Two bodies of masses $m$ and $3 m$ , moving with velocities $3 v$ and $v$ respectively along same direction, collide with each other. After collision they stick together and move with a velocity $V$ in the same direction. Then:

V = v

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V = \frac{3}{2} v

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V = 2 v

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V = \frac{4}{3} v

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Solution

The correct option is B $V = \frac{3}{2} v$
Here, the collision is inelastic. And law of conservation of momentum for inelastic collision is given by,

$m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) V$
$\Rightarrow V = \frac{m_{1} u_{1} + m_{2} u_{2}}{m_{1} + m_{2}}$
$\Rightarrow V = \frac{m \times 3 v + 3 m \times v}{m + 3 m}$
$\Rightarrow V = \frac{3 m v + 3 m v}{4 m}$
$\Rightarrow V = \frac{6 m v}{4 m} = \frac{3}{2} v$

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Two bodies of masses m and 3m, moving with velocities 3v and v respectively along same direction, collide with each other. After collision they stick together and move with a velocity V in the same direction. Then:

The correct option is B V=32vHere, the collision is inelastic. And law of conservation of momentum for inelastic collision is given by,m1u1+m2u2=(m1+m2)V⇒V=m1u1+m2u2m1+m2⇒V=m×3v+3m×vm+3m⇒V=3mv+3mv4m⇒V=6mv4m=32v

Two bodies of masses $m$ and $3 m$ , moving with velocities $3 v$ and $v$ respectively along same direction, collide with each other. After collision they stick together and move with a velocity $V$ in the same direction. Then: