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 $={u}_{1}=400m/s$

Initial velocity of the wooden block $={u}_{2}=0m/s$

To find = velocity acquired by the block.

Solution

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

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 the given value in the equation, 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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Updated on: 10-Oct-2022

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