A stone is dropped from the top of a tower of height 60 m. At the same time, another stone is thrown up from the foot of the tower with a velocity 20 m/s. Calculate the displacement of the centre of mass of the two stones at time of collision. (g=10 m/s^(2))

A stone is dropped from the top of a tower of height 60 m. At the same

1.	A stone is dropped from the top of a tower of height 60 m. At the same time, another stone is thrown up from the foot of the tower with a velocity 20 m/s. Calculate the displacement of the centre of mass of the two stones at time of collision. (g=10 m/s^(2))
Answer» SOLUTION :Initially the centre of mass of the TWO stones is at a height of 30 m from the ground. Acceleration of the centre of mass `= (mxxg+mxxg)/(m+m)`=g (downward) Let the stones colide after time t. At that time DISTANCE travelled by the first STONE `= (1)/(2) "gt"^(2)` and distance travelled by the second stone = `ut -(1)/(2)"gt"^(2)` Now, `(1)/(2)"gt"^(2)+ut-(1)/(2)"gt"^(2)=60` `:.""t=(60)/(u)=(60)/(20)=3 s` Initital speed of the centre of mass, `u_("cm") = (mxx0+mxx20)/(m+m)` = 10 m/s (upward) `:.` Displacement of the centre of mass `= u_("cm")xxt-(1)/(2)"gt"^(2)=10xx3-(1)/(2)xx10xx9=-15`m Therefore, the centre of mass will move downward by 15 m.