Question

a bullet of mass 10 g is moving with a velocity of 400m/s gets embedded in a freely suspended wooden block of mass 900 g . what is the velocity acquired by the block ?

Answer 1

the velocity acquired by block is 46.1m/s

Answer 2

Answer:

$\fbox{Solution}$

Here,Mass of the bullet ,

$m _{1} = 10g$

$= \frac{10}{1000} kg$

$= 0.01kg$

And,

Velocity of the bullet,

$v _{1} =$

400 m/s

So , Momentum of the bullet

$= m _{1} \times v _{1}$

= 0.01 × 400 kg.m/s............(1)

Now ,this bullet of mass 10g gets embedded into a wooden block of mass 900g .

So, the mass of wooden block along with the embedded Bullet will become

$900 + 10 = 910g. \: \: \: thus$

Mass of wooden block+ bullet,

$m _{2} = 900 + 10$

$= 910g$

$= \frac{910}{1000} kg$

$= 0.91kg$

And,

velocity of wooden block+ bullet,

$v _{2} = ?$

( to be calculated)

So,

Momentum of wooden block+ bullet

$= m _{2} \times v _{2}$

$= 0.91 \times v _{2} \: \: kg.m$

{ it is kg. m/s not only m} ....

.......(2)

Now according to the law of conservation of momentum,the two momenta as given by equation (1) and (2) should be equal

So,

$m _{1} \times v _{1} = m _{2} \times v _{2}$

or,

$0.01 \times 400 = 0.91 \times v _{2}$

$v _{2} = \frac{0.01 \times 400}{0.91}$

$\fbox{4.4 m/s }$

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