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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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?
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: [tex]m_1[/tex]
[tex] = 10g \: = \frac{10}{1000} kg \: = 0.01kg \\ \\ [/tex]
Mass of the wooden block: [tex]m_2[/tex]
[tex] = 900g = \frac{900}{1000}kg \: = 0.9kg[/tex]
Initial velocity of the bullet: [tex]u_1[/tex]
= [tex]\frac{400m}{s} [/tex]
Initial velocity of the wooden block: [tex]u_2[/tex]
= [tex] \frac{0m}{s} [/tex]
To find:
Solution:
Final Velocity : vm/s
Now,
According to the law of conservation of momentum:
[tex]\left({m}_{1}\times {u}_{1}\right)+\left({m}_{2}\times {u}_{2}\right)=\left({m}_{1}+{m}_{2}\right)\times v[/tex]
[tex]v=\frac{\left({m}_{1}\times {u}_{1}\right)+\left({m}_{2}\times {u}_{2}\right)}{\left({m}_{1}+{m}_{2}\right)}[/tex]
Substituting:
We get,
[tex]v=\frac{\left(0.01\times 400\right)+\left(0.9\times 0\right)}{\left(0.01+0.9\right)}[/tex]
[tex]v=\frac{4}{0.91}[/tex]
v=4.4m/s
Therefore, 4.4 m/s is the velocity acquired by the block.
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Answer:
Velocity =mass /acceleration