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Standard XII

Physics

Momentum

A particle of...

Question

A particle of mass M moves along the X-axis with speed $V_{0}$ and collides and sticks to a particle of mass $m$ moving with a speed $V_{0}$ along y-axis. The velocity of the combined particle after the collision :

\frac{M^i + m^j}{(M + m)} V_{0}

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\frac{m^i + M^j}{(M + m)} V_{0}

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(m^i + M^j) V_{0}

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Z e r o

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Solution

The correct option is A $\frac{M^i + m^j}{(M + m)} V_{0}$
Applying conservation of linear momentum,

$M V_{0}^i + m V_{0}^j = (M + m) v^{'}$

$v^{'} = \frac{M V_{0}^i + m V_{0}^j}{M + m}$

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A particle of mass M moves along the X-axis with speed V0 and collides and sticks to a particle of mass m moving with a speed V0 along y-axis. The velocity of the combined particle after the collision :

The correct option is A M^i+m^j(M+m)V0Applying conservation of linear momentum,MV0^i+mV0^j=(M+m)v′v′=MV0^i+mV0^jM+m

A particle of mass M moves along the X-axis with speed $V_{0}$ and collides and sticks to a particle of mass $m$ moving with a speed $V_{0}$ along y-axis. The velocity of the combined particle after the collision :

The correct option is A $\frac{M^i + m^j}{(M + m)} V_{0}$
Applying conservation of linear momentum,

$M V_{0}^i + m V_{0}^j = (M + m) v^{'}$

$v^{'} = \frac{M V_{0}^i + m V_{0}^j}{M + m}$