Three identical particle with velocities $v_{0}^i, - 3 v_{0}^j$ and $5 v_{0}^k$ collide successively with each other in such a way that they form a single particle. The velocity vector of resultant particle is

\frac{v_{0}}{3} (^i +^j +^k)

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\frac{v_{0}}{3} (^i -^j +^k)

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\frac{v_{0}}{3} (^i - 3^j +^k)

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\frac{v_{0}}{3} (^i - 3^j + 5^k)

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Solution

The correct option is D $\frac{v_{0}}{3} (^i - 3^j + 5^k)$
Since there is no external force acting on the system, momentum is conserved.
m: mass of each particle;
$\to v =$ Velocity of the combined particle;
$_{i} =_{f}$ ;
$∴ m v_{0}^i = m (- 3 v_{0}^j) + m (5 v_{0}^k) = 3 m \to v$ ;

$∴ \to v = \frac{v_{0}}{3} (^i - 3^i + 5^k)$ ;

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Three identical particle with velocities v0^i,−3v0^j and 5v0^k collide successively with each other in such a way that they form a single particle. The velocity vector of resultant particle is

The correct option is D v03(^i−3^j+5^k)Since there is no external force acting on the system, momentum is conserved.m: mass of each particle;→v=Velocity of the combined particle;→Pi=→Pf;∴mv0^i=m(−3v0^j)+m(5v0^k)=3m→v;∴→v=v03(^i−3^i+5^k);

Three identical particle with velocities $v_{0}^i, - 3 v_{0}^j$ and $5 v_{0}^k$ collide successively with each other in such a way that they form a single particle. The velocity vector of resultant particle is