Question

Two particles A and B of masses 1 kg and 2 kg respectively are kept 1 m apart and are released to move under mutual attraction. Find the speed of A when that of B is 3 cm/hour. What is the separation between the particles at this instant?

• A. v = 2 × 10^-5m/s , d = 0.31m.
• B. v = 10^-5m/s , d = 0.31m.
• C. v = 2 × 10^-5m/s , d = 0.81m
• D. v = 10^-5m/s , d = 0.41m.

Answer 1

The correct option is A

v = 2 × 10^-5m/s , d = 0.31m.

The linear momentum of the pair A + B is zero initially. As only mutual attraction is taken into account, which is internal when A + B is taken as the system, the linear momentum will remain zero. The particles move in opposite directions. If the speed of A is v when the speed of B is 3.6 cm/hour

(1kg)v = (2kg) $\left({10}^{-5}m/s\right)$

or v = $2\times {10}^{-5}m/s$.

The Potential energy of the pair is $-\frac{G{m}_A{m}_B}{R}$ with usual symbols.Initial potential energy

= $\frac{6.67\times {10}^{11}N-m^2/k{g}^2\times 2kg\times 1kg}{1m}$

= -13.34 $\times {10}^{-11}H$.

If the seperation at the given instant is d, using conservation of energy,

= -13.34 $\times {10}^{11}J$.

= $\frac{13.34\times {10}^{-11}J-m}{d}+\frac{1}{2}\left(2kg\right)({10}^{-5}m/s{)}^2+\frac{1}{2}(2kg\left)\right(2\times {10}^{-5}m/s{)}^2$

Solving this, d = 0.31m.