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A bullet of mass $20 g$ moving with the velocity of $200 m / s$ strikes and gets embedded into a stationary wooden block of mass $980 g$ . Find the velocity with which the block moves.

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5 m / s

5.5 m / s

4 m / s

4.5 m / s

answer is C.

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Detailed Solution

A bullet of mass

20 g

moving with the velocity of

200 m / s

strikes and gets embedded into a stationary wooden block of mass

980 g

. The block moves with a velocity of

4 m / s

.
Given,
Mass of the bullet

m_{1} = 20 g

= 0.02 kg

And velocity of the bullet

u_{1} = 200 m / s

Mass of the blocks

m_{2} = 980 g

= 0.98 kg

and
Velocity of the block

u_{2} = 0

Total initial momentum of bullet and wooden blocks,

p_{1} = m_{1} u_{1} + m_{2} u_{2}

= 0.02 \times 200 + 0

= 4 kg m s^{- 1}

⸫ Total final momentum of bullet and wooden block

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

= 4 kg m s^{- 1}

Velocity of block and bullet,

v = \frac{p_{2}}{m_{1} + m_{2}}

= \frac{4}{0.02 + 0.98}

= 4 m / s

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