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Two particles of masses 1.0 kg and 2.0 kg are placed at a separation of 50 cm. Assuming that the only foces acting on the particles are their mutual gravitation find the initial acceleration of hthe two particles. |
Answer» The force of gravitastion exerted by one particle on another is `F_1=(Gm_1m_2)/r^2` `=96.67xx10^-11(N-m^2)/(kg^2)(1.0kg)xx(2.0kg))/((0.5m)^2)` `=5.3xx10^-10N` The acceleration of 1.0 kg particle is `a_`=F/m_1=(5.3xx10^-10N)/(1.0kg)` `=5.3xx10^-10N ms^2-` This acceleration is towards the 2.0 kg particle. The acceleration of the 2.0 kg particle is `a_2=F/m_2=(5.3xx10^-10N)/(2.0kg)` `=2.65xx10^-10ms^-2` This acceleration is towards the 1.0 kg particle. |
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