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Q.

A $10 g$ bullet travelling at $200 m / s$ strikes and embeds itself in a $2 kg$ target that is initially at rest but is free to move. At what speed does the target move away?

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a

4 m / s

b

3 m / s

c

1 m / s

d

6 m / s

answer is C.

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Detailed Solution

When a

10 g

bullet travelling at

200 m / s

strikes and embeds itself in a

2 kg

target that is initially at rest, the target moves away at a rate of

1 m / s

Weight of the bullet (

m_{1})

is

10 g,

= 10 \times 10^{- 3} kg

= 0.01 kg

Bullet velocity

(v_{1})

is given as

200 m / s

The mass of the target

m_{2}

is

2 kg

Let

v_{2}

be the velocity of the target after being hit by the bullet.
According to the law of conservation of momentum:
When two objects collide with each other, the sum of their momentum remains conserved when no external force is applied on the bodies. Therefore,

m_{1} \times v_{1} = m_{2} \times v_{2}

\Rightarrow 0.01 \times 200 = 2 \times v_{2}

\Rightarrow v_{2} = \frac{0.01 \times 200}{2}

\Rightarrow v_{2} = 1 m / s

Therefore, the final velocity of the target is

1 m / s

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