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 900g. what is the velocity acquired by the block

Answer 1

Your answer is attached. Hope it helps.

Answer 2

The velocity acquired by the block is 4.4 m/s.

Step-by-step explanation:

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'.

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.