A $5 g m$ lump of clay, moving with a velocity of $10 c m / s$ towards east, collides head-on with another $2 g m$ lump of clay moving with $15 c m / s$ towards west. After collision, the two lumps stick together. The velocity of the compound lump will be-

5 cm/s towards east

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5 cm/s towards west

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2.88 cm/s towards east

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2.5 cm/s towards west

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Solution

The correct option is C 2.88 cm/s towards east
Given that,

$m_{1} = 5 g$

$m_{2} = 2 g$

$v_{1} = 10 c m / s$

$v_{2} = 15 m / s$

Now, according to the law of conservation

$m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v$

$5 \times 10 - 2 \times 15 = 7 v$

$v = \frac{20}{7}$

$v = 2.88 c m / s$

Hence, the velocity is $2.88 c m / s$ in east

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