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A bullet of mass $15 g$ moving with a velocity of $300 m / s$ in a horizontal direction gets embedded in a freely suspended wooden block of mass $1.2 kg$ . What is the velocity of the blocks after the bullet is embedded in it?

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1.5 m / s

- 15 m / s

3.7 m / s

None of the above

answer is C.

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Detailed Solution

A bullet of mass

15 g

moving with a velocity of

300 m / s

in a horizontal direction gets embedded in a freely suspended wooden block of mass

1.2 kg

. The velocity of the block after the bullet is embedded in it is

3.7 m / s

.
Given:
Mass of the bullet

m_{1} = 15 g = 0.015 kg

Mass of the block

m_{2} = 1.2 kg

Initial velocity of the bullet

u_{1} = 300 m / s

Initial velocity of the block

u_{2} = 0 m / s

According to the law of conservation of momentum,

m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) v

Where

m_{1}

m_{2}

u_{1}

u_{2}

and

v

are the mass of the bullet, mass of the block, speed of the bullet, speed of the block and the speed of the system after action respectively.
On substituting the given values we get the velocity of the block after the bullet is embedded in it as

\Rightarrow (0.015) (300) + (1.2) (0) = (0.015 + 1.2) v

\Rightarrow v = 4.5 / 1.215

\Rightarrow v = 3.7 m / s

Hence the velocity of the block is

3.7 m / s

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