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Question: A projectile is fixed with velocity v<sub>0</sub> at an angle 60<sup>0</sup> with horizontal. At top...

A projectile is fixed with velocity v0 at an angle 600 with horizontal. At top of its trajectory it explodes into three fragment of equal mass. First fragment retraces the path, second moves vertically upward with speed ′3v0′2\frac{'3v_{0}'}{2}. The speed of third fragment –

A

3v02\frac{3v_{0}}{2}

B

5v02\frac{5v_{0}}{2}

C

v0

D

2v0

Answer

5v02\frac{5v_{0}}{2}

Explanation

Solution

Let velocity of third fragment be v→\overrightarrow{v}

P→i{\overrightarrow{P}}_{i} = P→f{\overrightarrow{P}}_{f}

̃ 3m . v02i^\frac{v_{0}}{2}\widehat{i}= – mv02i^m\frac{v_{0}}{2}\widehat{i} + m . 3v02j^\frac{3v_{0}}{2}\widehat{j} + mv→m\overrightarrow{v}

̃ v = 4v02i^\frac{4v_{0}}{2}\widehat{i}– 3v02j^\frac{3v_{0}}{2}\widehat{j}

̃ v = 5v02\frac{5v_{0}}{2}