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A bullet of mass m traveling horizontally at a very high speed v embeds itself in a block of mass M that is sitting at rest on a nearly frictionless surface. What is the final speed vf of the block-bullet system after the bullet embeds itself in the block?

Answer :

Answer:

[tex]vf=\dfrac{m\ v}{(M+m)}[/tex]

Explanation:

Given that

mass of the bullet = m

Speed of the bullet = v

mass of the block = M

Initial speed of the block = 0

The final speed of the bullet and block combine = vf

As we know that ,if external force on the system is zero then the linear momentum of the system will remain conserve.

Pi= Pf

So by using linear momentum conservation

m v + M x 0 = (M+m) vf

m v = (M+m) vf

[tex]vf=\dfrac{m\ v}{(M+m)}[/tex]

Therefore the final spped of the system will be

[tex]vf=\dfrac{m\ v}{(M+m)}[/tex]

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