A lump of clay of mass $10 g$ travelling with a velocity $10 c m s e c^{- 1}$ , towards east collides head on with another lump of clay of mass $10 g$ travelling with velocity of $20 c m s e c^{- 1}$ towards west. If the two lumps coalesce after collision, what is its velocity if no external force acts on the system?

$15 c m s e c^{- 1}$ towards west

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$15 c m s e c^{- 1}$ towards east

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$5 c m s e c^{- 1}$ towards west.

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$5 c m s e c^{- 1}$ towards east.

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Solution

The correct option is D
$5 c m s e c^{- 1}$ towards east.

Step 1: Given data
Mass of lump of clay, $m_{1} = 10 g$
The velocity of the lump of clay before the collision, $u_{1} = 20 c m s e c^{- 1}$ , towards the west which is positive.
Mass of another lump of clay, $m_{2} = 10 g$
The velocity of another lump of clay, $u_{2} = 10 c m / s e c$ , towards east which is negative.
Step 2: Find the velocity of two lumps after the collision, $v$
Since both masses are coalesced after collision hence, applying the conservation of momentum,
Putting the given values,
$m_{1} u_{1} - m_{2} u_{2} = (m_{1} + m_{2}) v 10 \times 20 - 10 \times 10 = (10 + 10) v 200 - 100 = 20 v 100 = 20 v v = 5 c m s e c^{- 1} v = 5 c m s e c^{- 1}, t o w a r d s e a s t$
Hence, option $(D)$ is the correct option.

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A lump of clay of mass 10g travelling with a velocity 10cmsec-1, towards east collides head on with another lump of clay of mass 10g travelling with velocity of20cmsec-1 towards west. If the two lumps coalesce after collision, what is its velocity if no external force acts on the system?