Question

Two particles of mass $1\text{kg}$ and $2\text{kg}$ are placed at a separation of $50\text{cm}$. Assuming that the only forces acting on the particles is due to their mutual attraction, find the initial acceleration of the two particles.

• A. $5.3\times {10}^{-10}{\text{m/s}}^2,2.65\times {10}^{-10}{\text{m/s}}^2$
  
• B. $4.3\times {10}^{-10}{\text{m/s}}^2,2.65\times {10}^{-10}{\text{m/s}}^2$
• C. $6.3\times {10}^{-10}{\text{m/s}}^2,2.65\times {10}^{-10}{\text{m/s}}^2$
• D. $2.3\times {10}^{-10}{\text{m/s}}^2,2.25\times {10}^{-10}{\text{m/s}}^2$

Answer 1

The correct option is A $5.3\times {10}^{-10}{\text{m/s}}^2,2.65\times {10}^{-10}{\text{m/s}}^2$


The force due to mutual attraction exerted on one particle by the another one is
$F=\frac{G{m}_1{m}_2}{R^2}$
$F=5.3\times {10}^{-10}\text{N}$

Now we have to calculate acceleration of $1\text{kg}$
$a_1=\frac{F}{m_1}=5.3\times {10}^{-10}$ ${\text{m/s}}^2$

The acceleration is directed towards the $2\text{kg}$ particle.

The acceleration of $2\text{kg}$ particle is
$a_2=\frac{F}{m_2}=2.65\times {10}^{-10}$ $\text{m}/s^2$

The acceleration of $2\text{kg}$ particle will be directed towards the $1\text{kg}$ particle.