \frac{GM2M2}{r/2}\)
\(\text{Thus } F1 \times r^2 = GM_1M_1 \text{and}\)
F2 \(\times (r/2)^2 = GM_1M_1\)
\(\text{Since } GM1M1 = GM_1M_1\)
\(\implies F_1 \times r^2 = F2 \times (r/2)^2\)
\(\text{Therefore } 200 \times r^2 = F2 \times r^2/4\)
F2 = \(\frac{4 \times 200 \times r^2}{r^2} = 800N\)