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A particle of mass $m$ is projected with velocity $v$ at an angle $\theta$ with the horizontal. At its highest point, it explodes into two pieces of equal mass, one of the piece continue to move on the original trajectory, then the velocity of second piece is
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The correct answer is:
$v \cos \theta$
$v \cos \theta$
According to the law of conservation of linear momentum at the highest point,
$$
m v \cos \theta=\frac{m}{2}(v \cos \theta)+\frac{m}{2} v^{\prime}
$$
$\Rightarrow \quad m v \cos \theta-\frac{m}{2} v \cos \theta=\frac{m}{2} v^{\prime}$
$\Rightarrow \quad v^{\prime}=2\left(v \cos \theta-\frac{v \cos \theta}{2}\right)=v \cos \theta$
$$
m v \cos \theta=\frac{m}{2}(v \cos \theta)+\frac{m}{2} v^{\prime}
$$
$\Rightarrow \quad m v \cos \theta-\frac{m}{2} v \cos \theta=\frac{m}{2} v^{\prime}$
$\Rightarrow \quad v^{\prime}=2\left(v \cos \theta-\frac{v \cos \theta}{2}\right)=v \cos \theta$
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