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Question

# A bomb is projected with initial velocity 9.8 m/s making 15∘ angle with the horizontal. Bomb explodes into two parts in the mass ratio 1 : 2 at the point where vertical component of velocity is zero. If the heavier mass falls vertically down, how far on the ground from projection point will the lighter mass fall?

A
4.9 m
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B
19.6 m
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C
9.8 m
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D
2.45 m
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Solution

## The correct option is C 9.8 mLet R be the range of projected bomb and m be the total mass of bomb. Heavier mass (m1)=2m3 and lighter mass (m2)=m3 At the maximum height, velocity in the vertical direction will be zero. i.e vy=0 Since, Fext=0 in x- direction, so horizontal range attainer by COM will not change. According to question, heavier mass will fall vertically down i.e range is R/2. Let x be the range of the lighter mass. So, ¯x=m1x1+m2x2m1+m2 ⇒R=(2m3×R2)+(m3×x)2m3+m3 ⇒R=R3+x3 ⇒x=2R ⇒x=2×u2sin 2θg=2×9.8×9.89.8×12=9.8 m ⇒x=9.8 m

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