A particle of mass $m$ moving with speed $v$ towards east strikes another particle of same mass moving with same speed $v$ towards the north. After striking, the two particle fuse together. Determine the speed this new particle of mass $2 m$ will move in north-east direction?

v

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\frac{v}{2}

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\frac{v}{\sqrt{2}}

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v \sqrt{2}

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Solution

The correct option is C $\frac{v}{\sqrt{2}}$

Moment before and after collision is same.
$2 m \to u = m v ˆ 1 + m v ˆ j 2 \to u = v ˆ 1 + v ˆ j | \to v = | \frac{1}{2} \sqrt{v^{2} + v^{2}} = \frac{v}{\sqrt{2}}$

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A particle of mass m moving with speed v towards east strikes another particle of same mass moving with same speed v towards the north. After striking, the two particle fuse together. Determine the speed this new particle of mass 2 m will move in north-east direction?

The correct option is C v√2Moment before and after collision is same.2m→u=mvˆ1+mvˆj2→u=vˆ1+vˆj|→v=|12√v2+v2=v√2

A particle of mass $m$ moving with speed $v$ towards east strikes another particle of same mass moving with same speed $v$ towards the north. After striking, the two particle fuse together. Determine the speed this new particle of mass $2 m$ will move in north-east direction?

The correct option is C $\frac{v}{\sqrt{2}}$

Moment before and after collision is same.
$2 m \to u = m v ˆ 1 + m v ˆ j 2 \to u = v ˆ 1 + v ˆ j | \to v = | \frac{1}{2} \sqrt{v^{2} + v^{2}} = \frac{v}{\sqrt{2}}$