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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» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Initially the centre of mass of the <a href="https://interviewquestions.tuteehub.com/tag/two-714195" style="font-weight:bold;" target="_blank" title="Click to know more about TWO">TWO</a> stones is at a height of <a href="https://interviewquestions.tuteehub.com/tag/30-304807" style="font-weight:bold;" target="_blank" title="Click to know more about 30">30</a> m from the ground. Acceleration of the centre of mass <br/> `= (mxxg+mxxg)/(m+m)`=g (downward)<br/> Let the stones colide after time t. <br/> At that time <a href="https://interviewquestions.tuteehub.com/tag/distance-116" style="font-weight:bold;" target="_blank" title="Click to know more about DISTANCE">DISTANCE</a> travelled by the first <a href="https://interviewquestions.tuteehub.com/tag/stone-1228007" style="font-weight:bold;" target="_blank" title="Click to know more about STONE">STONE</a> `= (1)/(2) "gt"^(2)` <br/> and distance travelled by the second stone = `ut -(1)/(2)"gt"^(2)` <br/> Now, `(1)/(2)"gt"^(2)+ut-(1)/(2)"gt"^(2)=60` <br/> `:.""t=(60)/(u)=(60)/(20)=3 s` <br/> Initital speed of the centre of mass, <br/> `u_("cm") = (mxx0+mxx20)/(m+m)` = 10 m/s (upward) <br/> `:.` Displacement of the centre of mass <br/> `= u_("cm")xxt-(1)/(2)"gt"^(2)=10xx3-(1)/(2)xx10xx9=-15`m <br/> Therefore, the centre of mass will move downward by 15 m.</body></html>


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