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Question
A shell is fired vertically up. At some point it explodes in three parts with mass ratio 1:1:2 . One part falls at 10m left of projected point. Heavier part falls at 20 m right. What is the position of the third part?
A shell is fired vertically up. At some point it explodes in three parts with mass ratio 1:1:2 . One part falls at 10m left of projected point. Heavier part falls at 20 m right. What is the position of the third part?
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Solution
The correct option is C 30 m left
Center of mass will reach the same point of projection = \(x_{cm} = (0,0)
Position of lighter mass, m = (-10 , 0)
Position of higher mass, 2m = (20 , 0)
Position of another lighter mass, m = (x, 0)
∴ O= \frac {m(-30)+2m(20)+m(x)}{m+2m+m} \Rightarrow 30m= -mx\)
⇒ x=-30m
i.e. Position of third part, x = 30m left
30 m left
Center of mass will reach the same point of projection = \(x_{cm} = (0,0)
Position of lighter mass, m = (-10 , 0)
Position of higher mass, 2m = (20 , 0)
Position of another lighter mass, m = (x, 0)
∴ O= \frac {m(-30)+2m(20)+m(x)}{m+2m+m} \Rightarrow 30m= -mx\)
⇒ x=-30m
i.e. Position of third part, x = 30m left
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