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?
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Answer 1

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:

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

Therefore, 4.4 m/s is the velocity acquired by the block.

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Answer 2

Answer:

Velocity =mass /acceleration