Question

a bullet of mass 10 g moving with a velocity of 400m per s gets empledded in a free suspended wooden block of mass 900g velocity acquired by the block?​

Answer 1

Answer:

We are given that a bullet of mass 10 g moving with a velocity of 400 m/s gets embedded in a freely suspended wooden block of mass 900 g. Let the final velocity acquired by the block be 'v m/s'. So, v = 4.4 m/s which means this is the velocity acquired by the block

Explanation:

Let the final velocity acquired by the block be 'v m/s'.

The mass of the bullet = m_1 = 10 g = \frac{10}{1000} \text{ kg}

= 0.01 kg

The mass of the wooden block = m_2 = 900 g = \frac{900}{1000} \text{ kg}

= 0.9 kg

Also, the initial velocity of the bullet is given by = u_1 = 400 m/s

and the initial velocity of the wooden block is given by = u_2 = 0 m/s

Now, the momentum conservation of law states that;

Momentum before collision = Momentum after collision

(m_1\times u_1 )+ (m_2\times u_2)= (m_1+m_2)\times v

\text{v} = \frac{(m_1\times u_1 )+ (m_2\times u_2)}{(m_1+m_2)}

= \frac{(0.01\times 400 )+ (0.9\times 0)}{(0.01+0.9)}

= \frac{4}{0.91} = 4.4

So, v = 4.4 m/s which means this is the velocity acquired by the block.

Answer 2

$\huge{\underline{\underline {\mathtt{\purple{Hola}\pink{Mate}}}}}$

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 900 g. What is the velocity acquired by the block?

Given:

Mass of the bullet $m_1$

$= 10g \: = \frac{10}{1000} kg \: = 0.01kg$

Mass of the wooden block $m_2$

$= 900g = \frac{900}{1000}kg \: = 0.9kg$

Initial velocity of the bullet $u_1$

$\frac{400m}{s}$

Initial velocity of the wooden block $u_2$

$\frac{0m}{s}$

To find:

• Velocity acquired by the block

Solution

We know that,

Final Velocity = vm/s

Now,

According to the law of conservation of momentum:

Momentum before collision = Momentum after collision

$\left({m}_{1}\times {u}_{1}\right)+\left({m}_{2}\times {u}_{2}\right)=\left({m}_{1}+{m}_{2}\right)\times v$

$v=\frac{\left({m}_{1}\times {u}_{1}\right)+\left({m}_{2}\times {u}_{2}\right)}{\left({m}_{1}+{m}_{2}\right)}$

Substituting:

We get,

$v=\frac{\left(0.01\times 400\right)+\left(0.9\times 0\right)}{\left(0.01+0.9\right)}$

$v=\frac{4}{0.91}$

v = 4.4m/s

∴ 4.4 m/s is your answer

Hope it helps ya!!

Hope it helps ya!!Mark As Brainliest