Question

A bullet of mass 10 g 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 wooden block

Answer 1

Hi,

Here is your answer

Given:

Mass of bullet(m1)=10g=0.01kg

Mass of wooden block(m2)=900g=0.9kg

Initial velocity of bullet(u1)=400m/s

Initial velocity of wooden block(u2)=0m/s

To Find:

Final velocity of the system(v)

Solution:

By the law of conservation of momentum

m1u1+m2u2=(m1+m2)×v

Substituting the given datas, we get

0.01×400+0.9×0=(0.01+0.9)v

4/0.91=v

v=4.4m/s

Notes:

1.Momentum before collision is equal to the momentum after collision in an isolated system

2. Momentum is the product of mass and velocity. It's unit is kg m/s

Hope this helps you.

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,

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

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

$\fbox{4.4 m/s }$

Please everyone like it and please verify it...it took a lot of time to write and it is correct also...