Question

A bullet of mass 10 g moving with a velocity of
400 m/s gets embedded in a freely suspended
wooden block of mass 990 g. The velocity
acquired by the block will be​

Answer 1

The velocity  acquired by the block will be​ 4

M1(mass of bullet) = 10 g = 0.01kg

m2(mass of wooden block)= 990 g= 0.99 kg

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

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

After the collision, the block as well as the bullet move with a common velocity. Let this velocity be v.

By Law of Conservation of Momentum

m1u1 + m2u2 = m1v + m2v

=   0.01 × 400 + 0 × 0.99 = v(m1+m2)

=   4 = v(0.99 + 0.01)

=   4 = v

HOPE IT HELPS!!!

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...