A ball of mass 100g is dropped at a time t=0. A second ball of mass 200g is dropped from the same point at t=0.2s. The distance between the centre of mass of two balls and the release point at t=0.4s…

Question

A ball of mass 100g is dropped at a time t=0. A second ball of mass 200g is dropped from the same point at t=0.2s. The distance between the centre of mass of two balls and the release point at t=0.4s is (Assume, g=10m/s^2 )

Accepted Answer

Correct answer: A. (a) Height fallen by first ball in 0.4s x_1= 12gt^2= 12 10 (0.4 )^2=0.8m Height fallen by second ball in 0.2s x= 12gt^2= 12 10 (0.2 )^2=0.2m Position of centre of mass is given by y_ CM= m_1x_1+m_2x_2m_1+m_2= 100 0.8+200 0.2100+200=0.4m

$A ball of mass 100 g is dropped at a time t = 0 . A second ball of mass 200 g is dropped from the same point at t = 0.2 s . The distance between the centre of mass of two balls and the release point at t = 0.4 s is (Assume, g = 10 m / s^{2})$

Solution

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